solutions for exercises in chapter 1

Solutions for exercises in chapter 1

E1.1 Verify thatS0 → ¬S1 = 〈2, 3, 1, 4〉

and(S0 → S1)→ (¬S1 → ¬S0) = 〈2, 2, 3, 4, 2, 1, 4, 1, 3〉.

S0 → ¬S1 = 〈2〉S0 ¬S1

= 〈2〉〈3〉〈1〉S1

= 〈2, 3, 1, 4〉;

(S0 → S1)→ (¬S1 → ¬S0) = 〈2〉(S0 → S1)(¬S1 → ¬S0)

= 〈2〉〈2〉S0 S1 〈2〉¬S1 ¬S0

= 〈2, 2, 3, 4, 2〉〈1〉S1 〈1〉S0

= 〈2, 2, 3, 4, 2, 1, 4, 1, 3〉.

E1.2 Prove that there is a sentential formula of each positive integer length.

If m is a positive integer, then

〈

m−1 times︷︸︸︷

1, 1, . . . , 1, S0〉

is a formula of length m, it ism−1 times︷︸︸︷¬¬ · · · ¬ S0.

E1.3 Prove that m is the length of a sentential formula not involving ¬ iff m is odd.

Proof. ⇒: We prove by induction on ϕ that if ϕ is a sentential formula not involving¬, then the length of ϕ is odd. This is true of sentential variables, which have length 1.Suppose that it is true of ϕ and ψ, which have length 2m+1 and 2n+1 respectively. Thenϕ→ ψ, which is 〈1〉ϕψ, has length 1 + 2m+ 1 + 2n+ 1 = 2(m+ n+ 1) + 1, which isagain odd. This finishes the inductive proof.⇐. We construct formulas without ¬ with length any odd integer by induction. 〈S0〉

is a formula of length 1. If ϕ has been constructed of length 2m+ 1, then S0 → ϕ, whichis 〈1, S0〉

ϕ, has length 2m+ 3. This finishes the inductive construction.

E1.4 Prove that a truth table for a sentential formula involving n basic formulas has 2n

rows.

We prove this by induction on n. For n = 1, there are two rows. Assume that for n basicformulas there are 2n rows. Given n + 1 basic formulas, let ϕ be one of them. For theothers, by the inductive hypothesis there are 2n rows. For each such row there are twopossibilities, 0 or 1, for ϕ. So for the n+ 1 basic formulas there are 2n · 2 = 2n+1 rows.

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E1.5 Use the truth table method to show that the formula

(ϕ→ ψ)↔ (¬ϕ ∨ ψ)

is a tautology.

ϕ ψ ϕ→ ψ ¬ϕ ¬ϕ ∨ ψ (ϕ→ ψ)↔ (¬ϕ ∨ ψ)

1 1 1 0 1 1

1 0 0 0 0 1

0 1 1 1 1 1

0 0 1 1 1 1


[ϕ ∨ (ψ ∧ χ)]↔ [(ϕ ∨ ψ) ∧ (ϕ ∨ χ)]

is a tautology.

Let θ be the indicated formula.

ϕ ψ χ ϕ ∨ ψ ϕ ∨ χ (ϕ ∨ ψ) ∧ (ϕ ∨ χ) ψ ∧ χ ϕ ∨ (ψ ∧ χ) θ

1 1 1 1 1 1 1 1 1

1 1 0 1 1 1 0 1 1

1 0 1 1 1 1 0 1 1

1 0 0 1 1 1 0 1 1

0 1 1 1 1 1 1 1 1

0 1 0 1 0 0 0 0 1

0 0 1 0 1 0 0 0 1

0 0 0 0 0 0 0 0 1


(ϕ→ ψ)→ (ϕ→ ¬ψ)

is not a tautology. It is not necessary to work out the full truth table.

ϕ ψ ϕ→ ψ ¬ψ ϕ→ ¬ψ (ϕ→ ψ)→ (ϕ→ ¬ψ)

1 1 1 0 0 0

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E1.8 Determine whether or not the following is a tautology:

S0 → (S1 → (S2 → (S3 → S1))).

Suppose that f is an assignment making the indicated formula false; we work towards acontradiction. Thus

(1) S0[f ] = 1 and

(2) (S1 → (S2 → (S3 → S1)))[f ] = 0.

From (2) we get

(3) S1[f ] = 1 and

(4) (S2 → (S3 → S1))[f ] = 0.

From (4) we get

(5) S2[f ] = 1 and

(6) (S3 → S1)[f ] = 0.

From (6) we get S1[f ] = 0, contradicting (3).

E1.9 Determine whether or not the following is a tautology; an informal method is betterthan a truth table:

([(ϕ→ ψ)→ (¬χ→ ¬θ)]→ χ → τ)→ [(τ → ϕ)→ (θ → ϕ)].

Suppose that f is an assignment which makes the given formula false; we want to get acontradiction. Thus we have

(1) ([(ϕ→ ψ)→ (¬χ→ ¬θ)]→ χ → τ)[f ] = 1 and

(2) [(τ → ϕ)→ (θ → ϕ)][f ] = 0.

By (2) we have

(3) (τ → ϕ)[f ] = 1 and

(4) (θ→ ϕ)[f ] = 0.

By (4) we have

(5) θ[f ] = 1 and

(6) ϕ[f ] = 0.

By (3) and (6) we get

(7) τ [f ] = 0.

By (1) and (7) we get

(8) [(ϕ→ ψ)→ (¬χ→ ¬θ)]→ χ[f ] = 0.

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It follows that

(9) [(ϕ→ ψ)→ (¬χ→ ¬θ)][f ] = 1 and

(10) χ[f ] = 0.

Now by (6) we have

(11) (ϕ→ ψ)[f ] = 1,

and hence by (9),

(12) (¬χ→ ¬θ)[f ] = 1.

By (5) we have

(13) (¬θ)[f ] = 0,

and hence by (12),

(14) (¬χ)[f ] = 0.

This contradicts (10).

E1.10 Determine whether the following statements are logically consistent. If the contractis valid, then Horatio is liable. If Horation is liable, he will go bankrupt. Either Horatiowill go bankrupt or the bank will lend him money. However, the bank will definitely notlend him money.

Let S0 correspond to “the contract is valid”, S1 to “Horatio is liable”, S2 to “Horatio willgo bankrupt”, and S3 to “the bank will lend him money”. Then we want to see if there isan assignment of values which makes the following sentence true:

(S0 → S1) ∧ (S1 → S2) ∧ (S2 ∨ S3) ∧ ¬S3.

We can let f(0) = f(1) = f(2) = 1 and f(3) = 0, and this gives the sentence the value 1.

E1.11 Write out an actual proof for ψ ⊢ ¬ψ → ϕ. This can be done by following theproof of Lemma 1.9, expanding it using the proof of the deduction theorem.

Following the proof of Lemma 1.9, the following is a ψ,¬ψ-proof:

(a) ¬ψ(b) ¬ψ → (¬ϕ→ ¬ψ) (1)(c) ¬ϕ→ ¬ψ (a), (b), MP(d) (¬ϕ→ ¬ψ)→ (ψ → ϕ) (3)(e) ψ → ϕ (c), (d), MP(f) ψ(g) ϕ (e), (f), MP

Now applying the proof of the deduction theorem, the following is a ψ-proof:

(a) [¬ψ → [(¬ψ → ¬ψ)→ ¬ψ]]→ [[¬ψ → (¬ψ → ¬ψ)]→ (¬ψ → ¬ψ)] (2)

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(b) ¬ψ → [(¬ψ → ¬ψ)→ ¬ψ] (1)(c) [¬ψ → (¬ψ → ¬ψ)]→ (¬ψ → ¬ψ) (a), (b), MP(d) ¬ψ → (¬ψ → ¬ψ) (1)(e) ¬ψ → ¬ψ (c), (d), MP(f) [¬ψ → (¬ϕ→ ¬ψ)]→ [¬ψ → [¬ψ → (¬ϕ→ ¬ψ)]] (1)(g) ¬ψ → (¬ϕ→ ¬ψ) (1)(h) ¬ψ → [¬ψ → (¬ϕ→ ¬ψ)]] (f), (g), MP(i) [(¬ϕ→ ¬ψ)→ (ψ → ϕ)]→ [¬ψ → [(¬ϕ→ ¬ψ)→ (ψ → ϕ)]] (1)(j) (¬ϕ→ ¬ψ)→ (ψ → ϕ) (3)(k) ¬ψ → [(¬ϕ→ ¬ψ)→ (ψ → ϕ)] (i), (j), MP(l) [¬ψ → [(¬ϕ→ ¬ψ)→ (ψ → ϕ)]]→ [[¬ψ → (¬ϕ→ ¬ψ)]

→ [¬ψ → (ψ → ϕ)]] (2)(m) [¬ψ → (¬ϕ→ ¬ψ)]→ [¬ψ → (ψ → ϕ)] (k), (l), MP(n) ¬ψ → (ψ → ϕ) (g), (m), MP(o) ψ → (¬ψ → ψ) (1)(p) ψ(q) ¬ψ → ψ (o), (p), MP(r) [¬ψ → (ψ → ϕ)]→ [(¬ψ → ψ)→ (¬ψ → ϕ)] (2)(s) (¬ψ → ψ)→ (¬ψ → ϕ) (n), (r), MP(t) ¬ψ → ϕ (q), (s), MP

Solutions for exercises in Chapter 2

E2.1 Give an exact definition of a language for the structure (ω,<).

The quadruple (11, ∅, ∅, rnk), where rnk is the function with domain 11 such thatrnk(11) = 2.

E2.2 Give an exact definition of a language for the set A (no individual constants, func-tion symbols, or relation symbols).

The quadruple (∅, ∅, ∅, ∅). Note that the last ∅ is the empty function.

E2.3 Describe a term construction sequence which shows that + • v0v0v1 is a term in thelanguage for (R,+, ·, 0, 1, <).

〈v0, •v0v0, v1,+ • v0v0v1〉.

E2.4 In any first-order language, show that the sequence 〈v0, v0〉 is not a term. Hint: useProposition 2.2.

Suppose that 〈v0, v0〉 is a term. This contradicts Proposition 2.2(ii).

E2.5 In the language for (ω, S, 0,+, ·), show that the sequence 〈+, v0, v1, v2〉 is not a term.Here S(i) = i+ 1 for any i ∈ ω. Hint: use Proposition 2.2.

Suppose it is a term. By Proposition 2.2(ii)(c), there are terms σ, τ such that 〈+, v0, v1, v2〉is 〈+〉στ . Thus 〈v0, v1, v2〉 = στ . So the term v0 is an initial segment of the termσ. By Proposition 2.2(iii) it follows that v0 = σ. Hence 〈v1, v2〉 = τ . This contradictsProposition 2.2(ii).

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E2.6 Prove Proposition 2.5.

We show by complete induction on i that ϕi ∈ Γ for all i < m. So, suppose that i < mand ϕj ∈ Γ for all j < i. By the definition of formula construction sequence, we have thefollowing cases.

Case 1. ϕi is an atomic formula. Then ϕi ∈ Γ by (i).Case 2. There is a j < i such that ϕi is ¬ϕj . By the inductive hypothesis, ϕj ∈ Γ.

Hence by (ii), ϕi ∈ Γ.Case 2. There are j, k < i such that ϕi is ϕj → ϕk. By the inductive hypothesis,

ϕj ∈ Γ and ϕk ∈ Γ. Hence by (iii), ϕi ∈ Γ.Case 4. There exist j < i and k ∈ ω such that ϕi is ∀vkϕj . By the inductive

hypothesis, ϕj ∈ Γ. Hence by (iv), ϕi ∈ Γ.This completes the inductive proof.

E2.7 Show how the structure (ω, S, 0,+, ·) can be put in the general framework of struc-tures.

(ω, S, 0,+, ·) can be considered to be the structure (ω,Rel′, F cn′, Cn′) where Rel′ = ∅,Cn′ is the function with domain 8 such that Cn′(8) = 0, and Fcn′ is the function withdomain 6, 7, 9 such that Fcn′(6) = S, Fcn′(7) = +, and Fcn′(9) = ·.

E2.8 Prove that in the language for the structure (ω,+), a term has length m iff m isodd.

First we show by induction on terms that every term has odd length. This is true forvariables. Suppose that it is true for terms σ and τ . Then also σ + τ has odd length.Hence every term has odd length.

Second we prove by induction on m that for all m, there is a term of length 2m+ 1.A variable has length 1, so our assertion holds for m = 0. Assume that there is a term σof length 2m+ 1. Then σ + v0 has length 2m+ 3. This finishes the inductive proof.

E2.9 Give a formula ϕ in the language for (Q,+, ·) such that for any a : ω → Q,(Q,+, ·) |= ϕ[a] iff a0 = 1.

Let ϕ be the formula ∀v1[v0 · v1 = v1].

E2.10 Give a formula ϕ which holds in a structure, under any assignment, iff the structurehas at least 3 elements.

∃v0∃v1∃v2(¬(v0 = v1) ∧ ¬(v0 = v2) ∧ ¬(v1 = v2)).

E2.11 Give a formula ϕ which holds in a structure, under any assignment, iff the structurehas exactly 4 elements.

∃v0∃v1∃v2∃v3(¬(v0 = v1) ∧ ¬(v0 = v2) ∧ ¬(v0 = v3) ∧ ¬(v1 = v2) ∧ ¬(v1 = v3) ∧ ¬(v2 = v3)

∧ ∀v4(v0 = v4 ∨ v1 = v4 ∨ v2 = v4 ∨ v3 = v4)).

E2.12 Write a formula ϕ in the language for (ω,<) such that for any assignment a,(ω,<) |= ϕ[a] iff a0 < a1 and there are exactly two integers between a0 and a1.

v0 < v1 ∧ ∃v2∃v3[v0 < v2 ∧ v2 < v3 ∧ v3 < v1

∧ ∀v4[v0 < v4 ∧ v4 < v1 → v4 = v2 ∨ v4 = v3]].

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E2.13 Prove that the formula

v0 = v1 → (Rv0v2 → Rv1v2)

is universally valid, where R is a binary relation symbol.

Let A be a structure and a : ω → A an assignment. Suppose that A |= (v0 = v1)[a]. Then

a0 = a1. Also suppose that A |= Rv0v2[a]. Then (a0, a2) ∈ RA. Hence (a1, a2) ∈ RA.Hence A |= Rv1v2[a], as desired.

E2.14 Give an example showing that the formula

v0 = v1 → ∀v0(v0 = v1)

is not universally valid.

Consider the structure Adef= (ω,<), and let a : ω → ω be defined by a(i) = 0 for all i ∈ ω.

Then A |= (v0 = v1)[a]. Now A 6|= (v0 = v1)[a01] since 1 6= 0, so A 6|= ∀v0(v0 = v1)[a].

Therefore A 6|= (v0 = v1 → ∀v0(v0 = v1))[a].

E2.15 Prove that ∃v0∀v1ϕ→ ∀v1∃v0ϕ is universally valid.

Assume that a : ω → A and A |= ∃v0∀v1ϕ[a]. Choose u ∈ A so that A |= ∀v1ϕ[a0u]. In

order to show that A |= ∀v1∃v0ϕ[a], let w ∈ A be given. Then A |= ϕ01uw. It follows that

A |= ∃v0ϕ[u1w]. Hence A |= ∀v1∃v0ϕ[a], as desired.

Solutions to exercises in Chapter 3

E3.1 Do the case Rσ0 . . . σm−1 for some m-ary relation symbol and terms σ0, . . . , σm−1

in the proof of Theorem 3.1, (L3).

We are assuming that vi does not occur in Rσ0 . . . σm−1; hence it does not occur in anyterm σi.

A |= (Rσ0 . . . σm−1)[a] iff 〈σA0 (a), . . . , σRm−1(a)〉 ∈ RA

iff 〈σA0 (b), . . . , σRm−1(b)〉 ∈ RA

(by Proposition 2.4)

iff A |= (Rσ0 . . . σm−1)[b].

E3.2 Prove that (L6) is universally valid, in the proof of Theorem 3.1.

Assume that A |= (σ = τ)[a] and A |= (ρ = σ)[a]. Then σA(a) = τA(a) and ρA(a) = σA(a),

so ρA(a) = τA(a), hence A |= (ρ = τ)[a].

E3.3 Prove that (L8) is universally valid, in the proof of Theorem 3.1.

Assume that A |= (σ = τ)[a]. Then σA(a) = τA(a). Assume that

A |= (Rξ0 . . . ξi−1σξi+1 . . . ξm−1)[a]; hence

〈ξA0 (a), . . . , ξAi−1(a), σA(a), ξAi+1(a), . . . , ξ

Am−1(a)〉 ∈ RA; hence

〈ξA0 (a), . . . , ξAi−1(a), τA(a), ξAi+1(a), . . . , ξ

Am−1(a)〉 ∈ RA; hence

A |= (Rξ0 . . . ξi−1τξi+1 . . . ξm−1)[a];

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hence (L8) is universally valid.

E3.4 Finish the proof of Proposition 3.9.

If i = 0 then ϕ itself is the desired segment, unique by Proposition 2.6(iii). If i > 0 thenactually i > 1 so that ϕi is within ψ, and the inductive hypothesis applies.

E3.5 Finish the proof of Proposition 3.11.

Suppose inductively that ϕ is ¬ψ. Thus ϕ is 〈1〉ψ. It follows that i > 0, so that ϕiappears in ψ; then the inductive hypothesis applies.

Suppose inductively that ϕ is ψ → χ. Thus ϕ is 〈2〉ψχ. It follows that i > 0, sothat ϕi appears in ψ or χ; then the inductive hypothesis applies.

Finally, suppose that ϕ is ∀vsψ with ψ a formula and s ∈ ω. Thus ϕ is 〈4, 5(s+1)〉ψ.Hence i > 0. If i = 1, then 〈5(s+1)〉 is the desired segment, unique by Proposition 2.6(iii).Suppose that i > 1. So ϕi is an entry in ψ and hence by the inductive assumption, thereis a segment 〈ϕi, ϕi+1, . . . ϕm〉 which is a term; this is also a segment of ϕ, and it is uniqueby Proposition 2.6(iii).

E3.6 Indicate which occurrences of the variables are bound and which ones free for thefollowing formulas.

∃v0(v0 < v1) ∧ ∀v1(v0 = v1).v4 + v2 = v0 ∧ ∀v3(v0 = v1).∃v2(v4 + v2 = v0).

First formula: the first and second occurrences of v0 are bound, and the third one is free.The first occurrence of v1 is free, and the other two are bound.

Second formula: the occurrence of v3 is bound. All other occurrences of variables are free.

Third formula: the two occurrences of v2 are bound. The other occurrences of variablesare free.


Induction on ϕ. Suppose that ϕ is ρ = ξ. Then by Proposition 3.13, σ occurs in ρ orξ. Suppose that it occurs in ρ. Let ρ′ be obtained from ρ by replacing that occurrenceof σ by τ . Then ρ′ is a term by Proposition 3.14. Since ψ is ρ′ = ξ, ψ is a formula.The case in which σ occurs in ξ is similar. Now suppose that ϕ is Rη0 . . . ηm−1 withR an m-ary relation symbol and η0, . . . , ηm−1 are terms. Then the occurrence of σ iswithin some ηi. Let η′i be obtained from ηi by replacing that occurrence by τ . Now ψ isRη0 . . . ηi−1η

′i . . . ηm−1, so ψ is a formula.

Now suppose that the result holds for ϕ′, and ϕ is ¬ϕ′. Then σ occurs in ϕ′, so ifψ′ is obtained from ϕ′ by replacing the occurrence of σ by τ , then ψ′ is a formula by theinductive assumption. Since ψ is ¬ψ′ also ψ is a formula.

Next, suppose that the result holds for ϕ′ and ϕ′′, and ϕ is ϕ′ → ϕ′′. Then theoccurrence of σ is within ϕ′ or is within ϕ′′. If it is within ϕ′, let ψ′ be obtained from ϕ′

by replacing that occurrence of σ by τ . Then ψ′ is a formula by the inductive hypothesis.Since ψ is ψ′ → ϕ′′, also ψ is a formula. If the occurrence is within ϕ′′, let ψ′′ be obtained

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from ϕ′′ by replacing that occurrence of σ by τ . Then ψ′′ is a formula by the inductivehypothesis. Since ψ is ϕ′ → ψ′′, also ψ is a formula.

Finally, suppose that the result holds for ϕ′, and ϕ is ∀vkϕ′. If i = 1, then σ is vk,

and by hypothesis τ is some variable vl. Then ψ is ∀vlϕ′, which is a formula. If i > 1, then

σ occurs in ϕ′, so if ψ′ is obtained from ϕ′ by replacing the occurrence of σ by τ , then ψ′

is a formula by the inductive assumption. Since ψ is ∀vkψ′ also ψ is a formula.

E3.8 Indicate all free and bound occurrences of terms in the formula v0 = v1 + v1 →∃v2(v0 + v2 = v1).

v0 is free in both of its occcurrences.v1 is free in all three of its occurrences.v2 is bound in both of its occurrences.v1 + v1 is free in its occurrence.v0 + v2 is bound in its occurrence.

E3.9 Prove Proposition 3.17

Induction on ϕ. If ϕ is atomic, then ψ is equal to ϕ, and θ is equal to χ and hence isa formula. Suppose the result is true for ϕ′ and ϕ is ¬ϕ′. If ψ = ϕ, again the desiredconclusion is clear. Otherwise the occurrence of ψ is within the subformula ϕ′. If θ′ isobtained from ϕ′ by replacing that occurrence by χ, then θ′ is a formula by the inductivehypothesis. Since θ is ¬θ′, also θ is a formula.

Now suppose the result is true for ϕ′ and ϕ′′, and ϕ is ϕ′ → ϕ′′. If ψ = ϕ, again thedesired conclusion is clear. Otherwise the occurrence of ψ is within the subformula ϕ′ oris within the subformula ϕ′′. If it is within ϕ′ and θ′ is obtained from ϕ′ by replacing thatoccurrence by χ, then θ′ is a formula by the inductive hypothesis. Since θ is θ′ → ϕ′′, alsoθ is a formula. If it is within ϕ′′ and θ′′ is obtained from ϕ′′ by replacing that occurrenceby χ, then θ′′ is a formula by the inductive hypothesis. Since θ is ϕ′ → θ′′, also θ is aformula.

Finally, suppose the result is true for ϕ′ and ϕ is ∀viϕ′. If ψ = ϕ, again the desired

conclusion is clear. Otherwise the occurrence of ψ is within the subformula ϕ′. If θ′ isobtained from ϕ′ by replacing that occurrence by χ, then θ′ is a formula by the inductivehypothesis. Since θ is ∀viθ

′, also θ is a formula.

E3.10 Show that the condition in Lemma 3.15 that the resulting occurrence of τ is freeis necessary. Hint: use Theorem 3.2; describe a specific formula of the type in Proposition3.15, but with τ not free, such that the formula is not universally valid.

Consider the language for (ω, S), and the formula

v0 = v1 → (∃v1(Sv0 = v1)↔ ∃v1(Sv1 = v1)).

Taking an assignment a : ω → ω with a0 = a1 makes this sentence false; hence it is notprovable, by Theorem 3.2.

E3.11 Do the case of implication in the proof of Lemma 3.15.

Suppose inductively that ϕ is χ→ θ.

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Case 1. The occurrence of σ in ϕ is within χ. Let χ′ be obtained from χ by replacingthat occurrence by τ , such that that occurrence is free in ψ, hence free in χ′. By theinductive hypothesis, ⊢ σ = τ → (χ ↔ χ′). Since ψ is χ′ → θ, a tautology gives thedesired result.

Case 2. The occurrence of σ in ϕ is within θ. Let θ′ be obtained from θ by replacingthat occurrence by τ , such that that occurrence is free in ψ, hence free in θ′. By theinductive hypothesis, ⊢ σ = τ → (θ ↔ θ′). Since ψ is χ→ θ′, a tautology gives the desiredresult.

E3.12 Prove that the hypothesis of Theorem 3.25 is necessary.

Consider the formula∀v0∃v1(v0 < v1)→ ∃v1(v1 < v1).

This formula is not universally valid; it fails to hold in (ω,<), for example.


Proof. By definition, ∃vi¬ϕ is ¬∀vi¬¬ϕ. Now ⊢ ϕ ↔ ¬¬ϕ by a tautology. Henceusing generalization and (L2) we get ⊢ ∀viϕ ↔ ∀vi¬¬ϕ. Hence another tautology yields⊢ ¬∀viϕ↔ ¬∀vi¬¬ϕ, i.e., ⊢ ¬∀viϕ↔ ∃vi¬ϕ.


Proof. ¬∃viϕ is the formula ¬¬∀vi¬ϕ, so a simple tautology gives the result.


Proof. By Theorem 3.25 we have ⊢ ∀vi¬ϕ → Subfvi

σ (¬ϕ). Since clearly Subfvi

σ (¬ϕ)is the same as ¬Subfvi

σ ϕ, a tautology gives ⊢ Subfvi

σ ϕ→ ∃viϕ.


By Corollary 3.26 and Corollary 3.32.


Proof. ⊢ ¬ϕ↔ ∀vi¬ϕ. Now use a tautology.


Assume ⊢ ϕ↔ ψ. By a tautology, ⊢ ϕ→ ψ. Hence by generalization and (L2), ⊢ ∀viϕ→∀viψ. Similarly, ⊢ ∀viψ → ∀viϕ. The exercise follows by a tautology.

E3.19 Prove Proposition 3.42

Assume ⊢ ϕ ↔ ψ. By a tautology, ⊢ ¬ϕ ↔ ¬ψ. Hence by exercise E3.18, ⊢ ∀vi¬ϕ ↔∀vi¬ψ. Now a tautology gives the desired result.

E3.20 Prove that

⊢ ∀v0∀v1(v0 = v1)→ ∀v0(v0 = v1 ∨ v0 = v2).

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⊢ ∀v0∀v1(v0 = v1)→ v0 = v1; Cor. 3.26 twice, taut. (1)

⊢ ∀v1(v0 = v1)→ v0 = v2; Thm. 3.25 (2)

⊢ ∀v0∀v1(v0 = v1)→ v0 = v2; (2), Cor. 3.26, taut. (3)

⊢ ∀v0∀v1(v0 = v1)→ v0 = v1 ∨ v0 = v2; (1), (3), taut. (4)

⊢ ∀v0∀v0∀v1(v0 = v1)→ ∀v0(v0 = v1 ∨ v0 = v2); (4), (L2), taut. (5)

⊢ ∀v0∀v1(v0 = v1)→ ∀v0(v0 = v1 ∨ v0 = v2). (5), Prop. 3.27, taut.

E3.21 Prove that

⊢ ∃v0(¬v0 = v1 ∧ ¬v0 = v2)→ ∃v0∃v1(¬v0 = v1).

⊢ ¬∀v0(v0 = v1 ∨ v0 = v2)→ ¬∀v0∀v1(v0 = v1); E3.20, taut. (1)

⊢ ¬∀v0(v0 = v1 ∨ v0 = v2)↔ ∃v0¬(v0 = v1 ∨ v0 = v2); Prop. 3.29 (2)

⊢ ¬(v0 = v1 ∨ v0 = v2)↔ (¬(v0 = v1) ∧ ¬(v0 = v2)); taut. (3)

⊢ ∃v0¬(v0 = v1 ∨ v0 = v2)↔ ∃v0(¬(v0 = v1) ∧ ¬(v0 = v2)); (3), Prop. 3.42 (4)

⊢ ¬∀v0(v0 = v1 ∨ v0 = v2)↔ ∃v0(¬(v0 = v1) ∧ ¬(v0 = v2)); (2), (4), taut. (5)

⊢ ¬∀v1(v0 = v1)↔ ∃v1¬(v0 = v1); Prop. 3.29 (6)

⊢ ∃v0¬∀v1(v0 = v1)↔ ∃v0∃v1¬(v0 = v1); (6), Prop. 3.42 (7)

⊢ ¬∀v0∀v1(v0 = v1)↔ ∃v0¬∀v1(v0 = v1); Prop. 3.29 (8)

⊢ ¬∀v0∀v1(v0 = v1)↔ ∃v0∃v1¬(v0 = v1) (7), (8), taut. (9)

⊢ ∃v0(¬v0 = v1 ∧ ¬v0 = v2)→ ∃v0∃v1(¬v0 = v1). (1), (5), (9), taut.


E4.1 Suppose that Γ ⊢ ϕ → ψ, Γ ⊢ ϕ → ¬ψ, and Γ ⊢ ¬ϕ → ϕ. Prove that Γ isinconsistent.

The formula (¬ϕ → ϕ) → ϕ is a tautology. Hence by Lemma 3.3, Γ ⊢ (¬ϕ → ϕ) → ϕ.Since also Γ ⊢ ¬ϕ→ ϕ, it follows that Γ ⊢ ϕ. Hence Γ ⊢ ψ and Γ ⊢ ¬ψ. Hence by Lemma4.1, Γ is inconsistent.

E4.2 Let L be a language with just one non-logical constant, a binary relation symbol R.Let Γ consist of all sentences of the form ∃v1∀v0[Rv0v1 ↔ ϕ] with ϕ a formula with onlyv0 free. Show that Γ is inconsistent. Hint: take ϕ to be ¬Rv0v0.

By Theorem 3.25 we have

(1) Γ ⊢ ∀v0[Rv0v1 ↔ ¬Rv0v0]→ [Rv1v1 ↔ ¬Rv1v1].

Now [Rv1v1 ↔ ¬Rv1v1]→ ¬(v0 = v0) is a tautology, so from (1) we obtain

Γ ⊢ ∀v0[Rv0v1 ↔ ¬Rv0v0]→ ¬(v0 = v0);

11

then generalization gives

Γ ⊢ ∀v1[∀v0[Rv0v1 ↔ ¬Rv0v0]→ ¬(v0 = v0)].

Then by Proposition 3.37 we get

Γ ⊢ ∃v1∀v0[Rv0v1 ↔ ¬Rv0v0]→ ¬(v0 = v0).

But the hypothesis here is a member of Γ, so we get Γ ⊢ ¬(v0 = v0). Hence by Lemma4.1, Γ is inconsistent.

Alternate proof (due to a couple of students). Suppose that Γ is consistent. Bythe completeness theorem let A be a model of Γ. Taking ϕ to be ¬Rv0v0, we get A |=∃v1∀v0[Rv0v1 ↔ ¬Rv0v0]. Let a : ω → A be any assignment. Then by Proposition 2.8(iv)there is a b ∈ A such that A |= ∀v0[Rv0v1 ↔ ¬Rv0v0][a

1b]. By the definition of satisfaction

of ∀, it follows that for any c ∈ A we have A |= [Rv0v1 ↔ ¬Rv0v0][a0 1c b ]. Hence (c, b) ∈ RA

iff (c, b) /∈ RA, contradiction.

E4.3 Show that the first-order deduction theorem fails if the condition that ϕ is a sentenceis omitted. Hint: take Γ = ∅, let ϕ be the formula v0 = v1, and let ψ be the formula v0 = v2.

v0 = v1 ⊢ v0 = v1

v0 = v1 ⊢ ∀v1(v0 = v1)

v0 = v1 ⊢ ∀v1(v0 = v1)→ v0 = v2 by Theorem 3.25

v0 = v1 ⊢ v0 = v2.

On the other hand, let A be the structure with universe ω and define a = 〈0, 0, 1, 1, . . .〉.Clearly A 6|= [v0 = v1 → v0 = v2][a]. Hence 6⊢ v0 = v1 → v0 = v2 by Theorem 3.2.

E4.4 In the language for Adef= (ω, S, 0,+, ·), let τ be the term v0 + v1 · v2 and ν the

term v0 + v2. Let a be the sequence 〈0, 1, 2, . . .〉. Let ρ be obtained from τ by replacing theoccurrence of v1 by ν.

(a) Describe ρ as a sequence of integers.

(b) What is ρA(a)?

(c) What is νA(a)?(d) Describe the sequence a1

νA(a)as a sequence of integers.

(e) Verify that ρA(a) = τA(a1

νA(a)) (cf. Lemma 4.4.)

(a) ρ is v0 + (v0 + v2) · v2; as a sequence of integers it is 〈7, 5, 9, 7, 5, 15, 15〉.

(b) ρA(a) = 0 + (0 + 2) · 2 = 4.

(c) νA(a) = 0 + 2 = 2.(d) a1

νA(a)= 〈0, 2, 2, 3, . . .〉.

(e) ρA(a) = 4, as above; τA(a1

νA(a)) = 0 + 2 · 2 = 4.

12

E4.5 In the language for Adef= (ω, S, 0,+, ·), let ϕ be the formula ∀v0(v0 · v1 = v1), let ν

be the formula v1 + v1, and let a = 〈1, 0, 1, 0, . . .〉.(a) Describe Subfv1ν ϕ as a sequence of integers

(b) What is νA(a)?(c) Describe a1

νA(a)as a sequence of integers.

(d) Determine whether A |= Subfv1ν ϕ[a] or not.(e) Determine whether A |= ϕ[a1

νA(a)] or not.

(a) Subfv1ν ϕ is ∀v0(v0 · (v1 + v1) = v1 + v1; as a sequence of integers it is

〈4, 5, 3, 9, 5, 7, 10, 10, 7, 10, 10〉.

(b) νA(a) = (v1 + v1)A(〈1, 0, 1, 0, . . .〉) = 0 + 0 = 0.

(c) a1

νA(a)= 〈1, 0, 1, 0, . . .〉.

(d) A |= Subfv1ν ϕ[a] iff A |= [∀v0(v0 · (v1 +v1) = v1 +v1][〈1, 0, 1, 0, . . .〉] iff for all a ∈ ω,a · (0 + 0) = 0 + 0; this is true.

(e) A |= ϕ[a1

νA(a)] iff A |= [∀v0(v0 · v1 = v1][〈1, 0, 1, 0, . . .〉] iff for all a ∈ ω, a · 0 = 0;

this is true.

E4.6 Show that the condition in Lemma 4.6 that

no free occurrence of vi in ϕ is within a subformula of the form ∀vkµ with vk a variableoccurring in ν

is necessary for the conclusion of the lemma.

In the language for A = (ω, S, 0,+, ·), let ϕ be the formula ∃v1[Sv1 = v0], ν = v1,and a = 〈1, 1, . . .〉. Note that the condition on v0 fails. Now Subfv0v1ϕ is the formula

∃v1[Sv1 = v1], and there is no a ∈ ω such that Sa = a, and hence A 6|= Subfv0v1ϕ[a]. Also,

νA(a) = vA1 (a) = a1 = 1, and hence a0

νA(a)= 〈1, 1, . . .〉. Since S0 = 1, it follows that

A |= ϕ[a0

νA(a)].

E4.7 Let A be an L -structure, with L arbitrary. Define Γ = ϕ : ϕ is a sentence and

A |= ϕ[a] for some a : ω → A. Prove that Γ is complete and consistent.

Note by Lemma 4.4 that A |= ϕ[a] for some a : ω → A iff A |= ϕ[a] for every a : ω → A.Let ϕ be any sentence. Take any a : ω → A. If A |= ϕ[a], then ϕ ∈ Γ and hence Γ ⊢ ϕ.Suppose that A 6|= ϕ[a]. Then A |= ¬ϕ[a], hence ¬ϕ ∈ Γ, hence Γ ⊢ ¬ϕ.

This shows that Γ is complete. Suppose that Γ is not consistent. Then Γ ⊢ ¬(v0 = v0)by Lemma 4.1. Then Γ |= ¬(v0 = v0) by Theorem 3.2. Since A is a model of Γ, it is alsoa model of ¬(v0 = v0), contradiction.

E4.8 Call a set Γ strongly complete iff for every formula ϕ, Γ ⊢ ϕ or Γ ⊢ ¬ϕ. Prove thatif Γ is strongly complete, then Γ ⊢ ∀v0∀v1(v0 = v1).

Assume that Γ is strongly complete. Then Γ ⊢ v0 = v1 or Γ ⊢ ¬(v0 = v1). If Γ ⊢ v0 = v1,then by generalization, Γ ⊢ ∀v0∀v1(v0 = v1). Suppose that Γ ⊢ ¬(v0 = v1). Then by

13

generalization, Γ ⊢ ∀v0¬(v0 = v1). By Theorem 3.25, Γ ⊢ ∀v0¬(v0 = v1) → ¬(v1 = v1).Hence Γ ⊢ ¬(v1 = v1). But also Γ ⊢ v1 = v1 by Proposition 3.4, so Γ is inconsistent byLemma 4.1, and hence again Γ ⊢ ∀v0∀v1(v0 = v1).

E4.9 Prove that if Γ is rich, then for every term σ with no variables occurring in σ thereis an individual constant c such that Γ ⊢ σ = c.

By richness we have Γ ⊢ ∃v0(v0 = σ)→ c = σ for some individual constant c. Then using(L4) it follows that Γ ⊢ c = σ.

E4.10 Prove that if Γ is rich, then for every sentence ϕ there is a sentence ψ with noquantifiers in it such that Γ ⊢ ϕ↔ ψ.

We proceed by induction on the number m of symbols ¬, →, ∀ in ϕ. (More exactly, bythe number of the integers 1,2,4 that occur in the sequence ϕ.) If m = 0, then ϕ is atomicand we can take ψ = ϕ. Assume the result for m and suppose that ϕ has m + 1 integers1,2,4 in it. Then there are three possibilities. First, ϕ = ¬ϕ′. Let ψ′ be a quantifier-freesentence such that Γ ⊢ ϕ′ ↔ ψ′. Then Γ ⊢ ϕ ↔ ¬ψ′. Second, ϕ = (ϕ′ → ϕ′′). Choosequantifier-free sentences ψ′ and ψ′′ such that Γ ⊢ ϕ′ ↔ ψ′ and Γ ⊢ ϕ′′ ↔ ψ′′. ThenΓ ⊢ ϕ ↔ (ψ′ → ψ′′). Third, ϕ = ∀viϕ

′. By richness, let c be an individual constant suchthat Γ ⊢ ∃vi¬ϕ

′ → Subfvi

c ¬ϕ′. Then by Theorem 3.31 we get

(1) Γ ⊢ ∃vi¬ϕ′ ↔ Subfvi

c ¬ϕ′.

Now Subfvi

c ϕ′ has only m integers 1,2,4 in it, so by the inductive hypothesis there is a

sentence ψ with no quantifiers in it such that Γ ⊢ Subfvi

c ϕ′ ↔ ψ and hence

(2) Γ ⊢ Subfvi

c ¬ϕ′ ↔ ¬ψ.

From (1) and (2) and a tautology we get Γ ⊢ ¬∃vi¬ϕ′ ↔ ψ. Then by Proposition 3.31,

Γ ⊢ ∀viϕ′ ↔ ψ, finishing the inductive proof.

E4.11 Describe sentences in a language for ordering which say that < is a linear orderingand there are infinitely many elements. Prove that the resulting set Γ of sentences is notcomplete.

Let Γ consist of the following sentences:

¬∃v0(v0 < v0);

∀v0∀v1∀v2[v0 < v1 ∧ v1 < v2 → v0 < v2];

∀v0∀v1[v0 < v1 ∨ v0 = v1 ∨ v1 < v0];∧

i<j<n

¬(vi = vj) for every positive integer n.

The following sentence ϕ holds in (Q, <) but not in (ω,<):

∀v0∀v1[v0 < v1 → ∃v2(v0 < v2 ∧ v2 < v1)].

Since ϕ does not hold in (ω,<), we have Γ 6⊢ ϕ, by Theorem 4.2. But since ϕ holds in(Q, <), we also have Γ 6⊢ ¬ϕ by Theorem 4.2. So Γ is not complete.

14

E4.12 Prove that if a sentence ϕ holds in every infinite model of a set Γ of sentences,then there is an m ∈ ω such that it holds in every model of Γ with at least m elements.

Suppose that ϕ holds in every infinite model of a set Γ of sentences, but for every m ∈ ωthere is a model M of Γ with at least m elements such that ϕ does not hold in M . Let ∆be the following set:

Γ ∪

∧

i<j<n

¬(vi = vj) : n a positive integer

∪ ¬ϕ.

Our hypothesis implies that every finite subset ∆′ of ∆ has a model; for if m is themaximum of all n such that the above big conjunction is in ∆′, then the hypothesis yieldsa model of ∆′. By the compactness theorem we get a model N of ∆. Thus N is an infinitemodel of Γ in which ϕ does not hold, contradiction.

E4.13 Let L be the language of ordering. Prove that there is no set Γ of sentences whosemodels are exactly the well-ordering structures.

Suppose there is such a set. Let us expand the language L to a new one L ′ by adding aninfinite sequence cm, m ∈ ω, of individual constants. Then consider the following set Θ ofsentences: all members of Γ, plus all sentences cm+1 < cm for m ∈ ω. Clearly every finitesubset of Θ has a model, so let A = (A,<, ai)i<ω be a model of Θ itself. (Here ai is the0-ary function, i.e., element of A, corresponding to ci.) Then a0 > a1 > · · ·; so ai : i ∈ ωis a nonempty subset of A with no least element, contradiction.

E4.14 Suppose that Γ is a set of sentences, and ϕ is a sentence. Prove that if Γ |= ϕ,then ∆ |= ϕ for some finite ∆ ⊆ Γ.

We prove the contrapositive: Suppose that for every finite subset ∆ of Γ, ∆ 6|= ϕ. Thusevery finite subset of Γ∪¬ϕ has a model, so Γ∪¬ϕ has a model, proving that Γ 6|= ϕ.

E4.15 Suppose that f is a function mapping a set M into a set N . Let R = (a, b) :a, b ∈M and f(a) = f(b). Prove that R is an equivalence relation on M .

If a ∈ M , then f(a) = f(a), so (a, a) ∈ R. Thus R is reflexive on M . Suppose that(a, b) ∈ R. Then f(a) = f(b), so f(b) = f(a) and hence (b, a) ∈ R. Thus R is symmetric.Suppose that (a, b) ∈ R and (b, c) ∈ R. Then f(a) = f(b) and f(b) = f(c), so f(a) = f(c)and hence (a, c) ∈ R.

E4.16 Suppose that R is an equivalence relation on a set M . Prove that there is a functionf mapping M into some set N such that R = (a, b) : a, b ∈M and f(a) = f(b).

Let N be the collection of all equivalence classes under R. For each a ∈M let f(a) = [a]R.Then (a, b) ∈ R iff a, b ∈M and [a]R = [b]R iff a, b ∈M and f(a) = f(b).

E4.17 Let Γ be a set of sentences in a first-order language, and let ∆ be the collection ofall sentences holding in every model of Γ. Prove that ∆ = ϕ : ϕ is a sentence and Γ ⊢ ϕ.

For ⊆, suppose that ϕ ∈ ∆. To prove that Γ ⊢ ϕ we use the compactness theorem, provingthat Γ |= ϕ. Let A be any model of Γ. Since ϕ ∈ ∆, it follows that A is a model of Γ, asdesired.

15

For ⊇, suppose that ϕ is a sentence and Γ ⊢ ϕ. Then by the easy direction of thecompleteness theorem, Γ |= ϕ. That is, every model of Γ is a model of ϕ. Hence ϕ ∈ ∆.

Solutions, Chapter 6

E6.1 Prove that if f : A → B and 〈Ci : i ∈ I〉 is a system of subsets of A, then

f[⋃

i∈I Ci]

=⋃

i∈I f [Ci].

x ∈ f

[⋃

i∈I

Ci

]

iff x ∈ rng

(

f ⋃

i∈I

Ci

)

iff ∃y ∈⋃

i∈I

Ci[f(y) = x]

iff ∃i ∈ I∃y ∈ Ci[f(y) = x]

iff ∃i ∈ I[x ∈ rng(f Ci)]

iff ∃i ∈ I[x ∈ f [Ci]]

iff x ∈⋃

i∈I

f [Ci].

E6.2 Prove that if f : A → B and C,D ⊆ A, then f [C ∩ D] ⊆ f [C] ∩ f [D]. Give anexample showing that equality does not hold in general.

Take any x ∈ f [C ∩ D]. Choose y ∈ C ∩ D such that x = f(y). Since y ∈ C, we havex ∈ f [C]. Similarly, x ∈ f [D]. So x ∈ f [C] ∩ f [D]. Since x is arbitrary, this shows thatf [C ∩D] ⊆ f [C] ∩ f [D].

For the required example, let dmn(f) = a, b with a 6= b and with f(a) = a = f(b).Let C = a and D = b. Then C ∩D = ∅, so f [C ∩D] = ∅, while f [C] = a = f [D]and hence f [C] ∩ f [D] = a 6= ∅. So f [C ∩D] 6= f [C] ∩ f [D].

E6.3 Given f : A → B and C,D ⊆ A, compare f [C\D] and f [C]\f [D]: prove theinclusions (if any) which hold, and give counterexamples for the inclusions that fail tohold.

We claim that f [C]\f [D] ⊆ f [C\D]. For, suppose that x ∈ f [C]\f [D]. Choose c ∈ Csuch that x = f(c). Since x /∈ f [D], we have c /∈ D. So c ∈ C\D and hence x ∈ f [C\D],proving the claim.

The other inclusion does not hold. For, take the same f, C,D as for exercise E6.2.Then C\D = a and so f [C\D] 6= ∅. But f [C] = a = f [D], so f [C]\f [D] = ∅.

E6.4 Prove that if f : A → B and 〈Ci : i ∈ I〉 is a system of subsets of B, then

f−1[⋃

i∈I Ci]

=⋃

i∈I f−1[Ci].

For any b ∈ B we have

b ∈ f−1

[⋃

i∈I

Ci

]

iff f(b) ∈⋃

i∈I

Ci

iff ∃i ∈ I[f(b) ∈ Ci]

16

iff ∃i ∈ I[b ∈ f−1[Ci]]

iff b ∈⋃

i∈I

f−1[Ci].

E6.5 Prove that if f : A → B and 〈Ci : i ∈ I〉 is a system of subsets of B, then

f−1[⋂

i∈I Ci]

=⋂

i∈I f−1[Ci].

For any a,

a ∈ f−1

[⋂

i∈I

Ci

]

iff f(a) ∈⋂

i∈I

Ci

iff ∀i ∈ I[f(a) ∈ Ci]

iff ∀i ∈ I[a ∈ f−1[Ci]]

iff a ∈⋂

i∈I

f−1[Ci].

E6.6 Prove that if f : A→ B and C,D ⊆ B, then f−1[C\D] = f−1[C]\f−1[D].

For any a,

a ∈ f−1[C\D] iff f(a) ∈ C\D

iff f(a) ∈ C and f(a) /∈ D

iff a ∈ f−1[C] and a /∈ f−1[D]

iff a ∈ f−1[C]\f−1[D].

E6.7 Prove that if f : A→ B and C ⊆ A, then

b ∈ B : f−1[b] ⊆ C = B\f [A\C].

First suppose that b is in the left side; but suppose also, aiming for a contradiction, thatb ∈ f [A\C]. Say b = f(a), with a ∈ A\C. Then a ∈ f−1[b], so a ∈ C, contradiction.

Second, suppose that b is in the right side. Take any a ∈ f−1[b]. Then f(a) = b,and it follows that a ∈ C, as desired.

E6.8 For any sets A,B define AB = (A\B) ∪ (B\A); this is called the symmetricdifference of A and B. Prove that if A,B,C are given sets, then A(BC) = (AB)C.

Let D = A ∪B ∪ C, A′ = D\A, B′ = D\B, and C′ = D\C. Then

AB = (A ∩B′) ∪ (B ∩ A′);

(AB)′ = ((A ∩B′) ∪ (B ∩A′))′

= (A ∩B′)′ ∩ (B ∩ A′)′

= (A′ ∪B) ∩ (B′ ∪ A)

= (A′ ∩B′) ∪ (A ∩B).

17

These equations hold for any sets A,B. Now

A(BC) = (A ∩ (BC)′) ∪ ((BC) ∩ A′

= (A ∩ ((B′ ∩ C′) ∪ (B ∩ C))) ∪ (((B ∩ C′) ∪ (C ∩B′)) ∩A′)

= (A ∩B′ ∩ C′) ∪ (A ∩B ∩ C) ∪ (A′ ∩B ∩ C′) ∪ (A′ ∩B′ ∩ C).

This holds for any sets A,B,C. Hence

(AB)C = C(AB)

= (C ∩A′ ∩B′) ∪ (C ∩A ∩B) ∪ (C′ ∩ A ∩B′) ∪ (C′ ∩ A′ ∩B)

= A(BC).

E6.9 For any set A letIdA = 〈x, x〉 : x ∈ A.

Justify this definition on the basis of the axioms.

IdA = y ∈ A× A : ∃x ∈ A[y = 〈x, x〉].

E6.10 Suppose that f : A → B. Prove that f is surjective iff there is a g : B → A suchthat f g = IdB. Note: the axiom of choice might be needed.

⇐: given b ∈ B, we have b = (f g)(b) = f(g(b)); so f is surjective.⇒: Assume that f is surjective. Let

A = (b, a) : a ∈ A, f(a) = b : b ∈ B.

Each member of A is nonempty; for let x ∈ A . Choose b ∈ B such that x = (b, a) : a ∈A, f(a) = b. Choose a ∈ A such that f(a) = b. So (b, a) ∈ x.

The members of A are pairwise disjoint: suppose x, y ∈ A with x 6= y. Choose b, cso that x = (b, a) : a ∈ A, f(a) = b and y = (b, a) : a ∈ A, f(a) = c. If u ∈ x ∩ y, thenthere exist a, a′ ∈ A such that u = (b, a), f(a) = b, and also (u = (c, a′), f(a′) = c. So byTheorem 6.3, b = c. But then x = y, contradiction.

Now by the axiom of choice, let C have exactly one element in common with eachmember of A . Then define

g = (b, a) ∈ C : a ∈ A, f(a) = b.

Now g is a function. For, suppose that (b, a), (b, a′) ∈ g. Let x = (b, a′′) : a′′ ∈ A, f(a′′) =b. Then (b, a), (b, a′) ∈ C ∩ x, so (b, a) = (b, a′). Hence a = a′.

Clearly g ⊆ B × A. Next, dmn(g) = B, for suppose that b ∈ B. Choose x ∈C ∩ (b, a′′) : a′′ ∈ A, f(a′′) = b; say x = (b, a) with a ∈ A, f(a) = b. Then x ∈ g and sob ∈ dmn(g).

Thus g : B → A. Take any b ∈ B, and let g(b) = a. So (b, a) ∈ g and hence f(a) = b.So f g = IdB .

18

E6.11 Let A be a nonempty set. Suppose that f : A → B. Prove that f is injective iffthere is a g : B → A such that g f = IdA.

First suppose that f is injective. Fix a ∈ A, and let

g = f−1 ∪ (b, a) : b ∈ B\rng(f).

Then g is a function. In fact, suppose that (b, c), (b, d) ∈ g. If both are in f−1, then(c, b(, (d, b) ∈ f , so f(c) = b = f(d) and hence c = d since f is injective. If (b, c) ∈ f−1

and b ∈ B\rng(f), the (c, b) ∈ f , so b ∈ rng(f), contradiction. If (b, c), (b, d) /∈ f−1, thenc = d = a.

Clearly then g : B → A. For any a ∈ A we have (a, f(a)) ∈ f , hence (f(a), a) ∈ f−1 ⊆g, and so g(f(a)) = a.

Second, suppose that g : B → A and g f = IdA. Suppose that f(a) = f(a′). Thena = (g f)(a) = g(f(a)) = g(f(a′)) = (g f)(a′) = a′.

E6.12 Suppose that f : A→ B. Prove that f is a bijection iff there is a g : B → A suchthat f g = IdB and g f = IdA. Prove this without using the axiom of choice.

⇒: Assume that f is a bijection. By E6.11 there is a g : B → A such that g f = IdA.We claim that f g = IdB . Since f is a bijection, the relation f−1 is also a bijection. Nowfor any b ∈ B,

(f g)(b) = f(g(b)) = f(g(f(f−1(b)))) = f((g f)(f−1(b))) = f(f−1(b)) = b.

So f g = IdB , as desired.⇐: Assume that g is as indicated. Then f is injective, since f(a) = f(b) implies

that a = g(f(a)) = g(f(a′)) = a′. And f is surjective, since for a given b ∈ B we havef(g(b)) = b.

E6.13 For any sets R, S define

R|S = (x, z) : ∃y((x, y) ∈ R ∧ (y, z) ∈ S).

Justify this definition on the basis of the axioms.

R|S = (x, z) ∈ dmn(R)× rng(S) : ∃y((x, y) ∈ R ∧ (y, z) ∈ S).

E6.14 Suppose that f, g : A→ A. Prove that

(A×A)\[((A× A)\f)|((A×A)\g)]

is a function.

Suppose that (x, y), (x, z) are in the indicated set, with y 6= z. By symmetry say f(x) 6= y.Then (x, y) ∈ [(A × A)\f ], so it follows that (y, z) ∈ g, as otherwise (x, z) ∈ [((A ×A)\f)|((A×A)\g)]. Hence (y, y) /∈ g, so (x, y) ∈ [((A×A)\f)|((A×A)\g)], contradiction.

19

E6.15 Suppose that f : A→ B is a surjection, g : A→ C, and ∀x, y ∈ A[f(x) = f(y)→g(x) = g(y)]. Prove that there is a function h : B → C such that h f = g. Define h as aset of ordered pairs.

Let h = (f(a), g(a)) : a ∈ A. Then h is a function, for suppose that (x, y), (x, z) ∈ h.Choose a, a′ ∈ A so that x = f(a), y = g(a), x = f(a′), and y = g(a′). Thus f(a) = f(a′),so g(a) = g(a′), as desired.

Since f is a surjection it is clear that dmn(h) = B. Clearly rng(h) ⊆ C. So h : B → C.If a ∈ A, then (f(a), b(a)) ∈ h, hence h(f(a)) = g(a). This shows that h f = g.

E6.16 The statement

∀A ∈ A ∀B ∈ B(A ⊆ B) implies that⋃

A ⊆⋂

B

is slightly wrong. Fix it, and prove the result.

If A has a nonempty member and B is empty, the implication does not hold. Add thehypothesis B 6= ∅.

Suppose that a ∈⋃A and B ∈ B; we want to show that a ∈ B. Choose A ∈ A such

that a ∈ A. Since A ⊆ B, we have a ∈ B.

E6.17 Suppose that ∀A ∈ A ∃B ∈ B(A ⊆ B). Prove that⋃

A ⊆⋃

B.

Suppose that a ∈⋃

A ; we want to show that a ∈ B. Choose A ∈ A such that a ∈ A.Then choose B ∈ B such that A ⊆ B. Then a ∈ B. Hence a ∈

⋃B.

E6.18 The statement

∀A ∈ A ∃B ∈ B(B ⊆ A) implies that⋂

B ⊆⋂

A .

is slightly wrong. Fix it, and prove the result.

If A is empty and⋂

B is nonempty, the statement is false. Fix it by adding the hypothesisthat A is nonempty.

Suppose that b ∈⋂

B and A ∈ A ; we want to show that b ∈ A. Choose B ∈ B suchthat B ⊆ A. Now b ∈ B since b ∈

⋂B, so b ∈ A.

Solutions, exercises in Chapter 7

E7.1 Prove that if x is an ordinal, then x is transitive and (x, (y, z) ∈ x× x : y ∈ z) isa well-ordered set.

By definition, x is transitive. Let R = (y, z) ∈ x× x : y ∈ z). Obviously R is a relation.By definition, R ⊆ x× x. R is irreflexive on x by Theorem 7.5. R is transitive since x istransitive. R is linear on x by Theorem 7.7. The final well-ordering property follows fromTheorem 7.13.

E7.2 Assume that x is transitive and (x, (y, z) ∈ x × x : y ∈ z) is a well-ordered set.Prove that for all y, z ∈ x, either y = z or y ∈ z or z ∈ y.

This is obvious.

E7.3 Assume that x is transitve and for all y, z ∈ x, either y = z or y ∈ z or z ∈ y.Prove that for all y, if y ⊂ x and y is transitive, then y ∈ x. Hint: apply the foundationaxiom to x\y.

20

Assume the hypothesis, and suppose that y ⊂ x and y is transitive. Choose z ∈ x\y suchthat z ∩ (x\y) = ∅. If u ∈ y, then u ∈ x since y ⊂ x. So u, z ∈ x, so by hypothesis we haveu ∈ z, u = z, or z ∈ u. Now u 6= z since z /∈ y and u ∈ y. And z /∈ u, since z ∈ u wouldimply, because y is transitive and u ∈ y, that z ∈ y, which is not true. Hence u ∈ z. Thisis true for any u ∈ y. So y ⊆ z. Clearly also z ⊆ y, so y = z ∈ x.

E7.4 Assume that x is transitive and for all y, if y ⊂ x and y is transitive, then y ∈ x.Show that x is an ordinal. Hint: let y = z ∈ x : z is an ordinal, and get a contradictionfrom the assumption that y ⊂ x.

Assume the hypothesis. Let y = z ∈ x : z is an ordinal. So y ⊆ x. Suppose that y ⊂ x.Now y is transitive, for assume that z ∈ y. Thus z ∈ x and z is an ordinal. Suppose thatw ∈ z. Then w ∈ x since x is transitive, and w is an ordinal since z is an ordinal. Sow ∈ y. Thus, indeed, y is transitive. So by assumption y ∈ x. Now y is a transitive setof transitive sets, so y is an ordinal. It follows that y ∈ y, contradiction. This proves thatx = y. So x is a transitive set of transitive sets, and hence x is an ordinal.

E7.5 Show that if x is an ordinal, then the following two conditions hold:

(i) For all y ∈ x, either y ∪ y = x or y ∪ y ∈ x.(ii) For all y ⊆ x, either

⋃y = x or

⋃y ∈ x.

Assume that x is an ordinal. Then (i) holds by Proposition 7.10. Now suppose that y ⊆ x.If z ∈

⋃y, choose w ∈ y such that z ∈ w. Then also w ∈ x, so z ∈ x since x is transitive.

This shows that⋃y ⊆ x. By Proposition 7.3,

⋃y is an ordinal. Hence by Proposition 7.8,

⋃y ≤ x.

E7.6 Assume the two conditions of exercise E7.5. Show that x is an ordinal. Hint: Showthat there is an ordinal α not in x. Taking such an ordinal α, show that there is a leastβ ∈ α ∪ α such that β /∈ x. Work with such a β to show that x is an ordinal.

By Theorem 7.6 there is an ordinal α not in x. Then by Theorem 7.13 there is a leastβ ∈ α ∪ α such that β /∈ x. Now we have two possibilities:

Case 1. β =⋃β. Now β ⊆ x, so by (ii) second clause, since

⋃β = β /∈ x we have

x =⋃β, hence x is an ordinal, as desired.

Case 2. β = (⋃β) + 1. Thus

⋃β is an ordinal smaller than β, so it is in x. By (i),

since β =⋃β + 1 /∈ x we have x = (

⋃β) + 1, hence x is an ordinal.

Solutions, exercises in Chapter 8

E8.1 Give an example of A,R such that R is not well-founded on A and is not set-likeon A.

We take On and R, where R = (α, β) : α > β. As shown after 8.2, R is not set-like onOn. It is also not well-founded on On, since ω is a nonempty set of ordinals, but if m ∈ ωthen (m+′ 1, m) ∈ R, so that ω does not have an R-minimal element.

E8.2 Give an example of A,R such that R is not well-founded on A but is set-like onA. Give one example with R and A are proper classes, and one example where they aresets.

21

Both are sets: let A = ω and R = (m,n) : m,n ∈ ω and m > n. Then ω does not havean R-minimal element, since for any m ∈ ω we have (m+′ 1, m) ∈ R.

Both are proper classes: let A = On and let

R = (m,n) : m,n ∈ ω and m > n ∪ (α, β) : α < β.

E8.3 Give an example of A,R such that R is well-founded on A but is not set-like onA.

Let A = V and R = (a, ∅) : a ∈ V, a 6= ∅. Thus predAR

(∅) = V, so R is not set-likeon V. Now let X be a nonempty set. If X = ∅, then ∅ ∈ X and ∀a ∈ X [(a, ∅) /∈ R]. IfX 6= ∅, take any a ∈ X\∅. Then ∀b ∈ X [(b, a) /∈ R].

E8.4 Suppose that R is a class relation contained in A×A, x ∈ A, and v ∈ predAR∗(x).Prove by induction on n that if n ∈ ω\1, f is a function with domain n +′ 1, ∀i <n[(f(i), f(i+′ 1)) ∈ R] and f(n) = v, then f(0) ∈ predAR∗(x).

Suppose that R is a class relation contained in A × A, x ∈ A, and v ∈ predAR∗(x).

We take n = 1 in the condition to be proved. So, suppose that f is a function withdomain 2 such that ∀i < 1[(f(i), f(i +′ 1)) ∈ R] and f(1) = v. Thus (f(0), v) ∈ R, sof(0) ∈ predAR(v). By Lemma 8.3(ii), f(0) ∈ predAR∗(x).

Now suppose that if n ∈ ω\1, f is a function with domain n+′ 1, ∀i < n[(f(i), f(i+′

1)) ∈ R] and f(n) = v, then f(0) ∈ predAR∗(x). Suppose also now that f is a function

with domain n +′ 2, ∀i < n +′ 1[(f(i), f(i+′ 1)) ∈ R] and f(n +′ 1) = v. Define g withdomain n +′ 1 by setting g(i) = f(i +′ 1) for all i < n +′ 1. Then ∀i < n[(g(i), g(i +′

1)) = (f(i +′ 1), f(i +′ 2)) ∈ R] and g(n) = f(n +′ 1) = v. Hence by the inductiveassumption, f(1) = g(0) ∈ pred

AR∗(x). We also have (f(0), f(1)) ∈ R, so by Lemma8.3(ii), f(0) ∈ predAR∗(x).

E8.5 Suppose that R is a class relation contained in A×A, (u, v) ∈ R∗, and (v, w) ∈ R∗.Show that (u, w) ∈ R∗.

Assume that R is a class relation contained in A × A, (u, v) ∈ R∗, and (v, w) ∈ R∗.Since (u, v) ∈ R∗, there exist n ∈ ω\1 and a function f with domain n +′ 1 such that∀i < n[(f(i), f(i+′ 1)) ∈ R, f(0) = u, and f(n) = v. From exercise E8.4 it follows that(u, w) ∈ R∗.

E8.6 Give an example of a proper class X which has a proper class of ∈-minimal elements.

Let X = α : α ≥ 2. We claim that all elements of X are ∈-minimal. Suppose thatα, β ≥ 2 and α ∈ β. Then α = β, Since β ≥ 2 we have 0, 1 ∈ β, so 0 = α = 1,contradiction.

E8.7 Give an example of a proper class relation R contained in A ×A for some properclass A, and a class function G mapping A × V into V such that R is set-like on A

but not well-founded on A and there is no class function F mapping A into V such thatF(a) = G(a,F〈pred

AR(a)) for all a ∈ A.

Let A = On and

R = (m,n) : m,n ∈ ω and m > n ∪ (α, β) : ω ≤ α < β.

22

Thus R is a proper class relation contained in A ×A. Clearly R is set-like on On but itis not well-founded on On. Define G : On×V→ V by setting

G(α, a) =

a(α+′ 1) if α ∈ ω and a is a function with domain m ∈ ω : m > α,∅ otherwise.

Suppose that F : A → V is such that F(α) = G(α,F predOnR(α)) for all α ∈ On. Letf = F ω. Choose b ∈ rng(f) such that b ∩ rng(f) = ∅. Say b = f(m) with m ∈ ω. Now

f(m) = F(m) = G(m,F predOnR(m)) = G(m,F n : n ∈ ω, n > m)

= F(m+′ 1) = f(m+′ 1),

so that f(m+′ 1) ∈ f(m) ∩ rng(f), contradiction.

E8.8 Give an example of a proper class relation R contained in some A ×A for someproper class A and a class function G mapping A ×V into V such that R is set-like onA but not well-founded on A but still there is a class function F mapping A into V suchthat F(a) = G(a,F〈predAR(a)) for all a ∈ A.

Let A and R be as in exercise E8.7, but define G(α, a) = α for all α ∈ On and all a ∈ V.Then the function F : On→ V such that F(α) = α for all α ∈ On is as desired.

Solutions to exercises in chapter 9

E9.1 Let (A,<) be a well order. Suppose that B ⊂ A and ∀b ∈ B∀a ∈ A[a < b→ a ∈ B].Prove that there is an element a ∈ A such that B = b ∈ A : b < a.

Let a be the least element of A\B. We claim that a is as desired. For, if b ∈ B, then itcannot happen that a ≤ b, since this would imply that a ∈ B; so b < a. And if b < a, thenb ∈ B by the minimality of a.

E9.2 Let (A,<) be a well order. Suppose that B ⊂ A and ∀b ∈ B∀a ∈ A[a < b→ a ∈ B].Prove that (A,<) is not isomorphic to (B,<).

Suppose that f is such an isomorphism from (A,<) onto (B,<). By exercise E9.1, leta ∈ A be such that B = x ∈ A : x < a. By Proposition 9.11, a ≤ f(a), contradictingthe assumption that f maps into B.

E9.3 Suppose that f is a one-one function mapping an ordinal α onto a set A. Definea relation ≺ which is a subset of A × A such that (A,<) is a well-order and f is anisomorphism of (α,<) onto (A,≺).

Define ≺= (a, b) ∈ A × A : f−1(a) < f−1(b). We check that (A,<) is a well-order.If a ∈ A and a ≺ a, then f−1(a) < f−a(a), contradiction. So ≺ is irreflexive. Supposethat a ≺ b ≺ c. Then f−1(a) < f−1(b) < f−1(c), so f−1(a) < f−1(c) and hence a ≺ c.So ≺ is transitive. Now given a, b ∈ A, either f−1(a) < f−1(b) or f−1(a) = f−1(b) orf−1(b) < f−1(a), so a ≺ b or a = b or b ≺ a. Thus (A,≺) is a linear order. Finally,suppose that ∅ 6= X ⊆ A. Then ∅ 6= f−1[X ], so let ξ be the least element of f−1[X ].Then f(ξ) ∈ X . Suppose that b ∈ X . Then f−1(b) ∈ f−1[X ], so ξ ≤ f−1(b). Hence

23

f(ξ) b. This shows that f(ξ) is the ≺-least element of X . We have shown that (A,≺) isa well-order.

We are given that f is a bijection from α onto A. If ξ, η ∈ α and ξ < η, thenf(ξ) ≺ f(η). If f(ξ) ≺ f(η), then ξ < η. Thus f is an isomorphism.

E9.4 Prove that 1 +m = m+ 1 for any m ∈ ω.

(Ordinary) induction on m. 0 + 1 = 1 = 1 + 0 using Theorem 9.21(vi). Assume that1 +m = m+ 1. Then 1 + (m+ 1) = (1 +m) + 1 = (m+ 1) + 1.

E9.5 Prove that m+ n = n+m for any m,n ∈ ω.

With m fixed, induction on n. 0 + m = m = m + 0 using Theorem 9.21(vi). Assumethat m+ n = n+m. Then (n+ 1) +m = n+ (1 +m) = n+ (m + 1) (by exercise E9.4)= (n+m) + 1 = (m+ n) + 1 = m+ (n+ 1).

E9.6 Prove that ω ≤ α iff 1 + α = α.

First note that 1 + ω =⋃

m∈ω(1 +m) =⋃

m∈ω(m+ 1) = ω, using Theorem 9.21(vi).⇒: Assume that ω ≤ α. By Theorem 9.21(vii) let δ be such that ω + δ = α. Then

1 + α = 1 + (ω + δ) = (1 + ω) + δ = ω + δ = α.⇐: It suffices to show that if m < ω then 1 + m 6= m. This is true by Theorem

9.21(vi).

E9.7 For any ordinals α, β let

α⊕ β = (α× 0) ∪ (β × 1).

We define a relation ≺ as follows. For any x, y ∈ α ⊕ β, x ≺ y iff one of the followingthree conditions holds:

(i) There are ξ, η < α such that x = (ξ, 0), y = (η, 0), and ξ < η.(ii) There are ξ, η < β such that x = (ξ, 1), y = (η, 1), and ξ < η.(ii) There are ξ < α and η < β such that x = (ξ, 0) and y = (η, 1).

Prove that (α⊕ β,≺) is a well order which is isomorphic to α+ β.

Clearly ≺ is a well-order. We show by transfinite induction on β, with α fixed, that(α ⊕ β,≺) is order isomorphic to α + β. For β = 0 we have α + β = α + 0 = α, whileα ⊕ β = α ⊕ 0 = α × 0. Clearly ξ 7→ (ξ, 0) defines an order-isomorphism from α onto(α × 0,≺). So our result holds for β = 0. Assume it for β, and suppose that f is anorder-isomorphism from α + β onto (α ⊕ β,≺). Now the last element of α ⊕ (β + 1) is(β, 1), and the last element of α+ (β + 1) is α+ β, so the function

f ∪ (α+ β, (β, 1))

is an order-isomorphism from α+ (β + 1) onto α⊕ (β + 1).Now assume that β is a limit ordinal, and for each γ < β, the ordinal α + γ is

isomorphic to α ⊕ γ. For each such γ let fγ be the unique isomorphism from α + γ onto

24

α⊕ γ. Note that if γ < δ < β, then fδ γ is an isomorphism from α+ γ onto α⊕ γ; hencefδ γ = fγ . It follows that

⋃

γ<β

fγ

is an isomorphism from α+ β onto α⊕ β, finishing the inductive proof.

E9.8 Given ordinals α, β, we define the following relation ≺ on α× β:

(ξ, η) ≺ (ξ′, η′) iff ((ξ, η) and (ξ′, η′) are in α× β and:

η < η′, or (η = η′ and ξ < ξ′).

We may say that this is the anti-dictionary or anti-lexicographic order.Show that the set α × β under the anti-lexicographic order is a well order which is

isomorphic to α · β.

We may assume that α 6= 0. It is straightforward to check that ≺ is a well-order.Now we define, for any (ξ, η) ∈ α× β,

f(ξ, η) = α · η + ξ.

We claim that f is the desired order-isomorphism from α× β onto α · β. If (ξ, η) ∈ α× β,then

f(ξ, η) = α · η + ξ < α · η + α = α · (η + 1) ≤ α · β.

Thus f maps into α · β.To show that f is one-one, suppose that (ξ, η), (ξ′, η′) ∈ α× β and f(ξ, η) = f(ξ′, η′).

Then by Theorem 9.26, (ξ, η) = (ξ′, η′). So f is one-one.To show that f maps onto α · β, let γ < α · β. Choose ξ and η so that γ = α · η + ξ

with ξ < α. Now η < β, as otherwise

γ = α · η + ξ ≥ α · η ≥ α · β.

It follows that f(ξ, η) = α · η + ξ = γ. so f is onto.Finally, we show that the order is preserved. Suppose that (ξ, η) ≺ (ξ′, η′). Then one

of these cases holds:Case 1. η < η′. Then

f(ξ, η) = α · η + ξ < α · η + α = α · (η + 1) ≤ α · η′ ≤ α · η′ + ξ′ = f(ξ′, η′),

as desired.Case 2. η = η′ and ξ < ξ′. Then f(ξ, η) < f(ξ′, η′).

Now it follows that f is the desired isomorphism.

E9.9 Suppose that α and β are ordinals, with β 6= 0. We define

αβw = f ∈ αβ : ξ < α : f(ξ) 6= 0 is finite.

25

For f, g ∈ αβw we write f ≺ g iff f 6= g and f(ξ) < g(ξ) for the greatest ξ < α for whichf(ξ) 6= g(ξ).

Prove that (αβw,≺) is a well-order which is order-isomorphic to the ordinal exponentβα. (A set X is finite iff there is a bijection from some natural number onto X.)

If α = 0, then βα = 1, and αβw also has only one element, the empty function (= theemptyset). So, assume that α 6= 0. If β = 1, then αβw has only one member, namely thefunction with domain α whose value is always 0. This is clearly order-isomorphic to 1, asdesired. So, suppose that β > 1.

Now we define a function f mapping βα into αβw. Let f(0) be the member of αβw

which takes only the value 0. Now suppose that 0 < ε < βα. By Theorem 9.29 write

ε = βγ(0) · δ(0) + βγ(1) · δ(1) + · · ·+ βγ(m−1) · δ(m− 1),

where ε ≥ γ(0) > γ(1) > · · · > γ(m − 1) and 0 < δ(i) < β for each i < m. Note thatβγ(0) ≤ ε < βα, so γ(0) < α. Then we define, for any ζ < α,

(f(ε))(ζ) =

0 if ζ /∈ γ(0), . . . , γ(m− 1),δ(i) if ζ = γ(i) with i < m.

Clearly f(ε) ∈ αβw. To see that f maps onto αβw, suppose that x ∈ αβw. If x takes onlythe value 0, then f(0) = x. Suppose that x takes on some nonzero value. Let

ξ < α : x(ξ) 6= 0 = γ(0), γ(1), . . . , γ(m− 1),

where γ(0) > γ(1) > · · · > γ(m− 1). Let δ(i) = x(γ(i)) for each i < m, and let

ε = βγ(0) · δ(0) + βγ(1) · δ(1) + · · ·+ βγ(m−1) · δ(m− 1).

Clearly then f(ε) = x.Now we complete the proof by showing that for any ε, θ < βα, ε < θ iff f(ε) < f(θ).

This equivalence is clear if one of ε, θ is 0, so suppose that both are nonzero. Write

ε = βγ(0) · δ(0) + βγ(1) · δ(1) + · · ·+ βγ(m−1) · δ(m− 1),

where α ≥ γ(0) > γ(1) > · · · > γ(m− 1) and 0 < δ(i) < β for each i < m, and

θ = βγ′(0) · δ′(0) + βγ

′(1) · δ′(1) + · · ·+ βγ′(n−1) · δ′(n− 1),

where α ≥ γ′(0) > γ′(1) > · · · > γ′(n− 1) and 0 < δ′(i) < β for each i < n.By symmetry we may suppose that m ≤ n. Note that N(β,m, γ, δ), k(β,m, γ, δ) = ε,

N(β, n, γ′, δ′), and k(β, n, γ′, δ′) = θ. We now consider several possibilities.Case 1. ε = θ. Then clearly f(ε) = f(θ).Case 2. γ ⊆ γ′, δ ⊆ δ′, and m < n. Thus ε < θ. Also, γ′(m) is the largest ξ < α such

that (f(ε))(ξ) 6= (f(θ))(ξ), and (f(ε))(ξ) = 0 < δ′(m) = (f(θ))(γ′(m)), so f(ε) < f(θ).

26

Case 3. There is an i < m such that γ(j) = γ′(j) and δ(j) = δ′(j) for all j < i, whileγ(i) 6= γ′(i). By symmetry, say that γ(i) < γ′(i). Then we have ε < θ. Since γ′(i) is thelargest ξ < α such that (f(ε))(ξ) 6= (f(θ))(ξ), and (f(ε))(γ′(i)) = 0 < δ′(i) = (f(θ))(γ′(i)),we also have f(ε) < f(θ).

Case 4. There is an i < m such that γ(j) = γ′(j) and δ(j) = δ′(j) for all j < i, whileγ(i) = γ′(i) and δ(i) 6= δ′(i). By symmetry, say that δ(i) < δ′(i). Then we have ε < θ.Since γ(i) is the largest ξ < α such that (f(ε))(ξ) 6= (f(θ))(ξ), and (f(ε))(γ′(i)) = δ(i) <δ′(i) = (f(θ))(γ′(i)), we also have f(ε) < f(θ).

E9.10 Show that for every nonzero ordinal α there are only finitely many ordinals β suchthat α = γ · β for some γ.

Suppose there are infinitely many such β; let 〈βi : i ∈ ω〉 be a one-one enumeration ofinfinitely many of them. For each i ∈ ω let γi be such that α = γi · βi. Clearly βi < βjiff γj < γi. We define 〈ij : j ∈ ω〉 by recursion. Let i0 be such that βi0 is the smallestelement of βk : k ∈ ω. Having defined i0, . . . , is, let is+1 be such that βis+1

is the smallestelement of

βk : k ∈ ω\βik : k ≤ s

Clearly βi0 < βi1 < · · ·, and hence γi0 > γi1 > · · ·, contradiction.

E9.11 Prove that n(ωω) = ω(ωω) for every natural number n > 1.

Note that nω = ω by an easy argument. Hence

ω(ωω) = (nω)(ωω)

= n(ω·(ωω))

= n(ωω). by Theorem 9.32

E9.12 Show that the following conditions are equivalent for any ordinals α, β:(i) α+ β = β + α.(ii) There exist an ordinal γ and natural numbers k, l such that α = γ ·k and β = γ · l.

⇒: Assume that α + β = β + α. The desired conclusion is clear if α = 0 or β = 0, soassume that α, β 6= 0. Write α = ωδ · k + ε with δ ≤ α, 0 < k ∈ ω, and ε < ωδ, and writeβ = ωρ · l + σ with ρ ≤ β, 0 < l ∈ ω, and σ < ωρ. If δ < ρ, then

α+ β = β < β + α,

contradiction. A similar contradiction is reached if ρ < δ. So δ = ρ. Now

α+ β = ωδ · (k + l) + σ = β + α = ωδ · (k + l) + ε,

so σ = ε. Hence α = (ωδ + ε) · k and β = (ωδ + ε) · l, as desired.⇐: Obvious.

E9.13 Suppose that α < ωγ. Show that α+ β + ωγ = β + ωγ.

27

Suppose that α, β, γ are ordinals and α < ωγ . If also β < ωγ , then α+β < ωγ by Theorem9.31, and also by Theorem 9.31 α+ β + ωγ = ωγ and β + ωγ = ωγ .

Now suppose that ωγ ≤ β. Write β = ωγ · δ + ε with δ > 0 and ε < ωγ .

(1) α+ ωγ · ϕ = ωγ · ϕ for every positive ϕ.

We prove (1) by induction on ϕ. It is true for ϕ = 1 by Theorem 9.31. Assume that itholds for ϕ. Then

α+ ωγ · (ϕ+ 1) = α+ ωγ · ϕ+ ωγ = ωγ · ϕ+ ωγ = ωγ · (ϕ+ 1),

as desired. Finally, assume that ϕ is limit and (1) holds for all ψ < ϕ. Let F (ϕ) = α+ ϕfor all ϕ, and G(ϕ) = ωγ · ϕ. Both of these are normal functions. Hence

α+ ωγ · ϕ = F (G(ϕ)) =⋃

ψ<ϕ

F (G(ψ)) =⋃

ψ<ϕ

(α+ ωγ · ψ) =⋃

ψ<ϕ

(ωγ · ψ) = ωγ · ϕ,

finishing the inductive proof of (1).Now by (1) we have

α+ β + ωγ = α+ ωγ · δ + ε+ ωγ = ωγ · δ + ε+ ωγ = β + ωγ .

E9.14 Show that the following conditions are equivalent:(i) α is a limit ordinal(ii) α = ω · β for some β 6= 0.(iii) For every m ∈ ω\1 we have m · α = α, and α 6= 0.

(i)⇒(ii): Assume (i). By Theorem 9.26 write α = ω · β + n with n < ω. If β = 0, thenα = n, contradiction. If n 6= 0, then α = ω · β + (n− 1) + 1, contradiction.

(ii)⇒(iii): Assume (ii). By Theorem 9.23(iii), α 6= 0. Suppose that m ∈ ω\1. Thenm · ω =

⋃

n∈ω(m · n) = ω by Theorem 9.23(iii), so m · α = α.(iii)⇒(i): Assume (iii), but suppose that α = β + 1. Then α = 2 · α = 2 · (β + 1) =

2 · β + 2 > α, contradiction.

E9.15 Show that (α+ β) · γ ≤ α · γ + β · γ for any ordinals α, β, γ.

Assume that α, β.γ 6= 0. Write α = ωδ · k + ε with δ ≤ α, 0 6= k ∈ ω, ε < ωδ, andβ = ωρ · l + σ with ρ ≤ β, 0 6= l ∈ ω, σ < ωρ. Also, write γ = ω · ξ +m with m ∈ ω. Nowwe consider some cases.

Case 1. δ < ρ. Then α+ β = β, and the desired conclusion follows.Case 2. δ = ρ. Note that if m > 0, then

α ·m = ωδ · k ·m+ ε;

β ·m = ωδ · l ·m+ σ;

(α+ β) ·m = ωδ · (k + l) ·m+ σ.

28

If ξ = 0 it is then clear that (α+ β) · γ = α · γ + β · γ. Hence assume that ξ > 0. Then

α · γ = α · ω · ξ + α ·m

= ωδ+1 · ξ + α ·m;

β · γ = ωδ+1 · ξ + β ·m;

α · γ + β · γ = ωδ+1 · ξ · 2 + β ·m;

(α+ β) · γ = ωδ+1 · ξ + (α+ β) ·m,

and the desired conclusion is clear.

Case 3 ρ < δ. Then if m > 0 we have

α ·m = ωδ · k ·m+ ε;

β ·m = ωρ · l ·m+ σ;

α ·m+ β ·m = ωδ · k ·m+ ε+ ωρ · l ·m+ σ;

(α+ β) ·m = ωδ · k ·m+ ε+ ωρ · l + σ.

Hence the desired conclusion follows if ξ = 0. Assume now that ξ 6= 0. Then

α · γ = ωδ+1 · ξ + α ·m;

β · γ = ωρ+1 · ξ + β ·m;

α · γ + β · γ = ωδ+1 · ξ + α ·m+ ωρ+1 · ξ + β ·m;

(α+ β) · γ = ωδ+1 · ξ + ωδ · k ·m+ ε+ ωρ · l + σ.

Again the desired conclusion holds.Solutions to exercises in chapter 10

E10.1 Show that any vector space over a field has a basis (possibly infinite).

Let V be any vector space over F . Let A = X ⊆ V : X is linearly independent, partiallyordered by ⊆. Then A 6= ∅, since trivially ∅ ∈ A. Now suppose that B is a subset of Asimply ordered by ⊆. We claim that

⋃B ∈ A; this will verify the hypothesis of Zorn’s

lemma. Suppose that v1, . . . , vn ∈⋃B, a1, . . . , an ∈ F , and a1v1 + · · · + anvn = 0; we

want to show that all ai are 0. For each i = 1, . . . , n choose Xi ∈ B such that vi ∈ Xi.Now Xi : i = 1, . . . , n has a largest member Xj under ⊆, since B is simply ordered.[Easy proof by induction on n.] Clearly vi ∈ Xj for all i = 1, . . . , n. Since Xj is linearlyindependent, it follows that each ai = 0, as desired.

Now we apply Zorn’s lemma to obtain a maximal member Y of A under ⊆. We claimthat Y is a basis for V . Since Y is linearly independent, it suffices to show that Y spansV . Suppose that w ∈ V . If w ∈ Y , then obviously w is in the span of Y . Suppose thatw /∈ Y . Then Y ⊂ Y ∪ w so by the maximality of Y , Y ∪ w is linearly dependent.Hence there is a natural number n, distinct elements v1, . . . , vn ∈ Y ∪ w, and elementsa1, . . . , an ∈ F , not all 0, such that a1v1 + · · ·+anvn = 0. Since Y is linearly independent,

29

not all vi are in Y ; say that vj = w. Then again because Y is linearly independent, wemust have aj 6= 0. So

w =

(

−a1

ajv1

)

+ · · ·+

(

−aj−1

ajvj−1

)

+

(

−aj+1

ajvj+1

)

+ · · ·+

(

−anajvn

)

,

so that w is in the span of Y , as desired.

E10.2 A subset C of R is closed iff the following condition holds:

For every sequence f ∈ ωC, if f converges to a real number x, then x ∈ C.

Here to say that f converges to x means that

∀ε > 0∃M∀m ≥M [|fm − x| < ε].

Prove that if 〈Cm : m ∈ ω〉 is a sequence of nonempty closed subsets of R, ∀m ∈ ω∀x, y ∈Cm[|x − y| < 1/(m + 1)], and Cm ⊇ Cn for m < n, then

⋂

m∈ω Cm is nonempty. Hint:use the Cauchy convergence criterion.

Let c be a choice function for P(R)\∅. For each m ∈ ω let fm = c(Cm). We claim thatf is a Cauchy sequence, and hence it converges to some point x. For, let ε > 0 be given.Choose m ∈ ω such that 1

m+1 < ε. Then for any n, p ≥ m we have fn, fp ∈ Cm and hence

by a hypothesis of the exercise, |fn − fp| <1

m+1 < ε, as desired. Now for any m ∈ ω wehave fn ∈ Cm for all n ≥ m, and hence x ∈ Cm. Thus x ∈

⋂

m∈ω Cm.

E10.3 Prove that every nontrivial commutative ring with identity has a maximal ideal.Nontrivial means that 0 6= 1. Only very elementary definitions and facts are needed here;they can be found in most abstract algebra books. Hint: use Zorn’s lemma.

Let R be a nontrivial commutative ring with identity. Let A be the collection of all properideals, partially ordered under ⊂. Obviously A 6= ∅. Suppose that B is a nonempty subsetof A simply ordered by ⊂. Let I =

⋃B. We claim that I is a proper ideal, so that it is

an upper bound for B. In fact, if a, b ∈ I, choose J,K ∈ B such that a ∈ J and b ∈ K.Since B is simply ordered by ⊂, by symmetry say J ⊂ K. Then a, b ∈ K, hence a+ b anda − b are also in K, and hence they are in I too. Also, if a ∈ I and b ∈ R, then a · b ∈ Iby an even easier argument. Thus I is an ideal. Clearly 1 /∈ I, so I is proper.

By Zorn’s lemma, A has a maximal element L. Clearly L is a maximal ideal.

E10.4 A function g : R→ R is continuous at a ∈ R iff for every sequence f ∈ ωR whichconverges to a, the sequence g f converges to g(a). (See Exercise E10.2.) Show that g iscontinuous at a iff the following condition holds:

∀ε > 0∃δ > 0∀x ∈ R[|x− a| < δ → |g(x)− g(a)| < ε].

Hint: for →, argue by contradiction.

→: Suppose that g is continuous at a but the indicated condition fails. Thus

(∗) ∃ε > 0∀δ > 0∃x ∈ R[|x− a| < δ and |g(x)− g(a)| ≥ ε].

30

Let c be a choice function for R. For each m ∈ ω let

fm = c

x ∈ R

[

|x− a| <1

m+ 1and |g(x)− g(a)| ≥ ε

]

.

Then f converges to a. In fact, given ξ > 0, choose M such that 1M−1

< ξ. Then for

any m ≥ M , |fm − a| <1

m+1 ≤1

M−1 < ξ. Since f converges to a and g is continuousat a, it follows that g f converges to g(a). Hence we can choose N such that ∀n ≥N [|g(fm)− g(a)| < ε]. But by the definition of f , |g(fN)− g(a)| ≥ ε, contradiction.←: Assume the indicated condition, and suppose that f ∈ ωR converges to a. In

order to show that g f also converges to a, let ε > 0 be given. By the condition, chooseδ > 0 such that ∀x ∈ R[|x− a| < δ → |g(x)− g(a)| < ε]. Since f converges to a, chooseM such that ∀m ≥M [|fm − a| < δ]. Then for any m ≥M we have |g(fm)− g(a)| < ε, asdesired.

E10.5 Show by induction on m, without using the axiom of choice, that if m ∈ ω and〈Ai : i ∈ m〉 is a system of nonempty sets, then there is a function f with domain m suchthat f(i) ∈ Ai for all i ∈ m.

For m = 0, the system itself is empty, and the desired function f is the empty set.Now suppose that 〈Ai : i ∈ m + 1〉 is a system of nonempty sets, and we know our

result for a system of m nonempty sets. So, let f be a function with domain m such thatf(i) ∈ Ai for all i ∈ m. Pick a ∈ Am, and let g = f ∪ (m, a). Clearly g is as desired,completing the inductive proof.

E10.6 Using AC, prove the following, which is called the Principle of Dependent Choice(which is also weaker than the axiom of choice, but cannot be proved in ZF). If A is anonempty set, R is a relation, R ⊆ A×A, and for every a ∈ A there is a b ∈ A such thataRb, then there is a function f : ω → A such that f(i)Rf(i+ 1) for all i ∈ ω.

Let c be a choice function for nonempty subsets of A. We define f : ω → A by recursion,as follows. Fix a ∈ A. For any m ∈ ω let

f(m) =

a if m = 0,c(x ∈ A : f(n)Rx) if m = n+ 1 and x ∈ A : f(n)Rx 6= 0,a otherwise.

By induction, f(i)Rf(i+ 1) for every i ∈ ω, as desired.

E10.7 Show that the axiom of choice implies (1), where (1) is(1) If < is a partial ordering and ≺ is a simple ordering which is a subset of <, then

there is a maximal (under ⊆) simple ordering ≪ such that ≺ is a subset of ≪, which inturn is a subset of <.

LetA = ≪:≪ is a simple ordering and ≺⊆≪⊆<.

Note that A is nonempty, since ≺∈ A . We partially order A by inclusion. To check thehypothesis of Zorn’s lemma, suppose that B is a nonempty subset of A simply ordered by

31

inclusion. We claim that⋃

B ∈ A ; this is clear, by checking all the necessary conditions.For example,

⋃B is transitive since if (a, b), (b, c) ∈

⋃B, then there are R, S ∈ B with

(a, b) ∈ R and (b, c) ∈ S; by symmetry R ⊆ S, hence (a, b), (b, c) ∈ S, hence (a, c) ∈ S,hence (a, c) ∈

⋃B.

So we apply Zorn’s lemma to obtain a maximal member ≪ of A ; this is as desired.

E10.8 Prove that (1) implies (2). [Given sets A and B, define f < g iff f and g areone-one functions which are subsets of A ×B, and f ⊂ g. Apply (1) to < and the emptysimple ordering.] Here (2) is

(2) For any two sets A and B, either there is a one-one function mapping A into B orthere is a one-one function mapping B into A.

Following the hint, we get a maximal simple ordering ≺ such that ≺⊆<. Let f =⋃

(≺).Since ≺ is a simply ordered collection of one-one functions, it is clear that f is a one-onefunction. It suffices to show that dmn(f) = A or rng(f) = B. Suppose that this is nottrue, and choose a ∈ A\dmn(f) and b ∈ B\rng(f). Let g = f ∪ (a, b). Clearly g is aone-one function contained in A×B. Thus if we define ≺′ as an extension of ≺ with g ≺′ ffor all g in the domain of ≺, we get a proper extension of ≺, contradiction.

E10.9 Prove that (2) implies (3). [Easy] Here (3) is

(3) For any two nonempty sets A and B, either there is a function mapping A onto B orthere is a function mapping B onto A.

Assume (2), and let A and B be nonempty sets. By (2) and symmetry, say that f is aone-one function mapping A into B. Fix a ∈ A, and define g with domain B by setting,for each b ∈ B,

g(b) =

f−1(b) if b ∈ rng(f),a otherwise.

Clearly g maps B onto A, as desired.

E10.10 Show in ZF that for any set A there is an ordinal α such that there is no one-onefunction mapping α into A. Hint: consider all well-orderings contained in A× A.

Let X be the set of all well-orderings contained in A× A. Now each ≺∈ X is isomorphicto an ordinal β≺. Let α =

⋃

≺∈X(β≺ + 1). Suppose that f is a one-one function mappingα into A. Let ≺= (f(ξ), f(η)) : ξ < η. Then ≺ is a well-ordering contained in A × A,and so βX = α; consequently α ∈ α, contradiction.

E10.11 Prove that (3) implies the axiom of choice. [Show that any set A can be well-ordered, as follows. Use exercise E10.10 to find an ordinal α which cannot be mappedone-one into P(A). Show that if f : A → α maps onto α, then 〈f−1[β] : β < α〉 is aone-one function from α into P(A).]

We follow the hint. Suppose that f : A → α maps onto α. Let g = 〈f−1[β] : β < α〉.Clearly g maps α into P(A). Suppose that g(β) = g(γ). Thus f−1[β] = f−1[γ].Choose a ∈ A such that f(a) = β; this is possible because f maps onto α. thus a ∈f−1[β] = f−1[γ], so f(a) ∈ γ, hence β = f(a) = γ. So g is one-one, contradictingthe choice of α.

32

Now it follows from (3) that there is a function f mapping α onto A. Define a ≺ biff the least element of f−1[a] is less than the first element of f−1[b]. Clearly ≺ is alinear order on A. To show that it is a well-order, let B be a nonempty subset of A. Letβ be the least element of f−1[B]. Then f(β) is clearly the ≺-least element of B.

E10.12 Show that the axiom of choice implies (4). [Use Zorn’s lemma.] Here (4) is

(4) A family F of subsets of a set A has finite character if for all X ⊆ A, X ∈ F iff everyfinite subset of X is in F . Principle (4) says that every family of finite character has amaximal element under ⊆.

Let F , a nonempty family of subsets of A, have finite character. We consider F as apartially ordered set under inclusion. It is nonempty by assumption. Now suppose thatG is a nonempty subset of F linearly ordered by inclusion. To show that

⋃G ∈ F , it

suffices to show that every finite F subset of it is in F , by the definition of finite character.For each a ∈ F choose Xa ∈ G such that a ∈ Xa. Since G is linearly ordered by inclusion,choose a ∈ F such that Xb ⊆ Xa for all b ∈ F . Now Xa ∈ F since G ⊆ F , and F is afinite subset of Xa, so F ∈ F by the definition of finite character.

Thus we have verified the hypotheses of Zorn’s lemma, and it gives the desired maximalelement.

E10.13 Show that (4) implies (5). Here (5) is

(5) For any relation R there is a function f ⊆ R such that dmnR = dmn f .

[Given a relation R, let F consist of all functions contained in R.]

Taking F as indicated, we verify that F has finite character. It is obviously nonempty,since ∅ ∈ F . Of course, if f ∈ F , then every finite subset of f is in F . Now suppose thatf ⊆ R and every finite subset of f is in F . We just need to show that f is a function.Suppose that (a, b), (a, c) ∈ f . Then (a, b), (a, c) is a finite subset of f , and so it is in F ,which means that it is a function, and so b = c. Thus f is a function.

Now by (4), let f be a maximal member of F under inclusion. So, f is a functionincluded in R. Suppose that a ∈ dmn(R)\dmn(f). Choose b such that (a, b) ∈ R. Thenf ⊂ f ∪ (a, b) ∈ F , contradiction. Therefore, dmn(R) = dmn(f), as desired.

E10.14 Show that (5) implies the axiom of choice. [Given a family 〈Ai : i ∈ I〉 ofnonempty sets, let R = (i, x) : i ∈ I and x ∈ Ai.]

We follow the hint. Let f be a function such that dmn(f) = dmn(R). Thus dmn(f) = Iand f(i) ∈ Ai for all i ∈ I.


E12.1 Define sets A,B with |A| = |B| such that there is a one-one function f : A → Bwhich is not onto.

Let A = B = ω and define f(m) = m + 1 for all m ∈ ω. Then f is not onto, since0 /∈ rng(f). Suppose that f(m) = f(n) and m 6= n. Say m < n. By Proposition 4.10,m+ 1 ≤ n < n+ 1 = m+ 1, contradiction.

33

E12.2 Define sets A,B with |A| = |B| such that there is an onto function f : A → Bwhich is not one-one.

Let A = B = ω. Define f(m+ 1) = m for any m ∈ ω and f(0) = 0.

E12.3 Show that the restriction λ 6= 0 is necessary in Proposition 12.43(ix).

Let κ = λ = 0, µ = 0, ν = 1. Then κµ = 00 = 1 by Theorem 12.43(i), and κν = 01 = 0 byTheorem 12.43(ii).

E12.4 Let F : P(A) → P(A), and assume that for all X, Y ⊆ A, if X ⊆ Y , thenF (X) ⊆ F (Y ). Let A = X : X ⊆ A and X ⊆ F (X), and set X0 =

⋃

X∈AX. Then

X0 ⊆ F (X0).

For any Y ∈ A we have Y ⊆ X0, and hence Y ⊆ F (Y ) ⊆ F (X0)), so X0 ⊆ F (X0).

E12.5 Under the assumptions of exercise E12.4 we actually have X0 = F (X0).

By exercise E12.4,X0 ⊆ F (X0), so F (X0) ⊆ F (F (X0)), hence F (X0) ∈ A , hence F (X0) ⊆X0; together with exercise E12.4 this proves that X0 = F (X0).

E12.6 Suppose that f : A→ B is one-one and g : B → A is also one-one. For every X ⊆A let F (X) = A\g[B\f [X ]]. Show that for all X, Y ⊆ A, if X ⊆ Y then F (X) ⊆ F (Y ).

We have f [X ] ⊆ f [Y ], hence B\f [Y ] ⊆ B\f [X ], hence g[B\f [Y ]] ⊆ g[B\f [X ]], henceF (X) = A\g[B\f [X ]] ⊆ A\g[B\f [Y ]] = F (Y ).

E12.7 Prove the Cantor-Schroder-Bernstein theorem as follows. Assume that f and gare as in exercise E12.6, and choose F as in that exercise. Let X0 be as in exercise E12.4.Show that A\X0 ⊆ rng(g). Then define h : A→ B by setting, for any a ∈ A,

h(a) =

f(a) if a ∈ X0,g−1(a) if a ∈ A\X0.

Show that h is one-one and maps onto B.

A\X0 = A\F (X0) = g[B\f [X ]] ⊆ rng(g). Now note that h X0 maps X0 onto f [X0] andis one-one, and h (A\X0) maps A\X0 onto g−1[A\X0] = g−1[g[B\f [X0]]] = B\f [X0] andis one-one. So h is the union of two functions with disjoint domains and disjoint ranges,so h is a one-one function, and it maps A onto B.

E12.8 Show that if α and β are ordinals, then |α ∔ β| = |α| + |β|, where ∔ is ordinaladdition and + is cardinal addition.

This is immediate from Proposition 12.21.

E12.9 Show that if α and β are ordinals, then |α ⊙ β| = |α| · |β|, where ⊙ is ordinalmultiplication and · is cardinal multiplication.

This is immediate from Proposition 12.29.

E12.10 Show that if α and β are ordinals, 2 ≤ α, and ω ≤ β, then |·αβ | = |α| · |β|. Herethe dot to the left of the first exponent indicates that ordinal exponentiation is involved.

34

First note that |α| ≤ α ≤ ·αβ and |β| ≤ β ≤ ·αβ , so |α| · |β| ≤ |·αβ |. Hence it sufficesto prove the other direction, which we do by induction on β, starting with β = ω. First,β = ω: If α < ω, then |·αω| =

∣∣⋃

m∈ω·αm

∣∣ = ω = |α| · |ω|. If α ≥ ω, then

|·αω| = |⋃

m∈ω

·αm| ≤∑

m∈ω

|·αm| ≤∑

m∈ω

|α| ≤ ω · |α|,

as desired.Now we assume the result for β ≥ ω. Then

|·αβ+1| = |·αβ ⊙ α| = |·αβ | · |α| = |α| · |β| · |α| = |α| · |β|.

using the inductive hypothesis. Finally, if β is a limit ordinal > ω and the result is truefor all γ < β, then

|·αβ | =

∣∣∣∣∣∣

⋃

γ<β

·αγ

∣∣∣∣∣∣

=

∣∣∣∣∣∣

⋃

ω≤γ<β

·αγ

∣∣∣∣∣∣

≤∑

ω≤γ<β

|·αγ |

=∑

ω≤γ<β

|α| · |γ| ≤∑

ω≤γ<β

|α| · |β| ≤ |α| · |β| · |β| = |α| · |β|.

E12.11 Prove that if |A| ≤ |B| then |P(A)| ≤ |P(B)|.

Let f be a one-one function mapping A into B. For each X ∈P(A) let g(X) = f [X ]. Sog maps P(A) into P(B). We claim that g is one-one. For, suppose that X, Y ∈ P(A)and X 6= Y . Say by symmetry that x ∈ X\Y . Then f(x) ∈ f [X ] but f(x) /∈ f [Y ] byone-oneness.

E12.12 Prove the following general distributive law:

c∏

i∈I

∑

j∈Ji

κij =∑

f∈P

c∏

i∈I

κi,f(i),

where P =∏

i∈I Ji.

The left side is the number of elements of

(1)∏

i∈I

⋃

j∈Ji

(κij × j)

,

and the right side is the number of elements of

(2)⋃

f∈P

(c∏

i∈I

κi,f(i) × f

)

.

35

For each f ∈ P let Ff be a bijection from∏

i∈I κi,f(i) onto∏ci∈I κi,f(i). Now given x in

(1) we define G(x) in (2) as follows. For each i ∈ I we have xi ∈⋃

j∈Ji(κij × j), and so

there is a unique j ∈ Ji such that x ∈ κij × j; let fx(i) be this j. Thus fx ∈ P . Now1st(xi) ∈ κi,fx(i) for all i ∈ I, so 〈1st(xi) : i ∈ I〉 ∈

∏

i∈I κi,fx(i). Now we define

G(x) = (Ffx(〈1st(xi) : i ∈ I〉), fx).

Clearly G(x) is in (2).Suppose that G(x) = G(y). Now fx = 2nd(G(x)) = 2nd(G(y) = fy. Write fx = g.

Then for any i ∈ I,

1st(xi) = (F−1g (1st(G(x))))i

= (F−1g (1st(G(y))))i

= 1st(yi),

and 2nd(xi) = g(i) = 2nd(yi). So xi = yi. Hence x = y. So G is one-one.To show that G maps onto (2), suppose that z is a member of (2). Choose h ∈ P such

that z ∈ (∏ci∈I κi,h(i))× h. Now F−1

h (1st(z)) ∈∏

i∈I κi,h(i), so for each i ∈ I we can let

xi = ((F−1h (1st(z))i, h(i)).

Then x is in (1). Moreover, clearly fx = h. Then 1st(xi) = (F−1h (1st(z))i, hence 〈1st(xi) :

i ∈ I〉 = F−1h (1st(z)), and so

G(x) = (Ffx(〈1st(xi) : i ∈ I〉), fx)

= (Fh(〈1st(xi) : i ∈ I〉), h)

= (Fh(F−1h (1st(z))), h)

= (1st(z), 2nd(z))

= z,

as desired.

E12.13 Show that for any cardinal κ we have κ+ = α : α is an ordinal and |α| ≤ κ.

First suppose that α < κ+. Then |α| ≤ α, so |α| ≤ κ. Now suppose that |α| ≤ κ. Thusthere is a one-one function from α into κ. If κ+ ≤ α, then we could also get a one-onefunction from κ+ into κ, so κ+ = |κ+| ≤ |κ| = κ, contradiction. So α < κ+, as desired.

E12.14 For every infinite cardinal λ there is a cardinal κ > λ such that κλ = κ.

Let κ = 2λ. Then κλ = (2λ)λ = 2λ·λ = 2λ = κ.

E12.15 For every infinite cardinal λ there is a cardinal κ > λ such that κλ > κ.

Let λ = ℵα. Note that cf(ℵα+ω) = ω ≤ λ. Let κ = ℵα+ω. Then κλ > κ.

E12.16 Prove that for every n ∈ ω, and every infinite cardinal κ, ℵκn = 2κ · ℵn.

36

We prove this by induction on n. n = 0: ℵκ0 = 2κ = 2κ · ℵ0. Assume it for n. Then byHausdorff’s theorem,

ℵκn+1 = ((ℵn)+)κ = ℵκn · (ℵn)

+ = 2κ · ℵn · ℵn+1 = 2κ · ℵn+1.

E12.17 Prove that∏ci∈I(κi · λi) =

∏ci∈I κi ·

∏ci∈I λi.

We have

(1)c∏

i∈I

(κi · λi) =

∣∣∣∣∣

∏

i∈I

(κi · λi)

∣∣∣∣∣

and

(2)c∏

i∈I

κi ·c∏

i∈I

λi =

∣∣∣∣∣

(c∏

i∈I

κi

)

×

(c∏

i∈I

λi

)∣∣∣∣∣

For each i ∈ I let fi be a bijection from κi · λi onto κi × λi. Let g be a bijection from∏ci∈I κi onto

∏

i∈I κi, and let h be a bijection from∏ci∈I λi onto

∏

i∈I λi. Now we define a

function F from∏

i∈I(κi ·λi) to(∏c

i∈I κi)×(∏c

i∈I λi)

by setting, for any x ∈∏

i∈I(κi ·λi),

F (x) = (g−1(〈1st(fi(xi)) : i ∈ I〉), h−1(2nd(fi(xi)) : i ∈ I〉)).

This does map into(∏c

i∈I κi)×(∏c

i∈I λi), since for any i ∈ I we have xi ∈ κi · λi, hence

fi(xi) ∈ κi × λi; so 1st(fi(xi)) ∈ κi. Thus 〈1st(fi(xi)) : i ∈ I〉 ∈∏

i∈I κi and hence

g−1(〈1st(fi(xi)) : i ∈ I〉) ∈∏ci∈I κi. Similarly h−1(2nd(fi(xi)) : i ∈ I〉) ∈

∏ci∈I λi, so that

F (x) ∈(∏c

i∈I κi)×(∏c

i∈I λi).

F is one-one; for suppose that F (x) = F (y). Then for any i ∈ I,

1st(F (x)) = g−1(〈1st(fi(xi)) : i ∈ I〉);

g(1st(F (x)) = 〈1st(fi(xi)) : i ∈ I〉;

(g(1st(F (x))i = 1st(fi(xi)).

Similarly, (g(1st(F (y))i = 1st(fi(yi)). Then 1st(fi(xi)) = 1st(fi(yi)) follows from theassumption that F (x) = F (y). Similarly, 2nd(fi(xi)) = 2nd(fi(yi)). So fi(xi) = fi(yi),and hence xi = yi. This being true for all i ∈ I, we have x = y, as desired.

F maps onto; for suppose that z ∈(∏c

i∈I κi)×(∏c

i∈I λi). Then 1st(z) ∈

∏ci∈I κi, so

g(1st(z)) ∈∏

i∈I κi. Hence for any i ∈ I, (g(1st(z)))i ∈ κi. Similarly, (h(2nd(z))i ∈ λi. It

follows that ((g(1st(z)))i, (h(2nd(z))i) ∈ κi × λi, and hence

f−1i (((g(1st(z)))i, (h(2

nd(z))i)) ∈ κi · λi.

We let xi = f−1i (((g(1st(z)))i, (h(2

nd(z))i)). So x ∈∏

i∈I(κi · λi), and

fi(xi) = ((g(1st(z)))i, (h(2nd(z))i);

1st(fi(xi)) = (g(1st(z)))i;

〈1st(fi(xi)) : i ∈ I〉 = g(1st(z));

g−1(〈1st(fi(xi)) : i ∈ I〉) = 1st(z).

37

Similarly, h−1(〈2nd(fi(xi)) : i ∈ I〉) = 2nd(z). So F (x) = z.By (1) and (2) this completes the exercise.

E12.18 Prove that ℵℵ1ω = 2ℵ1 · ℵℵ0

ω .

Note that ℵω =∑

m∈ω ℵm, since ℵn ≤∑

m∈ω ℵm for each n ∈ ω, hence

ℵω ≤∑

m∈ω

ℵm ≤∑

m∈ω

ℵω = ω · ℵω = ℵω.

Hence by Theorem 12.41 we have ℵω <∏cm∈ω ℵm+1 ≤

∏cm∈ω ℵm. So

ℵℵ1ω ≤

(c∏

m∈ω

ℵm

)ℵ1

=c∏

m∈ω

ℵℵ1m

=c∏

m∈ω

(2ℵ1 · ℵm) by exercise E12.16

=

c∏

m∈ω

2ℵ1 ·c∏

m∈ω

ℵm by exercise E12.17

= 2ℵ1·ℵ0 ·c∏

m∈ω

ℵm

= 2ℵ1 ·c∏

m∈ω

ℵm

≤ 2ℵ1 ·c∏

m∈ω

ℵω

= 2ℵ1 · ℵℵ0ω

≤ ℵℵ1ω .

E12.19 Prove that ℵℵ0ω =

∏

n∈ω ℵn.

By the argument at the beginning of the solution of exercise E12.18, ℵω =∑

m∈ω ℵm <∏cm∈ω ℵm. Hence

ℵℵ0ω ≤

(c∏

m∈ω

ℵm

)ℵ0

=

c∏

m∈ω

ℵℵ0m =

c∏

m∈ω

(2ℵ0 ·ℵm) = 2ℵ0 ·c∏

m∈ω

ℵm =

c∏

m∈ω

ℵm ≤c∏

m∈ω

ℵω = ℵℵ0ω .

E12.20 Prove that for any infinite cardinal κ, (κ+)κ = 2κ.

By Hausdorff’s theorem, (κ+)κ = κκ · κ+ = 2κ · κ+ = 2κ.

38

E12.21 Show that if κ is an infinite cardinal and C is the collection of all cardinals lessthan κ, then |C| ≤ κ.

Let κ = ℵα. Thus C ⊆ ω ∪ ℵβ : β < α. Hence |C| ≤ ω + |α|. Now ω ≤ κ, and|α| ≤ α ≤ ℵα = κ since ℵ is a normal function. Hence |C| ≤ κ.

E12.22 Show that if κ is an infinite cardinal and C is the collection of all cardinals lessthan κ, then

2κ =

(∑

ν∈C

2ν

)cf(κ)

.

First suppose that κ is a successor cardinal λ+. Then

2κ ≤

(∑

ν∈C

2ν

)κ

=

(∑

ν∈C

2ν

)cf(κ)

≤

(∑

ν∈C

2λ

)κ

= (|C| · 2λ)κ ≤ (λ+ · 2λ)κ ≤ (2κ)κ = 2κ,

as desired.Now suppose that κ is a limit cardinal. Let 〈µξ : ξ < cf(κ)〉 be a strictly increasing

sequence of cardinals with supremum κ. Then

2κ = 2

∑

ξ<cf(κ)µξ =

c∏

ξ<cf(κ)

2µξ ≤

(∑

ν∈C

2λ

)cf(κ)

≤ (2κ)cf(κ) = 2κ.

E12.23 Prove that for any limit ordinal τ ,∏cξ<τ 2ℵξ = 2ℵτ .

2ℵτ = 2

∑

ξ<τℵξ =

c∏

ξ<τ

2ℵξ .

E12.24 Assume that κ is an infinite cardinal, and 2λ < κ for every cardinal λ < κ. Show

that 2κ = κcf(κ).

If κ is a successor cardinal, then cf(κ) = κ and the desired conclusion is clear. Supposethat κ is a limit cardinal. Let 〈µξ : ξ < cf(κ)〉 be a strictly increasing sequence of cardinalswith supremum κ. Then

2κ = 2

∑

ξ<cf(κ)µξ =

c∏

ξ<cf(κ)

2µξ ≤c∏

ξ<cf(κ)

κ = κcf(κ) ≤ κκ = 2κ.

E12.25 Suppose that λ is a singular cardinal, cf(λ) = ω, and 2κ < λ for every κ < λ.

Prove that 2λ = λω.

Let 〈κn : n ∈ ω〉 be a system of cardinals less than λ with supremum λ. Then

2λ = 2∑

n∈ωκn =

c∏

n∈ω

2κn ≤c∏

n∈ω

λ = λω ≤ λλ = 2λ.

39


E13.1 Let (A,+, ·,−, 0, 1) be a Boolean algebra. Show that (A,, ·, 0, 1) is a ring withidentity in which every element is idempotent. This means that x · x = x for all x.

Obviously is commutative, and it is associative by Proposition 13.11(iii). Clearly x0 =x for all x. Clearly xx = 0, so each element x has itself as additive inverse. Hence(A,, 0) is an abelian group.

Clearly · is associative. The distributive law holds by Proposition 13.11(ii). Clearlyx · 1 = x for all x, and clearly x · x = x for all x.

Hence (A,, ·, 0, 1) is a ring with identity in which every element is idempotent.

E13.2 Let (A,+, ·, 0, 1) be a ring with identity in which every element is idempotent.Show that A is a commutative ring, and (A,⊕, ·,−, 0, 1) is a Boolean algebra, where forany x, y ∈ A, x⊕ y = x+ y + xy and for any x ∈ A, −x = 1 + x. Hint: expand (x+ y)2.

x+ y = (x+ y)2 = x2 + xy + yx+ y2 = x+ xy + yx+ y, and hence 0 = xy + yx for anyx, y. Setting x = y, we get 0 = x + x, and so x is its own additive inverse. Then from0 = xy + yx we see that yx is the additive inverse of xy, hence xy = yx. Thus the ring iscommutative.

To show that (A,⊕, ·,−, 0, 1) is a Boolean algebra, we need to check all of the axioms.

(C): Clear.

(A): For any x, y, z,

x⊕ (y ⊕ z) = x+ (y ⊕ z) + x(y ⊕ z)

= x+ y + z + yz + x(y + z + yz)

= x+ y + z + yz + xy + xz + xyz;

Hence, using (C),

(x⊕ y)⊕ z = z ⊕ (x⊕ y)

= z + x+ y + xy + zy + zxy

= above.

(A′): obvious.

(C′): obvious.

(L):x⊕ xy = x+ xy + xxy = x+ xy + xy = x.

(L′):x(x⊕ y) = x(x+ y + xy) = xx+ xy + xxy = x+ xy + xy = x.

(D):

x(y ⊕ z) = x(y + z + yz) = xy + xz + xyz;

xy ⊕ xz = xy + xz + xyxz = xy + xz + xyz.

40

(D′): See Proposition 13.12.

(K): x+ (−x) = x+ 1 + x = 1.

(K’): x(1 + x) = x+ xx = x+ x = 0.

Thus we have a BA.

E13.3 Show that the processes described in exercises E2.1 and E2.2 are inverses of oneanother.

For each BA (A,+, ·,−, 0, 1) let R(A,+, ·,−, 0, 1) = (A,, ·, 0, 1) be the associated ring,and for each ring (A,+, ·, 0, 1) with identity in which every element is idempotent letB(A,+, ·, 0, 1) = (A,⊕, ·,−, 0, 1) be the associated Boolean algebra. We want to showthat R and B are inverses of each other.

First suppose that (A,+, ·,−, 0, 1) is a BA. Let R((A,+, ·,−, 0, 1)) = (A,, ·, 0, 1) bethe associated ring, and let B(R((A,+, ·,−, 0, 1)) = (A,⊕, ·,−′, 0, 1) be the BA associatedwith that ring; we want to show that + = ⊕ and − = −′. We have

x⊕ y = xy(x · y)

= x(y · −(x · y) + x · y · −y)

= x(y · −x)

= x · −(y · −x) + y · −x · −x

= x+ y · −x

= x+ y · x+ y · −x

= x+ y.

Also, −′x = 1x = −x.Second, suppose that (A,+, ·, 0, 1) is a ring with identity in which every element

is idempotent, let B((A,+, ·, 0, 1)) = (A,+′, ·,−′, 0, 1) be the associated BA, and letR(B((A,+, ·, 0, 1))) = (A,′, ·, 0, 1) be the ring associated with it. We want to showthat + = ′. We have

x′y = (x · −′y) +′ (y · −′x)

= (x · (1 + y)) +′ (y · (1 + x))

= (x+ xy) +′ (y + xy)

= x+ xy + y + xy + (x+ xy)(y + xy)

= x+ y + xy + xy + xy + xxy + xyy + xyxy

= x+ y + xy + xy + xy + xy + xy + xy

= x+ y.

E13.4 Prove that a filter F is an ultrafilter iff F is maximal among the set of all filtersG such that 0 /∈ G.

41

⇒: Assume that F is an ultrafilter. Hence by definition 0 /∈ F . Suppose that F ⊂ G withG a filter. Choose x ∈ G\F . Since x /∈ F it follows that −x ∈ F , and hence −x ∈ G. So0 = x · −x ∈ G. So F is maximal among the set of filters G such that 0 /∈ G.⇐: Suppose that F is maximal among the set of filters G such that 0 /∈ G. Suppose

that a ∈ A and a /∈ F . Let G = x ∈ A : a · y ≤ x for some y ∈ F. Then G is a filter onA. In fact, obviously conditions (1) and (2) hold. For (3), suppose that x, z ∈ G. Choosey, w ∈ F such that x = a · y and z = a ·w. Now y ·w ∈ F , and x · z = a · y ·w. So x · z ∈ G.Thus, indeed, G is a filter on A. Clearly also F ⊆ G. Clearly a ∈ G (taking y = 1), soF ⊂ G.

It follows by supposition that 0 ∈ G. Say 0 = a · y, with y ∈ F . then y ≤ −a, so−a ∈ F . Thus F is an ultrafilter.

E13.5 Prove that for any nonzero a ∈ A there is an ultrafilter F such that a ∈ F .

Let A = G : G is a filter in A, a ∈ G, and 0 /∈ G. We consider A as a partially orderedset under ⊆. To verify the hypothesis of Zorn’s lemma, suppose that B is a subset of A

linearly ordered by ⊆. Now x ∈ A : a ≤ x is clearly a member of A , so we may assumethat B is nonempty. Let H =

⋃B. Since B is nonempty, it is clear that a ∈ H. Suppose

that x ∈ H and x ≤ y. Choose G ∈ B such that x ∈ G. Then y ∈ G since G is a filter.So y ∈ H. Suppose that x, y ∈ H. Choose G,G′ ∈ B such that x ∈ G and y ∈ G′. Bysymmetry say G ⊆ G′. Then x, y ∈ G′, so x · y ∈ G′, hence x · y ∈ H. Thus we have shownthat H is a filter on A. Clearly 0 /∈ H. So H is a member of A which is an upper boundfor B.

Thus by Zorn’s lemma, A has a maximal member G. By exercise E13.4, G is asdesired.

E13.6 Prove that any BA is isomorphic to a field of sets. (Stone’s representation theorem)Hint: given a BA A, let X be the set of all ultrafilters on A and define f(a) = F ∈ X :a ∈ F.

Let X be the collection of all ultrafilters, and let F,G ∈ X .F ∈ f(−a) iff −a ∈ F iff a /∈ F , so f(−a) = X\f(a).Suppose that F ∈ f(a+ b). Then a+ b ∈ F . Suppose that F /∈ f(a). Then a /∈ F , so

−a ∈ F , hence −a · (a+ b) ∈ F . Since −a · (a+ b) ≤ b, also b ∈ F , so F ∈ f(b). This showsthat f(a + b) ⊆ f(a) ∪ f(b). On the other hand, if F ∈ f(a), then a ∈ F ; but a ≤ a + b,so also a+ b ∈ F ; hence F ∈ f(a+ b). Altogether this shows that f(a+ b) = f(a) ∪ f(b).

Suppose that a 6= b. Then ab 6= 0, so ab ∈ F for some ultrafilter F , by exerciseE13. Hence F ∈ f(ab) = [f(a)\f(b)]∪ [f(b)\f(a)], and so f(a) 6= f(b). So f is one-one.

E13.7 Suppose that F is an ultrafilter on a BA A. Let 2 be the two-element BA. (Thisis, up to isomorphism, the BA of all subsets of 1.) For any a ∈ A let

f(a) =

1 if a ∈ F ,0 if a /∈ F .

Show that f is a homomorphism of A into 2.

f(a · b) = 1 iff a · b ∈ F iff a, b ∈ F iff f(a) · f(b) = 1. Hence f(a · b) = f(a) · f(b).

42

f(−a) = 1 iff −a ∈ F iff a /∈ F iff f(a) = 0. Hence f(−a) = −f(a).f(a+ b) = f(−(−a · −b)) = −(−f(a) · −f(b)) = f(a) + f(b).f(0) = f(a · −a) = f(a) · −f(a) = 0.f(1) = f(a+−a) = f(a) +−f(a) = 1.

E13.8 (A knowledge of logic is assumed.) Suppose that L is a first-order language and Tis a set of sentences of L . Define ϕ ≡T ψ iff ϕ and ψ are sentences of L and T |= ϕ↔ ψ.Show that this is an equivalence relation on the set S of all sentences of L . Let A be thecollection of all equivalence classes under this equivalence relation. Show that there areoperations +, ·, − on A such that for any sentences ϕ, ψ,

[ϕ] + [ψ] = [ϕ ∨ ψ];

[ϕ] · [ψ] = [ϕ ∧ ψ];

−[ϕ] = [¬ϕ].

Finally, show that (A,+, ·,−, [∃v0(¬(v0 = v0))], [∃v0(v0 = v0)]) is a Boolean algebra.

≡T is reflexive: T |= ϕ↔ ϕ for any sentence ϕ.≡T is symmetric: If T |= ϕ↔ ψ, then T |= ψ ↔ ϕ.≡T is transitive: If T |= ϕ↔ ψ and T |= ψ ↔ χ, then T |= ϕ↔ χ.+ is well-defined: If T |= ϕ↔ ϕ′ and T |= ψ ↔ ψ′, then T |= (ϕ ∨ ψ)↔ (ϕ′ ∨ ψ′).· is well-defined: If T |= ϕ↔ ϕ′ and T |= ψ ↔ ψ′, then T |= (ϕ ∧ ψ)↔ (ϕ′ ∧ ψ′).− is well-defined: If T |= ϕ↔ ϕ′, then T |= ¬ϕ↔ ¬ϕ′.Finally, we need to check the axioms for BAs:

(A) holds since

[ϕ] + ([ψ] + [χ]) = [ϕ ∨ (ψ ∨ χ)] = [(ϕ ∨ ψ) ∨ χ] = ([ϕ] + [ψ]) + [χ];

(A′) holds since

[ϕ] · ([ψ] · [χ]) = [ϕ ∧ (ψ ∧ χ)] = [(ϕ ∧ ψ) ∧ χ] = ([ϕ] · [ψ]) · [χ];

(C) holds since[ϕ] + [ψ] = [ϕ ∨ ψ] = [ψ ∨ ϕ] = [ψ] + [ϕ];

(C′) holds since[ϕ] · [ψ] = [ϕ ∧ ψ] = [ψ ∧ ϕ] = [ψ] · [ϕ];

(L) holds since[ϕ] + [ϕ] · [ψ]) = [ϕ ∨ (ϕ ∧ ψ)] = [ϕ];

(L′) holds since[ϕ] · [ϕ] + [ψ]) = [ϕ ∧ (ϕ ∨ ψ)] = [ϕ];

(D) holds since

[ϕ] · ([ψ] + [χ]) = [ϕ ∧ (ψ ∨ χ)] = [(ϕ ∧ ψ) ∨ (ϕ ∧ χ)] = [ϕ] · [ψ] + [ϕ] · [χ];

43

for (D′) see Proposition 2.12; (K) holds since

[ϕ] +−[ϕ] = [ϕ ∨ ¬ϕ] = [∃v0(v0 = v0)];

(K′) holds since

[ϕ] · −[ϕ] = [ϕ ∧ ¬ϕ] = [∃v0(¬(v0 = v0))].

E13.9 (A knowledge of logic is assumed.) Show that every Boolean algebra is isomorphicto one obtained as in exercise E13.8. Hint: Let A be a Boolean algebra. Let L be thefirst-order language which has a unary relation symbol Ra for each a ∈ A. Let T be thefollowing set of sentences of L :

∀x∀y(x = y);

∀x[R−a(x)↔ ¬Ra(x)] for each a ∈ A;

∀x[Ra·b(x)↔ Ra(x) ∧Rb(x)] for all a, b ∈ A;

∀xR1(x).

We follow the hint, and consider ≡T . Define f(a) = [∀xRa(x)] for any a ∈ A. To showthat f preserves ·, suppose that a, b ∈ A. Note that

T |= ∀xRa·b(x)↔ ∀xRa(x) ∧ ∀xRb(x);

hence f(a · b) = f(a) · f(b).To proceed we need the following fact

(1) T |= ∀xϕ↔ ϕ for any variable x and any formula ϕ.

In fact, trivially T |= ∀xϕ→ ϕ, and T |= ϕ→ ∃xϕ. Since T |= x = y, clearly T |= ∃xϕ→∀xϕ. So (1) holds.

Now to show that f preserves −, suppose that a ∈ A. Then T |= ∀xR−a(x) ↔∀x¬Ra(x). By (1), T |= ∀x¬Ra(x) ↔ ¬Ra(x) and T |= ¬Ra(x) ↔ ¬∀xRa(x). Puttingthese statements together we have T |= ∀xR−a(x) ↔ ¬∀xRa(x), and it follows that fpreserves −.

To show that f is one-one, suppose that a, b ∈ A and a 6= b; say a · −b 6= 0. Let Fbe an ultrafilter on A such that a · −b ∈ F . We now define an L -structure A. Let A = 1.For each a ∈ A, let

RAa =

1 if a ∈ F ,0 otherwise.

Clearly A is a model of T . Also, A |= Ra·−b(x). It follows that [∀xRa·−b(x)] 6= [∃v0(¬(v0 =v0))], and so f(a) = [∀xRa(x)] 6= [∀xRb(x)] = f(b), as desired.

It remains only to show that f maps onto.

(2) For any formula ϕ there is an a ∈ A such that T |= ϕ↔ Ra(x).

Condition (2) is easily proved by induction on ϕ, using (1). Hence f is onto.

44

E13.10 Let A be the collection of all subsets X of Ydef= r ∈ Q : 0 ≤ r such that there

exist an m ∈ ω and a, b ∈ m(Y ∪ ∞) such that a0 < b0 < a1 < b1 < · · · < am−1 <bm−1 ≤ ∞ and

X = [a0, b0) ∪ [a1, b1) ∪ . . . ∪ [am−1, bm−1).

Note that ∅ ∈ A by taking m = 0, and Y ∈ A since Y = [0,∞).(i) Show that if X is as above, c, d ∈ Y ∪∞ with c < d, c ≤ a0, then X ∪ [c, d) ∈ A,

and c is the first element of X ∪ [c, d).(ii) Show that if X is as above and c, d ∈ Y ∪ ∞ with c < d, then X ∪ [c, d) ∈ A.(iii) Show that (A,∪,∩, \, ∅, Y ) is a Boolean algebra.

(i): Assume the hypothesis. If m = 0 the desired conclusion is clear, so suppose thatm > 0. We consider several cases.

Case 1. bm−1 ≤ d. Then X ∪ [c, d) = [c, d) ∈ A.Case 2. There is an i < m− 1 such that bi ≤ d < ai+1. Then

X ∪ [c, d) = [c, d) ∪ [ai+1, bi+1) ∪ . . . ∪ [am−1, bm−1) ∈ A.

Case 3. There is an i < m such that ai ≤ d < bi. Then

X ∪ [c, d) = [c, bi) ∪ [ai+1, bi+1) ∪ . . . ∪ [am−1, bm−1) ∈ A.

Case 4. d < a0. Then

X ∪ [c, d) = [c, d) ∪ [a0, b0) ∪ . . . ∪ [am−1, bm−1) ∈ A.

(ii): Again we consider several cases.Case 1. c ≤ a0. Then X ∪ [c, d) ∈ A by (i).Case 2. There is an i < m such that ai ≤ c ≤ bi. Let X ′ = [ai, bi)∪ . . .∪ [am−1, bm−1).

Then by (i) applied to X ′ and [ai, d) we get X ′ ∪ [ai, d) ∈ A, and ai is the least element ofX ′ ∪ [ai, d). Clearly

X ∪ [c, d) = [a0, b0) ∪ . . . ∪ [ai−1, bi−1) ∪X′ ∪ [ai, d) ∈ A.

Case 3. There is an i < m − 1 such that bi < c < ai+1. Then we can apply (i) to[ai+1, bi+1) ∪ . . . ∪ [am−1, bm−1) and [c, d) to get the desired result as in Case 2.

Case 4. c = bm−1. Then

X ∪ [c, d) = [a0, b0) ∪ . . . ∪ [am−1, d) ∈ A.

Case 5. bm−1 < c. This case is clear.(iii): From (ii) it is clear that A is closed under ∪. Now suppose that X is given as

above. To show that also Y \X ∈ A, we consider several cases.Case 1. m = 0. So X = ∅, and Y = [0,∞) ∈ A.Case 2. m > 0, 0 < a0, and bm−1 <∞. Then

Y \X = [0, a0) ∪ [b0, a1) ∪ . . . ∪ [bm−2, am−1) ∪ [bm−1,∞) ∈ A.

45

Case 3. m > 0, a0 = 0, and bm−1 <∞. Then

Y \X = [b0, a− 1) ∪ . . . ∪ [bm−2, am−1) ∪ [bm−1,∞) ∈ A.

Case 4. m > 0, 0 < a0, and bm−1 =∞. Then

Y \X = [0, a0) ∪ [b0, a1) ∪ . . . ∪ [bm−2, am−1) ∈ A.

Case 5. m > 0, 0 = a0, and bm−1 =∞. Then

Y \X = [b0, a1) ∪ . . . ∪ [bm−2, am−1) ∈ A.

Thus (iii) holds.

E13.11 (Continuing exercise E13.10) For each n ∈ ω let xn = [n, n+ 1), an interval inQ. Show that

∑

n∈ω x2n does not exist in A.

Suppose that the sum does exist. Let X =∑

n∈ω x2n, and assume that X is as in exerciseE13.10.

We claim that bm−1 = ∞. In fact, if bm−1 < ∞, then there is an m ∈ ω such thatbm−1 < 2m; then x2m = [2m, 2m+ 1) is disjoint from X according to the form of X , butx2m ≤ X by definition, contradiction. So our claim holds.

Now choose m ∈ ω so that am−1 < 2m+1. Then [2m+1, 2m+2)∩x2n = ∅ for all n,hence [2m+1, 2m+2)∩X = ∅. But [2m+1, 2m+2) ⊆ [am−1, bm−1) ⊆ X , contradiciton.

E13.12 Let A be the Boolean algebra of all subsets of some nonempty set X, under thenatural set-theoretic operations. Show that if 〈ai : i ∈ I〉 is a system of elements of A, then

∏

i∈I

(ai +−ai) = 1 =∑

ε∈I2

∏

i∈I

aε(i)i ,

where for any y, y1 = y and y0 = −y.

First note that the big products and sums are just the ordinary intersections and unions.Obviously ai + −ai = ai ∪ (X\ai) = X = 1, giving the first equality. Now suppose thatx ∈ X . We define

ε(i) =

1 if x ∈ ai,0 otherwise.

Clearly then x ∈ aε(i)i for each i ∈ I, and hence x is in the right side of the second equality,

as desired.

E13.13 Let M be the set of all finite functions f ⊆ ω × 2. For each f ∈M let

Uf = g ∈ ω2 : f ⊆ g.

Let A consist of all finite unions of sets Uf .(i) Show that A is a Boolean algebra under the set-theoretic operations.

46

(ii) For each i ∈ ω, let xi = U(i,1). Show that

ω2 =∏

i∈ω

(xi +−xi)

while ∑

ε∈ω2

∏

i∈ω

xε(i)i = ∅,

where for any y, y1 = y and y0 = −y.

(i): Obviously A is closed under ∪. Now suppose that a ∈ A; we want to show that(ω2)\a ∈ A. Say a =

⋃

f∈N Uf , where N is a finite subset of M . Let

P =

g ∈M : dmn(g) ⊆

⋃

f∈N

dmn(f) and ∀f ∈ N∃i ∈ dmn(g) ∩ dmn(f)[f(i) 6= g(i)]

.

Clearly P is a finite subset of M . We claim that (ω2)\a =⋃

g∈P Ug. First suppose thath ∈ (ω2)\a. Let g = h

⋃

f∈N dmn(f). So g ∈M and h ∈ Ug. We claim that g ∈ P . For,suppose that f ∈ N . Then Uf ⊆ a, so it follows that h /∈ Uf . So we can choose i ∈ dmn(f)such that f(i) 6= h(i). Clearly i ∈ dmn(g) and f(i) 6= g(i). This shows that g ∈ P , proving⊆ of our claim.

For ⊇, suppose that g ∈ P and h ∈ Ug. Suppose that h ∈ a. Choose f ∈ N suchthat h ∈ Uf , hence f ⊆ h. But g ⊆ H too, so there is an i ∈ dmn(g) ∩ dmn(f) such thatf(i) 6= g(i). But this means that f(i) 6= h(i), contradicting f ⊆ h. We have now shown(i).

Clearly xi ∪ ((ω2)\xi = ω2 for any i ∈ ω. Hence ω2 =∏

i∈ω(xi +−xi).

Now suppose that ε ∈ ω2; we want to show that∏

i∈ω xε(i)i = 0, i.e., that there is no

nonzero element a of A such that a ≤ xε(i)i for all i ∈ ω. suppose that a is such an element.

Then there is a g ∈ M such that Ug ⊆ a. Take any i /∈ dmn(g), and let h ∈ ω2 be any

function such that g ⊆ h and h(i) 6= ε(i). Then h ∈ Ug but h /∈ xε(i)i , contradiction.

E13.14 Suppose that (P,≤, 1) is a forcing order. Define

p ≡ q iff p, q ∈ P, p ≤ q, and q ≤ p.

Show that ≡ is an equivalence relation, and if Q is the collection of all ≡-classes, thenthere is a relation on Q such that for all p, q ∈ P , [p]≡ [q]≡ iff p ≤ q. Finally,show that (Q,) is a partial order, i.e., is reflexive on Q, transitive, and antisymmetric(q1 q2 q1 implies that q1 = q2); moreover, q ≤ [1] for all q ∈ Q.

Since ≤ is reflexive on P , clearly also ≡ is reflexive on P . Clearly ≡ is symmetric. Nowsuppose that p ≡ q ≡ r. Thus p ≤ q, q ≤ p, q ≤ r, and r ≤ q. Then p ≤ r and r ≤ p, sop ≡ r. So ≡ is an equivalence relation on P .

Let= (a, b) : ∃p, q ∈ P [p ≤ q, a = [p], and b = [q]].

47

Obviously then p ≤ q implies that [p] [q]. Now suppose that [p] [q]. Choose p′, q′ ∈ Psuch that p′ ≤ q′, [p] = [p′], and [q] = [q′]. Then p ≤ p′ and q′ ≤ q, so p ≤ q.

To show that is a partial order on Q, first suppose that a ∈ Q. Write a = [p]. Thenp ≤ p, so a a. Thus is reflexive on Q. Now suppose that a b c. Then there existp, q, q′, r such that p ≤ q, a = [p], b = [q], q′ ≤ r, b = [q′], and c = [r]. Then q ≤ q′ since[q] = [q′]. So p ≤ q ≤ q′ ≤ r, hence p ≤ r. So a = [q] [r] = c. This shows that istransitive. Finally, suppose that a b a. Then there exist p, q, q′, r such that p ≤ q,a = [p], b = [q], q′ ≤ r, b = [q′], and a = [r]. Then q ≤ q′ since [q] = [q′]. Also r ≤ p since[p] = [r], so q ≤ q′ ≤ r ≤ p, hence q ≤ p. But also p ≤ q, so a = [p] = [q] = b. So is apartial order. Clearly a ≤ [1] for all a ∈ Q.

E13.15 We say that (P,<) is a partial order in the second sense iff < is transitive andirreflexive. (Irreflexive means that for all p ∈ P , p 6< p.) Show that if (P,<) is a partialorder in the second sense and if we define by p q iff (p, q ∈ P and p < q or p = q),then A (P,) is a partial order. Furthermore, show that if (P,≤) is a partial order, andwe define p ≺ q by p ≺ q iff (p, q ∈ P , p ≤ q, and p 6= q), then B(P,≺) is a partial orderin the second sense.

Also prove that A and B are inverses of one another.

Clearly is reflexive on P . Now suppose that x y z. If x = y or y = z, then x zby supposition. If x < y < z, then x < z, and so x z. Thus is transitive. Supposethat x y x, but x 6= y. Then x < y < x, hence x < x, contradiction. So isantisymmetric. Hence (P,) is a partial order.

Now suppose that (P,≤) is a partial order, and define p ≺ q by p ≺ q iff (p, q ∈ P ,p ≤ q, and p 6= q). Clearly ≺ is irreflexive. Suppose that p ≺ q ≺ r. Then p ≤ q ≤ r, sop ≤ r. Suppose that p = r. Then p ≤ q ≤ p, so p = q by antisymmetry, contradiction.Thus p 6= r, and so p ≺ r. So (P,≺) is a partial order in the second sense.

Next, suppose that (P,<) is a partial order in the second sense, and let A (P,<) =(P,). Furthermore, let B(A (P,<)) = (P,<′). Then

p <′ q iff (p q and p 6= q) iff ((p < q or p = q) and p 6= q) iff p < q.

Thus B(A (P,<)) = (P,<).Finally, suppose that (P,≤) is a partial order. Let B(P,≤) = (P,≺), and let

A (B(P,≤)) = (P,≤′). Then

p ≤′ q iff (p ≺ q or p = q) iff ((p ≤ q and p 6= q) or p = q) iff p ≤ q.

E13.16 Show that if (P,≤, 1) is a forcing order and we define ≺ by p ≺ q iff (p, q ∈ P ,p ≤ q and q 6≤ p), then (P,≺) is a partial order in the second sense. Give an examplewhere this partial order is not isomorphic to the one derived from (P,≤, 1) by the procedureof exercise E13.14.

≺ is irreflexive, since x ≺ x would imply that x 6≤ x, a contradiction. For transitivity,suppose that x ≺ y ≺ z. Then x ≤ y and y ≤ z, so x ≤ z. Also, y 6≤ x and z 6≤ y. Supposethat z ≤ x. Then y ≤ z ≤ x and hence y ≤ x, contradiction. Hence z 6≤ x, and so x ≺ z.Thus (P,≺) is a partial order in the second sense.

48

For the example, letX be any infinite set, and let ≤ be X×X . Fix 1 ∈ X . So (X,≤, 1)is a quasiorder. The partial order constructed in exercise E13.14 has only one element,while the partial order of the present exercise has X , an infinite set, as its underlying set.Note that ≺ is empty.

E13.17 Prove that if (P,, 1) is a forcing order such that the mapping e from P intoRO(P ) is one-one, then (P,) is a partial order. Give an example of a forcing order suchthat e is not one-one. Give an example of an infinite forcing order Q such that e is notone-one, while for any p, q ∈ Q, p ≤ q iff e(p) ⊆ e(q).

Suppose that P is a forcing order such that e is one-one, and p ≤ q ≤ p. Then P ↓ p =P ↓ q, and hence e(p) = e(q). So p = q. Hence (P,≤) is a partial order.

For an example of a forcing order such that e is not one-one, take any simple orderingwith greatest element; see the remarks preceding Proposition 13.21.

For the final example, take any infinite set Q, and take the forcing order (Q,Q×Q, q)for any element q ∈ Q. So e is the constant function with value Q. For any p, q ∈ Q wehave p ≤ q and e(p) ⊆ e(q), so these statements are equivalent trivially.

E13.18 (Continuing E13.14.) Let P = (P,≤, 1) be a forcing order, and let Q = (Q,, [1])be as in exercise E13.14. Show that there is an isomorphism f of RO(P) onto RO(Q) suchthat f eP = eQ π, where π : P → Q is defined by π(p) = [p] for all p ∈ P .

We will apply Theorem 13.22. For any p ∈ P let j(p) = eQ(π(p)). Thus j : P → RO(Q).Suppose that 0 6= X ∈ RO(Q). By Theorem 13.20(i), choose q ∈ Q such that

eQ(q) ≤ X . Say q = [p]. Then j(p) = eQ(π(p)) ≤ X . So j[P ] is dense in RO(Q).Suppose that p, q ∈ P and p ≤ q. Then [p] [q], and so j(p) ≤ j(q) by Theorem

13.20(ii).Suppose that p, q ∈ P . If p 6⊥ q, choose r ≤ p, q. Then j(r) ≤ j(p), j(q), so j(p)∩j(q) 6=

∅. If j(p) ∩ j(q) 6= ∅, then by Theorem 13.20(iii), π(p) 6⊥ π(q). So there is an r ∈ Q suchthat r ≤ π(p), π(q). Say r = π(s). Then s ≤ p, q, so p 6⊥ q.

This verifies the hypotheses of Theorem 13.22, and the desired conclusion follows.Solutions to exercises in chapter 14

E14.1 Write out all the elements of Vα for α = 0, 1, 2, 3, 4.

V0 = ∅.

V1 = P(V0) = P(∅) = ∅ = 1.

V2 = P(V1) = P(∅) = ∅, ∅ = 2.

V3 = P(V2) = P(∅, ∅) = ∅, ∅, ∅, ∅, ∅. Note that V3 has four elements,but it is not equal to 4, since, for example, ∅ ∈ V3\4.

For V4, it helps to use the usual abbreviations for natural numbers. Thus V3 = 0, 1, 2, 1.We list out the subsets of V3 with 0,1,2,3,4 elements:

V4 = P(V3) = 0 0 elements; 1 of these

0, 1, 2, 1 1 element, 4 of these

49

0, 1, 0, 2, 0, 1, 1, 2, 1, 1, 2, 1 2 elements, 6 of these

0, 1, 2, 0, 1, 1, 0, 2, 1, 1, 2, 1 3 elements, 4 of these

0, 1, 2, 1 4 elements, 1 of these.

E14.2 Define by recursion

S(α) =⋃

β<α

P(S(β))

for every ordinal α. Prove that Vα = S(α) for every ordinal α.

First we apply the recursion theorem 8.7. Define G : On × V → V by setting, for anyordinal α and any set x,

G(α, x) =

⋃

β<αP(x(β)) if x is a function with domain α,∅ otherwise.

Then obtain F : On → V by Theorem 6.7: for any ordinal α, F(α) = G(α,F α). ThusF(α) =

⋃

β<αP(F(β)).

We prove that Vα = S(α) for all α by induction:

S(0) =⋃

β<0

P(S(β)) = ∅ = V0;

S(α+ 1) =⋃

β<α+1

P(S(β))

=⋃

β<α

P(S(β)) ∪P(S(α))

= Vα ∪P(Vα) (inductive hypothesis)

= Vα+1 (using Theorem 14.5(ii));

S(α) =⋃

β<α

P(S(β)) (with α limit)

=⋃

γ<α

⋃

β<γ

P(S(β))

=⋃

γ<α

Vγ (inductive hypothesis)

= Vα.

E14.3 Determine exactly the ranks of the following sets in terms of the ranks of the setsentering into their definitions. In some cases the rank is not completely determined by theranks of the constituents; in such cases, describe all possibilities.

(i) x (iv) x ∪ y (vii)⋃x (x) R−1

(ii) x, y (v) x ∩ y (viii) dmn(R)(iii) (x, y) (vi) x\y (ix) P(x)

50

(i): Let α = rank(x). Thus x ∈ Vα+1. Hence x ⊆ Vα+1, so x ∈P(Vα+1) = Vα+2. Sorank(x) ≤ α + 1. Suppose that rank(x) < α + 1. Then x ∈ Vα+1 = P(Vα), hencex ⊆ Vα, so x ∈ Vα, and so rank(x) < α, contradiction. Thus rank(x) = α+ 1.

(ii): Let α = rank(x) and β = rank(y). We claim that rank(x, y) = max(α, β) + 1.To prove this, by symmetry it suffices to assume that α ≤ β. Hence x, y ∈ Vβ+1, sox, y ⊆ Vβ+1, so x, y ∈ P(Vβ+1) = Vβ+2. Hence rank(x, y) ≤ β + 1. Suppose thatrank(x, y) < β + 1. Then x, y ∈ Vβ+1 = P(Vβ), so x, y ⊆ Vβ . Hence y ∈ Vβ ,contradiction.

(iii): rank((x, y)) = max(rank(x), rank(y)) + 2 by (i) and (ii).(iv): Let rank(x) = α and rank(y) = β. We claim that rank(x ∪ y) = max(α, β). To

prove this, by symmetry we may assume that α ≤ β. Thus x, y ∈ Vβ+1, so x, y ⊆ Vβ , hencex ∪ y ⊆ Vβ and so x ∪ y ∈ Vβ+1. Hence rank(x ∪ y) ≤ β. Suppose that rank(x ∪ y) < β.Then x ∪ y ∈ Vβ , so by exercise E14.2, x ∪ y ⊆ Vγ for some γ < β. So y ⊆ Vγ , hencey ∈ Vγ+1 ⊆ Vβ , contradiction.

(v): Let α and β be as in (iv). We claim that rank(x ∩ y) ≤ min(rank(x), rank(y)),and actually can be anything ≤ this minimum. To prove this, by symmetry assume thatα ≤ β. Then x ∩ y ⊆ x ⊆ Vα, so rank(x ∩ y) ≤ α.

For the second assertion, suppose that γ is any ordinal, and suppose that δ ≤ γ. Wedefine two sets x and y such that min(rank(x), rank(y)) = γ while rank(x ∩ y) = δ. Letx = δ∪γ and y = γ. Then rank(x) = γ+1, rank(y) = γ, and rank(x∩y) = rank(δ) = δ.

(vi): Let α = rank(x). We claim that rank(x\y) ≤ α, and it can be anything ≤ α. Infact, x\y ⊆ x ⊆ Vα, so x\y ∈ Vα+1, and so rank(x\y) ≤ α.

For the second assertion, let β ≤ α; we define x, y so that rank(x) = α whilerank(x\y) = β. Let x = α and y = α\β.

(vii): Let rank(x) = α. We claim that rank (⋃x) =

⋃α. (Thus it is α if α is limit or

0, and it is β if α = β + 1.) To prove this, first we show that⋃x ⊆ V⋃

α. For, suppose

that y ∈⋃x. Say y ∈ z ∈ x. Now x ∈ Vα+1, so x ⊆ Vα. Hence z ∈ Vα. Hence by exercise

E14.2, z ⊆ Vβ for some β < α. So y ∈ Vβ . Now β ⊆⋃α, so β ≤

⋃α. Hence y ∈ V⋃

α, as

desired.It follows that

⋃x ∈ V(

⋃α)+1, and hence rank (

⋃x) ≤

⋃α. Suppose that

⋃x ∈ V⋃α.

Then by exercise E14.2⋃x ⊆ Vγ for some γ <

⋃α. Say γ < δ < α. Then x ⊆ Vδ. In

fact, if y ∈ x, then y ⊆⋃x ⊆ Vγ , hence y ∈ Vγ+1 ⊆ Vδ. So, indeed, x ⊆ Vδ. Hence

x ∈ Vδ+1 ⊆ Vα, contradiction. This proves that rank (⋃x) =

⋃α.

(viii): Let α = rank(R). We claim, first of all, that rank(dmn(R)) ≤⋃⋃

α. For, takeany x ∈ dmn(R). Choose y such that (x, y) ∈ R. So x ∈ x ∈ (x, y) ∈ R; it follows thatx ∈

⋃⋃R. So dmn(R) ⊆

⋃⋃R, and so rank(dmn(R)) ≤ rank(

⋃⋃R) =

⋃⋃α by (vii).

Next we claim that if β ≤⋃⋃

α, then there is a set R such that rank(R) = α whilerank(dmn(R)) = β. To give the examples here, we consider two cases.

Case 1. β = 0. Then we take R = α. We use the easy fact that no ordinal is an orderedpair. [(a, b) has at most two elements, and is a nonempty set. So the only possibilities for(a, b) to be an ordinal are (a, b) = 1 or (a, b) = 2. Since (a, b) = a, a, b, neither caseis really possible, since the members of (a, b) are nonempty.]

Case 2. 0 < β. Let R = (ξ, 0) : ξ < β ∪ α. To see that this works, note thatdmn(R) = β = rank(β). But we also need to see that rank(R) ≤ α. For this we consider

51

several subcases.Subcase 2.1. α is a limit ordinal. Then

⋃⋃α = α, rank((ξ, 0) : ξ < β) ≤ α

Subcase 2.2. α = γ + 1 for some limit ordinal γ. Since β ≤ γ, clearly rank((ξ, 0) :ξ < β) ≤ γ < α

Subcase 2.3. α = γ + 2 for some limit ordinal γ. Similar to Subcase 2.2.Subcase 2.4. α = γ + n for some limit ordinal γ and some n ∈ ω\3. Then

β ≤ γ + n− 2, and so rank((ξ, 0) : ξ < β) ≤ γ + nSubcase 2.5. 0 < β ≤ α− 2, with α ∈ ω\3. Again clearly ok.

These are all of the possibilities.(ix): Let rank(x) = α. We claim that rank(P(x)) = α+1. Now x ∈ Vα+1, so x ⊆ Vα.

Hence y ⊆ Vα for every y ⊆ x. So P(x) ⊆ P(Vα) = Vα+1; hence P(x) ∈ Vα+2. Thisshows that rank(P(x)) ≤ α + 1. Suppose that P(x) ∈ Vα+1. Hence x ∈ P(x) ⊆ Vα, sox ∈ Vα, contradiction. Hence rank(P(x)) = α+ 1.

(x): We claim that for all ordinals α, β, ∃R[rank(R) = α and rank(R−1) = β] iff β ≤ αand one of the following conditions holds:

(1) β = γ + 3 for some ordinal γ.

(2) β is a limit ordinal.

To prove this, we first note that if a and b have ranks ξ, η respectively, then (a, b) hasrank max(ξ, η) + 2 by (iii). Hence if S is a collection of ordered pairs, then rank(S) =suprank(s)+1 : s ∈ S, and hence rank(S) is either a limit ordinal (if rank(s)+1 : s ∈ Sdoes not have a greatest element) or it is of the form γ + 3. It follows that if rank(R) = αand rank(R−1) = β, then β ≤ α and (1) or (2) holds.

Now suppose that β ≤ α. If β = γ + 3 for some γ, let R = (γ, γ) ∪ α; thenrank(R) = α and rank(R−1) = β. If β is a limit ordinal, let R = (γ, γ) : γ < β ∪α; thenrank(R) = α and rank(R−1) = β.

E14.4 Define xRy iff (x, 1) ∈ y. Show that R is well-founded and set-like on V.

Suppose that X is a nonempty set. Choose x ∈ X of smallest rank. Suppose that y ∈ Xand yRx. Thus (y, 1) ∈ x, so y ∈ y ∈ (y, 1) ∈ x, and hence rank(y) < rank(x),contradiction. Hence R is well-founded on V.

For any y ∈ V we have predVR(y) = x : (x, 1) ∈ y. Note that if (x, 1) ∈ y then x ∈x ∈ x, x, 1 = (x, 1) ∈ y, so x ∈

⋃⋃y. So pred

VR(y) = x ∈

⋃⋃y : (x, 1) ∈ y,

and hence predVR(y) is a set. Thus R is set-like on V.

E14.5 (Continuing E14.4) By recursion let y = (x, 1) : x ∈ y for any set y. Let F bethe Mostowski collapsing function for R,V in exercise E14.4. Prove that F(y) = y forevery set y.

First we show that the function ˇ exists. Let S = (x, y) : x ∈ y. So S is well-foundedand set-like on V. Define G : V ×V→ V by setting, for any sets y, f ,

G(y, f) =

(f(x), 1) : x ∈ y if f is a function with domain predVS(y),∅ otherwise.

52

Then let H be obtained from G by Theorem 5.7: for any set y, H(y) = G(y,H pred

VS(y)). Hence for any set y, H(y) = (H(x), 1) : x ∈ y, as desired.

Now for the main part of the exercise, suppose that it is not true. So there is a y suchthat F(y) 6= y. Let W = w ∈ trcl(y ∪ y) : F(w) 6= w. Thus y ∈ W , so W 6= ∅. By thefoundation axiom, choose z ∈W with z ∩W = ∅. Then

F(z) = F(s) : sRz (definition of F)

= F(s) : (s, 1) ∈ z (definition of R)

= F(t) : t ∈ z (definition of z)

= t : t ∈ z (since z ∩W = ∅)

= z,

contradiction.

E14.6 Define xRy iff x ∈ trcl(y). Show that R is well-founded and set like on V.

Clearly xRy implies that rank(x) < rank(trcl(y)) = rank(y), so R is well-founded. Forany a ∈ V, the class b ∈ V : bRa = b : b ∈ trcl(a) = trcl(a), a set.

E14.7 Let F be the Mostowski collapsing function for R,V. Show that F(x) = rank(x)for every set x.

Suppose not, and by the foundation axiom let x be a set such that F(x) 6= rank(x) whileF(y) = rank(y) for every y ∈ x. Note that F(x) is transitive: if u ∈ v ∈ F(x), choosey ∈ trcl(x) such that v = F (y). Then choose z ∈ trcl(y) such that u = F(z). Thenz ∈ trcl(x), and so u ∈ F(y). It follows that F(x) is an ordinal. Now

F(x) = F(y) : y ∈ V and yRx

= F(y) : y ∈ trcl(x)

= rank(y) : y ∈ trcl(x).

Now if y ∈ trcl(x), then rank(y) < rank(trcl(x)) = rank(x). So F(x) ≤ rank(x). Con-versely, if α ∈ rank(x), then α ∈ rank(trcl(x)) and hence α ≤ rank(y) for some y ∈ trcl(x).Since F(x) is an ordinal, it follows that α ∈ F(x).

Thus F(x) = rank(x), contradiction.

E14.8 Prove that if a is transitive, then rank(b) : b ∈ a is an ordinal.

Let S = rank(b) : b ∈ a, and let α be the least ordinal not in S. Thus α ⊆ S. Weclaim that α = S. Suppose not, and let β be the least member of S\α; so α < β. Thusthere is a b ∈ a such that β = rank(b). Now a is transitive, so if c ∈ b then c ∈ a andhence rank(c) ∈ S with rank(c) < β, so rank(c) < α. Hence β = supc∈b(rank(c) + 1) ⊆ α,contradiction.

E14.9 Show that for any set a we have rank(trcl(a)) = rank(a).

53

We use the notation in the proof of Theorem 14.6. First we prove that rank(dm) = rank(a)for every m ∈ ω, by induction on m. Obviously rank(d0) = rank(a). For the inductivestep we use Theorem 14.8:

rank(a) ≤ rank(dm+1)

= max(rank(dm), rank(⋃

dm))

= max(rank(a),⋃

b∈dm

rank(b))

≤ max(rank(a),⋃

b∈dm

rank(dm))

= max(rank(a), rank(a))

= rank(a).

Now rank(trcl(a)) = rank(a) by Exercise E14.3(vii).

E14.10 For any infinite cardinal κ, let H(κ) be the set of all x such that |trcl(x)| < κ.Prove that Vω = H(ω). (H(ω) is the collection of all hereditarily finite sets.) Hint:Vω ⊆ H(ω) is easy. For the other direction, suppose that x ∈ H(ω), let t = trcl(x), and letS = rank(y) : y ∈ t. Show that S is an ordinal.

First we show Vω ⊆ H(ω). Suppose that x ∈ Vω. Say x ∈ Vn with n ∈ ω. Then n = m+ 1for somem, and so x ⊆ Vm. Hence by Theorem 14.6, trcl(x) ⊆ Vm. So |trcl(x)| ≤ |Vm| < ω.It follows that x ∈ H(ω), as desired.

Conversely, suppose that x ∈ H(ω). Let t = trcl(x). So t is finite. Let S = rank(y) :y ∈ t. Thus S is finite, since t is. S is a set of ordinals; we claim that it actually isan ordinal. It suffices to show that S is transitive. Suppose that β ∈ α ∈ S; we mayassume that α is minimum with this property. Write α = rank(y) with y ∈ t. Now byTheorem 14.8(iv), rank(y) = supz∈y(rank(z) + 1). It follows that there is a z ∈ y suchthat β < rank(z) + 1. Now z ∈ t since t is transitive. Thus with γ = rank(z), we haveγ ∈ S, and hence we cannot have β < γ, as this would contradict the minimality of α. Soβ = γ ∈ S, proving our claim: S is a finite ordinal.

For each y ∈ t we have y ∈ VS+1, hence x ⊆ t ⊆ VS+1, so x ∈P(VS+1) = VS+2 ⊆ Vω,as desired.

E14.11 Which axioms of ZFC are true in On?

Extensionality holds, by Theorem 14.11.Comprehension fails. Let ϕ(x, z) be the formula ∃y(y ∈ x). Suppose that [∃w∀x(x ∈

w ↔ x ∈ 2 ∧ ∃y(y ∈ x))]On. So, choose an ordinal w such that ∀x ∈ On(x ∈ w ↔ x ∈2∧∃y ∈ On(y ∈ x)). Thus ∀x(x ∈ w↔ x ∈ 2∧x 6= ∅). So 1 ∈ w but 0 /∈ w, contradictingw being an ordinal.

Pairing holds, by Theorem 14.13.Union holds, by Theorem 14.14Power set holds, by Theorem 14.15. In fact, given an ordinal x, any subset of x which

is an ordinal is ≤ x itself, so we can take y = x.

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Replacement holds. To prove this, we use Theorem 14.16. So, suppose that ϕ is aformula with free variables among x, y, A, w1, . . . , wn, and suppose that A,w1, . . . , wn areordinals such that

∀x ∈ A∃!y[y ∈ On ∧ ϕOn(x, y, A, w1, . . . , wn)].

For each x ∈ A, let yx be the unique ordinal such that ϕOn(x, yx, A, w1, . . . , wn). Letα = supx∈A yx. Then clearly y ∈ On : ∃x ∈ AϕOn(x, y, A, w1, . . . , wn) ⊆ α, as desired.

Foundation holds, by Theorem 14.17.Infinity holds. This is clear if we write the axiom out:

∃x[∃y(y ∈ x ∧ ∀z(z /∈ y)∧

∀y ∈ x∃z ∈ x∀w(w ∈ z ↔ w ∈ y ∨ w = y)]

Clearly ω works for x.Axiom of choice. This holds trivially. Given an ordinal A , if A = 0, then the

conclusion holds if we take B to be any ordinal; and if A 6= 0, the hypothesis of theimplication in the axiom is false.

E14.12 Show that the power set operation is absolute for Vα for α limit.

We first show that the power set operation is defined in Vα. We take the followingformula as the official definition for this operation:

∀z ∈ y[z ⊆ x] ∧ ∀z[z ⊆ x→ z ∈ y].

Now given x ∈ Vα, we have rank(P(x)) = rank(x)+1 by the solution for exercise E14.3(ix).Hence P(x) ∈ Vα. Clearly y = P(x) satisfies the above formula. The uniqueness conditionis clear.

Since every subset of x is in Vα by Theorem 14.8(iii), clearly PVα(x) = P(x). SeeProposition 14.21.

E14.13 Let M be a countable transitive model of ZFC. Show that the power set operationis not absolute for M .

Note that ω ∈ M by Theorem 14.28. If the power set operation is absolute for M , thenPM (ω) = P(ω) ∈ M . But |P(ω)| = 2ω and M is transitive, so M is uncountable,contradiction.

E14.14 Show that Vω is a model of ZFC− Inf.

Extensionality: true by Theorem 14.11.Comprehension: assume that ϕ has free variables among x, z, w1, . . . , wn and we are

given z, w1, . . . , wn ∈ Vω. Let

A = x ∈ z : ϕVω (x, z, w1, . . . , wn).

Choose m ∈ ω so that z, w1, . . . , wn ∈ Vm. Then A ⊆ z ⊆ Vm, so A ∈ P(Vm) = Vm+1 ⊆Vω. So A ∈ Vω. Hence the comprehension axiom holds in Vω, by Theorem 14.12.

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Pairing: if x, y ∈ Vω, choose n such that x, y ∈ Vn. Then x, y ⊆ Vn, so x, y ∈P(Vn) = Vn+1 ⊆ Vω. So x, y ∈ Vω. By Theorem 14.13, the pairing axiom holds in Vω.

Union: if x ∈ Vω, choose n such that x ∈ Vn. Then⋃x ⊆ Vn, so

⋃x ∈ P(Vn) =

Vn+1 ⊆ Vω. By Theorem 14.14, the union axiom holds in Vω.Power set: if x ∈ Vω, choose m such that x ∈ Vn. Then y ⊆ Vn for each y ⊆ x, so

P(x) ⊆P(Vn) = Vn+1 ∈ Vn+1 ⊆ Vω. So the power set axiom holds by Theorem 14.15.Replacement: preparing to use Theorem 14.16, let ϕ be a formula with free variables

among x, y, A, w1, . . . , wn, suppose that A,w1, . . . , wn ∈ Vω, and also assume that

∀x ∈ A∃!y[y ∈ Vω ∧ ϕVω (x, y, A, w1, . . . , wn)].

Choose m ∈ ω such that A,w1, . . . , wn ∈ Vm. For each x ∈ A, let yx be the unique setsuch that

yx ∈ Vω ∧ ϕVω (x, yx, A, w1, . . . , wn)],

and let px ∈ ω be such that yx ∈ Vpx. Now A is finite, so there is a q ∈ ω such that m < q

and px < q for every x ∈ A. It follows that

y ∈ Vω : ∃x ∈ Aϕ(x, y, A, w1, . . . , wn) = yx : x ∈ A ⊆ Vq ∈ Vω,

as desired.Finally, foundation holds by Theorem 14.17.

E14.15 Show that the formula ∃x(x ∈ y) is not absolute for all nonempty sets, but it isabsolute for all nonempty transitive sets.

Let A = ∅. Then ∃x(x ∈ ∅) holds in V , but not in A, since there is no a ∈ A suchthat a ∈ ∅.

Now suppose that B is a nonempty transitive set, and y ∈ B. Then y has an elementiff it has an element in B, and so ∃x(x ∈ y) iff ∃x ∈ B(x ∈ y) iff (∃x(x ∈ y))B. So theformula is absolute for B. (Note that ∃x(x ∈ y) is not quite a ∆0 formula.)

E14.16 Show that the formula ∃z(x ∈ z) is not absolute for every nonempty transitiveset.

Take the transitive set 2. Then ∃z(1 ∈ z), but this does not hold in 2, since there is noz ∈ 2 such that 1 ∈ z.

E14.17 A formula is Σ1 iff it has the form ∃xϕ with ϕ a ∆0 formula; it is Π1 iff it hasthe form ∀xϕ with ϕ a ∆0 formula.

(i) Show that “X is countable” is equivalent on the basis of ZF to a Σ1 formula.(ii) Show that “α is a cardinal” is equivalent on the basis of ZF to a Π1 formula.

(i) Basically the following statement proves this:

X is countable iff ∃f [f is a one-one function and rng(f) ⊆ ω].

(ii) Basically the following statement proves this:

α is a cardinal iff ∀f [α is an ordinal and ∀β ∈ α(f : β → α→ f is not onto)].

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However, in each case we have more work to do, since the “insides” were not proved to be∆0 in this chapter; all we can really use are the definitions. For brevity, we say “is ∆0”instead of “is equivalent under ZF to a ∆0 formula”.

(1) “x is an ordinal” is ∆0. For,

x is an ordinal iff x is transitive and ∀y ∈ x[y is transitive].

(2) “x is a successor ordinal” is ∆0. For,

x is a successor ordinal iff x is an ordinal and ∃y ∈ x(x = y ∪ y)].

(3) “n is a natural number” is ∆0. For,

n is a natural number iff (n = ∅ or n is a successor ordinal)

and (∀y ∈ n[y = ∅ or y is a successor ordinal]).

(4) “a is an ordered pair” is ∆0. For,

a is an ordered pair iff ∃x ∈ a∃y ∈ a∃u ∈ x∃v ∈ y[x = u and y = u, v and a = x, y]

(5) “R is a relation” is ∆0. For,

R is a relation iff ∀a ∈ R[a is an ordered pair]

(6) If ϕ(a, f) is ∆0, then so is ∀a ∈⋃fϕ(a, f). For,

∀a ∈⋃

fϕ(a, f) iff ∀x ∈ f∀a ∈ xϕ(a, f).

(7) “f is a function” is ∆0. For,

f is a function iff f is a relation and ∀a ∈ f∀b ∈ f∀x ∈⋃a∀y ∈

⋃a∀z ∈

⋃b

[a = (x, y) and b = (x, z)→ y = z].

(8) “f is a one-one function” is ∆0. For,

f is a one-one function iff f is a function and ∀a ∈ f∀b ∈ f∀x ∈⋃a∀y ∈

⋃a∀z ∈

⋃b

[a = (x, y) and b = (z, y)→ x = z].

(9) “x is in the range of f” is ∆0. For,

x is in the range of f iff ∃a ∈ f∃y ∈⋃a[a = (y, x)].

(10) “x is in the domain of f” is ∆0. For,

x is in the domain of f iff ∃a ∈ f∃y ∈⋃a[a = (x, y)].

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(11) x = dmn(f) is ∆0. For,

x = dmn(f) iff ∀y ∈ x(y is in the domain of f) and

∀a ∈ f∀y ∈ a∀u ∈ a(u is in the domain of f implies that u ∈ x).

Now we can give the details on (i) and (ii):

X is countable iff ∃f [f is a one-one function and X = dmn(f)

and ∀a ∈ f∀x ∈⋃a[x is in the range of f implies that x is a natural number]].

α is a cardinal iff ∀f [α is an ordinal and

∀β ∈ α[f is a function and β = dmn(f) implies that ∃x ∈ α(x is not in the range of f)]

E14.18 Prove that if κ is an infinite cardinal, then H(κ) ⊆ Vκ.

Let a ∈ H(κ). Let α = rank(b) : b ∈ trcl(a). Thus α is an ordinal by exercise E14.8. Byexercise E14.9 and Theorem 14.8(iv) rank(a) = rank(trcl(a)) ≤ α+ 1. Now |trcl(a)| < κ,and 〈rank(b) : b ∈ trcl(a)〉 maps trcl(a) onto α, so |rank(a)| ≤ |trcl(a)| < κ. Hencerank(a) < κ, as desired.

E14.19 Prove that for κ regular, H(κ) = Vκ iff κ = ω or κ is inaccessible.

⇒: Assume that H(κ) = Vκ and κ is uncountable. Suppose that λ < κ ≤ 2λ. Thusλ ∈ Vλ+1, and so P(λ) ∈ Vλ+2 ⊆ Vκ. But |P(λ) = 2λ ≥ κ, so P(λ) /∈ Vκ.⇐: κ = ω implies that H(κ) = Vκ by exercise E14.14. Now suppose that κ is

inaccessible. By exercise E14.18 we have H(κ) ⊆ Vκ. Now suppose that S ∈ Vκ. Letα = rank(S). Then also α = rank(trcl(S)) by exercise E14.14. So trcl(S) ∈ Vα+1, hencetrcl(S) ⊆ Vα+1 and so |trcl(S)| ≤ |Vα+1| = iβ with ω + β = α + 1. Now β < κ andiβ < iκ = κ, so S ∈ H(κ).

E14.20 Assume that κ is an infinite cardinal. Prove the following:(a) H(κ) is transitive.(b) H(κ) ∩On = κ.(c) If x ∈ H(κ), then

⋃x ∈ H(κ).

(d) If x, y ∈ H(κ), then x, y ∈ H(κ).(e) If y ⊆ x ∈ H(κ), then y ∈ H(κ).(f) If κ is regular and x is any set, then x ∈ H(κ) iff x ⊆ H(κ) and |x| < κ.

(a): Suppose that x ∈ y ∈ H(κ). Then |trcl(y)| < κ. Moreover, x ∈ trcl(y), hencex ⊆ trcl(y), hence trcl(x) ⊆ trcl(y). So |trcl(x)| < κ.

(b): If α ∈ H(κ) ∩ON, then |α| = |trcl(α)| < κ, and so α < κ. Conversely, if α ∈ κ,then |trcl(α)| = |α| < κ, so α ∈ H(κ).

(c): If y ∈⋃x, then ∃z[y ∈ z ∈ x], hence ∃z[y ∈ z ∈ trcl(x)], hence y ∈ trcl(x). Thus

⋃x ⊆ trcl(x), hence trcl(

⋃x) ⊆ trcl(x). So

⋃x ∈ H(κ).

(d): Note that trcl(x) ∪ x is transitive and contains x. If b is any transitive setcontaining x, then x ∈ b, hence trcl(x) ⊆ b, so that trcl(x) ∪ x ⊆ b. This proves thattrcl(x) = trcl(x) ∪ x. Hence |trcl(x)| < κ.

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Similarly, trcl(x, y) = trcl(x) ∪ trcl(y), so |trcl(x, y)| < κ. So x, y ∈ H(κ).(e): obvious.(f): Suppose that κ is regular and x is any set. If x ∈ H(κ), then x ⊆ H(κ) by

(a), and |x| ≤ |trcl(x)| < κ. Now suppose that x ⊆ H(κ) and |x| < κ. Now clearlytrcl(x) = x ∪

⋃

y∈x trcl(y). Hence |trcl(x)| ≤ 1 +∑

y∈x |trcl(y)| < κ; so x ∈ H(κ).

E14.21 Show that if κ is regular and uncountable, then H(κ) is a model of all of the ZFCaxioms except possibly the power set axiom.

Extensionality: true since H(κ) is transitive.

Comprehension: using Theorem 14.12, it suffices to take a formula ϕ with free vari-ables among x, z, w1, . . . , wn, assume that z, w1, . . . , wn ∈ H(κ), and prove that x ∈z : ϕH(κ)(x, z, w1, . . . , wn) ∈ H(κ). This is true since the indicated set is a subset of z;hence its transitive closure is a subset of trcl(z), which has size less than κ.

Pairing: Given x, y ∈ H(κ), clearly x, y ∈ H(κ).

Union: given x ∈ H(κ), clearly⋃x ⊆ trcl(x), hence trcl(

⋃x) ⊆ trcl(x), and so

⋃x ∈

H(κ).Replacement: Suppose that ϕ is a formula with free variables among those in the list

x, y, A, w1, . . . , wn, A,w1, . . . , wn ∈ H(κ), and

∀x ∈ A∃!y[y ∈ H(κ) ∧ ϕH(κ)(x, y, A, w1, . . . , wn)].

So for each x ∈ A let yx ∈ H(κ) such that ϕH(κ)(x, y, A, w1, . . . , wn) holds. Let Y = yx :x ∈ A. Then trcl(Y ) =

⋃

x∈A trcl(yx), and this has size less than κ since κ is regular.

Clearly z ∈ H(κ) : ∃x ∈ AϕH(κ)(x, y, A, w1, . . . , wn) ⊆ Y .Infinity: obviously ω ∈ H(κ); see Theorem 14.27.Foundation: true since H(κ) is transitive.Axiom of choice: obvious.


E15.1 Let I and J be sets with I infinite and |J | > 1, and let P = (P,≤, ∅), where P isthe collection of all finite functions contained in I × J and ≤ is ⊇ restricted to P . Showthat P satisfies the condition of Lemma 15.2.

Suppose that p ∈ P . Pick any i ∈ I\dmn(p), and let j, k be distinct elements of J . Thenp ⊆ p ∪ (i, j), p ⊆ p ∪ (i, k), and these two extensions of p are incompatible.

E15.2 Show that if the condition in the hypothesis of Lemma 15.2 fails, then there is aP-generic filter G over M such that G ∈M , and G intersects every dense subset of P (notonly those in M). [Cf. Lemma 15.1.]

Let p be such that for all q, r, if q and r are ≤ p then they are compatible. Define

G = q : ∃r[r ≤ q and (r ≤ p or p ≤ r)].

We claim that G is as desired. Clearly G ∈M .

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Note that p ∈ G, by taking r = p.To check (1), suppose that q, r ∈ G; we want to find s ∈ G with s ≤ q, r. Choose

t ≤ q such that t ≤ p or p ≤ t, and choose u ≤ r such that u ≤ p or p ≤ u. If t ≤ p andu ≤ p, then by the choice of p there is a v ≤ t, u. Then v ≤ u ≤ p, and v ≤ t ≤ q, hencev ∈ G, and v ≤ q and v ≤ u ≤ r, as desired.

If t ≤ p and p ≤ u, then t ≤ u ≤ r, so t ∈ G and t ≤ q, r, as desired.If p ≤ t and u ≤ p, then u ≤ r implies that u ∈ G, and u ≤ p ≤ t ≤ q, as desired.Finally, if p ≤ t, u, then p ≤ q, r and p ∈ G, as desired. So (1) holds.(2) is obvious.Now suppose that D is dense. Choose q ∈ D such that q ≤ p. Then q ∈ G, as desired.

E15.3 Assume the hypothesis of Lemma 15.2. Show that there does not exist a P-genericfilter over M which intersects every dense subset of P (not only those which are in M).Hint: Take G generic, and show that p ∈ P : p /∈ G is dense. Thus in the definition ofgeneric filter, the condition on dense sets being in M is necessary.

Let D be the set indicated in the hint. Let p be any element of P , and choose incompatibleq, r ≤ p by the hypothesis of Lemma 15.2. Then it is not true that both q, r ∈ G, as desired;this checks that D is dense. Obviously G ∩D = ∅, proving the assertion of the exercise.

E15.4 Show that if P satisfies the condition of Lemma 15.2, then it has uncountably manydense subsets.

(Solution due to Josh Sanders) For each p ∈ P let p(0), p(1) be members of P such thatp(0), p(1) ≤ p and p(0) ⊥ p(1); thus p(0), p(1) < p. We now define an element pf foreach finite sequence f of 0s and 1s, by recursion on the domain of f . Let p∅ be anyelement of P . Having defined pf , let pf0 = pf (0) and pf1 = pf (1). For each f ∈ ω2let Kf = pfm : m ∈ ω. Now if f, g ∈ ω2 and f 6= g, choose m minimum such thatf(m) 6= g(m). Clearly then pf(m+1) ∈ Kf\Kg. Thus Kf 6= Kg for distinct f, g ∈ ω2. Foreach f ∈ ω2 let Df = P\Kf . So Df 6= Dg for f 6= g.

We claim that each Df is dense; this will prove the statement of the exercise. To seethis, take any q ∈ P . If q ∈ Df , then there is nothing to prove. Suppose that q /∈ Df .Thus q ∈ Kf . Say q = pfm. Let ε = 1 − f(m). Then p(fm)ε ∈ Df and p(fm)ε ≤ q, asdesired.

E15.5 Assume the hypothesis of Lemma 15.2. Show that there are 2ω filters which areP-generic over M .

Let D0, D1, . . . list all of the dense subsets of P which are in M . We now define elementsrf and pf in P for f a finite sequence of 0’s and 1’s, by recursion on the length of f . Letp∅ = 1 and choose r∅ ∈ D0 so that r∅ ≤ p∅. Now suppose that pf and rf have beendefined, with f having domain n ∈ ω. Choose pf0 and pf1 both ≤ rf so that pf0 ⊥ pf1.Then choose rf0 ≤ pf0 with rf0 ∈ Dn+1, and choose rf1 ≤ pf1 with rf1 ∈ Dn+1.

For any f ∈ ω2 let

Gf = q ∈ P : pfn ≤ q for some n ∈ ω.

Clearly Gf is P-generic; and there are 2ω such filters.

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E15.6 Let P = (1,≤, 1). Prove that the collection of all P-names is a proper class.

Clearly the following two facts hold with no assumption about P .

(1) If σ is a P -name, then so is (σ, 1).

(2) If A is a set of P -names, then (σ, 1) : σ ∈ A is a P -name.

Now let A be the collection of all P -names, and suppose that A is a set. By (2), also

τdef= (σ, 1) : σ ∈ A is a P -name, and by (1), (τ, 1) is a P -name. So σ

def= (τ, 1) ∈ A.

Thus

τ ∈ τ ∈ τ, τ, 1 = (τ, 1) ∈ (τ, 1) = σ ∈ σ ∈ σ, σ, 1 = (σ, 1) ∈ τ,

contradiction.

E15.7 Show that p σ = τ iff the following two conditions hold.(i) For every (ξ, q) ∈ σ and every r ≤ p, q one has r ξ ∈ τ .(ii) For every (ξ, q) ∈ τ and every r ≤ p, q one has r ξ ∈ σ.

First assume that p σ = τ . For (i), suppose that (ξ, q) ∈ σ and r ≤ p, q. Let G beP-generic over M with r ∈ G. Then also p, q ∈ G, so ξG ∈ σG and σG = τG. HenceξG ∈ τG, as desired. (ii) is similar.

Second assume the two conditions. To show that p σ = τ , let G be P-generic overM with p ∈ G. Suppose that x ∈ σG. Then there is a (ξ, q) ∈ σ such that q ∈ G andx = ξG. Choose r ∈ G such that r ≤ p, q. By (i) we have ξG ∈ τG. This shows thatσG ⊆ τG. The other inclusion is similar.

E15.8 Assume that P ∈ M , p, q ∈ P , and p ⊥ q. Show that τ ∈ MP : p τ = ∅ is aproper class in M .

In M we define members τα of MP by recursion, such that p τα = 0, and such thatthe ranks increase. let τ0 = 0. Having defined τα, let τα+1 = (τα, q). Note that if G isgeneric and p ∈ G, then q /∈ G, and so (τα+1)G = ∅; so p τα+1 = ∅. For λ a limit ordinal,let τλ = (τα, q) : α < λ. Clearly p τλ = 0.

E15.9 A forcing order is separative iff it is antisymmetric (p ≤ q ≤ p implies that p = q),and for all p, q, if p 6≤ q then there is an r ≤ p such that r ⊥ q. Show that the forcingorder of exercise E15.1 is separative.

If p ⊇ q ⊇ p, then p = q. Now suppose that p 6⊇ q. Choose (i, j) ∈ q\p. If i ∈ dmn(p),then p ⊥ q, as desired. If i /∈ dmn(p), let k ∈ J\j and define r = p ∪ (i, k). Thenr ⊇ p and r ⊥ q.

E15.10 Assume that P ∈ M is separative and p, q, r ∈ P . Prove that the following twoconditions are equivalent:

(i) p ((∅, q), r) = 1.(ii) p ≤ r and p ⊥ q.

⇒: Assume that

(1) p ((0, q), r) = 1.

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Suppose that p 6≤ r. By the definition of separative, choose s such that s ≤ p and s ⊥ r.Let G be P-generic over M with s ∈ G. Then ((0, q), r)G = 0, contradiction. Thusp ≤ r. Suppose that p and q are compatible; say t ≤ p, q. Let G be P-generic over Mwith t ∈ G. Then q ∈ G, so (0, q)G = 0. Also r ∈ G, so ((0, q), r)G = 0 6= 1,contradiction.⇐: Suppose that p ≤ r and p ⊥ q. Suppose that G is P-generic over M and p ∈ G.

Then q /∈ G, so (0, q)G = 0. Also, r ∈ G, so ((0, q), r)G = (0, q)G = 0 = 1, asdesired.

E15.11 Suppose that f : A→ M with f ∈ M [G]. Show that there is a B ∈ M such that

f : A→ B. Hint: let f = τG and B = b : ∃p ∈ P [p b ∈ rng(τ)].

In the hint, the definition of B takes place in M ; so B ∈M . Suppose that b is in the rangeof f . Thus bG = b ∈ rng(τG), so we can choose p ∈ B such that p b ∈ rng(τ). So b ∈ B,as desired.

E15.12 Assume that P ∈ M and α is a cardinal of M . Then for any P-generic G overM the following conditions are equivalent:

(1) For all B ∈M , αB ∩M = αB ∩M [G].(2) αM ∩M = αM ∩M [G].

(1)⇒(2): Assume (1). Obviously ⊆ in (2) holds. Now suppose that f ∈ αM ∩M [G]. Byexercise E15.11 choose B ∈M such that f : α→ B. So by (1), f ∈M , as desired.

(2)⇒(1): Assume (2). Then ⊆ in (1) is clear. Suppose that f ∈ αB ∩M [G]. Thenf ∈ αM ∩M [G] since M is transitive, so f ∈M by (2).

E15.13 Suppose that P ∈ M is a forcing order satisfying the condition of Lemma 15.2.Assume that

M = M0 ⊆M1 ⊆M2 ⊆ · · · ⊆Mn ⊆ · · · (n ∈ ω),

where Mn+1 = Mn[Gn] for some Gn which is P-generic over Mn, for each n ∈ ω. Showthat the power set axiom fails in

⋃

n∈ωMn.

Assume that R =⋃

n∈ωMn does satisfy the power set axiom. Then R |= ∃y∀z(z ⊆P → z ∈ y). Choose y ∈ R so that R |= ∀z(z ⊆ P → z ∈ y). Say y ∈ Mn. ThenR |= Gn ⊆ P → z ∈ y. By absoluteness, R |= Gn ⊆ P . So R |= Gn ∈ y, henceGn ∈ y ∈Mn. This contradicts Lemma 15.2.

E15.14 Prove that the following conditions are equivalent:

[[ϕ(σ0, . . . , σm−1)↔ ψ(σ0, . . . , σm−1)]] = 1

[[ϕ(σ0, . . . , σm−1)]] = [[ψ(σ0, . . . , σm−1)]].

We omit the parameters σ0, . . . , σm−1. First assume that [[ϕ↔ ψ]] = 1. Then

0 = −[[ϕ↔ ψ]]

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= −[[(ϕ→ ψ) ∧ (ψ → ϕ)]]

= −[[¬(ϕ ∧ ¬ψ) ∧ ¬(ψ ∧ ¬ϕ)]]

= −([[¬(ϕ ∧ ¬ψ)]] · [[¬(ψ ∧ ¬ϕ)]]

= −(−[[ϕ ∧ ¬ψ) · −[[ψ ∧ ¬ϕ]])

= [[ϕ ∧ ¬ψ]] + [[ψ ∧ ¬ϕ]]

= ([[ϕ]] · −[[ψ]]) + ([[ψ]] · −[[ψ]].

It follows that [[ϕ]] = [[ψ]].

Conversely, if [[ϕ]] = [[ψ]], then we can reverse the above equations to get−[[ϕ↔ ψ]] = 0,so that [[ϕ↔ ψ]] = 1.

E15.15 Prove that [[σ = τ ]] · [[τ = ρ]] ≤ [[σ = ρ]].

Suppose not. Then [[σ = τ ]] · [[τ = ρ]] · −[[σ = ρ]] 6= 0. By Theorem 9.20(i) choose p sothat e(p) ≤ [[σ = τ ]] · [[τ = ρ]] · −[[σ = ρ]]. Then (p ∗ σ = τ)M , (p ∗ τ = ρ)M , andp ∗ (¬σ = ρ)M . Let G be P-generic over M . Then by Theorem 15.19, σG = τG, τG = ρG,and σG 6= ρG, contradiction.

E15.16 Prove that if ZFC |= ϕ then [[ϕ]] = 1, for any sentence ϕ.

Suppose that [[ϕ]] 6= 1. Thus −[[ϕ]] 6= 0, so by Theorem 13.20(i) choose p so that e(p) ≤−[[ϕ]] = [[¬ϕ]]. Thus (p ∗ ¬ϕ)M . Let G be P-generic over M . Then by Theorem 15.19 wehave ¬ϕM [G].


E16.1 Show that Fn(ω1, 2, ω1) preserves cardinals ≥ ω2.

Clearly |Fn(ω1, 2, ω1)| = ω1, so Fn(ω1, 2, ω1) is ω2-cc. Hence the result follows by Propo-sition 16.5.

E16.2 A system 〈Ai : i ∈ I〉 of sets is an indexed ∆-system iff there is a set r (againcalled the root such that Ai ∩ Aj = r for all distinct i, j ∈ I. Note that in an indexedsystem 〈Ai : i ∈ I〉 it is possible to have distinct i, j ∈ I such that Ai = Aj; in fact, all ofthe Ai’s could be equal, in which case the system is already an indexed ∆-system.

Prove that if κ is an uncountable regular cardinal and 〈Ai : i ∈ I〉 is a system of finitesets with |I| ≥ κ, then there is a J ∈ [I]κ such that 〈Ai : i ∈ J〉 is an indexed ∆-system.

We may assume that |I| = κ. Define i ≡ j iff i, j ∈ I and Ai = Aj . Thus ≡ is anequivalence relation on I. If some equivalence class K has κ elements, then 〈Ai : i ∈ K〉is an indexed ∆ system with |K| = κ, as desired; the kernel is Ai for any i ∈ K. Supposethat every equivalence class has size less than κ. Then there are κ equivalence classes. LetJ ⊆ I have one element from each equivalence class, and let A = Aj : j ∈ J. ThenA is a collection of finite sets, and |A | = κ. Hence by Theorem 16.6 let B ∈ [A ]κ be a∆-system. For each B ∈ B there is an iB ∈ J such that AiB = B. Let K = iB : B ∈ B.Then 〈Aj : j ∈ K〉 is an indexed ∆-system, and |K| = κ since the function i is clearlyone-one.

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E16.3 Here we work only in ZFC (or in a fixed model of it). Suppose that (X,<) is a linearorder. Let P be the set of all pairs (p, n) such that n ∈ ω and p ⊆ X×n is a finite function.Define (p, n) ≤ (q,m) iff m ≤ n, dmn(q) ⊆ dmn(p), ∀x ∈ dmn(q)[p(x) ∩m = q(x), and

∀x, y ∈ dmn(q), if x < y then p(x)\p(y) ⊆ m.

Show that P has ccc.

Suppose that A is an uncountable subset of P . By the ∆-system theorem, we may assumethat 〈dmn(p) : (p, n) ∈ A 〉 is a ∆-system, say with root r. We may also assume thatp r = q r whenever (p, n), (q,m) ∈ A . Now suppose that (p, n), (q,m) ∈ A . Let s bethe maximum of m and n. Clearly p∪ q is a function, and so (p∪ q, s) ∈ P . We claim that(p∪q, s) ≤ (p, n), (q,m), as desired. By symmetry it suffices to show that (p∪q, s) ≤ (q,m).Suppose that x ∈ dmn(q). Then (p ∪ q)(x) ∩m = q(x) ∩m = q(x). If x, y ∈ dmn(q) andx < y, then (p ∪ q)(x)\(p ∪ q)(y) = q(x)\q(y) ⊆ m.

E16.4 Continuing exercise E16.3, suppose that we are working in a c.t.m. M of ZFC.Let G be P-generic over M . For each x ∈ X let

ax =⋃

p(x) : (p, n) ∈ G for some n ∈ ω, with x ∈ dmn(p).

Thus ax ⊆ ω. Show that if x < y, then ax\ay is finite.For each z ∈ X let Dz = (p, n) : z ∈ dmn(p). Given any (q,m) ∈ P , if z /∈ dmn(q)

clearly (q ∪ (z, 0), m) ∈ P , (q ∪ (z, 0), m) ∈ Dz, and (q ∪ (z, 0), m) ≤ (q,m). So Dzis dense.

Choose (p, n) ∈ Dx∩G and (q,m) ∈ Dy ∩G. Say (p, n), (q,m) ≥ (r, s) ∈ G. We claimthen that ax\ay ⊆ s. Let i ∈ ax\ay. Say i ∈ u(x) with (u, t) ∈ G and x ∈ dmnu. Say(u, t), (r, s) ≥ (v, z) ∈ G. Thus v(x)\v(y) ⊆ s. Now i ∈ u(x), so i ∈ v(x). Also, i /∈ v(y)since i /∈ ay. So i ∈ s, as desired.

E16.5 Continuing exercises E16.3 and E16.4, show that if x < y, then ay\ax is infinite.Hint: for each i < ω let

Ei = (p, n) : x, y ∈ dmn(p) and |p(y)\p(x)| ≥ i,

and show that Ei is dense.

For each i < ω let

Ei = (p, n) : x, y ∈ dmn(p) and |p(y)\p(x)| ≥ i.

We claim that Ei is dense. Let (q, n) be given. Wlog x, y ∈ dmn(q). Say dmn(q) is

u0 < · · · < uj = x < · · · < um−1.

Let dmn(r) = dmn(q), r(ut) = q(ut) for t ≤ j; choose w > n with |w − n| = i, and letr(ut) = q(ut) ∪ (w\n) for j < t. Then (q, n) ≥ (r, w) ∈ Ei, as desired.

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Now for any i ∈ ω we show that |ay\ax| ≥ i. Choose (p, n) ∈ Ei ∩ G. We claimthat p(y)\p(x) ⊆ ay\ax (as desired). Let j ∈ p(y)\p(x). So j ∈ ay, and j < n sincep(y) ⊆ n. Suppose that j ∈ ax. Say y ∈ q(x), with (q, v) ∈ G and x ∈ dmn(q). Say(p, n), (q, v) ≥ (r, s) ∈ G. Then j ∈ r(x) since j ∈ q(x). Hence j ∈ r(x) = p(x) ∩ n, soj ∈ p(x), contradiction.

E16.6 Define a set A of finite sets with |A | = ω while there is no ∆-system B ∈ [A ]ω.

Let A = ω. Suppose that B ∈ [A ]ω is a ∆-system, say with kernel r. Choose m ∈ ω withn < m for all n ∈ r. Then m ∩ (m+ 1) = m 6= r, contradiction.

E16.7 Let κ be singular. Define a set A of finite sets with |A | = κ while there is no∆-system B ∈ [A ]κ.

Suppose that κ is uncountable and singular. Let 〈λξ : ξ < cf(κ)〉 be a strictly in-creasing continuous sequence of cardinals with supremum κ. Let A = λξ, α : ξ <cf(κ), λξ < α < λξ+1. Clearly |A | = κ. Suppose that B ∈ [A ]κ is a ∆-system, say withkernel G. If G = ∅, then for each ξ < cf(κ), B has at most one member in [λξ, λξ+1), andso |B| ≤ cf(κ) < κ, contradiction. Let α ∈ G. Say λξ ≤ α < λξ+1. Now B has somemember F not in [λξ, λξ+1), as otherwise |B| ≤ λξ+1. Then G 6⊆ F , contradiction.


E17.1 Show that Theorem 17.2 does not extend to ω1. Hint: consider ω1×Q and ω∗1×Q,

both with the lexicographic order, where ω∗1 is ω1 under the reverse order (α <∗ β iff

β < α).

Clearly each of these linear orders is dense with no first or last element. In fact, concerningthe first linear order, suppose that (ξ, r) < (η, s). If ξ < η, then (ξ, r) < (ξ, r+ 1) < (η, s).If ξ = η, then r < s, and so

(ξ, r) <

(

ξ,r + s

2

)

< (ξ, s).

Thus ω1 ×Q is dense. It does not have a first or last element, since if (ξ, r) is given, then(0, r − 1) < (ξ, r) < (ξ, r+ 1). The other order is treated similarly.

We claim that these two linear orders are not isomorphic. Suppose to the contrarythat f is an isomorphism of ω1 ×Q onto ω∗

1 ×Q. For all ξ < ω1 let f(ξ, 0) = (α(ξ), q(ξ)).Now if ξ < η < ω1, then α(ξ) ≥ α(η). Hence

(1) There is a ρ < ω1 such that for all ξ ∈ [ρ, ω1) we have α(ξ) = α(ρ).

In fact, suppose not. Thus for every ρ < ω1 there is a ξ ∈ (ρ, ω1) such that α(ξ) < α(ρ).Define 〈ρn : n ∈ ω〉 by recursion as follows. Let ρ0 = 0. If ρn has been defined, chooseρn+1 > ρn such that α(ρn+1) < α(ρn). Then 〈α(ρn) : n ∈ ω〉 is a strictly decreasingsequence of ordinals, contradiction.

Thus (1) holds. Now 〈q(ξ) : ρ ≤ ξ < ω1〉 is a strictly increasing sequence of rationals,contradicting the fact that |Q| = ω.

E17.2 For any infinite cardinal κ, consider κ2 under the lexicographic order, as for Hα.Show that it is a complete linear order.

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Let A ⊆ κ2; we want to find a lub for A. We define f ∈ κ2 by recursion, as follows: forany α < κ,

f(α) =

1 if there is a g ∈ A such that g α = f α and g(α) = 1,0 otherwise.

Suppose that g ∈ A and f < g. Then f χ(f, g) = g χ(f, g), f(χ(f, g)) = 0, andg(χ(f, g)) = 1. This contradicts the definition of f . (We are using χ(f, g) in the same wayas for Hα.) Thus g ≤ f for all g ∈ A.

Suppose that h is an upper bound for A and h < f . Thus h χ(h, f) = f χ(h, f),h(χ(h, f)) = 0, and f(χ(h, f)) = 1. By the definition of f , there is a g ∈ A such thatχ(g, f) = χ(h, f) and g(χ(g, f) = 1. But then χ(g, h) = χ(h, f) too, and it follows thath < g. This contradicts h being an upper bound for A.

Thus f is the desired lub of A.

E17.3 Suppose that κ and λ are cardinals, with ω ≤ λ ≤ κ. Let µ be minimum such thatκ < λµ. Take the lexicographic order on µλ, as for Hα. Show that this gives a dense linearorder of size λµ with a dense subset of size κ.

Note that obviously µ ≤ κ. Clearly µλ is a linear order. Let

D = f ∈ µλ : there is a ξ < µ such that f(ξ) = 1 and f(η) = 0 for all η ∈ (ξ, µ).

Clearly |D| ≤ κ. We can show that µλ is dense, and D is dense in it, by finding an elementof D between any two elements f < g of µλ. For brevity let α = χ(f, g). Now define h by:

h(β) =

f(β) if β ≤ α;f(α+ 1) + 1 if β = α+ 1;1 if β = α+ 2;0 if β ∈ (α+ 2, µ).

Clearly h is as desired.If |D| < κ, we can simply add κ elements to it to satisfy the requirement that |D| = κ.

E17.4 Show that P(ω) under ⊆ contains a chain of size 2ω. Hint: remember that|ω| = |Q|.

Let f : ω → Q be a bijection. For each real number r, let ar = f−1[s ∈ Q : s < r].Clearly r < t implies that ar ⊆ at; and in fact ar ⊂ at since there is a rational numberu such that r < u < t, and so f−1(u) ∈ at\ar. Now a is an isomorphic embedding byProposition 4.14.

E17.5 A subset S of a linear order L is weakly dense iff for all a, b ∈ L, if a < b thenthere is an s ∈ S such that a ≤ s ≤ b. Show that the following conditions are equivalentfor any cardinals κ, λ such that ω ≤ κ ≤ λ:

(i) There is a linear order of size λ with a weakly dense subset of size κ.(ii) P(κ) has a chain of size λ.

⇒: Clearly we may assume that κ < λ. Assume that L is a linear order with a weaklydense subset D of size κ. Let f be a bijection from D to κ. For each r ∈ L\D let

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ar = f [d ∈ D : d < r]. thus ar ∈ P(κ), and ar ⊆ as if r, s ∈ L\D with r < s. Also,by the weak denseness of D, there is a d ∈ D such that r ≤ d ≤ s. Since r, s ∈ L\D, wehave r < d < s, and so d ∈ as\ar. Thus r < s implies that ar ⊂ as. The other implicationfollows from this one. Note that |L\D| = λ.⇐: Again we may assume that κ < λ. Let L be a chain in P(κ) of size λ. For each

α < κ let

xα =⋃

a : a ∈ L, α /∈ a.

(1) If α, β < κ, then xα ⊆ xβ or xβ ⊆ xα.

For, suppose that γ ∈ xα\xβ and δ ∈ xβ\xα. Say γ ∈ a ∈ L with α /∈ a and δ ∈ b ∈ Lwith β /∈ b. Also, β ∈ a and α ∈ b, since γ /∈ xβ and δ /∈ xα. Say by symmetry a ⊆ b.Then β ∈ b, contradiction.

(2) If α ∈ κ and a ∈ L, then a ⊆ xα or xα ⊆ a.

For, suppose that α ∈ κ and a ∈ L. If α /∈ a, then a ⊆ xα. Suppose that α ∈ a. Ifα /∈ b ∈ L, we then must have b ⊂ a, since a and b are comparable. Thus xα ⊆ a in thiscase. So (2) holds.

By (1) and (2), the set Mdef= L∪xα : α ∈ κ is a chain. Its size is clearly λ. We claim

that xα : α < κ is weakly dense in it, which will finish the proof (expanding xα : α < κto a set of size κ if necessary). For, suppose that a, b ∈ L and a ⊂ b. Choose α ∈ b\a.Then clearly a ⊆ xα. Also, xα ⊆ b by the proof of (2).

E17.6 Suppose that Li is a linear order with at least two elements, for each i ∈ ω. Let∏

i∈ω Li have the lexicographic order. Show that it is not a well-order.

For each i ∈ ω let ai < bi be elements of Li. For each j ∈ ω define f j by setting, for eachi ∈ ω,

f j(i) =

ai if i ≤ j or j + 1 < i;bi if i = j + 1.

Then f0 > f1 > · · ·.

E17.7 Suppose that L is a ccc dense linear order. Show that L has a dense subset of size≤ ω1. Hint: let ≺ be a well-order of L, and let

N = p ∈ L : there is an open set U in L such that p is the ≺-first element of U,

and show that N is dense in L and has size at most ω1.

We may assume that |L| ≥ ω1. Let ≺ be a well-order of L, and define

N = p ∈ L : there is an open U ⊆ L such that p is the ≺-first element of U.

Then N is dense in L. For, if a, b ∈ L and a < b, let p be the ≺-first element of (a, b).Then p ∈ N and a < p < b, as desired.

Thus it suffices to show that |N | ≤ ω1. Suppose, to the contrary, that |N | > ω1.

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For each p ∈ N let

Ip =⋃

(a, b) : a, b ∈ L, a < b, and p is the ≺-first element of (a, b).

Then

(1) For each p ∈ N , the set Ip is a nonempty convex open subset of L.

In fact, it is clearly nonempty and open. If u < w < v with u, v ∈ Ip, choose a, b, c, d ∈ Lsuch that u ∈ (a, b), v ∈ (c, d), and p is the ≺-first element of both (a, b) and (c, d). Clearlythen p is the ≺-first element of (min(a, c),max(b, d)), and w ∈ (min(a, c),max(b, d)), sow ∈ Ip. Thus (1) holds.

(2) If p ≺ q, then Ip ∩ Iq = ∅ or Iq ⊂ Ip.

In fact, suppose that Ip ∩ Iq 6= ∅; say x ∈ Ip ∩ Iq. Choose a, b, c, d ∈ L such that x ∈ (a, b),p is the ≺-first element of (a, b), x ∈ (c, d), and q is the ≺-first element of (c, d). Sincex ∈ (a, b) ∩ (c, d), it follows that (a, b) ∪ (c, d) = (min(a, c),max(b, d)), and p is the ≺-firstelement of this interval. Thus

(min(a, c),max(b, d)) ⊆ Ip.

Now let t be any member of Iq. Say that t ∈ (e, f), with q the ≺-first element of (e, f).Now q is also the ≺-first element of (c, d), so (c, d) ∪ (e, f) = (min(c, e),max(d, f)), andq is the ≺-first element of this interval. q is also a member of (min(a, c),max(b, d)), so(min(a, c),max(b, d)) ∪ (min(c, e),max(d, f)) = (min(a, c, e),max(b, d, f)), and p is the ≺-first element of this interval. Since t is in this interval, it follows that t ∈ Ip. This proves(2).

Let 〈pα : α < ω2〉 list the first ω2 elements of N under ≺. Let M = Ip : p ∈ N. Nowwe construct by recursion two sequences 〈Aα : α < ω1〉 and 〈βα : α < ω1〉. Let β0 = 0,and let A0 be a subset of ω2 with 0 ∈ A0 such that Ipγ

: γ ∈ A0 is a maximal collectionof pairwise disjoint members of M . Thus A0 is countable. Suppose now that we haveconstructed Aγ and βγ for all γ < α, where α < ω1, so that each Aγ is a countable subsetof ω2. Let βα be any ordinal less than ω2 and greater than each ordinal in

⋃

γ<αAγ . Thenlet Aα be maximal subject to the following conditions:

(3) Aα ⊆ [βα, ω2).

(4) βα ∈ Aα.

(5) Ipγ: γ ∈ Aα is pairwise disjoint.

This finishes the construction. Now take any δ greater than each ordinal in⋃

α<ω1Aα,

and with pε < pδ for all ε ∈⋃

α<ω1Aα. For each α < ω1 there is a γα ∈ Aα such

that Ipδ∩ Ipγα

6= ∅. Hence by (2) we have Ipδ⊂ Ipγα

. Hence if α < β < ω1 thenIpγβ

⊂ Ipγα. For brevity let q(α) = pγα

for each α < ω1. So Iq(β) ⊂ Iq(α) if α < β < ω1.

Now for each limit ordinal µ < ω1 choose ξ(µ), η(µ), ρ(µ) such that ξ(µ) ∈ Iq(µ)\Iq(µ+1),η(µ) ∈ Iq(µ+1)\Iq(µ+2), and ρ(µ) ∈ Iq(µ+2)\Iq(µ+3). Then ξ(µ), η(µ), ρ(µ) are distinctelements all in Iq(µ)\Iq(µ+3). Hence two of them, say εµ and θµ, with εµ < θµ, are on the

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same side of Iq(µ+3), say the left side. It follows that 〈(εµ, θµ) : µ limit < ω1〉 is a systemof pairwise disjoint sets, contradiction.

E17.8 Let 〈Li : i ∈ I〉 be a system of linear orders, with I itself an ordered set. Show thatif each Li is dense without first or last elements, then also

∑

i∈I Li is dense without firstor last elements.

Suppose that (i, a) < (j, b). If i < j, let c be an element of Li such that a < c. Then(i, a) < (i, c) < (j, b). If i = j, let c be an element of Li such that a < c < b. Then(i, a) < (i, c) < (i, b). Thus

∑

i∈I Li is dense. Given (i, a), choose c, d ∈ Li with c < a < d.Then (i, c) < (i, a) < (i, d). Thus

∑

i∈I Li is does not have a first or last element.

E17.9 Let κ be any infinite cardinal number. Let L0 be a linear order similar to ω∗+ω+1;specifically, let it consist of a copy of Z followed by one element a greater than every integer,and let L1 be a linear order similar to ω∗ + ω + 2; say it consists of a copy of Z followedby two elements a < b greater than every integer. For any f ∈ κ2 let

Mf =∑

α<κ

Lf(α).

Show that if f, g ∈ κ2 then Mf and Mg are not isomorphic.Conclude that there are exactly 2κ linear orders of size κ up to isomorphism.

Let f, g ∈ κ2. We assume that Mf is isomorphic to Mg and show that f = g. Let F be anisomorphism of Mf onto Mg. Clearly

(1) For all x ∈Mf ∪Mg the following conditions are equivalent:(a) x does not have an immediate predecessor.(b) x = (a, ξ) for some ξ < κ.

Now for any x ∈ Mf , x has an immediate predecessor iff F (x) has an immediate prede-cessor, as is easily seen. We claim then that F (a, ξ) = (a, ξ) for all ξ < κ. We prove thisby transfinite induction. Suppose that F (a, η) = (a, η) for all η < ξ. Now F (a, ξ) does nothave an immediate predecessor, so by (1) it has the form (a, ρ) for some ρ. We cannot haveρ < ξ, since this would contradict F being one-one, by the supposition. If ξ < ρ, then wewould have F−1(a, ξ) < F−1(a, ρ) = (a, ξ), again contradicting the inductive assumption.Thus F (a, ξ) = (a, ξ), finishing the inductive proof.

Next we claim

(2) For any x ∈Mf the following conditions are equivalent:(a) x does not have an immediate predecessor, but it has an immediate successor y

which in turn does not have an immediate successor.(b) x = (a, ξ) for some ξ such that f(ξ) = 1.

This is obvious, and a similar condition for Mg holds.Now the property given in (2)(a) is preserved under isomorphisms, so by the above,

for any ξ < κ,

f(ξ) = 1 iff (a, ξ) satisfies (2)(a)

iff F (a, ξ) satisfies (2)(a)

iff g(ξ) = 1.

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Thus f = g, as desired.The required conclusion in the exercise is clear.

E17.10 Let κ be an uncountable cardinal. Let L0 be a linear order similar to η+1+η ·ω∗1 ;

specifically consisting of a copy of the rational numbers in the interval (0, 1] followed byQ× ω1, where Q× ω1 is ordered as follows: (r, α) < (s, β) iff α > β, or α = β and r < s.Let L1 be a linear order similar to η · ω1 + 1 + η · ω∗

1 ; specifically, we take L1 to be the set

(q, α, 0) : q ∈ Q, α < ω1 ∪ (0, 0, 1) ∪ (q, α, 2) : q ∈ Q, α < ω1,

with the following ordering:

(q, α, 0) < (r, β, 0) iff α < β, or α = β and q < r;

(q, α, 0) < (0, 0, 1) < (r, β, 2) for all relevant q, r, α, β;

(q, α, 2) < (r, β, 2) iff α > β, or α = β and q < r.

For each f ∈ κ2 let

Mf =∑

α<κ

Lf(α).

Show that each Mf is a dense linear order without first or last elements, and if f, g ∈ κ2and f 6= g, then Mf and Mg are not isomorphic.

Conclude that for κ uncountable there are exactly 2κ dense linear orders without firstor last elements, of size κ, up to isomorphism.

For L0, if q is an element in the initial (0, 1] part, then q−1 < q; so L0 has no least element.If (q, α) is an element in the second part, then (q, α) < (q + 1, 0); so L0 has no greatestelement. For denseness, suppose that x < y in L0. If both are in the first part, clearlythere is a z such that x < z < y. Suppose that x is in the first part and y = (q, α) is inthe second part. Then x < (q − 1, α) < y. Finally, suppose that x = (r, β) and y = (s, γ)are both in the second part. If β > γ, then x < (r + 1, β) < y. If β = γ, then r < s andthe desired element is clear.

For L1, given any element (q, α, 0) in the first part, we have (q − 1, α, 0) < (q, α, 0),so L1 does not have a least element. If (q, α, 2) is any element in the third part, then(q, α, 2) < (q+ 1, α, 2), so L2 does not have a greatest element. Suppose that x < y in L1.We can consider several cases.

Case 1. x = (q, α, 0), y = (r, β, 0), and α < β. Then x < (q + 1, α, 0) < y.Case 2. x = (q, α, 0), y = (r, α, 0). Then q < r; so with q < t < r we have

x < (t, α, 0) < y.Case 3. x = (q, α, 0), y = (0, 0, 1). Then x < (q + 1, α, 0) < y.Case 4. x = (q, α, 0), y = (r, β, 2). Then x < (0, 0, 1) < y.Case 5. x = (0, 0, 1), y = (q, α, 2). Then x < (q − 1, α, 2) < y.Case 6. x = (q, α, 2), y = (r, β, 2), and α > β. Then x < (q + 1, α, 2) < y.Case 7. x = (q, α, 2), y = (r, α, 2). Thus q < r. Then with q < t < r we have

x < (t, α, 2) < y.

Thus L1 is dense.

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Before beginning the main part of the exercise, we note the following facts.

(1) Each q in the first part of L0, except for the final 1, has character (ω, ω).

(2) The final 1 has character (ω, ω1). A strictly decreasing sequence with limit 1 is

〈(0, α) : α < ω1〉.

(3) Every element in the second part of L0 has character (ω, ω).

(4) Every element in the first part of L1 has character (ω, ω).

(5) (0, 0, 1) has character (ω1, ω1).

In fact, a strictly increasing sequence with limit (0, 0, 1) is

〈(0, α, 0) : α < ω1〉.

A strictly decreasing sequence with limit (0, 0, 1) is

〈(0, α, 2) : α < ω1〉.

Now we suppose that f ∈ κ2. We give some properties of Mf :

(6) The character of an element (x, ξ) of Mf is equal to the character of x in Lf(ξ).

This is true because Lf(ξ) is a convex set in Mf , i.e., if u < v < w with u, w ∈ Lf(ξ), thenv ∈ Lf(ξ). Also, the fact that Lf(ξ)) does not have a least or greatest element is needed tosee (6).

(7) For each ξ < κ there is a unique element of Mf of the form (xξ, ξ) which has character(ω, ω1) if f(ξ) = 0 and has character (ω1, ω1) if f(ξ) = 1; all other elements of the form(y, ξ) have character (ω, ω).

Now we can treat the main part of the exercise. Suppose that f, g ∈ κ2 and Mf isisomorphic to Mg; say F is an isomorphism. The sequence

〈F (xξ, ξ) : ξ < κ〉,

with xξ given in (7), is an increasing sequence of elements ofMg such that all other elementsof Mg have character (ω, ω). In Mg there is only one sequence of order type κ consistingof elements which do not have character (ω, ω), by (7) for Mg. Hence F (xξ, ξ) = (yξ, ξ),where yξ is defined for Mg like xξ was for Mf in (7). But xξ and yξ then have the samecharacters, and so f(ξ) = g(ξ). Thus f = g.

The final statement of the exercise is clear.Solutions to exercises in Chapter 18

E18.1 Let κ be an uncountable regular cardinal, and suppose that there is a κ-Aronszajntree. Show that there is one which is a normal subtree of <κ2. Hint: for each α < κ let gαbe an injection of Levα(T ) into |Levα(T )|2 and glue these maps together.

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Let T be a κ-Aronszajn tree. By Theorem 18.7 we may assume that it is well-pruned. Fors ∈ T and α < ht(s, T ) let sα be the unique element of height α below s. For each α < κ,let gα be an injection from Levα(T ) into |Levα(T )|2.

We define by recursion sequences 〈µα : α < κ〉 and 〈Fα : α < κ〉. Let µ0 = |Lev0(T )|,and let F0 = g0. Now suppose that µα, Fα have been defined so that the followingconditions hold:

(1α) µα < κ.

(2α) Fα is a function with domain⋃

β≤α Levβ(T ).

(3α) for all β, γ < α, if β < γ then µβ ≤ µγ and Fβ ⊆ Fγ .

(Clearly these conditions hold for α = 0.) Now let µα+1 = µα+ |Levα+1(T )| (ordinal addi-tion). Let Fα+1 be the extension of Fα such that for every t ∈ Levα(T ), every immediatesuccessor s of t, and every β < µα+1,

(Fα+1(s))(β) =

(Fα(t))(β) if β < µα,(gα(s))(ξ) if β = µα + ξ with ξ < |Levα+1(T )|.

Clearly (1α+1)–(3α+1) hold.Now suppose that α is a limit ordinal and µβ , Fβ have been defined for all β < α so

that (1β)–(3β) hold. Let ν =⋃

β<α µβ , G =⋃

β<α Fβ , and set µα = ν + |Levα(T )|. LetFα be the extension of G such that for every s ∈ Levα(T ) and every β < µα,

(Fα(s))(β) =

(G(sγ))(β) if β < µγ with γ < α,(gα(s))(ξ) if β = ν + ξ with ξ < |Levα(T )|.

Clearly (1α)–(3α) hold.Let H =

⋃

α<κ Fα. Clearly

(4) If s ∈ Levα(T ), then H(s) ∈ µα2,

(5) If u < s then H(u) ⊂ H(s).

We prove (5) by induction on the level of s. It is vacuously true for level 0. Now supposeinductively that s has level α+ 1. Say that t is the immediate predecessor of s. Let γ bethe level of u. Then for any β < µγ we have

(H(s))(β) = (Fα+1(s))(β) = (Fα(t))(β) = (H(t))(β) = (H(u))(β).

Finally, suppose inductively that s has limit level α. Then for any β < γ we have

(H(s))(β) = (Fγ(sγ))(β) = (H(u))(β).

Hence (5) holds.

(6) If s, t ∈ T have the same height and s 6= t, then H(s) 6= H(t).

We prove (6) by induction on the common height α of s and t. If α = 0 the conclusion isclear since g0 is one-one. Suppose inductively that they both have height α+ 1. Let s′, t′

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be their immediate predecessors. If s′ 6= t′, then H(s′) 6= H(t′), so H(s) 6= H(t) by (5).Suppose that s′ = t′. Then H(s) 6= H(t) since gα is one-one. Finally, suppose inductivelythat α is limit. Then H(s) 6= H(t) since gα is one-one. So (6) holds.

Now let T ′ = h ∈ <κ2 : h ⊆ H(s) for some s ∈ T. We claim that T ′ is as desired.Clearly it is a normal subtree of <κ2. Now consider any α < κ. Choose β minimum suchthat α ≤ µβ .

(7) If h ∈ T ′ with dmn(h) = α, then there is an s ∈ T of height β such that h ⊆ H(s).

In fact, choose t ∈ T such that h ⊆ H(t). Then dmn(H(t)) ≥ α ≥ µβ , so t has height ≥ β.Let s ∈ T of height β with s ≤ t. Then H(s), h ⊆ H(t), so h ⊆ H(s), as desired.

It follows from (7) that each level of T ′ has size less than κ. From (5) and (7) itfollows that T ′ does not have a chain of size κ.

E18.2 Do exercise E18.1 for κ-Suslin trees.

We use the same construction as in exercise 18.1. Thus our new tree T ′ does not have anychain of size κ. Suppose that A is an antichain of size κ. For each a ∈ A choose sa ∈ Tsuch that a ⊆ H(sa). Since T is a Suslin tree, choose distinct a, b ∈ A such that sa and sbare comparable. Say sa ≤ sb. Then H(sa) ⊆ H(sb). Since a, b ⊆ H(sb), it follows that aand b are comparable, contradiction.

E18.3 Suppose that T and T ′ are κ-Aronszajn trees. Define an order < on T × T ′ by(s, s′) < (t, t′) iff s < t and s′ < t′. Show that T × T ′ is not a tree.

Clearly T × T ′ is a partial order. Let x0 < x1 < x2 in T and y0 < y1 < y2 in T ′. Then(x1, y0) < (x2, y2) and (x0, y1) < (x2, y2), but (x1, y0) and (x0, y1) are incomparable.

E18.4 Suppose that T and T ′ are κ-Aronszajn trees. Let

T ×′ T ′ =⋃

α<κ

Levα(T )× Levα(T ′);

(s, s′) < (t, t′) iff (s, s′), (t, t′) ∈ T ×′ T ′, s < t, and s′ < t′;

Show that (T ×′ T ′, <) is a κ-Aronszajn tree.

Clearly T×′T ′ is a partial order. Suppose that (s, t) ∈ T×′T ′. Say that s and t have heightα, and let 〈aξ : σ < α〉 and 〈bξ : σ < α〉 are the systems of predecessors of s, t respectively.Clearly 〈(aξ, bξ) : ξ < α〉 is the system of predecessors of (s, t). So we have a tree. Eachlevel α has size |Levα(T ) × Levα(T ′)| < κ; so T ×′ T ′ is a κ-tree. If 〈(xξ, yξ) : ξ < β〉 isa chain in T ×′ T ′, then 〈xξ : ξ < β〉 is a chain in T , and so β < κ. Thus T ×′ T ′ is aκ-Aronszajn tree.

E18.5 Assume that κ is regular and uncountable. Suppose that T is a κ-Suslin tree. Withthe order on T ×′ T given in exercise E18.4, show that T ×′ T is not a κ-Suslin tree. Hint:first show that for every α < κ there is an element s of T at level α such that there areincomparable t, u > s.

First we claim

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(1) For every α < κ there is an s of height α such that there are incomparable t, u > s.

In fact, otherwise we get an α < κ such that for every s of height α, the set t ∈ T : s < tis a chain. Since all chains have size less than κ, for each s of level α choose βs greaterthan the level of all t with s < t. Let γ < κ be such that βs < γ for all s of level α.Let β be any element of T of level γ, and let s be its predecessor at level α. But γ > βs,contradiction. Thus (1) holds.

Now we construct elements sα, tα, uα for α < κ by recursion. Suppose that they havebeen constructed for all β < α. Let

γ = max(supht(tβ , T ) : β < α, supht(uβ, T ) : β < α) + 1.

By (1), choose sα of height γ with incomparable tα, uα > sα of some common level > γ.We claim that 〈(tα, uα) : α < κ〉 is an antichain in T ×′ T . For, suppose that β < α

and (tβ, uβ) < (tα, uα). So tβ < tα and uβ < uα. Since tβ , sα < tα, we have tβ < sα(since sα is at a higher level than tβ). Similarly, uβ < sα. So tβ and uβ are comparable,contradiction.

E18.6 A tree T is everywhere branching iff every t ∈ T has at least two immediatesuccessors. Show that every everywhere branching tree has at least 2ω branches.

We define a branch bf for every f ∈ ω2 by defining elements ah for every h ∈ <ω2 byrecursion on dmn(h). Let a∅ be a root of the tree. Suppose that ah has been defined forevery h ∈ n2. For each h ∈ n2, let ah0 and ah1 be two immediate successors of ah. Thisfinishes the definition of the ah’s. Now let bf be an extension of 〈afn : n ∈ ω〉 to a branch.Clearly this is as desired.

E18.7 Show that the hypothesis that all levels are finite is necessary in Konig’s theorem.

For each n ∈ ω let fn ∈n+1ω be any function such that fn(0) = n, and let T be the tree

consisting of all g ∈ <ωω such that g ⊆ fn for some n. Then two elements g, h ∈ T arecomparable iff they are both contained in the same fn. If C is a maximal chain in T , itmust be a subset of some fn, and hence is finite.

E18.8 Show that if κ is singular with cf(κ) = ω, then there is no κ-Aronszajn tree withall levels finite.

Let 〈αn : n ∈ ω〉 be a strictly increasing sequence of ordinals with supremum κ. Supposethat T is a κ-Aronszajn tree with all levels finite. Define

T ′ = t ∈ T : there is an n ∈ ω such that t has height αn.

Then T ′, with the order induced by T , is a tree of height ω with all levels finite. Henceby Konig’s theorem it has an infinite branch B. Let B′ = t ∈ T : t ≤ s for some s ∈ T ′.Then B′ is a branch in T of size κ, contradiction.

E18.9 Prove that if κ is singular and there is a cf(κ)-Aronszajn tree, then there is aκ-Aronszajn tree with all levels of power less than cf(κ).

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Let T be a cf(κ)-Aronszajn tree. Let 〈µα : α < cf(κ)〉 be a strictly increasing continuoussequence of cardinals with supremum κ, and with µ0 = 0. We define

T ′ = (t, β) : there is an α < cf(κ) such that t ∈ Levα(T ) and µα ≤ β < µα+1;

(t, β) < (t′, β′) iff t < t′, or t = t′ and β < β′.

Clearly this gives a partial order. To show that it is a tree, suppose that (t, β) ∈ T ′. Wedefine a function f from β into the set of predecessors of (t, β) as follows. Let ht(t) = α.Suppose that γ < β. then there is a δ such that µδ ≤ γ < µδ+1. Clearly δ ≤ α. We definef(γ) = (t′, γ), where t′ is the predecessor of t at level δ. Clearly then f(γ) < (t, β). Clearlyalso f is order preserving and maps onto the set of all predecessors of (t, β). Thus T ′ is atree. For each β < κ, say with µα ≤ β < µβ+1 we have

Levβ(T′) = (t, β) : t ∈ Levα(T ).

Thus each level of T ′ has size less than cf(κ). If B is a branch of size κ, then for eachβ < κ it has an element of height at least β, and hence for each α < cf(κ) it has an elementwhose first coordinate has height at least α. These first coordinates are linearly ordered.This contradicts T being a cf(κ)-Aronszajn tree.

Thus T ′ is as desired.

E18.10 Show that for every infinite cardinal κ there is an eventually branching tree T ofheight κ such that for every subset S of T , if S is a tree under the order induced by T andevery element of S has at least two immediate successors, then S has height ω.

The idea is to put a copy of <ω2 on top of longer and longer chains. More precisely, define

T = (α, ξ, ∅) : α < κ, ξ < α ∪ (α, α, f) : α < κ, f ∈ <ω2;

(α, ξ, f) < (β, η, g) iff α = β and either ξ < η, or ξ = η = α and f ⊂ g.

Clearly T is a tree. The height of an element (α, ξ, ∅) is ξ, and the height of an element(α, α, f) is α+ n, where f ∈ n2. In particular, T has height κ.

Now suppose that S is as indicated in the exercise, and take any element (α, ξ, f) ofS. (α, ξ, f) is a root of S iff α = ξ and f = ∅. It follows that all the non-root elements ofS have the form (α, α, f), and so in S the height of every element is finite.

E18.11 Show that if κ is an uncountable regular cardinal and T is a κ-Aronszajn tree, thenT has a subset S such that under the order induced by T , S is a well-pruned κ-Aronszajntree in which every element has at least two immediate successors.

By Theorem 18.7, we may assume that T is well-pruned. Now we construct a strictlyincreasing sequence 〈αξ : ξ < κ〉 of ordinals less than κ. Let α0 = 0. Suppose that αξhas been defined. Now T is eventually branching. (See the remark before Theorem 18.7.)Hence for each t ∈ Levαξ

(T ) there is an ordinal βt > αξ such that t has at least twosuccessors at level βt. Let αξ+1 be any ordinal less than κ such that βt < αξ+1 for allt ∈ Levαξ

(T ). Note by the well-prunedness condition, each t ∈ Levαξhas at least two

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successors at level αξ+1. Finally, suppose that η is a limit ordinal less than κ, and αξ hasbeen constructed for all ξ < η. Let αη = supξ<η αξ.

Let S =⋃

ξ<κ Levαξ(T ). Clearly S is as desired.


E19.1 Assume that κ is an uncountable regular cardinal and 〈Aα : α < κ〉 is a sequenceof subsets of κ. Let D = α<κAα. Prove the following:

(i) For all α < κ, the set D\Aα is nonstationary.(ii) Suppose that E ⊆ κ and for every α < κ, the set E\Aα is nonstationary. Show

that E\D is nonstationary.

(i): For any β < κ,

D\Aβ = α < κ : ∀γ < α(α ∈ Aγ) and α /∈ Aβ

⊆ α < κ : α ≤ β;

Hence D\Aβ is nonstationary.(ii): For each β < κ let Cβ be club in κ such that (E\Aβ) ∩ Cβ = ∅. Let F be the

diagonal intersection of the Cβ ’s; thus

F = γ < κ : ∀α < γ(γ ∈ Cα).

Thus F is club. We claim that F ∩ (E\D) = ∅ (as desired). For, suppose that γ ∈F ∩ (E\D). Since γ /∈ D, there is a β < γ such that γ /∈ Aβ. Since γ ∈ F , we have γ ∈ Cβ .Since (E\Aβ) ∩ Cβ = ∅, this is a contradiction.

E19.2 Let κ > ω be regular. Show that there is a sequence 〈Sα : α < κ〉 of stationarysubsets of κ such that Sβ ⊆ Sα whenever α < β < κ, and α<κSα = 0. Hint: useTheorem 19.12.

By Theorem 19.12, let 〈Aα : α < κ〉 be pairwise disjoint stationary subsets of κ. ThenAα\(α + 1) is also stationary. Let Sα =

⋃

β>α(Aβ\(β + 1)). Then Sα is stationary andα < β implies that Sβ ⊆ Sα. Let D be the diagonal intersection of the Sα’s:

D = γ < κ : ∀α < γ(γ ∈ Sα).

Thus 0 ∈ D. Suppose that 0 6= γ ∈ D. Then 0 < γ, so γ ∈ S0. Hence there is a β > 0 suchthat γ ∈ Aβ\(β + 1). Hence β < γ. So γ ∈ Sβ . Choose δ > β such that γ ∈ Aδ\(δ + 1).So γ ∈ Aβ ∩Aδ = ∅, contradiction.

E19.3 Suppose that κ is uncountable and regular, and for each limit ordinal α < κ we aregiven a function fα ∈

ωα. Suppose that S is a stationary subset of κ. Let n ∈ ω. Showthat there exist a t ∈ nκ and a stationary S′ ⊆ S such that for all α ∈ S′, fα n = t.

We define sequences 〈S0, S1, . . . , Sn〉 of stationary subsets of S and 〈β0, . . . , βn−1〉 of ordi-nals less than κ. Let S0 = S. Suppose that Si has been defined, i < n. Let g(α) = fα(i) forall α ∈ Si. Then g is a regressive function, and hence there exist a stationary subset Si+1

of Si and an ordinal βi such that g(α) = βi for all α ∈ Si+1. This finishes the construction.

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If α ∈ Sn, then for any i < n we have α ∈ Si+1, and hence fα(i) = βi. Hence we canlet t(i) = βi for all i < n, and the property of the exercise holds.

E19.4 Suppose that cf(κ) > ω, C ⊆ κ is club of order type cf(κ), and 〈cβ : β < cf(κ)〉 isthe strictly increasing enumeration of C. Let X ⊆ κ. Show that X is stationary in κ iffβ < cf(κ) : cβ ∈ X is stationary in cf(κ).

Assume X ⊆ κ. Let X ′ = β < cf(κ) : cβ ∈ X.⇒: Assume that X is stationary in κ. We want to show that X ′ is stationary in

cf(κ). Let D′ be club in cf(κ). Define D = cβ : β ∈ D′. We claim that D is club inκ. For closure, suppose that α < κ is a limit ordinal and D ∩ α is unbounded in α. Letγ =

⋃β ∈ D′ : cβ < α.

(1) γ < cf(κ).

For, since C is unbounded in κ, there is a β < cf(κ) such that α < cβ . Now if ε ∈ D′ andcε < α we have cε < cβ , hence ε < β. Therefore γ ≤ β < cf(κ), proving (1).

(2) γ is a limit ordinal, and D′ ∩ γ is unbounded in γ.

For, suppose that δ < γ. Choose β ∈ D′ such that cβ < α and δ < β. Since D ∩ α isunbounded in α, choose cε ∈ D ∩α such that cβ < cε. Then δ < β < ε ≤ γ. Since β ∈ D′,this proves (2).

(3) cγ = α.

For, suppose that δ < cγ . Now cγ =⋃

β<γ cβ by (2), so there is a β < γ such thatδ < cβ . By the definition of γ, there is a β′ ∈ D′ such that β < β′ and cβ′ < α. Thusδ < cβ < cβ′ < α, so δ < α. This proves that cγ ≤ α. On the other hand, suppose thatδ < α. Since D ∩ α is unbounded in α, there is a β ∈ D′ such that δ < cβ < α. Thusβ ≤ γ, so δ < cγ . This proves that α ≤ cγ , and finishes the proof of (3).

Now by (2) we have γ ∈ D′, and hence (3) yields α ∈ D, as desired; we have nowproved that D is closed in κ.

To show that D is unbounded in κ, let α < κ be arbitrary. Choose β < cf(κ) suchthat α < cβ . Since D′ is unbounded in cf(κ), choose β′ ∈ D′ such that β < β′. Thusα < cβ < cβ′ ∈ D, as desired.

So we have shown that D is club in κ. Since X is stationary in κ, choose β ∈ D′ suchthat cβ ∈ X . thus β ∈ D′ ∩X ′, as desired.

⇐: Assume that X ′ is stationary in cf(κ). Let D be club in κ, and let D′ = β <cf(κ) : cβ ∈ D. We claim that D′ is club in cf(κ). To show that it is closed, supposethat α < cf(κ) is a limit ordinal and D′ ∩ α is unbounded in α. We claim that D ∩ cαis unbounded in cα. For, suppose that γ < cα. Then there is a β < α such that γ < cβ .Choose δ ∈ D′ ∩ α such that β < δ; this is possible since D′ ∩ α is unbounded in α. Thusγ < cβ < cδ ∈ D ∩ cα, as desired. So D′ is closed in cf(κ). To show that it is unbounded,suppose that α < cf(κ). Now C ∩D is club in κ, so there is a β such that cα < cβ ∈ D.So α < β ∈ D′. This shows that D′ is unbounded in cf(κ). Hence D′ is club in cf(κ).

Choose β ∈ D′ ∩X ′. Then cβ ∈ D ∩X , as desired.

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E19.5 Suppose that κ is regular and uncountable, and S ⊆ κ is stationary. Also, supposethat every α ∈ S is an uncountable regular cardinal. Show that

Tdef= α ∈ S : S ∩ α is non-stationary in α

is stationary in κ. Hint: given a club C in κ, let C′ be the set of all limit points of C andlet α be the least element of C′ ∩ S; show that α ∈ T ∩ C.

Assume the notation of the exercise. Let C be club in κ; we want to show that T ∩C 6= ∅.Let C′ be the set of all limit points of C, i.e., the set of all limit ordinals α ∈ κ such thatC ∩ α is unbounded in α. Clearly C′ ⊆ C, and C′ is club in κ. Since S is stationary inκ, let α be the least element of S ∩ C′. Clearly C′ ∩ α is closed in α; we claim that it isalso unbounded in α. For, suppose that β < α. Now C ∩ α is unbounded in α, so we canconstruct a sequence 〈γi : i < ω〉 of members of C such that β < γ0 < γ1 < · · · < α. Letδ = supi∈ω γi. Then δ ∈ C′, and δ < α since α is uncountable and regular. So C′ ∩ α isclub in α. Now S∩C′ ∩α = ∅ by the minimality of α, so C′ ∩α is a club in α which showsthat S ∩ α is non-stationary in α. So α ∈ T ∩ C, as desired.

E19.6 Suppose that κ is uncountable and regular, and κ ≤ |A|. Suppose that C is aclosed subset of [A]<κ and D is a directed subset of C with |D| < κ. (Directed means thatif x, y ∈ D then there is a z ∈ D such that x ∪ y ⊆ z.) Show that

⋃D ∈ C. Hint: use

induction on |D|.

Proof. If D is finite, then⋃D ∈ D; so

⋃D ∈ C. Suppose that |D| = ω; say D =

xn : n ∈ ω. For each n ∈ ω choose yn ∈ D so that xm : m ≤ n ∪ ym : m < n ⊆ yn.Then

⋃

n∈ω yn ∈ C since C is closed, and⋃D =

⋃

n∈ω yn.Now suppose inductively that |D| > ω. Let |D| = κ and write D = xα : α < κ.

For all y, z ∈ D let f(y, z) ∈ D be such that y, z ⊆ f(y, z). We now define 〈Eα : α < κ〉by recursion. Suppose that Eβ has been defined for all β < α so that Eβ ⊆ D, Eβis directed, and |Eβ| ≤ |β| + ω. Let F0 = xα ∪

⋃

β<αEβ . So |F0| ≤ |α| + ω. LetFn+1 = Fn ∪ f(x, y) : x, y ∈ Fn. Then |Fn+1 ≤ |α| + ω. Let Eα =

⋃

n∈ω Fn. ThenEα ⊆ D, Eα is directed, and |Eα| ≤ |α|+ ω.

By the inductive hypothesis, yαdef=⋃Eα is in C, and yα ⊆ yβ for α < β. Hence

⋃D =

⋃

α<κ yα ∈ C.

E19.7 Let κ be uncountable and regular, and κ ≤ |A|. If f : [A]<ω → [A]<κ let Cf =x ∈ [A]<κ : ∀s ∈ [x]<ω[f(s) ⊆ x]. Show that Cf is club in [A]<κ.

Suppose that xξ ∈ Cf for all ξ < α, with α < κ and xξ ⊆ xη for ξ < η. Clearly⋃

ξ<α xξ ∈

Cf if α is a successor ordinal. Suppose that α is a limit ordinal. Take any s ∈[⋃

ξ<α xξ

]<ω

.

Then there is a ξ < α such that s ∈ [xξ]<ω, and hence f(s) ∈ xξ ⊆

⋃

η<ζ xη. Thus Cf isclosed.

To show that Cf is unbounded, let y ∈ [A]<κ. Define z0 = y and zn+1 = zn ∪ f(s) :s ∈ [zn]<ω. By induction, zn ∈ [A]<κ for all n ∈ ω. Now

⋃

n∈ω zn ∈ Cf , showing that Cfis unbounded.

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E19.8 (Continuing Exercise E19.7) Let κ be uncountable and regular, and κ ≤ |A|. Let Dbe club in [A]<κ. Show that there is an f : [A]<ω → [A]<κ such that Cf ⊆ D. Hint: showthat there is an f : [A]<ω → D such that ∀e ∈ [A]<ω[e ⊆ f(e)] and ∀e1, e2 ∈ [A]<ω[e1 ⊆e2 → f(e1) ⊆ f(e2)].

We claim that there is an f : [A]<ω → D such that ∀e ∈ [A]<ω[e ⊆ f(e)] and ∀e1, e2 ∈[A]<ω[e1 ⊆ e2 → f(e1) ⊆ f(e2)]. We define f by induction on |e|. Let f(∅) be anymember of D. Suppose that f(e) has been defined for all e ∈ [A]<ω such that |e| < m, andsuppose that e ∈ [A]<ω with |e| = m. Let f(e) be a member of D such that e ⊆ f(e) andf(e\a) ⊆ f(e) for all a ∈ e. Clearly f is as desired.

Now we show that Cf ⊆ D. Let x ∈ Cf . Note that f(e) : e ∈ [x]<ω is directed andhas union x. Hence x ∈ D by Exercise 19.6.

E19.9 Let κ be uncountable and regular, κ ≤ |A|, and A ⊆ B. If Y ∈ [A]<κ, let

Y B = x ∈ [B]<κ : x∩A ∈ Y . Show that if Y is club in [A]<κ, then Y B is club in [B]<κ.

Assume the hypotheses. Suppose that α < κ and 〈xξ : ξ < α〉 is a sequence of membersof Y B with xξ ⊆ xη for ξ < η. Then xξ ∩ A ∈ Y for all ξ < α, and xξ ∩ A ⊆ xη ∩ A forξ < η. Hence

⋃

ξ<α(xξ ∩ A) ∈ Y . So⋃

ξ<α xξ ∈ YB. Thus Y B is closed.

To show that Y B is unbounded, let b ∈ [B]<κ. Then b∩A ∈ [A]<κ, so there is a c ∈ Ysuch that b∩A ⊆ c. Then b ⊆ c∪ (b\A), and (c∪ (b\A))∩A = c ∈ Y . So c∪ (b\A) ∈ Y B.

E19.10 Let κ be uncountable and regular, κ ≤ |A|, and A ⊆ B. If Y ∈ [B]<κ, letY A = y ∩ A : y ∈ Y . Show that if Y is stationary in [B]<κ then Y A is stationaryin [A]<κ.

Assume the hypotheses. Suppose that C is club in [A]<κ. Then by Exercise E19.9, CB isclub in [B]<κ. Choose y ∈ Y ∩ CB . Then y ∩A ∈ (Y A) ∩ C.

E19.11 With κ,A,B as in exercise E19.9, suppose that f : [B]<ω → [B]<κ. For eache ∈ [A]<ω define

x0(e) = e;

xn+1(e) = xn(e) ∪ f(s) : s ∈ [xn(e)]<ω;

w(e) =⋃

n∈ω

xn(e).

Also, for each y ∈ [A]<κ let v(y) =⋃w(e) : e ∈ [y]<ω.

Prove that w(e) ∈ Cf for all e ∈ [A]<ω and v(y) ∈ Cf for all y ∈ [A]<κ.

If z ∈ [w(e)]<ω, then z ∈ [xn(e)]<ω for some n, and hence f(z) ∈ xn+1(e) ⊆ w(e). Thus

w(e) ∈ Cf . Note that if e1 ⊆ e2 then xn(e1) ⊆ xn(e2) for all n (by induction), and sow(e1) ⊆ w(e2). Now suppose that z ∈ [v(y)]<ω. Then there is a finite F ⊆ [y]<ω such thatz ⊆

⋃

e∈F w(e). Let e′ =⋃

e∈F e. Then⋃

e∈F w(e) ⊆ w(e′). So z ∈ [w(e′)]<ω. It followsfrom the first part of this solution that f(z) ∈ w(e′) ⊆ v(y). Thus v(y) ∈ Cf .

E19.12 With κ,A,B as in exercise E19.9, suppose that S is stationary in [A]<κ. Show

that SB is stationary in [B]<κ. Hint: use exercises E19.8 and E19.11.

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Assume the hypotheses. Suppose that D is club in [B]<κ. By Exercise 19.8 there is anf : [B]<ω → [B]<κ such that Cf ⊆ D. For any e ∈ [A]<ω let g(e) = w(e) ∩ A. We claimthat Cf A = Cg. Suppose that y ∈ Cf , so that y∩A ∈ Cf A. To show that y∩A ∈ Cg,let e ∈ [y ∩A]<ω. Then xn(e) ⊆ y by induction on n. Since xn(e) = e, it is true for n = 0.Suppose that xn(e) ⊆ y. If s ∈ [xn(e)]

<ω, then s ∈ [y]<ω, so f(s) ⊆ y since y ∈ Cf . Hencexn+1(e) ⊆ y. It follows that w(e) ⊆ y, and so g(e) ⊆ y ∩ A. This shows that y ∩ A ∈ Cg,and proves that Cf A ⊆ Cg. Now suppose that y ∈ Cg. Hence ∀e ∈ [y]<ω[g(e) ⊆ y]. Weclaim that v(y)∩A = y. For, if e ∈ [y]<ω, then g(e) ⊆ y, i.e., w(e)∩A ⊆ y. So v(y)∩A ⊆ y.If a ∈ y, then a ∈ w(a) ⊆ v(y); so v(y) ∩ A = y. This shows that Cg ⊆ Cf A. ThusCg = Cf A.

Choose z ∈ Cg ∩ S. Then z ∈ (Cf A) ∩ S, so there is a y ∈ Cf such that z = y ∩A.Thus y ∩ A ∈ S, so y ∈ SB ∩D.


E20.1 Suppose that κω > κ. Show that there is a family A of subsets of κ, each of sizeω, with |A | = κ+ and the intersection of any two members of A is finite.

Let K = <ωκ. Thus |K| = κ. Let F be a bijection from K to κ. For each f ∈ ωκ let

Xf = F [f m : m ∈ ω].

Clearly each Xf has size ω. If f, g ∈ ωκ and f 6= g, choose p ∈ ω such that f(p) 6= g(p).Then

Xf ∩Xg = F [f m : m ∈ ω] ∩ F [g m : m ∈ ω]

= F [f m : m ∈ ω ∩ g m : m ∈ ω]

⊆ F [f m : m ≤ p]

and hence Xf ∩Xg is finite. Since κω ≥ κ+, the desired result follows.

E20.2 Suppose that κ is any infinite cardinal, and λ is minimum such that κλ > κ. Showthat there is a family A of subsets of κ, each of size λ, with the intersection of any twomembers of A being of size less than λ, and with |A | = λ+.

Let K = <λκ. Thus |K| = κ. Let F be a bijection from K to κ. For each f ∈ λκ let

Xf = F [f m : m ∈ λ].

Clearly each Xf has size λ. If f, g ∈ λκ and f 6= g, choose p ∈ λ such that f(p) 6= g(p).Then

Xf ∩Xg = F [f m : m ∈ λ] ∩ F [g m : m ∈ λ]

= F [f m : m ∈ λ ∩ g m : m ∈ λ]

⊆ F [f m : m ≤ p]

and hence Xf ∩Xg has size less than λ. Since κλ > κ, the desired result follows.

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E20.3 Suppose that κ is uncountable and regular. Show that there is a family A of subsetsof κ, each of size κ with the intersection of any two members of A of size less than κ, andwith |A | = κ+. Hint: (1) show that there is a partition of κ into κ subsets, each of sizeκ; (2) Use Zorn’s lemma to start from (1) and produce a maximal almost disjoint set; (3)Use a diagonal construction to show that the resulting family must have size > κ.

First of all, recall that κ can be partitioned into κ sets, each of size κ. Namely, if f :κ × κ → κ is a bijection, let Xα = f [(α, β) : β < κ]; then clearly 〈Xα : α < κ〉 is asclaimed.

Thus we can apply Zorn’s lemma to get a maximal collection A ⊆ [κ]κ such that themembers of A are pairwise almost disjoint, and |A | ≥ κ.

Hence we just have to get a contradiction from the assumption that |A | = κ. Makingthis assumption, let 〈Yα : α < κ〉 be a one-one enumeration of A . Note that for any α < κ,

Yα\⋃

β<α

Yβ = Yα\⋃

β<α

(Yα ∩ Yβ)

has size κ. This enables us to define by recursion a sequence 〈zα : α < κ〉 like this: havingdefined zβ for all β < α, choose

zα ∈ Yα\

zβ : β < α ∪⋃

β<α

Yβ

.

Then Zdef= zα : α < κ is a set of size κ, and for any α < κ,

Z ∩ Yα ⊆ zβ : β ≤ α,

so that |Z ∩ Yα| < κ. This contradicts the maximality of A .

E20.4 Prove that if F is an uncountable family of finite functions each with range ⊆ ω,then there are distinct f, g ∈ F such that f ∪ g is a function.

We apply Theorem 20.4 to the indexed system 〈dmn(f) : f ∈ F 〉 and get an uncountablesubset G of F such that 〈dmn(f) : f ∈ G 〉 is an indexed ∆-system; say that dmn(f) ∩dmn(g) = D for all distinct f, g ∈ G . Then

G =⋃

h∈Dω

f ∈ G : f D = h;

since the index set Dω is countable and G is uncountable, there exist an h ∈ Dω for whichthere are two distinct f, g ∈ G such that f D = g D = h. Then f ∪ g is a function.

E20.5 (Double ∆-system theorem) Suppose that κ is a singular cardinal with cf(κ) > ω.Let 〈λα : α < cf(κ)〉 be a strictly increasing sequence of successor cardinals with supremumκ, with cf(κ) < λ0, and such that for each α < cf(κ) we have (

∑

β<α λβ)+ ≤ λα. Suppose

that 〈Aξ : ξ < κ〉 is a system of finite sets. Then there exist a set Γ ∈ [cf(κ)]cf(κ), a

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sequence 〈Ξα : α ∈ Γ〉 of subsets of κ, a sequence 〈Fα : α ∈ Γ〉 of finite sets, and a finiteset G, such that the following conditions hold:

(i) 〈Ξα : α ∈ Γ〉 is a pairwise disjoint system, and |Ξα| = λα for every α ∈ Γ.(ii) 〈Aξ : ξ ∈ Ξα〉 is a ∆-system with root Fα for every α ∈ Γ.(iii) 〈Fα : α ∈ Γ〉 is a ∆-system with root G.(iv) If ξ ∈ Ξα, η ∈ Ξβ , and α 6= β, then Aξ ∩ Aη = G.

Let κ =⋃

α<cf(κ) Ξ′α where the Ξ′

α’s are pairwise disjoint, with |Ξ′α| = λα for every

α < cf(κ). For each α < cf(κ) let Ξ′′α ∈ [Ξ′

α]λα be such that 〈Aη : η ∈ Ξ′′α〉 is a ∆-system,

say with root Fα. Choose Γ ∈ [cf(κ)]cf(κ) such that 〈Fα : α ∈ Γ〉 is a ∆-system, say withroot G. For each α ∈ Γ let

Bα =⋃

⋃

ξ∈Ξ′′β

Aξ : β ∈ Γ, β < α

.

We claim

(1) |Bα| < λα.

In fact,

|Bα| ≤∑

β<αβ∈Γ

∣∣∣∣∣∣

⋃

ξ∈Ξ′′β

Aξ

∣∣∣∣∣∣

≤∑

β<αβ∈Γ

λβ < λα.

So (1) holds. Now for any α ∈ Γ,

Ξ′′α =

⋃

J∈[Bα]<ω

ξ ∈ Ξ′′α : Aξ ∩Bα = J,

so by (1) there is a Cα ∈ [Bα]<ω such that Ξ′′′α

def= ξ ∈ Ξ′′

α : Aξ ∩ Bα = Cα has size λα.Note that Cα ⊆ Fα, since for distinct ξ, η ∈ Ξ′′′

α we have Cα = Aξ ∩ Aη ∩Bα ⊆ Fα. Nextnote that 〈Aξ\Fα : ξ ∈ Ξ′′′

α 〉 is a system of pairwise disjoint sets; hence for each β ∈ Γ, theset ξ ∈ Ξ′′′

α : (Aξ\Fα) ∩ Fβ 6= 0 is finite. Since |Γ| = cf(κ) < λα, it follows that the set

Ξαdef= Ξ′′′

α \ξ ∈ Ξ′′α : (Aξ\Fα) ∩ Fβ 6= ∅ for some β ∈ Γ.

has size λα.Now we can verify the conditions of the exercise. Conditions (i)–(iii) are clear. Now

suppose that ξ ∈ Ξα, η ∈ Ξβ , and α 6= β. Say β < α. Suppose that γ ∈ Aξ ∩ Aη\G; wewant to get a contradiction. Since Fα ∩ Fβ = G, we have two possibilities.

Case 1. γ /∈ Fα. But γ ∈ Aξ ∩Bα = Cα ⊆ Fα, contradiction.Case 2. γ ∈ Fα\Fβ . Thus γ ∈ (Aη\Fβ) ∩ Fα, contradicting the definition of Ξβ .

E20.6 Suppose that F is a collection of countable functions, each with range ⊆ 2ω, andwith |F | = (2ω)+. Show that there are distinct f, g ∈ F such that f ∪ g is a function.

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Let κ = ω1, λ = (2ω)+, and apply Theorem 20.4 with 〈Ai : i ∈ I〉 replaced by 〈dmn(f) :f ∈ F 〉. We get J ∈ [F ]λ such that 〈dmn(f) : f ∈ J〉 is an indexed ∆-system, say withroot r. Now

J =⋃

h:r→2ω

f ∈ J : f r = h,

and |r(2ω)| = 2ω, so there is an h : r → 2ω such that Kdef= f ∈ J : f r = h has size

(2ω)+. For any two f, g ∈ K, the set f ∪ g is a function.

E20.7 For any infinite cardinal κ, any linear order of size at least (2κ)+ has a subset oforder type κ+ or one similar to (κ+, >).

Let L be a linear order of size (2κ)+, and let ≺ be a well-order of L. Define f : [L]2 → 2by setting, for any a, b ∈ [L]2, say with a < b,

f(a, b) =

0 if a ≺ b,1 if b ≺ a.

By the Erdos-Rado theorem, i.e., Corollary 8.7, let A ∈ [L]κ+

such that A is homogeneousfor f . If f takes the value 0 on [A]2, then A is well-ordered under <, and since its size isκ+, it has a subset of order type κ+. Similarly if f takes the value 1 on [A]2.

E20.8 For any infinite cardinal κ, any tree of size at least (2κ)+ has a branch or anantichain of size at least κ+.

Suppose that T is a tree of size at least (2κ)+. Let S be a subset of T of size (2κ)+. Definef : [S]2 → 2 by setting, for distinct s, t ∈ S,

f(s, t) =

1 if s and t are comparable,0 otherwise.

By the Erdos-Rado theorem, let X ⊆ S be homogeneous for f of size κ+. So if f hasconstant value 1 on [X ]2, then X is a chain of size κ+, hence extends to a branch of sizeat least κ+, while if f has constant value 0 on [X ]2, then X is an antichain of size κ+.

E20.9 Any uncountable tree either has an uncountable branch or an infinite antichain.

Suppose that T is an uncountable tree. Let S ∈ [T ]ℵ1 . We define f : [S]2 → 2 by setting,for any distinct s, t ∈ S,

f(s, t) =

1 if s and t are comparable,0 otherwise.

Then the desired conclusion follows from the Dushnik-Miller theorem.

E20.10 Suppose that m is a positive integer. Show that any infinite set X of positiveintegers contains an infinite subset Y such that one of the following conditions holds:

(i) The members of Y are pairwise relatively prime.(ii) There is a prime p < m such that for any two a, b ∈ Y , p divides a− b.

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(iii) If a, b are distinct members of Y , then a, b are not relatively prime, but thesmallest prime divisor of a− b is at least equal to m.

Let p0, . . . , pi−1 list all of the primes < m, in order. Thus i = 0 if m = 1. Definef : [X ]2 → i+ 2 by setting, for any distinct x, y ∈ X ,

f(x, y) =

i if x and y are relatively prime,j if j < i and pj is the smallest prime dividing x− y,i+ 1 otherwise.

Applying Ramsey’s theorem in the form

ω → (ω, . . . , ω︸︷︷︸

i+1 times

)2,

we get an infinite homogeneous subset Y of X . If f [[Y ]2] = i, then any two membersof Y are relatively prime. If f [[Y ]2] = j with j < i, then pj divides x − y for any twomembers x, y of Y . If f [[Y ]2] = i + 1, then for any two members x, y of Y , the leastprime dividing x− y is at least as big as m.

E20.11 Suppose that X is an infinite set, and (X,<) and (X,≺) are two well-orderingsof X. Show that there is an infinite subset Y of X such that for all y, z ∈ Y , y < z iffy ≺ z.

Define f : [X ]2 → 2 by setting, for any distinct x, y ∈ X , say with x < y,

f(x, y) =

1 if x ≺ y0 otherwise.

By Ramsey’s theorem, let Y be an infinite subset of X which is homogeneous for f . Iff [[Y ]2 = 0, then x < y implies x ≻ y for all distinct x, y ∈ Y . Now since Y is infiniteand < is a well-order of Y , the order type of Y under < is an infinite ordinal. Hence thereis a system 〈yn : n ∈ ω〉 of elements of Y such that y0 < y1 < · · ·. Hence y0 ≻ y1 ≻ · · ·,contradicting the fact that ≺ is a well-order.

Hence f [[Y ]2 = 1. This means that for any distinct x, y ∈ Y we have x < y iff x ≺ y,as desired.

E20.12 Let S be an infinite set of points in the plane. Show that S has an infinite subsetT such that all members of T are on the same line, or else no three distinct points of Tare collinear.

Define f : [S]3 → 2 by

f(s, t, u) =

1 if s, t, u are on a line,0 otherwise.

Let T be an infinite subset of S homogeneous for f . If f [[T ]2] = 1, then all points of Tare on a line. If f [[T ]2] = 0, then no three points of T are on a line.

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E20.13 We consider the following variation of the arrow relation. For cardinals κ, λ, µ, ν,we define

κ→ [λ]µν

to mean that for every function f : [κ]µ → ν there exist an α < ν and a Γ ∈ [κ]λ suchthat f [[Γ]µ] ⊆ ν\α. In coloring terminology, we color the µ-element subsets of κ with νcolors, and then there is a set which is anti-homogeneous for f , in the sense that there isa color α and a subset of size λ all of whose µ-element subsets do not get the color α.

Prove that for any infinite cardinal κ,

κ 6→ [κ]κ2κ .

Hint: (1) Show that there is an enumeration 〈Xα : α < 2κ〉 of [κ]κ in which every memberof [κ]κ is repeated 2κ times. (2) Show that |[κ]κ| = 2κ. (3) Show that there is a one-one〈Yα : α < 2κ〉 such that Yα ∈ [Xα]κ for all α < 2κ. (4) Define f : [κ]κ → 2κ so that for allα < 2κ one has

f(Yα) = o.t.(β < α : Xβ = Xα).

We follow the hint.

(1) There is an enumeration 〈Xα : α < 2κ〉 of [κ]κ in which every member of [κ]κ isrepeated 2κ times.

In fact, let f : 2κ → [κ]κ be a surjection and let g : 2κ × 2κ → 2κ be a bijection. For eachα < 2κ let Xα = f(1st(g−1(α))). Then for all α, β < 2κ,

Xg(α,β) = f(1st(g−1(g(α, β)))) = f(1st(α, β)) = f(α),

and (1) follows.

(2) |[κ]κ| = 2κ.

To prove (2), let f : κ× κ→ κ be a bijection. For each g ∈ κ2 let

Zg =⋃

α<κ,

g(α)=1

f [κ× α].

Then |Zg| = κ provided that g is not identically 0, and Zg 6= Zh if g 6= h. (If g(α) 6= h(α),say g(α) = 1 and h(α) = 0; then f [κ × α] ⊆ Zg but f [κ × α] ∩ Zh = ∅.) Thus2κ ≤ |[κ]κ|+ 1 ≤ 2κ + 1 with cardinal addition, and so (2) follows.

(3) There is a one-one 〈Yα : α < 2κ〉 such that Yα ∈ [Xα]κ for all α < 2κ.

We construct Yα by recursion. If Yβ has been constructed for all β < α, where α < 2κ,choose Yα ∈ [Xα]κ\Yβ : β < α; this is possible by (2). So (3) holds.

Now we define f : [κ]κ → 2κ so that for all α < 2κ one has

f(Yα) = o.t.(β < α : Xβ = Xα).

85

This defines f on Yα : α < 2κ, and it can be extended to all of [κ]κ in any fashion.Now we show that f is the desired counterexample. For, suppose that β < 2κ, Γ ∈ [κ]κ,

and f [[Γ]κ] ⊆ 2κ\β. Choose α < 2κ such that Xα = Γ and γ < α : Xγ = Γ has ordertype β. Then Yα ∈ [Γ]κ and f(Yα) = β, contradiction.


E21.1 Assume MA(κ). Suppose that X is a compact Hausdorff space, and any pairwisedisjoint collection of open sets in X is countable. Suppose that Uα is dense open in X foreach α < κ. Show that

⋂

α<κ Uα 6= ∅.

Let P consist of all nonempty open subsets of X , with ⊆ as the order. Note that forU, V ∈ P , U is compatible with V iff U ∩ V 6= ∅. Hence ccc holds for P . For each α < κlet Dα = p ∈ P : p ⊆ Uα. We claim that Dα is dense in the sense of P . For, supposethat p ∈ P . Since Uα is (topologically) dense, we have p ∩ Uα 6= ∅. By regularity of thespace there is a nonempty open set q such that q ⊆ p ∩ Uα. Thus q ∈ Dα and q ⊆ p, asdesired.

So, we apply MA(κ) and obtain a filter G intersecting each Dα. Because G is a filter,it has the fip as a collection of open sets. Hence by compactness,

⋂

p∈G p 6= ∅. For anyα < κ there is a p ∈ G ∩Dα, and hence p ⊆ Uα. This implies that

⋂

p∈G p ⊆⋂

α<κ Uα.

E21.2 A partial order P is said to have ω1 as a precaliber iff for every system 〈pα : α < ω1〉of elements of P there is an X ∈ [ω1]

ω1 such that for every finite subset F of X there is aq ∈ P such that q ≤ pα for all α ∈ F .

Show that MA(ω1) implies that every ccc partial order P has ω1 as a precaliber.Hint: for each α < ω1 let

Wα = q ∈ P : ∃β > α(q and pα are compatible).

Show that there is an α < ω1 such that Wα = Wβ for all β > α, and apply MA(ω1) to Wα.

Let 〈pα : α < ω1〉 be a system of elements of P ; we want to come up with a set X asindicated. For each α < ω1 let

Wα = q ∈ P : ∃β > α(q and pα are compatible).

Clearly if α < β < ω1 then Wβ ⊆Wα. Now we claim

(1) ∃α∀β ∈ (α, ω1)[Wα = Wβ ].

In fact, otherwise we get a strictly increasing sequence 〈αξ : ξ < ω1〉 of ordinals such thatWαξ+1

⊂ Wαξfor all ξ < ω1. Choose qξ ∈ Wαξ

\Wαξ+1for all ξ < ω1. Then there is an

ordinal βξ such that αξ < βξ ≤ αξ+1 and qξ and pβξare compatible; say rξ ≤ qξ, pβξ

. Weclaim that rξ and rη are incompatible for ξ < η < ω1 (contradicting ccc for P ). For, ifs ≤ rξ, rη, then qξ and pβη

are compatible, and hence qξ ∈Wαξ+1, contradiction. Thus (1)

holds.We are going to apply MA(ω1) to Wα. The dense sets are as follows. For each

β ∈ (α, ω1), letDβ = q ∈Wα : ∃γ ∈ (β, ω1)[q ≤ pγ ].

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To prove density, suppose that r ∈Wα. Then, since Wα = Wβ it follows that r and pγ arecompatible for some γ > β, as desired.

So, let G be a filter on Wα intersecting each set Dβ . It follows that there exist astrictly increasing sequence 〈βξ : ξ < ω1〉 and a sequence 〈qξ : ξ < ω1〉 such that qξ ≤ pβξ

with qξ ∈ G for all ξ < ω1. Clearly then pβξ: ξ < ω1 has the desired property.

E21.3 Call a topological space X ccc iff every collection of pairwise disjoint open sets inX is countable. Show that

∏

i∈I Xi is ccc iff ∀F ∈ [I]<ω[∏

i∈F Xi is ccc]. Hint: use the∆-system theorem.

⇒: Suppose that∏

i∈I Xi is ccc and F ∈ [I]<ω. Also suppose that A is a pairwise disjointcollection of open sets in

∏

i∈F Xi. Then

A′ def

=

x ∈∏

i∈I

Xi : 〈xi : i ∈ T 〉 ∈ U

: U ∈ A

is a collection of pairwise disjoint open sets in∏

i∈I Xi, and hence A ′ is countable. Clearlythis implies that A is countable.⇐: Suppose that ∀F ∈ [I]<ω[

∏

i∈F Xi is ccc], and B is a collection of pairwise disjointopen sets in

∏

i∈I Xi. Suppose that B is uncountable; we want to get a contradiction. Wemay assume that each U ∈ B is basic open, which means that there exist a finite set FUand an open set VU in

∏

i∈F Xi such that U = x ∈∏

i∈I Xi : 〈xi : i ∈ FU 〉 ∈ VU.

E21.4 Assuming MA(ω1), show that any product of ccc spaces is ccc.

By exercise E21.3 it suffices to show that any product of two ccc spacesX, Y is ccc. Supposethat A is an uncountable collection of pairwise disjoint open subsets of X × Y ; we wantto get a contradiction. We may assume that each member of A has the form U × V , withU open in X and V open in Y . Let 〈Uα × Vα : α < ω1〉 be a one-one enumeration of asubset of A . Let P be the partially ordered set consisting of all nonempty open subsets ofX , ordered by ⊆. Thus by exercise 21.2, P has ω1 as a precaliber. Hence let M ∈ [ω1]

ω1

be such that for every finite subset F of M there is a V ∈ P such that V ⊆ Uα for allα ∈ F . Take any distinct α, β ∈M . Then Uα ∩ Uβ 6= ∅, while (Uα × Vα) ∩ (Uβ × Vβ) = ∅,so Vα ∩ Vβ = ∅. This contradicts ccc for Y .

E21.5 Assume MA(ω1). Suppose that P and Q are ccc partially ordered sets. Define ≤on P ×Q by setting (a, b) ≤ (c, d) iff a ≤ c and b ≤ d. Show that < is a ccc partial orderon P ×Q. Hint: use exercise E21.2.

Suppose that 〈(pα, qα) : α < ω1〉 is a system of elements of P ×Q; we want to find distinctα, β < ω1 such that (pα, qα) and (pβ, qβ) are compatible. By exercise 21.2, let Γ ∈ [ω1]

ω1

be such that for every finite subset F of Γ there is an r ∈ P such that r ≤ pα for all α ∈ F .Since Q has ccc, there exist distinct α, β ∈ Γ such that qα and qβ are compatible. Also,pα and pβ are compatible. So (pα, qα) and (pβ , qβ) are compatible.

E21.6 We define <∗ on ωω by setting f <∗ g iff f, g ∈ ωω and ∃n∀m > n(f(m) < g(m).Suppose that MA(κ) holds and F ∈ [ωω]κ. Show that there is a g ∈ ωω such that f <∗ g

87

for all f ∈ F . Hint: let P be the set of all pairs (p, F ) such that p is a finite functionmapping a subset of ω into ω and F is a finite subset of F . Define (p, F ) ≤ (q, G) iffq ⊆ p, G ⊆ F , and

∀f ∈ G∀n ∈ dmn(p)\dmn(q)[p(n) > f(n)].

P is ccc: assume that X is an uncountable subset of P. There are only countably manyfinite functions from ω to ω, so there are distinct (p, F ), (p,G) ∈ P. Then (p, F ∪ G) ≤(p, F ), (p,G). So X is not an incompatible set.

For each h ∈ F let Dh = (p, F ) ∈ P : h ∈ F. Then Dh is dense, since for any(p, F ) ∈P we have (p, F ∪ h) ≤ (p, F ).

For each n ∈ ω let En = (p, F ) : n ∈ dmn(p). Then En is dense. For, suppose that(p, F ) ∈ P and n /∈ dmn(p). Choose m ∈ ω greater than each member of f(n) : f ∈ F.Then clearly (p ∪ (n,m), F ) ≤ (p, F ).

Let G be a filter on P intersecting each Dh and En. Let g =⋃

(p,F )∈F p. Then g is a

function since G is a filter. Moreover, g ∈ ωω since G ∩ En 6= ∅ for all n. Now let f ∈ F .Choose (p, F ) ∈ G ∩Df . Thus p ∈ F . We claim that if m is greater than each member ofdmn(p) then g(m) > f(m) (as desired).

Since m ∈ dmn(g), choose (q,H) ∈ G such that m ∈ dmn(q). Choose (r,K) ∈ Gwith (r,K) ≤ (p, F ), (q,H). Then m ∈ dmn(q) ⊆ dmn(r), so m ∈ dmn(r). Hencem ∈ dmn(r)\dmn(p), and (r,K) ≤ (p, F ), so g(m) = r(m) > f(m).

E21.7 Let B ⊆ [ω]ω be almost disjoint of size κ, with ω ≤ κ < 2ω. Let A ⊆ B with A

countable. Assume MA(κ). Show that there is a d ⊆ ω such that |d∩x| < ω for all x ∈ A ,and |x\d| < ω for all x ∈ B\A . Hint: Let 〈ai : i ∈ ω〉 enumerate A . Let

P = (s, F,m) : s ∈ [ω]<ω, F ∈ [B\A ]<ω, and m ∈ ω;

(s′, F ′, m′) ≤ (s, F,m) iff s ⊆ s′, F ⊆ F ′, m ≤ m′, and

∀x ∈ F

[(

x\⋃

i∈m

ai

)

∩ s′ ⊆ s

]

.

Show that P satisfies ccc. To apply MA(κ), one needs various dense sets. The mostcomplicated is defined as follows. Let D = (s, F,m, i, n) : (s, F,m) ∈ P, i < m, andn ∈ ai\s. Clearly |D | = κ. For each (s, F,m, i, n) ∈ D let

E(s,F,m,i,n) = (s′, F ′, m′) ∈ P : (s, F,m) and (s′, F ′, m′) are incompatible

or (s′, F ′, m′) ≤ (s, F,m) and n ∈ s′.

P is ccc: assume that X is an uncountable subset of P. There are only countably manyfinite functions from ω to ω, so there are distinct (p, F ), (p,G) ∈ P. Then (p, F ∪ G) ≤(p, F ), (p,G). So X is not an incompatible set.

For each x ∈ B\A let Dx = (s, F,m) ∈ P : x ∈ F. Then Dx is dense, since clearly(s, F ∪ x, m) ≤ (s, F,m) for any (s, F,m) ∈ P.

Let D = (s, F,m, i, n) : (s, F,m) ∈ P, i < m, and n ∈ ai\s. Clearly |D | = κ. Foreach (s, F,m, i, n) ∈ D let

E(s,F,m,i,n) = (s′, F ′, m′) ∈ P : (s, F,m) and (s′, F ′, m′) are incompatible

or (s′, F ′, m′) ≤ (s, F,m) and n ∈ s′.

88

Now E(s,F,m,i,n) is dense for each (s, F,m, i, n) ∈ D . For, suppose that (s′, F ′, m′) isgiven. We may assume that (s, F,m) and (s′, F ′, m′) are compatible; say (s′′, F ′′, m′′) ≤(s, F,m), (s′, F ′, m′). We may assume that n /∈ s′′. We claim that (s′′ ∪ n, F ′′, m′′) ≤(s′′, F ′′, m′′) (as desired). This is true since for any x ∈ F ′′ we have n /∈ (x\

⋃

j∈m′′ aj), byvirtue of n ∈ ai and i ∈ m ≤ m′′.

Next, for any i < ω let Hi = (s, F,m) ∈ P : i < m. Then Hi is dense, since(s, F, i+ 1) ≤ (s, F,m) for any m ≤ i.

Now let G be a filter on P intersecting all of these dense sets. Let d =⋃

(s,F,m)∈G s.

Take any x ∈ B\A , and choose (s, F,m) ∈ G ∩ Dx; so x ∈ F . We claim that x ∩ d ⊆⋃

i∈m(x∩ai)∪s, so that x∩d is finite. For, suppose that n ∈ x∩d. choose (s′, F ′, m′) ∈ Gsuch that n ∈ s′. Take (s′′, F ′′, m′′) ∈ G with (s′′, F ′′, m′′) ≤ (s, F,m), (s′, F ′, m′). Thenn ∈ s′′. Since (x\

⋃

i∈m ai) ∩ s′′ ⊆ s and n ∈ x ∩ s′′, it follows that n ∈

⋃

i∈m ai ∪ s, asdesired.

Next, take any i < ω. Choose (s, F,m) ∈ G∩Hi. Thus i < m. We claim that ai\s ⊆ d,so that a\d is finite. To prove this, take any n ∈ ai\s. Then (s, F,m, i, n) ∈ D , so wecan choose (s′, F ′, m′) ∈ G∩E(s,F,m,i,n). Since (s, F,m) and (x′, F ′, m′) are compatible aselements of G, it follows that (s′, F ′, m′) ≤ (s, F,m) and n ∈ s′. Thus n ∈ d, as desired.This proves the claim.

E21.8 [The condition that A is countable is needed in exercise E21.7.] Show that thereexist A ,B such that B is an almost disjoint family of infinite subsets of ω, A ⊆ B,|A | = |B\A | = ω1, and there does not exist a d ⊆ ω such that |x\d| < ω for all x ∈ A ,and |x ∩ d| < ω for all x ∈ B\A . Hint: construct A = aα : α < ω1 and B\A = bα :α < ω1 by constructing aα, bα inductively, making sure that the elements are infinite andpairwise almost disjoint, and also aα ∩ bα = ∅, while for α 6= β we have aα ∩ bβ 6= ∅.

First we show that the hint works. Suppose that we have constructed 〈aα : α < ω1〉 and〈bα : α < ω1〉, so that they are infinite and pairwise almost disjoint, with the additionalindicated property. Suppose that d exists as indicated. Wlog ∀α < ω1(aα\d = F andbα ∩ d = G). Choose m ∈ a0 ∩ b1. If m ∈ d, then m ∈ b1 ∩ d = G ⊆ b0, so m ∈ a0 ∩ b0,contradiction. If m /∈ d, then m ∈ a0\d = F ⊆ a1, so m ∈ a1 ∩ b1, contradiction.

Now we do the construction indicated in the hint. Let 〈ci : i < ω〉 be a system ofpairwise disjoint infinite subsets of ω. Let 〈ci,j : j < ω〉 be a one-one enumeration of ci.Then we define for each m ∈ ω

am = c2m ∪ c2n+1,0 : n < m;

bm = c2m+1 ∪ c2n,0 : n < m.

Clearly this defines infinite pairwise almost disjoint subsets of ω, am∩bm = ∅ for all m ∈ ω,and for n < m we have c2n,0 ∈ an ∩ bm and c2n+1,0 ∈ am ∩ bn.

Now suppose that aβ and bβ have been defined for all β < α, with ω ≤ α < ω1. Letf be a function from ω onto α. By recursion, choose

xm ∈ bf(m)\

(⋃

n<m

bf(n) ∪⋃

n<m

af(n) ∪ xn : n < m ∪ yn : n < m

)

;

89

ym ∈ af(m)\

(⋃

n<m

bf(n) ∪⋃

n<m

af(n) ∪ xn : n ≤ m ∪ yn : n < m

)

.

Let aα = xm : m ∈ ω and bα = ym : m ∈ ω. Then aα ∩ bα = ∅ and

xm ∈ aα ∩ bf(m) ⊆ x0, . . . , xm;

ym ∈ bα ∩ af(m) ⊆ y0, . . . , ym;

aα ∩ af(m) ⊆ x0, . . . , xm;

bα ∩ bf(m) ⊆ y0, . . . , ym.

Hence the construction is complete.

E21.9 Suppose that A is a family of infinite subsets of ω such that⋂F is infinite for

every finite subset F of A . Suppose that |A | ≤ κ. Assuming MA(κ), show that there isan infinite X ⊆ ω such that X\A is finite for every A ∈ A . Hint: use Theorem 21.5.

Let A ′ = X ⊆ ω : ω\X ∈ A , and let B = ω. Clearly the hypothesis of Theorem 21.5holds for A ′ and B. Hence by 21.5 there is a d ⊆ ω such that |d∩X | < ω for all X ∈ A ′

and |d| = |d ∩ ω| = ω. Clearly d is as desired.

E21.10 Show that MA(κ) is equivalent to MA(κ) restricted to ccc partial orders of car-dinality ≤ κ. Hint: Assume the indicated special form of MA(κ), and assume given a cccpartially ordered set P and a family D of at most κ dense sets in P ; we want to find afilter on P intersecting each member of D. We introduce some operations on P . For eachD ∈ D define fD : P → P by setting, for each p ∈ P , fD(p) to be some element of Dwhich is ≤ p. Also we define g : P × P → P by setting, for all p, q ∈ P ,

g(p, q) =

p if p and q are incompatible,r with r ≤ p, q if there is such an r.

Here, as in the definition of fD, we are implicitly using the axiom of choice; for g, wechoose any r of the indicated form.

We may assume that D 6= ∅. Choose D ∈ D, and choose s ∈ D. Now let Q be theintersection of all subsets of P which have s as a member and are closed under all of theoperations fD and g. We take the order on Q to be the order induced from P . Apply thespecial form to Q.

We assume the indicated special form of MA(κ), and assume given a ccc partially orderedset P and a family D of at most κ dense sets in P ; we want to find a filter on P intersectingeach member of D . We introduce some operations on P . For each D ∈ D define fD : P →P by setting, for each p ∈ P , fD(p) to be some element of D which is ≤ p. Also we defineg : P × P → P by setting, for all p, q ∈ P ,

g(p, q) =

p if p and q are incompatible,r with r ≤ p, q if there is such an r.

90

Here, as in the definition of fD, we are implicitly using the axiom of choice; for g, wechoose any r of the indicated form.

We may assume that D 6= ∅. Choose D ∈ D , and choose s ∈ D. Now let Q be theintersection of all subsets of P which have s as a member and are closed under all of theoperations fD and g. We take the order on Q to be the order induced from P .

(1) |Q| ≤ κ.

To prove this, we give an alternative definition of Q. Define

R0 = s;

Rn+1 = Rn ∪ g(a, b) : a, b ∈ Rn ∪ fD(a) : D ∈ D and a ∈ Rn.

Clearly⋃

n∈ω Rn = Q. By induction, |Rn| ≤ κ for all n ∈ ω, and hence |Q| ≤ κ, as desiredin (1).

We also need to check that Q is ccc. Suppose that X is a collection of pairwiseincompatible elements ofQ. Then these elements are also incompatible in P , since x, y ∈ Xwith x, y compatible in P implies that g(x, y) ≤ x, y and g(x, y) ∈ Q, so that x, y arecompatible in Q. It follows that X is countable. So Q is ccc.

Now if D ∈ D , then D ∩Q is dense in Q. In fact, take any q ∈ Q. Then fD(q) ∈ Qand fD(q) ≤ q, as desired.

Now we can apply our special case of MA(κ) to Q and D ∩Q : D ∈ D; we obtain afilter G on Q such that G ∩D ∩Q 6= ∅ for all D ∈ D . Let

G′ = p ∈ P : q ≤ p for some q ∈ G.

We claim that G′ is the desired filter on P intersecting each D ∈ D .

Clearly if p ∈ G′ and p ≤ r, then r ∈ G′.

Suppose that p1, p2 ∈ G′. Choose q1, q2 ∈ G such that qi ≤ p1 for each i = 1, 2. Then

there is an r ∈ G such that r ≤ q1, q2. Then r ∈ G′ and r ≤ p1, p2. So G′ is a filter on P .

Now take any D ∈ D . Then as proved above, D ∩ Q is dense in Q. It follows thatG ∩D ∩Q 6= ∅; say q ∈ G ∩D ∩Q. Then q ∈ G′ ∩D, as desired.

E21.11 Define x ⊂∗ y iff x, y ⊆ ω, x\y is finite, and y\x is infinite. Assume MA(κ),and suppose that L,<) is a linear ordering of size at most κ. Show that there is a system〈ax : x ∈ L〉 of infinite subsets of ω such that for all x, y ∈ L, x < y iff ax ⊂

∗ ay. Hint: letP consist of all pairs (p, n) such that n ∈ ω, p is a function whose domain is a finite subsetof L, and ∀x ∈ dmn(p)[p(x) ⊆ n]. Define (p, n) ≤ (q,m) iff m ≤ n, dmn(q) ⊆ dmn(p),∀x ∈ dmn(q)[p(x) ∩m = q(x)], and ∀x, y ∈ dmn(q)[x < y → p(x)\p(y) ⊆ m].

First note that it is enough to do the construction so that ∀x, y ∈ L[x < y → ax ⊂∗ ay].

In fact, knowing this, if ax ⊂∗ ay, then we must have x < y, as otherwise y ≤ x and hence

ay ⊂∗ ax or ay = ax, both of which are ruled out by ax ⊂

∗ ay.

Now we show that P has ccc. Suppose that A is an uncountable subset of P . By theindexed ∆-system theorem, Theorem 16.4, there is an uncountable subset B of A such that

91

〈dmn(p) : (p, n) ∈ B〉 is an indexed ∆-system. Say M ∈ [L]<ω with dmn(p)∩dmn(q) = Mfor any distinct (p, n), (q,m) ∈ B. Now let Q =Mω. Then

B =⋃

f∈Q

(p, n) ∈ B : ∀x ∈M [p(x) = f(x)].

Since Q is countable, there exist a C ∈ [B]<ω and an f ∈ Q such that ∀(p, n) ∈ C ∀x ∈M [p(x) = f(x)]. Now take two distinct members (p, n) and (q,m) of C . Say m ≤ n. Wenow define a function r. dmn(r) = dmn(p) ∪ dmn(q). For any x ∈ dmn(r),

r(x) =

p(x) if x ∈ dmn(p),q(x) if x ∈ dmn(q)\dmn(p).

Clearly (r, n) ∈ P . We claim (r, n) ≤ (p,m), (q, n); this will show that P has ccc. First weshow that (r, n) ≤ (p,m). We have m ≤ n and dmn(p) ⊆ dmn(r). Take any x ∈ dmn(p).Then r(x) ∩ m = p(x) ∩ m = p(x). Suppose that x, y ∈ dmn(p) and x < y. Thenr(x)\r(y) = p(x)\p(y) ⊆ m. Therefore (r, n) ≤ (p,m). Second we show that (r, n) ≤ (q, n).We have n ≤ n and dmn(q) ⊆ dmn(r). Take any x ∈ dmn(q). If also x ∈ dmn(p) thenx ∈M , and r(x)∩n = p(x)∩n = f(x)∩n = q(x)∩n = q(x). Suppose that x, y ∈ dmn(q)and x < y.

Case 1. x, y ∈ dmn(p). Then r(x)\r(y) = p(x)\p(y) ⊆ m ≤ n.Case 2. x ∈ dmn(p) and y /∈ dmn(p). Then x ∈ M , and so r(x)\r(y) = p(x)\q(y) =

f(x)\q(y) = q(x)\q(y) ⊆ n.Case 3. x /∈ dmn(p) and y ∈ dmn(p). Similar to Case 2.Case 4. x, y /∈ dmn(p). The conclusion is clear.

This finishes the proof that P has ccc.Now after defining certain dense sets we are going to take a filter G with respect to

them and then define

ax =⋃

(p,n)∈G

x∈dmn(p)

p(x)

for each x ∈ L.To show that we can define ax for each x ∈ L we consider Dx

def= (p, n) ∈ P : x ∈

dmn(p). Then Dx is dense, since given (p, n) ∈ P with x /∈ dmn(p), we may assume thatn > 0, and we have (p∪ (x, 0), n) ∈ P and (p∪ (x, 0), n) ≤ (p, n), as is easily checked.

To show that for a given x ∈ L the set ax is infinite, we consider for each i ∈ ω theset

Eixdef= (p, n) : x ∈ dmn(p) and i < n and p(x) 6⊆ i.

To show that this set is dense, let (q,m) ∈ P . By the argument for Dx we may assume thatx ∈ dmn(q). Let n = max(i+ 1, m). Let p have domain dmn(q), and for each y ∈ dmn(q)let p(y) = q(y) ∪ n. Clearly (p, n+ 1) ∈ P . We have m < n+ 1 and dmn(q) = dmn(p).For any y ∈ dmn(q) we have p(y)∩m = q(y). Now suppose that y, z ∈ dmn(q) and y < z.Then p(y)\p(z) = q(y)\q(z) ⊆ m. So (p, n+ 1) ≤ (q,m), and clearly (p, n+ 1) ∈ Eix.

92

Next, for any x, y ∈ L such that x < y and any i ∈ ω let

Fixy = (p, n) ∈ P : x, y ∈ dmn(p) and ∃j ≥ i[j ∈ p(y)\p(x)].

To show that this set is dense, let (q,m) be given. Wlog x, y ∈ dmn(q). Let n = max(i+1, m). Let dmn(p) = dmn(q), and for z ∈ dmn(q) let

p(z) =

q(z) if z 6= y,q(z) ∪ n if y ≤ z.

Clearly (p, n + 1) ∈ P . Since p(x) = q(x) ⊆ m and n ≥ m, we have n ∈ p(y)\p(x).If z ∈ dmn(q), then p(z) ∩ m = q(z). Suppose that u, v ∈ dmn(q) and u < v. Thenp(u)\p(v) ⊆ m; this is only questionable if y ≤ u, and this case is clear. Thus (p, n+ 1) ≤(q,m).

Now we take a filter G with respect to all of these dense sets. So we can define ax forx ∈ L as above, and each ax is infinite. Now suppose that x, y ∈ L and x < y.

Using Dx and Dy , let (p, n) ∈ G with x, y ∈ dmn(p). We claim that ax\ay ⊆ n. Weprove that ax\n ⊆ ay. Suppose that i ∈ ax\n. Choose (q,m) ∈ G such that x ∈ dmn(q)and i ∈ q(x). Then choose (r, s) ∈ G such that (r, s) ≤ (p, n), (q,m). Now i ∈ q(x), soi ∈ r(x). Since x < y and x, y ∈ dmn(p), we have r(x)\r(y) ⊆ n. Now i ≥ n, so it followsthat i ∈ r(y). Hence i ∈ ay, as desired.

It remains only to show that ay\ax is infinite. Suppose it is finite; say ay\ax ⊆ iwith i ∈ ω. Choose (p, n) ∈ Fixy ∩ G. Thus x, y ∈ dmn(p) and there is a j ≥ i such thatj ∈ p(y)\p(x). Thus j ∈ ay. Since ay\i ⊆ ax, we have j ∈ ax. Choose (q,m) ∈ G suchthat x ∈ dmn(q) and j ∈ q(x). Choose (r, s) ≤ (p, n), (q,m). Then j ∈ q(x), so j ∈ r(x).Now r(x) ∩ n = p(x), so j ∈ p(x), contradiction.

E21.12 If A ,B are nonempty countable subsets of [ω]ω and a ⊆∗ b whenever a ∈ A andb ∈ B, then there is a c ∈ [ω]ω such that a ⊆∗ c ⊆∗ b whenever a ∈ A and b ∈ B.

Write A = an : n ∈ ω and B = bn : n ∈ ω. Let

c =⋃

n∈ω

⋃

m≤n

am

∩⋂

m≤n

bm

.

Now suppose that p ∈ ω. Then

ap\c =⋂

n∈ω

ap ∩

⋂

m≤n

(ω\am) ∪⋃

m≤n

(ω\bm)

=⋂

n<p

ap ∩

⋂

m≤n

(ω\am) ∪⋃

m≤n

(ω\bm)

∩⋂

n≥p

ap ∩

⋂

m≤n

(ω\am) ∪⋃

m≤n

(ω\bm)

93

=⋂

n<p

ap ∩

⋂

m≤n

(ω\am) ∪⋃

m≤n

(ω\bm)

∩⋂

n≥p

ap ∩⋃

m≤n

(ω\bm)

⊆ ap ∩⋃

m≤p

(ω\bm),

and this last set is finite.Furthermore,

c\bp =⋃

n<p

⋃

m≤n

am

∩⋂

m≤n

bm ∩ (ω\bp)

⊆

(⋃

m<p

am

)

\bp,

and this last set is finite.The set c is infinite, as otherwise a0 = (a0 ∩ c) ∪ (a0\c) would be finite.

E21.13 Suppose that A is a nonempty countable family of members of [ω]ω, and ∀a, b ∈A [a ⊆∗ b or b ⊆∗ a]. Also suppose that ∀a ∈ A [a ⊂∗ d], where d ∈ [ω]ω. Then there is ac ∈ [ω]ω such that ∀a ∈ A [a ⊆∗ c ⊂∗ d].

If ∃a ∈ A ∀b ∈ A [b ⊆∗ a], then the conclusion is obvious. So suppose that no such a exists.Then there is a sequence 〈an : n ∈ ω〉 of elements of A such that an ⊂

∗ am for n < m, andthe sequence is cofinal in A in the ⊆∗-sense. Let C = a0∪am+1\am : m ∈ ω∪ω\d.Then C is an almost disjoint family, except that possibly ω\d is finite. By Theorem 11.1and Corollary 11.6, let e ⊆ ω be infinite and almost disjoint from each member of C . Letc = d\e. Then for any n ∈ ω,

an+1\c = (an+1\d) ∪ (an+1 ∩ e)

⊆ (an+1\d) ∪

⋃

i≤n

(ai+1\ai) ∪ a0

∩ e,

and the last set is finite. Thus an+1 ⊆∗ c, hence b ⊆∗ c for all b ∈ A .

Since c ⊆ d, we have c ⊆∗ d. Also, d\c = d ∩ e, and this is infinite since e\d is finite.Thus c ⊂∗ d

Note that c is infinite, since a ⊆∗ c for all a ∈ A .

E21.14 If a, b ∈ [ω]ω and a ⊂∗ b, then there is a c ∈ [ω]ω such that a ⊂∗ c ⊂∗ b.

Write b\a = d ∪ e with d, e infinite and disjoint. Let c = a ∪ d.

94

E21.15 Suppose that A and B are nonempty countable subsets of [ω]ω, ∀x, y ∈ A [x ⊆∗ yor y ⊆∗ x], ∀x, y ∈ B[x ⊆∗ y or y ⊆∗ x], and ∀x ∈ A ∀y ∈ B[a ⊂∗ b]. Then there is ac ∈ [ω]ω such that a ⊂∗ c ⊂∗ b for all a ∈ A and b ∈ B.

By Exercise 21.12 choose d ⊆ ω such that ∀a ∈ A ∀b ∈ B[a ⊆∗ d ⊆∗ b]. Thus either∀a ∈ A [a ⊂∗ d] or ∀b ∈ B[d ⊂∗ b].

Case 1. ∀a ∈ A [a ⊂∗ d]. By Exercise 21.13 choose e ⊆ ω such that ∀a ∈ A [a ⊆∗

e ⊂∗ d]. By Exercise 11.14 choose c ∈ [ω]ω such that e ⊂∗ c ⊂∗ d.Case 2. ∀b ∈ B[d ⊂∗ b]. Then ∀b ∈ B[(ω\b) ⊂∗ (ω\d)]. By exercise E21.13 choose

e ⊆ ω such that ∀b ∈ B[(ω\b) ⊆∗ e ⊂∗ (ω\d)]. By exercise E21.14 choose c ⊆ ω such thate ⊂∗ c ⊂∗ (ω\d). Then ∀a ∈ A ∀b ∈ B[a ⊂∗ (ω\c) ⊂∗ b].

E21.16 Suppose that am ∈ [ω]ω for all m ∈ ω, am ⊂∗ an whenever m < n ∈ ω, b ∈ [ω]ω,

and am ⊂∗ b for all m ∈ ω. Then there is a c ∈ [ω]ω such that ∀m ∈ ω[am ⊂

∗ c ⊂∗ b] andc is near to an : n ∈ ω.

By Exercise 21.15 choose d ⊆ ω such that ∀m ∈ ω[an ⊂∗ d ⊂∗ b]. Now for each

m ∈ ω,⋃

i≤m(am ∩ am+1) is finite, and am+1\⋃

i≤m ai = am+1\⋃

i≤m(am+1 ∩ ai),so am+1\

⋃

i≤m ai is infinite. Choose em ⊆ am+1\⋃

i≤m ai such that |em| = m. Letc = d\

⋃

m∈ω em. Thus c ⊆∗ d ⊂∗ b.If n ∈ ω, then

an\c = (an\d) ∪⋃

m∈ω

(an ∩ em) = (an\d) ∪⋃

m<n

(an ∩ em),

and this last set is finite. Hence an ⊆∗ c. Since n is arbitrary, it follows that an ⊂

∗ c forall n ∈ ω.

Also for any m ∈ ω we have am+1\c ⊇ am+1 ∩ em = em, and so |am+1\c| ≥ m. Itfollows that for any n ∈ ω, am : am\c ⊆ n ⊆ a0, . . . , an. So c is near to am : m ∈ ω.

E21.17 Suppose that A ⊆ [ω]ω, ∀x, y ∈ A [x ⊂∗ y or y ⊂∗ x], b ∈ [ω]ω, ∀x ∈ A [x ⊂∗ b],and ∀a ∈ A [b is near to d ∈ A : d ⊂∗ a].

Then there is a c ∈ [ω]ω such that ∀a ∈ A [a ⊂∗ c ⊂∗ b] and c is near to A .

Proof. We consider several cases.Case 1. ∃a ∈ A ∀d ∈ A [d ⊆∗ a]. By Exercise 21.14, choose c such that a ⊂∗ c ⊂∗ b.

Choose n ∈ ω such that c\b ⊆ n. Then for any m ∈ ω and any d ∈ A , if d\c ⊆ m thend\b ⊆ (d\c) ∪ (c\b) ⊆ max(m,n). Hence

d ∈ A : d\c ⊆ m ⊆ a ∪ d ∈ A : d ⊂∗ a and d\b ⊆ max(m,n),

and the later set is finite, since b is near to d ∈ A : d ⊂∗ a. Thus c is as desired.Case 2. ∀a ∈ A ∃d ∈ A [a ⊂∗ d] and b is near to A . By Exercise 21.15, choose c so

that ∀a ∈ A [a ⊂∗ c ⊂∗ b]. Choose n ∈ ω such that c\b ⊆ n. Then for any m ∈ ω and anyd ∈ A , if d\c ⊆ m then d\b ⊆ (d\c) ∪ (c\b) ⊆ max(m,n). Hence

d ∈ A : d\c ⊆ m ⊆ a ∪ d ∈ A : d\b ⊆ max(m,n),

95

and the later set is finite, since b is near to A . Thus c is as desired.Case 3. ∀a ∈ A ∃d ∈ A [a ⊂∗ d] and b is not near to A . For each m ∈ ω let

Bm = a ∈ A : a\b ⊆ m. Since b is not near to A , choose m so that Bm is infinite.Note that p < q → Bp ⊆ Bq. Hence Bn is infinite for every n ≥ m. Now we claim

(1) ∀n ≥ m∀a ∈ A ∃d ∈ Bn[a ⊆∗ d].

In fact, otherwise we get n ≥ m and a ∈ A such that ∀d ∈ Bn[d ⊂∗ a]. Now b is near tod ∈ A : d ⊂∗ a by a hypothesis of the lemma, so d ∈ A : d ⊂∗ a and d\b ⊆ n is finite.But Bn ⊆ d ∈ A : d ⊂∗ a and d\b ⊆ n, contradiction. So (1) holds.

Next we claim

(2) ∀n ≥ m∀d ∈ Bn[e ∈ Bn : e ⊂∗ d is finite].

In fact, suppose that n ≥ m, d ∈ Bn and e ∈ Bn : e ⊂∗ d is infinite. Since b isnear to a ∈ A : a ⊂∗ d, the set a ∈ A : a ⊂∗ d and a\b ⊆ n is finite. Bute ∈ Bn : e ⊂∗ d ⊆ a ∈ A : a ⊂∗ d and a\b ⊆ n, contradiction. So (2) holds.

From (2) it follows that Bn has order type ω under ⊂∗, for each n ≥ m. Now clearlyA =

⋃

p∈ω Bp, so A is countable.Now by Exercise 21.16, choose cm such that ∀d ∈ Bm[d ⊂∗ cm ⊂

∗ b] and cm is nearto Bm. By (1), a ⊂∗ cm for each a ∈ A . Now suppose that n ≥ m and cn has beendefined so that a ⊂∗ cn for each a ∈ A . Again by Exercise 21.16 choose cn+1 such that∀d ∈ Bn+1[d ⊂

∗ cn+1 ⊂∗ cn] and cn+1 is near to Bn+1. Thus we have

∀a ∈ A [a ⊂∗ · · · ⊂∗ cn+1 ⊂∗ cn ⊂

∗ · · · ⊂∗ cm ⊂∗ b].

By Exercise 21.15, choose d so that ∀a ∈ A ∀n ≥ m[a ⊂∗ d ⊂∗ cn]. We claim that d isnear to A , completing the proof. For, let n ∈ ω. Let p = max(m,n), and choose q ≥ psuch that d\cp ⊆ q. Then

a ∈ A : a\d ⊆ n ⊆ a ∈ A : a\d ⊆ p

= a ∈ Bp : a\d ⊆ p

⊆ a ∈ Bp : a\cp ⊆ q,

where the last inclusion holds since a\cp = (a\d)∪ (d\cp). The last set is finite since cp isnear to Bp, as desired.

E21.18 (The Hausdorff gap) There exist sequences 〈aα : α < ω1〉 and 〈bα : α < ω1〉 ofmembers of [ω]ω such that ∀α, β < ω1[α < β → aα ⊂

∗ aβ and bβ ⊂∗ bα], ∀α, β < ω1[aα ⊂

∗

bβ], and there does not exist a c ⊆ ω such that ∀α < ω1[aα ⊂∗ c and c ⊂∗ bα].

We construct by recursion aα, bα ⊆ ω for α < ω1 so that aα ⊂∗ bα, α < β → aα ⊂

∗ aβ andbβ ⊂

∗ bα, and for all α < ω1, bβ is near to aα : α < β.Let a0 = ∅, b0 = ω. Suppose that aα and bα have been constructed for all α < β

so that aα ⊂∗ bα, α < γ < β → aα ⊂

∗ aγ and bγ ⊂∗ bβ, and α < β → bα is near to

aγ : γ < α. By exercise 21.15 choose c such that ∀α < β[aα ⊂∗ c ⊂∗ bα]. Suppose that

96

α < β. We claim that c is near to aγ : γ < α. In fact, suppose that m ∈ ω. Choosen ≥ m such that c\bα ⊆ n. Now for any γ < α we have aγ\bα ⊆ (aγ\c) ∪ (c\bα), so

aγ : γ < α and aγ\c ⊆ m ⊆ aγ : γ < α and aγ\bα ⊆ n,

and the latter set is finite since bα is near to aγ : γ < α. Thus indeed c is near toaγ : γ < α. Now by exercise E21.17 there is a bβ such that ∀α < β[aα ⊂

∗ bβ ⊂∗ c] and

bβ is near to aa : α < β. By exercise E21.15 choose aβ so that ∀α < β[aα ⊂∗ aβ ⊂

∗ bβ ].This finishes the construction.

Now suppose that d ⊆ ω and ∀α < ω1[aα ⊂∗ d ⊂∗ bα]. Now ω1 =

⋃

m∈ωα < ω1 :aα\d ⊆ m, so we can choose m ∈ ω such that |α < ω1 : aα\d ⊆ m| = ω1. Hencethere is an α < ω1 such that β < α : aβ\d ⊆ m is infinite. Choose p ≥ m such thatd\bα ⊆ p. Now aβ\bα ⊆ (aβ\d) ∪ (d\bα), so β < α : aβ\d ⊆ m ⊆ β < α : aβ\bα ⊆ p,contradicting bα near to aβ : β < α.


E22.1 Let κ be an uncountable regular cardinal. We define S < T iff S and T arestationary subsets of κ and the following two conditions hold:

(1) α ∈ T : cf(α) ≤ ω is nonstationary in κ.(2) α ∈ T : S ∩ α is nonstationary in α) is nonstationary in κ.

Prove that if ω < λ < µ < κ, all these cardinals regular, then Eκλ < Eκµ, where

Eκλ = α < κ : cf(α) = λ,

and similarly for Eκµ.

First of all, α ∈ Eκµ : cf(α) ≤ ω is empty, so of course it is nonstationary in κ.For (2), let C = (µ, κ). We claim that

α ∈ Eκµ : Eκλ ∩ α is nonstationary in α ∩ C = ∅;

this will prove (2). In fact, suppose that α is in the indicated intersection. Let D be clubin α such that Eκλ ∩D = ∅. Now α ∈ Eκµ , so cf(α) = µ. Define αξ for all ξ < λ as follows.Let α0 be the least member of D. If αξ ∈ D has been defined, take any member αξ+1

of D greater than αξ. If ξ is limit less than λ, let αξ =⋃

η<ξ αη. Then αξ ∈ D becauseD is closed. Now let β =

⋃

ξ<λ αξ. Then β ∈ D since D is closed, and cf(β) = λ. Soβ ∈ Eκλ ∩D, contradiction.

E22.2 Continuing exercise E22.1: Assume that κ is uncountable and regular. Show thatthe relation < is transitive.

Suppose that A < B < C. Then by definition

(1) α ∈ C : cf(α) ≤ ω is nonstationary in κ.

(2) α ∈ B : α ∩ A is nonstationary is nonstationary in κ.

(3) α ∈ C : α ∩B is nonstationary is nonstationary in κ.

97

We want to show

α ∈ C : α ∩A is nonstationary is nonstationary.

Our assumptions give us clubs M,N in κ such that

α ∈ B : α ∩A is nonstationary ∩M = ∅ and

α ∈ C : α ∩B is nonstationary ∩N = ∅.

Let M ′ be the set of all limits of members of M ; so also M ′ is club in κ. Now it sufficesto show that

α ∈ C : α ∩A is nonstationary ∩M ′ ∩N = ∅.

So, suppose that α ∈ C ∩M ′ ∩N ; we show that α ∩A is stationary in α. To this end, letP be club in α, and let P ′ be the set of all of its limit points. Now α ∈ C ∩N , so α∩B isstationary. Since α ∈M ′, it follows that α ∩M is club in α. So M ∩ P ′ is club in α, andso we can choose β ∈ α ∩B ∩M ∩ P ′. Now β ∈ B ∩M , so β ∩A is stationary in β. Sinceβ ∈ P ′, it follows that P ∩ β is club in β. So β ∩A ∩ P 6= ∅, hence A ∩ P 6= ∅, as desired.

E22.3 If κ is an uncountable regular cardinal and S is a stationary subset of κ, we define

Tr(S) = α < κ : cf(α) > ω and S ∩ α is stationary in α.

Suppose that A,B are stationary subsets of an uncountable regular cardinal κ and A < B.Show that Tr(A) is stationary.

Assume the conditions of the exercise. Thus by definition, α ∈ B : cf(α) ≤ ω isnonstationary in κ, and also α ∈ B : A ∩ α is non-stationary in α is non-stationaryin κ. Hence there is a club C in κ such that C ∩ α ∈ B : cf(α) ≤ ω = ∅ and alsoC ∩ α ∈ B : A ∩ α is non-stationary in α = ∅. Thus B ∩ C ⊆ Tr(A), and it follows thatTr(A) is stationary in κ.

E22.4 (Real-valued measurable cardinals) We describe a special kind of measure. A mea-sure on a set S is a function µ : P(S)→ [0,∞) satisfying the following conditions:

(1) µ(∅) = 0 and µ(S) = 1.

(2) If µ(s) = 0 for all s ∈ S,

(3) If 〈Xi : i ∈ ω〉 is a system of pairwise disjoint subsets of S, then µ(⋃

i∈ω Xi) =∑

i∈ω µ(Xi). (The Xi’s are not necessarily nonempty.)

Let κ be an infinite cardinal. Then µ is κ-additive iff for every system 〈Xα : α < γ〉 ofnonempty pairwise disjoint sets, wich γ < κ, we have

µ

(⋃

α<γ

Xα

)

=∑

α<γ

µ(Xα).

98

Here this sum (where the index set γ might be uncountable), is understood to be

supF⊆γ,

F finite

∑

α∈F

µ(Xα).

We say that an uncountable cardinal κ is real-valued measurable iff there is a κ-additivemeasure on κ. Show that every measurable cardinal is real-valued measurable. Hint: let µtake on only the values 0 and 1.

Suppose that κ is measurable. Thus κ is uncountable, and there is a κ-complete nonprin-cipal ultrafilter U on κ. Now for any X ⊆ κ we define

µ(X) =

1 if X ∈ U ,0 otherwise.

Conditions (1) and (2) in the definition of measure are clear. We can check (3) and κ-additivity simultaneously, by assuming that 〈Xα : α < β〉 is a system of pairwise disjointsubsets of κ, with β < κ. If µ(

⋃

α<β Xα) = 0, clearly µ(Xα) = 0 for all α < β, and so

µ

(⋃

α<γ

Xα

)

=∑

α<γ

µ(Xα).

Suppose that µ(⋃

α<β Xα) = 1. Thus⋃

α<β Xα ∈ U . If Xα /∈ U for all α < β, thenκ\Xα ∈ U for all α < β, and hence by κ-completeness,

κ\

⋃

α<β

Xα

=⋂

α<β

(κ\Xα) ∈ U,

contradiction. Hence Xα ∈ U for some α < β. There can be only one such α, since ifγ 6= α and Xγ ∈ U , then ∅ = Xα ∩Xγ ∈ U , contradiction. Hence again

µ

(⋃

α<γ

Xα

)

=∑

α<γ

µ(Xα).

E22.5 Suppose that µ is a measure on a set S. A subset A of S is a µ-atom iff µ(A) > 0and for every X ⊆ A, either µ(X) = 0 or µ(X) = µ(A). Show that if κ is a real-valued measurable cardinal, µ is a κ-additive measure on κ, and A ⊆ κ is a µ-atom, thenX ⊆ A : µ(X) = µ(A) is a κ-complete nonprincipal ultrafilter on A. Conclude that κ isa measurable cardinal if there exist such µ and A.

Let F be the indicated set. Obviously A ∈ F . Suppose that X ∈ F and X ⊆ Y ⊆ A.Then

µ(A) = µ(X ∪ (Y \X)∪ (A\Y )) = µ(X)+µ(Y \X)+µ(A\Y ) = µ(A)+µ(Y \X)+µ(A\Y ),

99

and so µ(A\Y ) = 0. Hence µ(A) = µ((A\Y ∪ Y ) = µ(A\Y ) + µ(Y ) = µ(Y ). So Y ∈ F .Now suppose that Y, Z ∈ F . Then

µ(A) = µ(Y ) = µ(Y ∩ Z) + µ(Y \Z) and

µ(A) = µ(Z) = µ(Y ∩ Z) + µ(Z\Y ).

It follows that µ(Y \Z) = µ(Z\Y ). If µ(Y \Z) = µ(A), then also µ(Z\Y ) = µ(A), andhence

2µ(A) = µ(Y \Z) + µ(Z\Y ) = µ((Y \Z) ∪ (Z\Y )) ≤ µ(A),

contradiction. So µ(Y \Z) = 0, and hence µ(A) = µ(Y ∩ Z). It follows that Y ∩ Z ∈ F .So, F is a filter.

Clearly ∅ /∈ F , so F is proper.If X ⊆ A, then µ(A) = µ(X) + µ(A\X), and hence µ(X) = µ(A) or µ(A\X) = µ(A).

So X ∈ F or A\X ∈ F . Thus F is an ultrafilter.Finally, for κ-completeness, suppose that A ∈ [F ]<κ Suppose that

⋂A /∈ F . Then

A\⋂

A ∈ F . Let 〈Xα : α < λ〉 be an enumeration of A . For each α < λ let Yα =⋂

β<αXβ\Xα.

(1)⋃

α<λ

Yα =⋃

α<λ

(A\Xα)

In fact, ⊆ is clear. Suppose that ξ ∈⋃

α<λ(A\Xα), and choose α < λ minimum such thatξ ∈ (A\Xα). Then ξ ∈ Yα. So (1) holds.

Clearly the Yα’s are pairwise disjoint. So from (1) we get

µ(A) = µ(

A\⋂

A

)

= µ

(⋃

α<λ

(A\Xα)

)

= µ

(⋃

α<λ

Yα

)

=∑

α<λ

µ(Yα),

and hence there is a α < λ such that µ(Yα) = 1. Hence µ(A\Xα) = µ(A) also, contradic-tion.

Hence F is κ-complete.Since all members of F have size κ by κ-completeness and nonprincipality, it follows

that |A| = κ. So κ is a measurable cardinal.

E22.6 Prove that if κ is real-valued measurable then either κ is measurable or κ ≤ 2ω.Hint: if there do not exist any µ-atoms, construct a binary tree of height at most ω1.

100

Let µ be a κ-additive measure on κ. By exercise E22.5, if there is a µ-atom, then κ ismeasurable. So, suppose that there does not exist any µ-atoms. We construct a treeunder ⊃ by constructing the levels Lα, as follows. L0 = κ. Suppose that Lα has beenconstructed, and that it is a nonempty collection of subsets of κ each of positive measure.For each X ∈ Lα let YX be a subset of X such that 0 < µ(YX) < µ(X); such a set existssince X is not a µ-atom. Then we define

Lα+1 = YX , X\YX : X ∈ Lα.

If α is a limit ordinal and Lβ has been constructed for every β < α, then we define

Lα =

⋂

β<α

Zβ : Zβ ∈ Lβ for all β < α and µ

⋂

β<α

Zβ

> 0

,

except that if Lα = ∅ the construction stops.Clearly this gives a tree. Let α be the least ordinal such that Lα is not defined. So α

is a limit ordinal.

(1) α ≤ ω1, and in fact, if 〈Zβ : α < γ〉 is a branch of the tree, thus with Zβ ⊂ Zδ ifδ < β < γ, then γ is countable.

In fact, we have µ(Zβ\Zβ+1) > 0 for every β < γ, and the sets Zβ\Zβ+1 are pairwisedisjoint. If γ ≥ ω1, then

γ =⋃

n∈ω

β < γ : µ(Zβ\Zβ+1) >1

n+ 1

,

and hence there would be an n ∈ ω such that

β < γ : µ(Zβ\Zβ+1) >1

n+ 1

is uncountable, which is not possible. So (1) holds.Similarly each level of our tree is countable. It follows that the tree has at most 2ω

branches.Let B be the collection of all branches in this tree, and for each B ∈ B let WB =

⋂

X∈B X . Let C = WB : B ∈ B\∅. Now clearly |C | ≤ 2ω, and C consists of measure0 sets.

(2) κ =⋃

C .

In fact, if α ∈ κ, then B = X ∈ T : α ∈ X is a branch, and so α ∈WB .From (2) it follows that κ ≤ 2ω, since the measure µ is κ-additive and µ(κ) = 1. In

fact, 2ω < κ would imply by (2) that µ(κ) = 0, contradiction.

E22.7 Let κ be a regular uncountable cardinal. Show that the diagonal intersection of thesystem 〈(α+ 1, κ) : α < κ〉 is the set of all limit ordinals less than κ.

101

For any β ∈ κ,

β ∈ α<κ(α+ 1, κ) iff ∀α < β[β ∈ (α+ 1, κ)]

iff ∀α < β[α+ 1 < β]

iff β is a limit ordinal.

E22.8 Let F be a filter on a regular uncountable cardinal κ. We say that F is normaliff it is closed under diagonal intersections. Suppose that F is normal, and (α, κ) ∈ F forevery α < κ. Show that every club of κ is in F . Hint: use exercise E22.7.

Let C be a club, and let 〈αξ : ξ < κ〉 be the strictly increasing enumeration of C, and letD be the set of all limit ordinals less than κ. By exercise E22.7 suffices to show that

D ∩ξ<κ(αξ, κ) ⊆ C.

So, take any β ∈ D ∩ξ<κ(αξ, κ). Thus β is a limit ordinal, and ∀ξ < β[β ∈ (αξ, κ)], i.e.,∀ξ < β[αξ < β]. Now ξ ≤ αξ for all ξ, so C ∩ β is unbounded in β. Hence β ∈ C.

E22.9 Let F be a proper filter on a regular uncountable cardinal κ. Show that the followingconditions are equivalent.

(i) F is normal

(ii) For any S0 ⊆ κ, if κ\S0 /∈ F and f is a regressive function defined on S0, thenthere is an S ⊆ S0 with κ\S /∈ F and f is constant on S.

(i)⇒(ii): Assume (i), and suppose that S0 ⊆ κ, κ\S0 /∈ F , and f is a regressive functionon S0. Suppose that the conclusion fails. Then for every γ < κ we have κ\f−1[γ] ∈ F ,as otherwise we could take S = f−1[γ]. By (i), take β ∈ γ<κ(κ\f

−1[γ]). Then∀γ < β[β ∈ κ\f−1[γ]; in particular, β ∈ κ\f−1[f(β)], contradiction.

(ii)⇒(i): Assume (ii), and suppose that 〈aα : α < κ〉 is a system of members ofF . Suppose that α<κaα /∈ F . Now ∀α ∈ κ\α<κaα∃β < α[α /∈ aβ ]. This gives us aregressive function f defined on κ\α<κaα such that for every α in that set, α /∈ af(α).Hence by (ii) choose S ⊆ κ\α<κaα such that f is constant on S, say with value γ, withκ\S /∈ F . Since aγ ∈ F , we have aγ 6⊆ κ\S. Choose β ∈ aγ ∩ S. Then β /∈ af(β) gives acontradiction.

E22.10 A probability measure on a set S is a real-valued function µ with domain P(S)having the following properties:

(i) µ(∅) = 0 and µ(S) = 1.

(ii) If X ⊆ Y , then µ(X) ≤ µ(Y ).

(iii) µ(a) = 0 for all a ∈ S.

(iv) If 〈Xn : n ∈ ω〉 is a system of pairwise disjoint sets, then µ(⋃

n∈ωXn) =∑

n∈ω µ(Xn). (Some of the sets Xn might be empty.)

Prove that there does not exist a probability measure on ω1. Hint: consider an Ulammatrix.

102

Suppose that µ is a probability measure on ω1. Let f = 〈fρ : ρ < ω1〉 be a family ofinjections fρ : ρ → ω. Define the function A : ω × ω1 → P(ω1) by setting, for any ξ < ωand α < ω1,

Aξα = ρ ∈ ω1\(α+ 1) : fρ(α) = ξ.

Take any α < ω1. Since⋃

n∈ω Anα = ω1\(α+1), Anα∩A

mα = ∅ for α 6= β, and µ(ω1\(α+1)) =

1, choose n(α) ∈ ω such that ϕ(An(α)α ) > 0. Then there exist M ∈ [ω1]

ω1 and m ∈ ω suchthat n(α) = m for every α ∈M . Then 〈Amα : α ∈M〉 is a system of pairwise disjoint setseach of positive measure, contradiction.

E22.11 Show that if κ is a measurable cardinal, then there is a normal κ-complete non-principal ultrafilter on κ. Hint: Let D be a κ-complete nonprincipal ultrafilter on κ. Definef ≡ g iff f, g ∈ κκ and α < κ : f(α) = g(α) ∈ D. Show that ≡ is an equivalence relationon κκ. Show that there is a relation ≺ on the collection of all ≡-classes such that for allf, g ∈ κκ, [f ] ≺ [g] iff α < κ : f(α) < g(α) ∈ D. Here for any function h ∈ κκ weuse [h] for the equivalence class of h under ≡. Show that ≺ makes the collection of allequivalence classes into a well-order. Show that there is a ≺ smallest equivalence class xsuch that ∀f ∈ x∀γ < κ[α < κ : γ < f(α) ∈ D. Let E = X ⊆ κ : f−1[X ] ∈ D. Showthat E satisfies the requirements of the exercise.

≡ is reflexive: κ = α < κ : f(α) = f(α), hence f ≡ f .≡ is symmetric: Assume that f ≡ g. Thus α < κ : f(α) = g(α) ∈ D. Hence

α < κ : g(α) = f(α) = α < κ : f(α) = g(α) ∈ D. Hence g ≡ f .≡ is transitive: Assume that f ≡ g ≡ h. Thus α < κ : f(α) = g(α) ∈ D and

α < κ : g(α) = h(α) ∈ D. Hence

α < κ : f(α) = g(α) ∩ α < κ : g(α) = h(α) ∈ D;

since

α < κ : f(α) = g(α) ∩ α < κ : g(α) = h(α) ⊆ α < κ : f(α) = h(α),

we get α < κ : f(α) = h(α) ∈ D, so f ≡ h.Now define

x ≺ y iff ∃f, g[x = [f ] and y = [g] and α < κ : f(α) < g(α) ∈ D].

(1) ∀f, g ∈ κκ[[f ] ≺ [g] iff α < κ : f(ϕ) < g(α) ∈ D].

In fact,⇐ is immediate from the definition. Now suppose that [f ] ≺ [g]. Choose f ′, g′ ∈ κκsuch that [f ] = [f ′], [g] = [g′], and α < κ : f ′(α) < g′(α) ∈ D. Then

α < κ : f(α) = f ′(α)∩α < κ : f ′(α) < g′(α) ∩ α < κ : g(α) = g′(α)

⊆ α < κ : f(α) < g(α);

the left side is in D, hence also the right side is in D, so α < κ : f(α) < g(α) ∈ D. Thus(1) holds.

103

≺ is irreflexive: α < κ : f(α) < f(α) = ∅ /∈ D, so [f ] 6≺ [f ].≺ is transitive: Assume that [f ] ≺ [g] ≺ [h]. Then

α < κ : f(α) < g(α) ∩ α < κ : g(α) < h(α) ⊆ α < κ : f(α) < h(α);

the left side is in D, hence also the right side is in D, so [f ] ≺ [h].≺ is a linear order: Suppose that f, g ∈ κκ are such that [f ] 6= [g] and [f ] 6≺ [g]. Now

κ = α < κ : f(α) < g(α) ∪ α < κ : f(α) = g(α ∪ α < κ : g(α) < f(α);

The first two sets are not in D, so the third one is in D, and hence [g] ≺ [f ].≺ is a well-order: Suppose not. Then we get a sequence 〈fm : m ∈ ω〉 of members of

κκ such that [fm+1] ≺ [fm] for all m ∈ ω. Thus α < κ : fm+1(α) < fm(α) ∈ D for allm ∈ ω. It follows that

⋂

m∈ω

α < κ : fm+1(α) < fm(α) ∈ D;

taking any element α in this intersection, we get . . . fm+1(α) < fm(α) . . ., contradiction.Now let k(α) = α for all α < κ. Then for any γ < κ we have

α < κ : γ < k(α) = α < κ : γ < α = κ\(γ + 1) ∈ D.

It follows that we can take the smallest equivalence class [f ] such that for any γ < κ wehave α < κ : γ < f(α) ∈ D. Now we let E = X ⊆ κ : f−1[X ] ∈ D. We claim that Eis as desired in the exercise.∅ /∈ E: This is true since f−1[∅] = ∅ /∈ D.If X ⊆ Y ⊆ κ and X ∈ E, then Y ∈ E: In fact, assume that X ⊆ Y ⊆ κ and X ∈ E.

Then f−1[X ] ⊆ f−1[Y ] and f−1[X ] ∈ D, so f−1[y] ∈ D, so that Y ∈ E.If X, Y ∈ E, then X ∩ Y ∈ E: In fact, f−1[X ∩ Y ] = f−1[X ]∩ f−1[y], so this is clear.If X ⊆ κ, then X ∈ E or (κ\X) ∈ E: For, suppose that X /∈ E. Then f−1[X ] /∈ D,

so f−1[κ\X ] = (κ\f−1[X ]) ∈ D, and hence (κ\X) ∈ E.E is nonprincipal: for any α < κ we have β < κ : α < f(β) ∈ D, and β < κ : α <

f(β) ⊆ β < κ : α 6= f(β), so β < κ : α 6= f(β) ∈ D, hence β < κ : α = f(β) /∈ D,hence f−1[α] /∈ D and so α /∈ E.

E is κ-complete: Suppose that 〈Xα : α < β〉 is a system of subsets of κ, withβ < κ and with [Xα] ∈ E for all α < β. Thus f−1[Xα] ∈ D for all α < β. Since

f−1[⋂

α<β Xα

]

=⋂

α<β f−1[Xα] ∈ D, it follows that

⋂

α<β Xα ∈ E.

E is normal: We apply exercise 22.9. Suppose that S0 ∈ E and g is regressive on S0.Note that f−1[S0] ∈ D. Let h = g f . Then for any α ∈ f−1[S0] we have h(α) < f(α),so that [h] ≺ [f ]. By the definition of f it then follows that there is a γ < κ such thatα < κ : γ < h(α) /∈ D. Hence α < κ : h(α) ≤ γ ∈ D. Now

α < κ : h(α) ≤ γ =⋃

δ≤γ

α < κ : h(α) = δ,

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and so there is a δ ≤ γ such that α < κ : h(α) = δ ∈ D. Now α < κ : h(α) = δ =h−1[δ] = f−1[g−1[δ], so g−1[δ] ∈ E. This checks the condition of exercise 22.9.


E23.1 In the ordering <L determine the first four sets and their order. Hint: use Lemma23.2

Recall that L is well-ordered by ordering each set Lα, placing the elements of Lα+1 afterall of the elements of Lα. We have L0 = ∅. By 23.23(viii) we have L1 = V1 = ∅,L2 = V2 = ∅, ∅, L3 = V3 = ∅, ∅, ∅, ∅, ∅. Thus the first four elements of L

are ∅, ∅, ∅, ∅, ∅. We have ∅ <L ∅ and ∅ <L both ∅ and ∅, ∅. Wejust need to determine the relative order of ∅ and ∅, ∅.

(1) Df(L2, ∅, 1) = P(1L2).

For, clearly ⊆ holds. We can see ⊇ by applying Lemma 23.2:

For ϕ equal to v0 = v0 we get 1L2 ∈ Df(L2, 1).

For ϕ equal to ¬(v0 = v0) we get ∅ ∈ Df(L2, 1).

For ϕ equal to ∃v1(v0 ∈ v1) we get 〈∅〉 ∈ Df(L2, 1).

For ϕ equal to ∃v0(v0 ∈ v1) we get 〈∅〉 ∈ Df(L2, 1).

This proves (1).Now L2 = x ∈ L2 : ∅〈x〉 ∈ 1L2 and ∅ = x ∈ L2 : ∅〈x〉 ∈ 〈∅〉. It follows

that in determining the order of our two elements we have n(L2) = n(∅) = 0. Itfollows that s(L2) = s(∅) = ∅. Now we need to determine R(L2) and R(∅). Since1L2 = Diag=(∅, 1, 0, 0), we have 1L2 ∈ Df

′(0, L2, ∅, 1). Clearly ∅ /∈ Df′(0, L2, ∅, 1), so

it follows that 1L2 <4L21 〈∅〉. Hence L2 preceeds ∅ in the order <L.Thus the first four sets are, in order, ∅, ∅, ∅, ∅, and ∅.

E23.2 Suppose that M is a nonempty transitive class satisfying the comprehension ax-ioms, and also ∀x ⊆M∃y ∈M[x ⊆ y]. Show that M is a model of ZF.

Extensionality holds by Theorem 10.11.Foundation holds by Theorem 10.17Comprehension is given.Pairing: given x, y ∈ M, we have x, y ⊆ M, hence there is a z ∈ M such that

x, y ⊆ z. So pairing holds by Theorem 10.13Union: Suppose that x ∈ M. Then

⋃x ⊆ M, so choose z ∈ M such that

⋃x ⊆ z.

Hence union holds by Theorem 10.14Power set: Clear by Theorem 10.15 and the hypothesis.Replacement: We apply Theorem 10.16 Assume that A,w1, . . . , wn ∈M and

∀x ∈ A∃!y ∈MϕM(x, y, A, w1, . . . , wn).

Letz = y ∈M : ∃x ∈ AϕM(x, y, A, w1, . . . , wn).

105

Then choose t ∈M such that z ⊆ t. So replacement holds.Infinity: By induction we show that m ∈M for all m ∈ ω. To show that ∅ ∈M, take

any a ∈M. By comprehension in M choose z ∈M such that ∀x[x ∈ z ↔ x ∈ a ∧ x 6= x].So z = ∅, as desired. Suppose m ∈ M. By the pairing axiom in M applied to m,m,choose a set a ∈M such that m ∈ a. Then by the compehension axiom in M take b ∈M

such that ∀x[x ∈ b ↔ x ∈ a ∧ x = m]. Thus b = m. By the pairing axiom in M

choose c ∈M such that m, m ∈ c. By the union axiom in bf M choose d ∈M such that∀Y ∀x[x ∈ Y ∈ c → x ∈ d]. So m, m ⊆ d. By the comprehension axiom in M choosee ∈M such that ∀x[x ∈ e↔ x ∈ d ∧ (x ∈ m∨ x = m)]. Hence e = m∪ m. This finishesthe induction. So ω ⊆ M. Hence by hypothesis there is a y ∈M such that ω ⊆ y. Nowby comprehension in M let z ∈M be such that ∀x[x ∈ z ↔ x ∈ y ∧ x is a finite ordinal].Thus z = ω. Hence infinity holds in M by Theorem 10.27.

E23.3 Show that if M is a transitive proper class model of ZF, then ∀x ⊆M∃y ∈M[x ⊆y].

Suppose that x ⊆M. Choose α such that x ⊆ Vα. Thus x ⊆ Vα ∩M.

(1) If α ∈M, then Vα ∩M ∈M.

This holds by Theorem 10.31(ii). By (1), it suffices to show that Ord ⊆M.Let ϕ(x, y) be the formula “rank(x) = y”. So ϕ is absolute for M by Theorem

10.30(iv). Given an ordinal β, choose x ∈ M such that rank(x) > β; this is possiblebecause M is a proper class. Now M |= ∀u∃vϕ(u, v), so choose y ∈M such that ϕM(x, y).Thus ϕ(x, y), so y is an ordinal greater than β. Since M is transitive, β ∈M.

E23.4 Show that for every ordinal α > ω, |Lα| = |Vα| iff α = iα.

Assume that α > ω; write α = ω+β. By Lemma 23.24, |Lα| = |α|. By Theorem 10.10(ii),|Vα| = iβ . So |Lα| = |Vα| iff |α| = iβ .

Now suppose that α = iα. Now if γ < α, then ω, γ < α since α is an uncountablecardinal. Hence ω + γ < α. Clearly then α = ω + α, and so α = β. It follows that

|Lα| = |α| = iα = iβ = |Vα|.

Now suppose that α 6= iα. If α is countable, then

|Lα| = |α| ≤ ω < iβ = |Vα|,

as desired. Suppose that α ≥ ω1. Then β ≥ ω1. Write β = ω1 + γ. Then

α = ω + β = ω + ω1 + γ = ω1 + γ = β.

Hence |Lα| = |α| ≤ α < iα = iβ = |Vα|, as desired.

E23.5 Assume V = L and α > ω. Then Lα = Vα iff α = iα.

⇒: by exercise E23.4⇐: First we claim

106

(*) (V = L) Vβ ⊆ Liβfor every ordinal β.

We prove this by induction on β. It is obvious for β = 0, and the inductive step with βlimit is obvious. Now assume it for β. Then using Theorem 23.32

Vβ+1 = P(Vβ) ⊆P(Liβ) ⊆ Li

+β

= Liβ+1,

finishing the inductive proof.Now assume that α = iα. Then Vα ⊆ Liα

= Lα ⊆ Vα using Theorem 23.23(vii).

E23.6 Assume V = L and prove that Lκ = H(κ) for every infinite cardinal κ.

First, Lω = Vω = H(ω). Now suppose that κ is uncountable and regular. Take any x ∈ Lκ.Choose α < κ such that x ∈ Lα. We may assume that α is infinite. Since Lα is transitive,we also have trcl(x) ⊆ Lα. Hence |trcl(x)| ≤ |Lα| = |α| < κ. So x ∈ H(κ). Thus wehave shown that Lκ ⊆ H(κ). Suppose that H(κ) 6⊆ Lκ. By the foundation axiom, choosex ∈ H(κ)\Lκ such that x ⊆ Lκ. Since |x| < κ and κ is regular, choose β < κ such thatx ⊆ Lβ . Hence x ∈ Lβ+ ⊆ Lκ by Theorem 23.32. This is a contradiction. So Lκ = H(κ).This finishes the case when κ is regular.

Now suppose that κ is singular. Then

Lκ =⋃

α<κ

Lα+ =⋃

α<κ

H(α+) = H(κ).

E23.7 Show that if ϕ(y1, . . . , yn, x) is a formula with at most the indicated variables free,then

∀α1, . . . , αn∀a[∀x[ϕ(α1, . . . , αn, x)↔ x = a]→ a ∈ OD].

Also show that ∅ ∈ OD.

Fix α1, . . . , αn and assume that ∀x[ϕ(α1, . . . , αn, x) ↔ x = a]. By Corollary 10.39 fixβ > max(α1, . . . , αn, rank(a)) so that ϕ is absolute for Vβ . Let

R = z ∈ n+1Vβ : ϕ(z0, . . . , zn).

Let s = 〈α1, . . . αn〉 ∈nβ. Then

∀x ∈ Vβ [s〈x〉 ∈ R↔ x = a].

By the absoluteness of ϕ,

R = s ∈ n+1Vβ : ϕ(s0, . . . , sn)Vβ ,

so R ∈ Df(Vβ , ∅, n+ 1) by Lemma 23.2. Hence a ∈ OD.Now for the last part of the exercise, we apply the first part to the formula y = x and

the ordinal 0. This gives

∀a[∀x[0 = x↔ x = a]→ a ∈ OD],

107

which means that 0 ∈ OD.

E23.8 We define s ⊳ t iff s, t ∈ <ωON and one of the following holds:(i) s = ∅ and t 6= ∅;(ii) s, t 6= ∅ and max(rng(s)) < max(rng(t));(iii) s, t 6= ∅ and max(rng(s)) = max(rng(t)) and dmn(s) < dmn(t);(iv) s, t 6= ∅ and max(rng(s)) = max(rng(t)) and dmn(s) = dmn(t) and ∃k ∈

dmn(s)[s k = t k and s(k) < t(k)].

Prove the following:

(v) ⊳ well-orders <ωON.(vi) ∀t ∈ <ωON[s : s ⊳ t is a set].(vii) For every infinite ordinal α we have |<ωα| = |α|.(viii) For every uncountable cardinal κ, the set <ωκ is well-ordered by ⊳ in order type

κ and is an initial segment of <ωON.(ix) <ωω is well-ordered by ⊳ in order type ω2.

(v): Clearly ⊳ is irreflexive, and ∀s, t ∈ <ωON[s 6= t→ s ⊳ t or t ⊳ s]. Now suppose thats ⊳ t ⊳ w.

Case 1. s = ∅. Then t, w 6= ∅, and so s ⊳ w.Case 2. s 6= ∅. Then clearly t, w 6= ∅ and max(rng(s)) ≤ max(rng(t)) ≤ max(rng(w)).

Subcase 2.1. max(rng(s)) < max(rng(t)). Then max(rng(s)) < max(rng(w)), andhence s ⊳ w.

Subcase 2.2. max(rng(s)) = max(rng(t)). Thus dmn(s) ≤ dmn(t).Subsubcase 2.2.1. max(rng(t)) < max(rng(w). max(rng(s)) < max(rng(w))

follows, and so s ⊳ w.Subsubcase 2.2.2. max(rng(t)) = max(rng(w)). So dmn(t) ≤ dmn(w).

Subsubsubcase 2.2.2.1. dmn(s) < dmn(t) or dmn(t) < dmn(w). Thendmn(s) < dmn(w) and hence s ⊳ w.

Subsubsubcase 2.2.2.2. dmn(s) = dmn(t) = dmn(w). Then there existk, l ∈ dmn(s) such that s k = t k, s(k) < t(k), t l = w l, and t(l) < w(l).

Subsubsubsubcase 2.2.2.2.1. k < l. Then s k = w k and s(k) < t(k) =w(k), so s ⊳ w.

Subsubsubsubcase 2.2.2.2.2. k = l. Then s k = w k and s(k) < t(k) <w(k), so s ⊳ w.

Subsubsubsubcase 2.2.2.2.3. l < k. Then s l = w l and s(l) = t(l) =w(l), so s ⊳ w.

This completes the proof that ⊳ is a linear order. Now suppose that A is a nonemptysubset of <ωON. If ∅ ∈ A, then it is the least element of A. So, suppose that ∅ /∈ A. LetB = max(rng(s)) : s ∈ A, and let α be the least member of B. Let C = dmn(s) : s ∈A,max(rng(s)) = α. Thus C 6= ∅, so let m be the least member of C. Let D = s(0) :s ∈ A,max(rng(s)) = α, dmn(s) = m, and let γ0 be the least member of D. Suppose thatγj has been defined for all j ≤ i so that i+ 1 < m and

Edef= s ∈ A : max(rng(s)) = α, dmn(s) = β, s (i+ 1) = 〈γ0, . . . , γi〉

108

is nonempty. Let F = s(i + 1) : s ∈ E, and let γi+1 be the least element of F . Byconstruction, there is an s ∈ A such that max(rng(s)) = α, dmn(s) = m, and s(i) = γifor all i < m. We claim that s is the least element of A. For, take any t ∈ A with s 6= t.Then max(rng(s)) = α ≤ max(rng(t)); if < holds here, then s ⊳ t, as desired. So suppose= holds. Then dmn(s) = m ≤ dmn(t). It < holds here, then s ⊳ t, as desired. So assumethat = holds. Let i ≤ m be maximum such that s i = t i. Then i < m since s 6= t.Then by construction s(i) < t(i). Hence s ⊳ t, as desired. Therefore, ⊳ is a well-order.

(vi): If t = ∅, then s ∈ <ωON : s ⊳ t = ∅. Now suppose that t 6= ∅. Clearlys ∈ <ωON : s ⊳ t ⊆ |<ω(α+ 1), where α = max(rng(t)).

(vii): We have∣∣<ωα

∣∣ ≤

∑

k∈ω

|kα| =∑

k∈ω

|α|k = |α|.

Also, ξ 7→ (0, ξ) is a one-one function from α into 1(α + 1) ⊆ <ω(α + 1), so equalityholds.

(viii): Assume that κ is an uncountable cardinal. We verify the second statementfirst. Suppose that s, t ∈ <ωON and s ⊳ t ∈ <ωκ. If s = ∅, then s ∈ <ωκ. If s 6= ∅, thenmax(rng(s)) ≤ max(rng(t)) ∈ κ; so max(rng(s)) ∈ κ and hence s ∈ <ωκ.

Now |<ωκ| = κ by (vii), so κ ≤ |<ωκ|. If s ∈ <ωκ, then t ∈ <ωOn : t ⊳ s ⊆<ω(max(rng(s))), and so by (vii), |t ∈ <ωOn : t ⊳ s| ≤ |max(ω, rng(s))| < κ. Hence bythe preceding paragraph, the order type of <ωκ is κ.

(ix): For each m ∈ ω let Am = s ∈ <ωω : s 6= ∅ and max(rng(s)) = m. Thus∀m ∈ ω∀s ∈ Am∀t ∈ Am+1[s ⊳ t]. Moreover, <ωω = ∅ ∪

⋃

m∈ω Am. For m ∈ ω and n apositive integer, let Bmn = s ∈ Am : dmn(s) = n. So ∀m ∈ ω∀n ∈ ω\1∀s ∈ Bmn∀t ∈Bm,n+1[s ⊳ t]. Each Bmn is finite. Hence Am has order type ω and <ωω has order typeω2.

E23.9 Prove that for any m ∈ ω and any set A, Df(A, ∅, n) = En(m,A, n) : m ∈ ω.

We prove ⊇ by complete induction on m. Suppose that En(m′, A, n) ∈ Df(A, ∅, n) when-ever m′ < m. If m = 0, then En(m,A, n) = ∅ ∈ Df(A, ∅, n) using Lemma 23.2. Supposethat m 6= 0, and write m = 2i · 3j · 5k · r with r not divisible by 2,3, or 5. If r = 1, k = 0,and i, j < n, then En(m,A, n) = Diag∈(A, n, i, j) ∈ Df ′(0, A, ∅, n) ⊆ Df(A, ∅, n). If r = 1,k = 1, and i, j < n, then En(m,A, n) = Diag=(A, n, i, j) ∈ Df

′(0, A, ∅, n) ⊆ Df(A, ∅, n). Ifr = 1 and k = 2, then i < m, hence by the inductive assumption En(i, A, n) ∈ Df(A, ∅, n),and clearly then En(m,A, n) = nA\En(i, A, n) ∈ Df(A, ∅, n). If r = 1 and k = 3, theni, j < m, hence by the inductive assumption En(i, A, n),En(i, A, n) ∈ Df(A, ∅, n), andclearly then En(m,A, n) = En(i, A, n)∩En(j, A, n) ∈ Df(A, ∅, n). If r = 1 and k = 4, theni < m, hence by the inductive assumption En(i, A, n+ 1) ∈ Df(A, ∅, n), and clearly thenEn(m,A, n) = Proj(A,En(i, A, n+ 1), n) ∈ Df(A, ∅, n). Finally, in any other case, againEn(m, i, n) = ∅ ∈ Df(A, ∅, n) by Lemma 23.2. This finishes the proof of ⊇.

For ⊆, we prove by induction on k that Df ′(k, A, ∅, n) ⊆ En(m,A, n) : m ∈ ω. Fork = 0, take any x ∈ Df

′(k, A, ∅, n). We have three possibilities.Case 1. x = Rel(A, ∅, n, i). Then x = ∅ = En(0, A, n).Case 2. x = Diag∈(A, n, i, j). If not(i, j < n), then x = ∅ = En(0, A, n). If i, j < n,

then x = En(2i · 3j, A, n).

109

Case 3. x = Diag=(A, n, i, j). If not(i, j < n), then x = ∅ = En(0, A, n). If i, j < n,then x = En(2i · 3j · 5, A, n).

Now suppose that Df ′(k, A, ∅, n) ⊆ En(m,A, n) : m ∈ ω. Take any x ∈ Df ′(k +1, A, ∅, n). If x ∈ Df

′(k, A, ∅, n), then the inductive hypothesis gives the desired conclusion.Suppose that x /∈ Df ′(k, A, ∅, n). Then we have three possibilities.

Case 1. There is an R ∈ Df′(k, A, ∅, n) such that x = nA\R. By the inductive

hypothesis choose i so that R = En(i, A, n). Then x = En(2i · 52, A, n).Case 2. There are R, S ∈ Df

′(k, A, ∅, n) such that x = R ∩ S. By the inductivehypothesis choose i, j so that R = En(i, A, n) and S = En(j, A, n). Then x = En(2i · 3j ·53, A, n).

Case 3. There is an R ∈ Df′(k, A, ∅, n + 1) such that x = Proj(A,R, n). By the

inductive hypothesis choose i so that R = En(i, A, n+ 1). Then x = En(2i · 54, A, n).

This completes the inductive proof.

E23.10 Prove that if ϕ(x0, . . . , xn−1) is a formula with free variables among x0, . . . , xn−1,then there is an m ∈ ω such that for every set A,

s ∈ nA : ϕA(s(0), . . . , s(n− 1)) = En(m,A, n).

We proceed by induction on the number of quantifiers in ϕ, and within that, by inductionon formulas in the usual sense. For brevity let S(ϕ) be the set s ∈ nA : ϕA(s(0), . . . , s(n−1)). Then

S(xi ∈ xj) = Diag∈(A, n, i, j) = En(2i · 3j , A, n);

S(xi = xj) = Diag=(A, n, i, j) = En(2i · 3j · 5, A, n);

if S(ψ) = En(i, A, n), then

S(¬ψ) = nA\S(ψ) = En(2i · 52, A, n);

if S(ψ) = En(i, A, n) and S(χ) = En(j, A, n), then

S(ψ ∧ χ) = S(ψ) ∩ S(χ) = En(2l · 3j · 53, A, n).

The inductive step from ψ to ∃yψ requires more care. By a change of bound variable weobtain a formula χ such that ∀yψ is logically equivalent to ∀xnχ, with the free variablesof χ among x0, . . . , xn. Hence with S(χ) = En(i, A, n+ 1),

S(∃yψ) = s ∈ nA : ∃xnχ(s(0), . . . , s(n− 1), xn)

= Proj(A, S(χ), n+ 1) = En(2i · 54, A, n).

E23.11 By exercise E23.8, for each uncountable cardinal κ there is an isomorphism fκfrom (κ,∈) onto (<ωκ,⊳). Then fκ ⊆ fλ for κ < λ. It follows that there is a functionEnon mapping ON onto <ωON such that α < β iff Enon(α) ⊳ Enon(β).

110

Now we define a class function Enod with domain ON, as follows. For any ordinal γ,

Enod(γ) =

a if there exist s, β,m, n such that Enon(γ) = s〈β, n,m〉with m,n ∈ ω, β ∈ ON, s ∈ <ωβ, dmn(s) = n, and∀x ∈ Vβ[s

x ∈ En(m,Vβ, n+ 1)↔ x = a],0 otherwise.

Prove that OD = Enod(γ) : γ ∈ ON.

First suppose that γ ∈ ON. If Enod(γ) = 0, then Enod(γ) ∈ OD by the second partof exercise E23.7. Suppose that Enod(γ) = a by the first part of the definition of Enod.Choose s, β,m, n accordingly. Now En(m,Vβ, n + 1) ∈ Df(A, ∅, n) by Exercise 23.9. Soa ∈ OD by definition.

Conversely, suppose that a ∈ OD. Choose β, n, s, R accordingly. By Exercise 23.9choose m so that R = En(m,Vβ, n+1). Let γ be such that Enon(γ) = s〈β,m, n〉. Clearlythen Enod(γ) = a.

E23.12 Now we define HOD = x ∈ OD : trcl(x) ⊆ OD.Prove that ON ⊆ HOD and HOD is transitive.

Let α ∈ ON. Now α is transitive, so trcl(α) = α. So it suffices to show that α ∈ OD. Weapply exercise E23.7 to the formula y = x to obtain ∀x[α = x ↔ x = α] → α ∈ OD. Soα ∈ OD.

If y ∈ x ∈ HOD, then trcl(x) ⊆ OD. Since y ∈ trcl(x), it follows that y ∈ OD, andalso trcl(y) ⊆ trcl(x), so trcl(y) ⊆ OD. Hence y ∈ HOD.

E23.13 Show that (Vα ∩HOD) ∈ HOD for every ordinal α.

Take any ordinal α. It suffices to show that (Vα ∩HOD) ∈ OD. Applying exercise E23.7to the formula x = Vy ∩HOD, with n = 1, α in place of α1, and Vα ∩HOD in place ofa, we have

∀x[x = Vα ∩HOD↔ x = Vα ∩HOD]→ (Vα ∩HOD) ∈ OD,

and the desired result follows.

E23.14 Prove without using the axiom of choice that HOD is a model of ZFC.

Extensionality: holds since HOD is transitive.Comprehension axioms: by Theorem 10.12 it suffices to take a formula ϕ with free

variables among x, z, w1, . . . , wn, assume that z, w1, . . . , wn ∈ HOD, and prove that

adef= x ∈ z : ϕHODϕ(x, z, w1, . . . , wn) ∈ HOD.

By exercise E23.11, choose ordinals α, β1, . . . , βn such that z = Enod(α) and wi = Enod(βi)for i = 1, . . . , n. Let ψ(y0, . . . , yn, v) be the formula

v = x ∈ Enod(y0) : ϕHOD(x,Enod(y1), . . . ,Enod(yn)).

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Then we apply exercise E23.7, with ψ(y1, . . . , yn, v) in place of ϕ(y1, . . . , yn, x) to obtain

∀v[ψ(α, β1, . . . , βn, v)↔ v = a]→ a ∈ OD.

Now ψ(α, β1, . . . , βn, v) is the formula

v = x ∈ Enod(α) : ϕHOD(x,Enod(β1), . . . ,Enod(βn)),

which is the same asv = x ∈ z : ϕHOD(x, w1, . . . , wn),

or simply v = a. So a ∈ OD. Since a ⊆ HOD, we get a ∈ HOD.Pairing: Given x, y ∈ HOD, take α > max(rank(x), rank(y)). Then x, y ∈ Vα ∩

HOD ∈ HOD by exercise E23.13. So Theorem 10.13 applies.Union: Given x ∈ HOD, take α > rank(x). Then

⋃x ∈ Vα ∩HOD ∈ HOD; apply

Theorem 10.14Power set: Given x ∈ HOD, write x = Enod(α). Apply exercise E23.7 to the formula

x = P(Enod(y)); we get

∀w[w = P(Enod(α))↔ w = P(Enod(α))]→P(Enod(α)) ∈ OD,

and hence P(x) = P(Enod(α)) ∈ OD. Now P(x) ⊆ HOD, so P(x) ∈ HOD. Hencepower set holds by Theorem 10.15

Replacement: to apply Theorem 10.16, let ϕ be a formula with free variables amongx, y, A, w1, . . . , wn, take any A,w1, . . . , wn ∈ HOD, and assume that

∀x ∈ A∃!y[y ∈ HOD ∧ ϕHOD(x, y, A, w1, . . . , wn)].

Choose α, β1, . . . , βn such that A = Enod(α) and wi = Enod(βi) for i = 1, . . . , n. Bythe replacement axiom there is a set Z such that for all x ∈ A there is a y ∈ Z suchthat y ∈ HOD ∧ ϕHOD(x, y, A, w1, . . . , wn). Let a = y ∈ Z : y ∈ HOD ∧ ∃x ∈AϕHOD(x, y, A, w1, . . . , wn). Then

a = y ∈ HOD : ∃x ∈ A[ϕHOD(x, y, A, w1, . . . , wn)].

In fact, ⊆ is clear. Conversely, suppose y ∈ HOD ∧ ∃x ∈ A[ϕHOD(x, y, A, w1, . . . , wn)].Take such an x. Choose y′ ∈ Z such that y′ ∈ HOD∧ϕHOD(x, y′, A, w1, . . . , wn). By theuniqueness condition, y = y′. Hence y ∈ a, as desired.

Since a ⊆ HOD and HOD is transitive, it follows that trcl(a) ⊆ HOD, and sotrcl(a) ⊆ OD. Hence it suffices to show that a ∈ OD. Now let ψ(v, s1, . . . , sn, z) be theformula

z = y ∈ HOD : ∃x ∈ Enod(v)[ϕHOD(x, y,Enod(v),Enod(s1), . . . ,Enod(sn))].

Now we apply exercise E23.7 to the formula ψ(v, s1, . . . , sn, z) in place of ϕ, and take theordinals α, β1, . . . , βn; this gives

∀z[ψ(α, β1, . . . , βn, z)↔ z = a]→ a ∈ OD;

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so

∀z[z = y ∈ HOD : ∃x ∈ Enod(α)

[ϕHOD(x, y,Enod(α),Enod(β1), . . . ,Enod(βn)↔ z = a]

→ a ∈ OD,

giving

∀z[z = y ∈ HOD : ∃x ∈ A[ϕHOD(x, y, A, w1, . . . , wn) ↔ z = a]→ a ∈ OD.

Recalling the definition of a, it follows that a ∈ OD.Foundation: holds since HOD is transitive.Infinity: holds since ω ∈ HOD by exercise E23.12; see Theorem 10.27Axiom of Choice: given A ∈ HOD, we want to find a relation R ∈ HOD which

well-orders A. By absoluteness (Proposition 10.32), this suffices. Let

R = (x, y) ∈ A× A : ∃ξ[x = Enod(ξ) ∧ ∀η ≤ ξ[y 6= Enod(η)]].

Clearly R well-orders A. To see that R ∈ HOD we will apply exercise E23.7 again. Chooseα so that A = Enod(α). Let ϕ(y, x) be the formula

x = (u, v) ∈ Enod(y)× Enod(y) : ∃ξ[u = Enod(ξ) ∧ ∀η ≤ ξ[v 6= Enod(η)].

Applying exercise E23.7, we get

∀x[ϕ(α, x)↔ x = R]→ R ∈ OD.

Now ψ(α, x) is

x = (u, v) ∈ A× A : ∃ξ[u = Enod(ξ) ∧ ∀η ≤ ξ[v 6= Enod(η)],

i.e., it is x = R. So we conclude that R ∈ OD. It remains only to check that R ⊆ HOD.If (u, v) ∈ R, then clearly u, v ∈ HOD. Hence by Theorem 10.28, (u, v) ∈ HOD.

E23.15 Show that the axiom of choice holds in L(B) iff trcl(A) can be well-ordered inL(B).

The direction⇐ is obvious. For⇒, we just need to modify a usual proof of AC in L. Notethat since C = ∅ in the definition of L(B), we have Rel(A, ∅, n, i) = ∅. By the proof ofLemma 23.1, we can simply ignore Rel(A, ∅, n, i).

Let ≺ be a well-order of trcl(B). Let A be any set, and n any natural number. Foreach R ∈ Diag∈(A, n, i, j) : i, j < n, let (Ch(0, A, n, R),Ch(1, A, n, R)) be the smallestpair (i, j), in the lexicographic order of ω×ω such that i, j < n and R = Diag∈(A, n, i, j).Now we define

R <0An S iff (Ch(0, A, n, R),Ch(1, A, n, R))<lex Ch(0, A, n, S),Ch(1, A, n, S)).

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Clearly this is a well-order of Diag∈(A, n, i, j) : i, j < n.In a very analogous way we can define a well-order <1An of Diag=(A, n, i, j) : i, j <

n.Now we can define a well-order <2An of

Diag∈(A, n, i, j) : i, j < n ∪ Diag=(A, n, i, j) : i, j < n

as follows. For any R, S in this union,

R <2An S iff R, S ∈ Diag∈(A, n, i, j) : i, j < n and R <0An S

or R ∈ Diag∈(A, n, i, j) : i, j < n, S /∈ Diag∈(A, n, i, j) : i, j < n

or R, S /∈ Diag∈(A, n, i, j) : i, j < n and R <1An S.

For the next few constructions, suppose that X and A are sets, n ∈ ω, and we are given awell-ordering < of X . Then we well-order nA\R : R ∈ X by setting

S ≺0,A,n,<,X T iff ∃S′, T ′ ∈ X [S′ < T ′ and S = nA\S′, T = nA\T ′].

We well-order R ∪ S : R, S ∈ X as follows. Suppose that U, V ∈ R ∪ S : R, S ∈ X.Let (R, S) be lexicographically smallest in X ×X (using <) such that U = R ∪ S, and let(R′, S′) be lexicographically smallest in X × X (using <) such that V = R′ ∪ S′. ThenU <1,A,n,<,X V iff (R, S) <lex (R′, S′).

We well-order proj(A,R, n) : R ∈ X as follows. Suppose that U, V ∈ proj(A,R, n) :R ∈ X. Let R be <-minimum in X such that U = proj(A,R, n), and let S be <-minimumin X such that V = proj(A, S, n). Then U <2,A,n,<,X V iff R < S.

Next, for any set A and any k, n ∈ ω we define a well-order <3kAn of Df ′(k, A, n) byinduction on k. Let <30An be <2An. Assume that <3kAn has been defined for all n ∈ ω,and let R, S ∈ Df ′(k + 1, A, n). Then we define R <3(k+1)An S iff one of the followingconditions holds:

(1) R, S ∈ Df ′(k, A, n) and R <3kAn S

(2) R ∈ Df ′(k, A, n) and S /∈ Df ′(k, A, n).

(3) R, S /∈ Df ′(k, A, n), R, S ∈ nA\T : T ∈ Df ′(k, A, n), and

R ≺0,A,n,<3kAn,Df ′(k,A,n) S.

(4) R, S /∈ Df ′(k, A, n), R ∈ nA\T : T ∈ Df ′(k, A, n), and S /∈ nA\T : T ∈Df ′(k, A, n).

(5) R, S /∈ Df ′(k, A, n), R, S /∈ nA\T : T ∈ Df ′(k, A, n), R, S ∈ T ∪ U : T, U ∈Df ′(k, A, n), and

R ≺1,A,n,<3kAn,Df ′(k,A,n) S.

(6) R, S /∈ Df ′(k, A, n), R, S /∈ nA\T : T ∈ Df ′(k, A, n), R ∈ T ∪ U : T, U ∈Df ′(k, A, n), and S /∈ T ∪ U : T, U ∈ Df ′(k, A, n).

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(7) R, S /∈ Df ′(k, A, n), R, S /∈ nA\T : T ∈ Df ′(k, A, n), R, S /∈ T ∪ U : T, U ∈Df ′(k, A, n), and

R ≺2,A,n,<3kA(n+1),Df ′(k,A,n) S.

Finally, for any set A and any natural number n, we well-order Df(A, n) as follows. LetR, S ∈ Df(A, n). Let k be minimum such that R ∈ Df ′(k, A, n), and let l be minimumsuch that S ∈ Df ′(l, A, n). Then we define

R <4An S iff k < l, or k = l and R <3kAn S.

Now we define a well-ordering <B5α of LB∅α by recursion. First of all, <B50=≺, the given

well-ordering of trcl(B). If α is a limit ordinal, then for any x, y ∈ LB∅α we define

x <B5α y iff ρ(x) < ρ(y) ∨ [ρ(x) = ρ(y) and x <B5ρ(x) y].

Clearly this is a well-order of LB∅α .

Now suppose that a well-order <B5α of LB∅α has been defined. Then for each n ∈ ω

we define the lexicographic order <B6nα on nLB∅α : for any x, y ∈ nL

B∅α ,

x <B6nα y iff ∃k < n[x k = y k and x(k) <B5α y(k)].

Clearly this is a well-order of nLB∅α . Now for any X ∈ LB∅α+ 1 = D(L

B∅α ), let n(X)

be the least natural number n such that

∃s ∈ nLB∅α ∃R ∈ Df(LB∅

α , n+ 1)[X = x ∈ LB∅α : s〈x〉 ∈ R].

Then let s(X) be the least member of n(X)LB∅α (under the well-order <B6n(X)α) such

that∃R ∈ Df(LB∅

α , n(X) + 1)[X = x ∈ LB∅α : s(X)〈x〉 ∈ R].

Then let R(X) be the least member of Df(LB∅α , n+ 1) (under <

B4LB∅α (n+1)

) such that

X = x ∈ LB∅α : s(X)〈x〉 ∈ R(X)].

Finally, for any X, Y ∈ LB∅α+ 1 we define X <B5(α+1) Y iff one of the followingconditions holds:

(i) X, Y ∈ LB∅α and X <B5α Y .

(ii) X ∈ LB∅α and Y /∈ L

B∅α .

(iii) X, Y /∈ LB∅α and one of the following conditions holds:

(a) n(X) < n(Y ).(b) n(X) = n(Y ) and s(X) <B6n(X)α s(Y ).(c) n(X) = n(Y ) and s(X) = s(Y ) and R(X) <

4LB∅α (n+1)

R(Y ).

Clearly this gives a well-order of LB∅α+1 .

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We denote the union of all the well-orders <B5α for α ∈ On by <BL. Under V = L(B) itis a well-ordering of the universe.

E23.16 Recall from elementary set theory the following definition of the standard well-ordering of On×On:

(α, β) ≺ (γ, δ) iff (α ∪ β < γ ∪ δ)

or (α ∪ β = γ ∪ δ and α < γ)

or (α ∪ β = γ ∪ δ and α = γ and β < δ).

Prove that ≺ is absolute for transitive class models of ZF.Now define ∆ : On→ On×On by recursion as follows:

∆(0) = (0, 0);

∆(α+ 1) =

(β, γ + 1) if ∆(α) = (β, γ) and γ < β,(0, β + 1) if ∆(α) = (β, γ) and γ = β,(β + 1, γ) if ∆(α) = (β, γ) and β + 1 < γ,(γ, 0) if ∆(α) = (β, γ) and β + 1 = γ;

∆(α) =≺ -least(β, γ) such that ∀δ < α[∆(δ) ≺ (β, γ)] if α is limit.

Prove:(1) If α < β, then ∆(α) ≺ ∆(β).(2) ∆ maps onto On×On.(3) ∆ is absolute for transitive class models of ZF.(4) ∆−1 is absolute for transitive class models of ZF.

Clearly ≺ is absolute, by the basic absoluteness results.For (1), fix α; we go by induction on β. Assume that the implication “α < β implies

that ∆(α) ≺ ∆(β)” holds for all β < γ; we prove it for γ. If γ ≤ α, the implicationvacuously holds. Now suppose that α < γ. If γ = β + 1, then α ≤ β, and hence∆(α) ∆(β) either trivially if α = β, or by the inductive hypothesis if α < β. Clearly∆(β) < ∆(β + 1), so ∆(α) < ∆(γ). The case of γ limit is clear.

For (2), suppose to the contrary that rng(∆) ⊂ On × On, and let (β, γ) be the ≺-least pair of ordinals not in rng(∆). Then the following list of all possiblities gives acontradiction in each case:

(1) β = γ = 0.(2) β = δ + 1, γ = 0; then (β, γ) is the immediate successor of (δ, δ + 1).(3) β limit, γ = 0; then (β, γ) is the least upper bound of (δ, β) : δ < β.(4) β = 0, γ = δ + 1; then (β, γ) is the immediate successor of (δ, δ).(5) β = ε+ 1 < δ + 1 = γ; then (β, γ) is the immediate successor of (ε, γ).(6) β = δ + 1 = γ; then (β, γ) is the immediate successor of (δ + 1, δ).(7) β = ε+ 1 > δ + 1 = γ; then (β, γ) is the immediate successor of (β, δ).(8) β limit, β < γ = δ + 1; then (β, γ) is the lub of (δ, γ) : δ < β.(9) β limit, β > δ + 1 = γ; then (β, γ) is the immediate successor of (β, δ).(10) β = 0, γ limit; then (β, γ) is the lub of (δ, 0) : δ < γ.

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(11) β = δ + 1 < γ with γ limit; then (β, γ) is the immediate successor of (δ, γ).(12) β = δ + 1 > γ with γ limit; then (β, γ) is the lub of (β, ε) : ε < γ.(13) β and γ are limit, with β < γ. Then (β, γ) is the lub of (δ, γ) : δ < β.(14) β = γ limit; then (β, γ) is the lub of (δ, β) : δ < β.(15) β and γ are limit, γ < β; then (β, γ) is the lub of (β, δ) : δ < γ.

(3) is clear by the theorem about absoluteness of recursive definitions.(4) holds on general grounds; the inverse of a one-one absolute function is clearly

absolute.

E23.17 Suppose that M is a transitive class model of ZFC, and every set of ordinals isin M . Show that M = V . Hint: take any set X. Let κ = |trcl(X)|, and let f : κ →trcl(X) be a bijection. Define αEβ iff α, β < κ and f(α) ∈ f(β). Use exercise E23.16to show that E ∈M . Take the Mostowski collapse of (κ,E) in M , and infer that X ∈M .

We follow the hint. Let ∆ be as in exercise E23.16. Then ∆−1[E] is a set of ordinals, soby supposition it is in M . It follows that E itself is in M . Now we work in M . Clearly Eis well-founded, since ∈ is, and obviously E is set-like. Hence we can apply the Mostowskicollapse; let G be the Mostowski collapsing function.

Now working outside M , we claim that G = f (hence f ∈ M). Since E is clearlywell-founded (outside M too), we can suppose that G 6= f and take the E-least α ∈ κ suchthat G(α) 6= f(α). Then for any x ∈M ,

x ∈ G(α) iff ∃β ∈ κ[x = G(β) and βEα (definition of G)

iff ∃β ∈ κ[x = f(β) and βEα (minimality of α)

iff ∃β ∈ κ[x = f(β) and f(β) ∈ f(α) (definition of E)

iff x ∈ f(α) (transitivity of rng(f)));

Thus G(α) = f(α), contradiction.Hence our claim holds: G = f . So f ∈ M . Since X ∈ trcl(X) = rng(f), and

rng(f) ∈M by absoluteness, if follows that X ∈M .

E23.18 Show that if X ⊆ ω1 then CH holds in L(X). Hint: show that if A ∈ P(ω) inL(X), then there are α, β < ω1 such that X ∈ Lα(X ∩ β).

We apply a reflection principle, Theorem 10.41. Let ϕ1 be enough of ZFC to prove that“x ∈ Ly(z)” is absolute for transitive models of ϕ1. Let ϕ2 be the extensionality axiom,let ϕ3 be the formula “x ∈ Ly(z)”, and let ϕ4 be the formula “x is an ordinal”. NowA ∈ Lα(X) for some ordinal α. By that reflection principle, let B be a countable setsuch that A, α,X, ω1 ⊆ B ⊆ L(X) and ϕ1, ϕ2, ϕ3ϕ4 are absolute for B, V . Let C bethe Mostowski collapse of B, with collapsing function G. Then (A ∈ Lα(X))B, since ϕ3

is absolute for B, V . Hence also (G(A) ∈ LG(α)(G(X))C , since G is an isomorphism. SoG(A) ∈ LG(α))G(X) since C, being isomorphic to B, is a model of ϕ1. Now A ⊆ ω, soclearly G(A) = A. Let γ be the least ordinal not in C. An easy argument using thetransitivity of C shows that C ∩Ord = γ.

(1) If δ ∈ γ, then G(δ) = δ.

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In fact, suppose that δ ∈ γ is ∈-minimal such that G(δ) 6= δ. If ε ∈ G(δ), choose β ∈ δ suchthat ε = G(β). Since β ∈ δ, we have G(β) = β; so ε = β ∈ δ. This shows that G(δ) ⊆ δ.If β ∈ δ, then G(β) = β and G(β) ∈ G(δ); so β ∈ G(δ). This shows that δ ⊆ G(δ). soG(δ) = δ, contradiction. Hence (1) holds.

Now α is an ordinal, so (α is an ordinal)B since ϕ4 is absolute for B, V . Hence (G(α) isan ordinal)C since G is an isomorphism. Hence G(α) is an ordinal, by absoluteness sinceC is transitive. Morover, C is countable and G(α) ∈ C, so G(α) is countable. This is partof the desired conclusion. Here is the other part:

G(X) = X ∩ γ.

For, suppose that y ∈ G(X). Choose β ∈ X such that y = G(β). Now δdef= G(ω1) is an

ordinal since B is a model of ϕ4. Since G(X) ⊆ G(ω1), it follows that β < δ < γ. So by(1) we have y = G(β) = β ∈ X ∩ γ.

On the other hand, if β ∈ X ∩ γ, then by (1), β = G(β) ∈ G(X). This finishes theproof of the statement of the hint.

It follows that in L(X) we have P(ω) ⊆⋃

α,β<ω1Lα(X ∩ β). Now it is clear (by

induction) that |Lα(X ∩ β)| ≤ max(ω, |α|, |β|), so CH follows.

E23.19 Show that if X ⊆ ω1 then GCH holds in L(X).

By induction it is clear that |Lα(X)| ≤ max(α, ω, |X |) for any ordinal α. Next, we claim

(*) There is a sentence ϕ which is a finite conjunction of members of ZF+V = L(X) suchthat

ZFC ⊢ ∀M [M transitive ∧ ϕM →M = Lo(M)(X)].

Proof. Let ϕ be a conjunction of V = L(X) together with enough of ZF to prove that〈Lα(X) : α ∈ On〉 is absolute, and also enough to prove that there is no largest ordinal.Then for any transitive set M , if ϕM , then o(M) is a limit ordinal, (∀x(x ∈ L(X)))M andhence M = L(X)M , and

M = L(X)M = x ∈M : (∃α(x ∈ L(α)(X)))M =⋃

α∈M

Lα(X) = Lo(M)(X),

as desired in (*).

(**) If V = L(X), then for every infinite ordinal α we have P(Lα(X)) ⊆ Lω1∪α+ .

For, let ϕ be as in (*). Assume that V = L(X) and α is an infinite ordinal. Take any A ∈P(Lα(X)). Let Y = Lα(X)∪A. Clearly X is transitive. Now |Y | ≤ |ω1∪α|. Now by areflection theorem, let M be a transitive set such that X ⊆M , |M | = |α|, and ϕM ↔ ϕV .But ϕV actually holds, so ϕM holds. Hence M = Lo(M) by (**). Now o(M) = M ∩ On,and |M | = |ω1 ∪ α|, so o(M) < ω1 ∪ α

+. Hence A ∈ X ⊆M = Lo(M)(X) ⊆ Lω1∪α+(X).Clearly (**) and exercise E23.18 imply GCH.


E24.1 Show that fin(ω, ω1) collapses ω1 to ω, but preserves cardinals ≥ ω2.

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For collapsing, see the beginning of Chapter 12.Clearly fin(ω, ω1) has size ω1, and hence satisfies ω2-cc. So the second assertion follows

from Proposition 12.5.

E24.2 Suppose that κ is an uncountable regular cardinal of M , and P ∈ M is a κ-ccquasi-order. Assume that C is club in κ, with C ∈ M [G]. Show that there is a C′ ⊆ Csuch that C′ ∈ M and C′ is club in κ. Hint: in M [G] let f : κ → κ be such that∀α < κ[α < f(α) ∈ C]. Apply Theorem 12.4.

By Theorem 12.4 let F ∈ M be such that F : κ→P(κ), f(α) ∈ F (α) for all α < κ, and(|F (α)| < κ)M for all α < κ. We now define g : κ→ κ by recursion. Let g(0) = sup(F (0)).If g(α) has been defined, let g(α + 1) = max(g(α) + 1, sup(F (g(α)))). For β limit, letg(β) =

⋃

α<β g(α).

(1) If β is limit, then g(β) ∈ C.

For, it suffices to show that g(β) ∩ C is unbounded in g(β), since g(β) is clearly a limitordinal. Suppose that γ < g(β). Choose α < β such that γ < g(α). Then γ < g(α) <f(g(α)) ∈ F (g(α)). Now f(g(α)) ≤ sup(F (g(α))) ≤ g(α + 1) < g(β). So we haveγ < f(g(α)) < g(β) with f(g(α)) ∈ C, as desired.

It follows that C′ def= g(β) : β limit is as desired.

E24.3 Suppose that κ is an uncountable regular cardinal of M , and P ∈ M is a κ-ccquasi-order. Assume that S ∈ M is stationary in κ, in the sense of M . Show that itremains stationary in M [G].

Let C ∈ M [G] be club in κ. By exercise E24.2, let C′ ∈ M be club in κ with C′ ⊆ C.Then ∅ 6= C′ ∩ S ⊆ C ∩ S.

E24.4 Suppose that κ is an uncountable regular cardinal of M , and P ∈M is a κ-closedquasi-order. Assume that S ∈ M is stationary in κ, in the sense of M . Show that itremains stationary in M [G].

Suppose that P is κ-closed, C is club in κ, C ∈ M [G], and S ∩ C = ∅. Let f ∈ M [G] bethe strictly increasing enumeration of C. Say τG = f . Choose q ∈ P such that

q τ : κ→ κ is strictly increasing, continuous, and such that ∀α ∈ κ[τ(α) /∈ S]

Now by induction define in M sequences 〈pα : α < κ〉, 〈zα : α < κ〉 such that each pα ∈ P ,each zα ∈ κ\S, p0 = q, pα τ(α) = zα, and pβ ≤ pα for α < β < κ; this is possible byκ-closure. We claim that rng(z) is club in κ with rng(z) ∩ S = ∅ (contradiction). Clearlyrng(z) ∩ S = ∅.

(1) If α < β < κ, then zα < zβ .

In fact,q ∀γ < κ∀δ < κ[γ < δ → τ(γ) < τ(δ)],

sopβ τ(α) < τ(β) ∧ τ(α) = zα ∧ τ(β) = zβ ;

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hence pβ zα < zβ , so zα < zβ , as desired in (1).

(2) If α < κ is limit, then zα =⋃

β<α zβ .

In fact,

q ∀γ < κ

γ limit → τ(γ) =⋃

δ<γ

τ(δ)

;

Hence pα forces this too. Let H be P-generic over M with pα ∈ H, and let g = τH . Nowpα τ(δ) = zδ for each δ ≤ α, so zα = g(α) =

⋃

δ<α g(δ) =⋃

δ<α zδ, as desired in (2).Now it follows that rng(z) is club in κ.

E24.5 Prove that if ZFC is consistent, then so is ZFC + GCH + ¬(V = L).

Let M be a c.t.m. of ZFC + GCH; it exists by the theory of constuctibility. Let P =fin(ω1, 2), and let G be P-generic over M . By Lemma 11.2, G /∈M . By Lemma 11.13, Mand M [G] have the same ordinals, so, since L is absolute, LM = LM [G] ⊆M , and so M [G]is not a model of V = L.

For GCH, let λ be any cardinal of M . Note that our quasi-order is ccc, so thatcardinals are preserved. Then by theorem 24.4,

λ+ ≤ (2λ)M [G] ≤ (ωλ1 )M = (2λ)M = (λ+)M = (λ+)M [G].

Solutions to exercises for Chapter 25

E25.1 Show that for any infinite cardinal κ, the partial order fn(κ, 2, ω) is isomorphic tofn(κ× ω, 2, ω).

Let f : κ×ω → κ be a bijection. Now we define F : fn(κ, 2, ω)→ fn(κ×ω, 2, ω) as follows.Let a ∈ Fn(κ, 2, ω). So a is a finite function contained in κ× 2. Let F (a) = a f−1. ThusF (a) is a function whose domain is

x ∈ dmn(f−1) : f−1(x) ∈ dmn(a) = f [dmn(a)],

and this is a finite subset of κ × ω. Obviously F (a) is a function which is a subset of(κ× ω)× 2. So F (a) ∈ Fn(κ× ω, 2, ω).

If a, b ∈ fin(κ, 2, ω), clearly a ⊆ b iff F (a) ⊆ F (b). In particular, F is one-one. ClearlyF (∅) = ∅.

So it remains only to show that F maps onto fin(κ× ω, 2, ω). Let b ∈ fin(κ× ω, 2, ω).Then clearly b f ∈ fin(κ, 2) and F (b f) = b f f−1 = b, as desired.

E25.2 Prove that if P and Q are isomorphic forcing orders, then RO(P) and RO(Q) areisomorphic Boolean algebras.

This is abstract generalized nonsense.Let f be an isomorphism from P onto Q, and let eP and eQ be the embeddings of P

into RO(P) and of Q into RO(Q) given in Chapter 9. Then the following conditions areclear:

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(1) (eQ f)[P ] is dense in RO(P).

(2) For all p, q ∈ P , if p ≤ q then eQ(f(p))) ≤ eQ(f(q)).

(3) For all p, q ∈ P , p ⊥ q iff eP(f(p)) · eP(f(q)) = 0.

Hence it follows from Theorem 9.22 that there is an isomorphism g from RO(P) into RO(Q)such that g eP = eQ f .

By symmetry we get an isomorphism h from RO(Q) into RO(P) such that h eQ =eP f−1.

Hence h g eP = h eQ f = eP f−1 f = eP. It follows from Theorem 9.22 thath g is the identity on RO(P).

Similarly, g h is the identity on RO(Q). So g is the desired isomorphism.

E25.3 Give an example of non-isomorphic forcing orders P and Q such that RO(P) andRO(Q) are isomorphic Boolean algebras.

Let P be fn(ω, 2, ω), let A = RO(P), and let Q = (A\0,≥, 1). For each a ∈ Q letj(a) = a. Then the conditions of Theorem 9.22 are clear, and so RO(Q) is isomorphic toA. So we just need to show that P and Q are not isomorphic. Suppose to the contrarythat f is an isomorphism from P onto Q. Let p = (0, 0). Now p 6= ∅, so f(p) 6= 1, andhence −f(p) 6= 0, so that −f(p) ∈ Q. Choose q ∈ P such that f(q) = −f(p). Let r beany member of fin(ω, 2) such that q ⊂ r. Hence r < q, and so f(r) < f(q) = −f(p), andso f(p) < −f(r). Since f(r) 6= 1, we have −f(r) 6= 0, so we can choose s ∈ P such thatf(s) = −f(r). Thus f(p) < f(s), so p < s. This means that s ⊂ p. Since p is a singleton,it follows that s = ∅, hence 1 = f(s) = −f(r), so f(r) = 0, contradiction.

E25.4 For any system 〈Pi : i ∈ I〉, we define the weak product∏wi∈I Pi as follows: the

underlying set is f ∈∏

i∈I Pi : i ∈ I : f(i) 6= 1 is finite, with f ≤ g iff f(i) ≤Pig(i)

for all i ∈ I. Prove that for any infinite cardinal κ, the forcing order fin(κ, 2) is isomorphicto∏wα<κ Pα, where each Pα is equal to fin(ω, 2).

Let F be a bijection from κ × ω onto κ. Now for each f ∈ fin(κ, 2) we define G(f) ∈∏

α<κ fin(ω, 2) as follows. For any α < κ, (G(f))α has domain i ∈ ω : F (α, i) ∈ dmn(f).Clearly this is a finite set. For any i ∈ dmn((G(f))α we set ((G(f))α)(i) = f(F (α, i)).Clearly α < κ : dmn(G(f))α 6= ∅ is finite, so G(f) ∈

∏wα<κ fin(ω, 2). Clearly f ⊆ g

implies that G(f) ≥ G(g).Conversely, for each x ∈

∏wα<κ fin(ω, 2) we define

H(x) = (α, i) : ∃β < κ∃j < ω[α = F (β, j), j ∈ dmn(xβ), and xβ(j) = i].

Now H(x) is a function. For, suppose that (α, i), (α, k) ∈ H(x) Then with α = F (b, j) wemust have j ∈ dmn(xβ) and xβ(j) = i and xβ(j) = k, so that i = k. The domain of H(x)is

F (β, j) : j ∈ dmn(xβ);

there are only finitely many β such that xβ 6= ∅, and for each β the set dmn(xβ) is finite,so H(x) is finite. So H(x) ∈ fin(κ, 2). Clearly x ≤ y implies that H(x) ⊇ H(y).

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Now suppose that f ∈ fin(κ, 2). we claim that H(G(f)) = f . For, if β < κ and j < ω,then

F (β, j) ∈ dmn(H(G(f)) iff j ∈ dmn((G(f))β

iff F (β, j) ∈ dmn(f),

and for any F (β, j) ∈ dmn(f),

(H(G(f)))(F (β, j)) = (G(f))β(j) = f(F (β, j)).

Hence H(G(f)) = f , as claimed.Finally, let x ∈

∏wα<κ fin(ω, 2); we claim that G(H(x)) = x. (This completes the

solution.) For, if β < κ and j ∈ ω, then

j ∈ dmn((G(H(x))β) iff F (β, j) ∈ dmn(H(x))

iff j ∈ dmn(xβ),

and for any j ∈ dmn(xβ) we have

(G(H(x))β(j) = (H(x))(F (β, j)) = xβ(j).

Thus G(H(x)) = x.

E25.5 We expand the language of set theory by adding an individual constant ∅. AnUrelement is an object a such that a 6= ∅ but a does not have any elements. (Plural isUrelemente.) A set is an object x which is either ∅ or has an element. Both of these arejust definitions, formally like this:

Ur(a)↔ a 6= ∅ ∧ ∀x(x /∈ a);

Set(x)↔ x = ∅ ∨ ∃y(y ∈ x).

Now we let ZFU be the following set of axioms in this language:

All the axioms of ZF except extensionality and foundation.∀x[¬(x ∈ ∅)].∀x, y[Set(x) ∧ Set(y) ∧ ∀z(z ∈ x↔ z ∈ y)→ x = y].∀x[Set(x) ∧ x 6= ∅ → ∃y ∈ x∀z(z ∈ x→ z /∈ y)].

We also reformulate the axiom of choice for ZFU; it is the following statement:

∀A Set(A ) ∧ ∀x ∈ A [Set(x) ∧ x 6= ∅]

∧ ∀x ∈ A ∀y ∈ A [x 6= y → ∀z[¬(z ∈ x ∧ z ∈ y)]]

→ ∃B∀x ∈ A ∃!y(y ∈ x ∧ y ∈ B).

We let ZFCU be all of these axioms.

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One can adapt most of elementary set theory to use these axioms; browsing through thefirst few chapters should convince one of this.

In this exercise, give a new definition of ordinal.Also, show that if we add the axiom ¬∃a[Ur(a)] we get a theory equivalent to ZF.

We define x is an ordinal to mean that x is a transitive set of transitive sets, just as before.But note that “set” is taken in the new sense; so an ordinal, by definition, does not haveany Urelemente as elements.

Suppose that we add ¬∃a[Ur(a)] to these new axioms. Since Set(a) is just the negationof this, we are now assuming that everything is a set. So the new extensionality axiomreduces to the old one, and the new foundation axiom reduces to the old one. The first newaxiom implies that ∅ is the same as the empty set given by the comprehension axiom. Thusall of the new axioms are derivable in ZF (including ¬∃a[Ur(a)]), and ZF is a consequenceof ¬∃a[Ur(a)] plus the new axioms plus the part of ZF allowed in the new framework.

E25.10

(1) If γ < β, then Wγ ⊆ Wβ.

An obvious induction on β, with γ fixed, proves (1).

(2) If x ∈ y ∈Wβ\U , then x ∈Wβ .

We prove this by induction on β. It vacuously holds for β = 0. Now assume that it holdsfor β, and x ∈ y ∈Wβ+1\U . If y ∈Wβ , then x ∈Wβ ⊆Wβ+1 by the induction hypothesis.Otherwise, y ∈P(Wβ), hence y ⊆Wβ , and obviously x ∈Wβ. Finally, the case β limit isclear.

(3) Wβ ∩ Vα = ∅ for all β.

We prove this by induction on β. It is clear for β = 0. Assume it for β, and suppose thatx ∈ Wβ+1. If x ∈ Wβ , then x /∈ Vα by the inductive hypothesis. Otherwise, ∅ 6= x ∈P(Wα). Choose y ∈ x. Now x ⊆Wβ , so y ∈Wβ . By the inductive hypothesis, y /∈ Vα, soalso x /∈ Vα. This takes care of the successor case, and the limit case in the induction isclear.

(4) Wβ = x ∈W : rankW (x) < β.

In fact, first suppose that x ∈Wβ . Let γ = rankW (x). If γ = −1, obviously rankW (x) < β.Suppose that γ ≥ 0. Then x /∈Wγ , so γ < β. This proves ⊆.

Now suppose that x ∈W and γdef= rankW (x) < β. Then x ∈Wγ+1 ⊆Wβ , as desired.

(5) If x, y ∈ W and x ∈ y, then rankW (x) < rankW (y).

In fact, assume the hypotheses. Then y /∈ U , as otherwise x ∈ Vα, contradicting (3). Hence

βdef= rankW (y) > −1. So y ∈ Wβ+1\Wβ , and hence y ∈ P(Wβ). So y ⊆ Wβ and hence

x ∈Wβ . Therefore rankW (x) < β, as desired.

(6) If x ∈W\U , then rankW (x) = supy∈x(rankW (y) + 1).

For, suppose that x ∈ W\U . Then x 6= ∅ by (3). If y ∈ x, then y ∈ W by (2), and hencerankW (y) < rankW (x) by (5). Hence ≥ holds. Let γ be the right side of (6). If y ∈ x, then

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rankW (y) < γ, and so y ∈ WrankW (y)+1 ⊆ Wγ . This shows that x ⊆ Wγ . So x ∈ P(Wγ).Since also x 6= ∅, as noted above, it follows that x ∈ Wγ+1, and hence rankW (x) ≤ γ,giving ≤.

(7) If x ∈W , then rank(x) = α+ 1 + rankW (x).

We prove (7) by ∈-induction. So, suppose that the implication is true for all members ofx, with x ∈W . If x ∈ U the conclusion is clear. Suppose that x /∈ U . Then

α+ 1 + rankW (x) = α+ 1 + supy∈x

(rankW (y) + 1) by (6)

= supy∈x

(α+ 1 + rankW (y) + 1)

= supy∈x

(rank(y) + 1) by the inductive hypothesis

= rank(x).

Here the last step uses a property of rank found in the M6730 notes.

(8) If a ∈ W\U , then W ∩ a 6= ∅.

In fact, by induction if a ∈Wβ\U then W ∩ a 6= ∅, so (8) holds.

(9) For any a ∈W we have UrW (a) iff a ∈ U\Z.

In fact, suppose that UrW (a). Thus a 6= Z and x /∈ a for all x ∈ W . Then a ∈ U by (8).Thus a ∈ U\Z. Conversely, if a ∈ U\Z, then there is no x ∈ W such that x ∈ a, forevery member of a is in Vα, and (3) applies. Hence UrW (a).

(10) For any a ∈W we have SetW (a) iff a /∈ U\Z.

This is clear from (9).

E25.11 The first axiom obviously holds, since every member of Z is in Vα; so (3) applies.

(11) New extensionality holds.

For, suppose that x, y ∈ W , SetM (x), SetM (y), and for all z ∈ W , z ∈ x iff z ∈ y. Thenby (10), x, y /∈ U . Suppose that z ∈ x. Then by (2), z ∈ W , and hence z ∈ y. Similarlyy ⊆ x. So x = y. This proves (11).

(12) New foundation holds.

For, suppose that x ∈ W , SetW (x), and x 6= Z. Then there is a y ∈ W such that y ∈ x.Choose such a y of smallest W -rank. Suppose that z ∈ W and z ∈ x. Then z /∈ y by (5)and the definition of y.

(13) Comprehension holds.

For, let ϕ be a formula with free variables among x, z, w1, . . . , wn, and suppose thatz, w1, . . . , wn ∈ W . If z ∈ U , let y = Z. Then for any x ∈ W , x /∈ y, and also it isnot true that x ∈ z ∧ ϕW , so the desired equivalence holds. Now suppose that z /∈ U .Let y′ = x ∈ z : ϕW . If y′ = ∅, then clearly again y = Z works for our equivalence. If

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y′ 6= ∅, let y = y′. Say z ∈ Wβ. If x ∈ y, then x ∈ z ∈ Wβ\U , and so x ∈ Wβ by (2). Soy ∈P(Wβ)\∅, and hence y ∈Wβ+1. Clearly

∀x ∈W [x ∈ y ↔ x ∈ z ∧ ϕW ]

as desired. So (13) holds.

(14) Pairing holds.

For, given x, y ∈ W , choose β so that x, y ∈ Wβ . Then x, y ∈ P(Wβ)\∅, so x, y ∈Wβ+1 and the desired conclusion follows.

(15) Union holds.

For, let A ∈W be given. We define

A =

(⋃

A ) ∩W if (⋃

A ) ∩W 6= ∅,Z otherwise.

We claim that A ∈ W . This is clear if (⋃

A ) ∩W = ∅, so suppose that (⋃

A ) ∩W 6= ∅.For each a ∈ A choose βa such that a ∈ Wβa

. Let γ = supa∈A βa. Then A ⊆ Wγ andA 6= ∅, so A ∈Wγ+1. Thus, indeed, A ∈W .

Now suppose that x, Y ∈ W and x ∈ Y ∈ A . So x ∈⋃

A ∩W , so x ∈ A, as desired.

(16) Power set holds.

To prove this, note that this axiom involves the defined notion ⊆. We claim:

(17) For any x, y ∈W , (x ⊆ y)W iff x ∈ U or x ⊆ y.

In fact, assume that x, y ∈ W . First suppose that (x ⊆ y)W and x /∈ U . Suppose thatz ∈ x. Then z ∈W by (2), and so z ∈ y since (x ⊆ y)W . Thus x ⊆ y.

Conversely, if x ∈ U , then (x ⊆ y)W by (3). Suppose that x ⊆ y. Then obviously(x ⊆ y)W .

Thus (17) holds.Now for the power set axiom, let x be given. Define y = U ∪(P(x)∩W ) Now suppose

that z ∈ W and (z ⊆ x)W . By (17) there are two possibilities. If z ∈ U , obviously z ∈ y.If z ⊆ x, again obviously z ∈ y. So (16) is proved.

(18) Infinity holds.

To prove (18), we first define by recursion a sequence 〈un : n ∈ ω〉 of members of W . Letu0 = Z and un+1 = (un ∪ un)

W . Let Ω = rngu. Now each un is in W , so there is a βnsuch that un ∈ Wβn

. Let γ = supn∈ω βn. Thus Ω ⊆ Wγ . Also, obviously Ω 6= ∅. HenceΩ ∈Wγ+1. Clearly Ω is what is needed in the infinity axiom.

(19) Replacement holds.

For, suppose that ϕ is a formula with free variables among x, y, A, w1, . . . , wn, assume thatA,w1, . . . , wn ∈ W , and suppose that (∀x ∈ A∃!yϕ)W . Written out more fully, this lastsupposition is:

∀x ∈W [x ∈ A→ ∃y ∈W [ϕ(x, y, A, w1, . . . , wn)W∧

∀z ∈W [ϕ(x, z, A, w1, . . . , wn)W → y = z]]].

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Thus ∀x ∈ A ∩W∃!y[y ∈ W ∧ ϕW ]. Hence by the replacement axiom we obtain a set Y ′

such that ∀x ∈ A ∩W∃y ∈ Y ′[y ∈W ∧ ϕW ]. Let

u = (x, y) : x ∈ A ∩W and y ∈ Y ′ ∩W and ϕW .

Thus u is a function with domain A ∩W . Define

Y =

rng(u) if rng(u) 6= ∅,Z otherwise.

We claim that Y ∈ W . This is clear if rng(u) = ∅, so suppose that rng(u) 6= ∅. Foreach y ∈ rng(u) choose βu such that y ∈ Wβu

. Let γ =⋃

y∈rng(u) βu. Clearly then

rng(u) ∈P(W + γ)\∅, so rng(u) ∈Wγ+1. This proves that Y ∈W .If x ∈ A ∩W , then u(x) ∈ Y and ϕ(x, u(x), A, w1, . . . , wn)

W , as desired.

E25.12 We extend f by recursion, denoting extension by f+. For any a ∈W ,

f+(a) =

f(a) if a ∈ U\Z,Z if a = Z,x : ∃b ∈ a[x = f+(b)] if a /∈ U .

Note that if a ∈ W\U , then a ⊆ W , by (2). To show that f is one-one and onto, itsuffices by symmetry to show that f+ (f−1)+ is the identity on W . We prove thatf+((f−1)+(a)) = a for all a ∈W , by induction on a. This is clear if a ∈ U . Now supposethat a ∈W\U . Then

f+((f−1)+(a)) = f+(x : ∃b ∈ a[x = (f−1)+(b)])

= y : ∃z ∈ x : ∃b ∈ a[x = (f−1)+(b)][y = f+(z)]

= y : ∃b ∈ a[y = f+((f−1)+(b))]

= y : ∃b ∈ a[y = b] (induction hypothesis)

= a.

So f : W → W is a bijection, and it takes Z to Z. If a, b ∈ W and a ∈ b, obviouslyf+(a) ∈ f+(b). Conversely, suppose that f+(a) ∈ f+(b). Then a = (f−1)+(f+(a)) ∈(f−1)+(f+(b)) = b. Thus f is the desired isomorphism.

E25.13 We define by recursion

D0 = a;

Dn+1 = Dn ∪ b ∈W : b ∈ c ∈ d for some c ∈W ∩Dn;

E =⋃

n∈ω

Dn.

By induction, Dn ⊆ W for each n ∈ ω, and hence E ⊆ W . Clearly a ⊆ E. Hence E 6= ∅,and it follows easily, by an argument used several times above, that E ∈W . Now suppose

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that b, c ∈ W and b ∈ c ∈ E. Choose n such that c ∈ Dn. It follows that b ∈ Dn+1 ⊆ E.Thus E is W -transitive.

For the second part of the problem, it suffices to prove that if T is the W -transitiveclosure of a, then f+[T ] is the W -transitive closure of f+(a). First, f+(a) ⊆ f+[T ]. For,let x ∈ f+(a). Then we can write x = f+(b) with b ∈ a ∩ W . Hence b ∈ T , and sox = f+(b) ∈ f+[T ]. So f+(a) ⊆ f+[T ]. Now suppose that S is any W -transitive set suchthat f+(a) ⊆ S. Let R = x : x ∈ W and f+(x) ∈ S. Now a ⊆ R. For, if y ∈ a, thenf+(y) ∈ f+(a) ⊆ S, and hence y ∈ R.

Suppose that F ∈ W is W -transitive and a ⊆ F . By an easy induction, Dn ⊆ F forevery n ∈ ω, and hence E ⊆ F .

E25.14 (i): Let a ∈ U . Then the W -transitive closure of a is clearly just a itself. Ifa = Z, clearly a is symmetric. If a 6= Z, then letting F = a in the definition shows thata is symmetric.

(ii): Since a is symmetric, let F be a finite subset of U\Z such that g+(a) = a forevery permutation g of U\Z which is the identity on F . Let G = f [F ]. So G is a finitesubset of U\Z. If g is a permutation of U\Z which is the identity on G, Then forany u ∈ F we have (f−1 g f)(u) = f−1(g(f(u)) = f−1(f(u)) = u. So f−1 g f is theidentity on F , and so (f−1)+(g+(f+(a))) = a = f−1(f(a)), and hence g+(f+(a)) = f+(a).This shows that f+(a) is symmetric.

(iii): By E25.13 and (ii), f+ maps H into H. By considering f−1, it is clear thatf+ H is a permutation of H. By E25.12 it is an automorphism.

(iv): We prove this by induction on ϕ. It is obvious for atomic formulas, and the in-duction hypothesis for ∨ is clear. Suppose that (¬ϕ)H holds. If ϕH(f+(v0), . . . , f

+(vn−1))holds, applying the inductive hypothesis to f−1 gives a contradiction. Finally, supposethat (∃xϕ(x, v0, . . . , vn−1))

H holds. Choose x ∈ H such that ϕH(x, v0, . . . , vn−1). ThenϕH(f+(x), f+(v0), . . . , f

+(vn−1)), hence (∃xϕ(f+(x), f+(v0), . . . , f+(vn−1))

H .

(v): We go through all of the axioms. First note that if a is hereditarily symmetricand not in U , then its elements are also hereditarily symmetric. Hence Ur and Set areabsolute for H,W .

• ∀x[¬(x ∈ Z)]. This holds since H ⊆W .

• New extensionality. This holds by the initial remark in (ii).

• New foundation. This also holds by that initial remark.

• Comprehension. Let ϕ be a formula with free variables among x, z, w1, . . . , wn, andsuppose that z, w1, . . . , wn ∈ H. Let y = x ∈ z : ϕH. Note that every member of z is inH, so ϕH makes sense here. Clearly it suffices to show that y ∈ H. Since each member of yis inH, it suffices to show that y is symmetric. Let F,G1, . . . , Gn be finite subsets of U\Zsuch that for every permutation f of U\Z, if f F is the identity then f+(z) = z, andfor each i = 1, . . . , n, if f Gi is the identity then f+(wi) = wi. Now take any permutationf of U\Z which is the identity on F ∪ G1 ∪ . . . ∪Gn. We claim that f+(y) = y, whichwill show that y is symmetric. Take any u ∈ f+(y). Then we can write u = f+(x) withx ∈ y. So x ∈ z and ϕH holds. Hence by (iv), also ϕH(f+(x), f+(z), f+(w1), . . . , f

+(wn))

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holds. Since f+(z) = z and f+(wi) = wi for each i, it follows that u = f+(x) ∈ y. Thusf+(y) ⊆ y. Hence y = (f−1)+(f+(y)) ⊆ f+(y) and so y = f+(y), as desired.

• Pairing. If x, y ∈ H, clearly x, y ∈ H.

• Union. Given A ∈ H, let A = x ∈W : ∃Y ∈ A [x ∈ Y ]. Clearly A is as desired.

• Power set. Given x ∈ H, let

y = z : ∀w ∈W [w ∈ z → w ∈ x.

Clearly y is as desired.

• Infinity. By induction it is clear that each un, defined in the proof of infinity in W givenabove, is in H. Hence so is Ω defined there, and so infinity holds.

• Replacement. This is clear by the proof of replacement in W .

E25.15 (a): Clear, by induction on m.(b): Again clear, by induction on n, with m fixed.(c): We follow the hint. Note that certain simple notions like unordered and ordered

pairs, relations, and functions, are absolute in the usual sense. Also note, by induction,that g+(i) = i for every permutation g of U\Z and every i ∈ ω.

Choose a finite subset F of U\Z such that g+(f) = f for any permutation g ofU\Z which is the identity on F . Clearly 〈f(i) : i ∈ ω〉 is a one-one function, by (b)and absoluteness. Choose u ∈ f(i) : i ∈ ω\F , and choose v ∈ U\F with u 6= v. Sayu = f(i). Let g be a permutation of U\Z which is the identity on F and takes u to v.Now (i, u) ∈ f , so

(i, v) = (i, g(u)) = (g+(i), g+(f(i))) ∈ g+(f) = f.

Hence f(i) = v, contradiction.

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solutions for exercises in chapter 1

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