Chapter II. Cardinal arithmetic

Homework problem 5. (ZF) Show that if ω � P(X) then in fact P(ω) � P(X).

In view of Homework problem 5, one may wonder whether, if X is Dedekind-finite but P(X)is Dedekind-infinite, then it follows that there is an infinite Dedekind-finite set Y such thatP(Y ) � X.This is not the case. To exhibit a counterexample, it is enough to know that it is consistentto have an infinite Dedekind finite set X that is the countable union of finite sets (in fact,sets of size 2). Notice that ω is a surjective image of X, so P(X) is Dedekind-infinite.Suppose that P(Y ) � X. Then certainly Y � X, so Y is a countable union of finite sets Yn.If Y is infinite then Yn 6= ∅ for infinitely many values of n. But then ω is also a surjectiveimage of Y , so ω (and in fact P (ω)) injects into P(Y ) and therefore into X, contradiction.

From now on, we assume the axiom of choice unless stated otherwise.

1 Preliminaries

We use κ, λ, . . . to denote cardinals (initial ordinals), usually infinite.

As usual, κ + λ is the cardinality of the disjoint union of two sets, one of size κ, and oneof size λ, κ+ λ = |κ t λ|. In fact, we can take κ+ λ = |κ+ λ|, where the latter + denotesordinal addition.

The advantage of doing this is that we can generalize it to infinite sums.

Definition 1 Let γ be an ordinal and (αi : i < γ) a sequence of ordinals. Then∑

i<γ αiis the ordinal resulting from concatenating the αi in the given order; formally, we considerthe disjoint union

⋃i∈γ αi × {i} and well-order it by setting (β, i) < (δ, j) iff i < j or else

i = j and β < δ.

It is straightforward to check that this is indeed a well-ordering.

Definition 2 Let I be a set and let (κi : i ∈ I) be an I-indexed family of cardinals. Then∑i∈I

κi =∣∣∣∣∑a<|I|

κπ(a)

∣∣∣∣where π : |I| → I is a bijection, and the latter sum is the infinitary addition of ordinalsdefined above.

One easily checks that this is well-defined (i.e., the same number is obtained for all bijectionsπ), and coincides with the cardinality of the disjoint union of the κi.

On the other hand, without choice, there is no reasonable way of defining∑

i∈I mi, even ifwe make the sum dependent on the specific index set I. For example, it is consistent to

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have an uncountable set that is the countable union of sets of size two. The issue is thateven if there are bijections between sets Ai and Bi for each i ∈ I, it is not necessarily thecase that there is a function that to each i assigns some such bijection, and it is indeedpossible that there is no bijection between the resulting sets

⋃i∈I Ai and

⋃i∈I Bi.

Homework problem 6. It is consistent with ZF that R is a countable union of count-able sets. It is also consistent with ZF that there are infinite Dedekind-finite subsets of R.However, show that these two statements cannot hold simultaneously.

The first result shows that cardinal addition trivializes in the presence of choice.

Lemma 3 Let I be a nonempty set. If the κi are nonzero, and either I or supi∈I κi isinfinite, then

∑i∈I κi = max{|I|, supi∈I κi}. In particular, κ+λ = max{κ, λ} if at least one

of κ, λ is infinite.

Proof: First we argue the case for two cardinals: Let m = max{κ, λ}. Then m ≤ κ+ λ ≤|κ × λ| ≤ |m × m| = m, where we use the fact that there is an injection A t B → A × Bwhenever at least one of A,B is infinite, and the fact that α and α× α have the same sizefor any infinite ordinal α, as shown in Lemma 21 of Chapter I.

(Let me remark that m + n ≤ m× n holds in the absence of choice if both m, n are at leasttwo. It fails in the generality above, since if m is infinite and Dedekind-finite, m+1 > m×1.)

The result is now easy: Clearly,∑

i∈I κi ≥ |I| and∑

i∈I κi ≥ κj for any j ∈ I. Also, ifm = supi∈I κi, then

∑i∈I κi ≤

∑i m = |m× I|. The result follows from the above and the

Schroder-Bernstein theorem. �

Unlike addition, multiplication is significantly more difficult.

Definition 4 Let I be a set and let (κi : i ∈ I) be an I-indexed family of cardinals. Then∏i∈I

κi =∣∣∣∣∏i∈I

κi

∣∣∣∣where the latter

∏denotes the product of the sets κi, the set of all (choice) functions

f : I →⋃i κi such that for all i ∈ I, f(i) ∈ κi.

Clearly, a product is zero is one of its factors is zero, and a finite product has the size ofthe Cartesian product of the sets involved.

One easily checks that if Ai ∼ Bi for all i ∈ I, then∏iAi ∼

∏iBi.

Just as with addition, without choice there is no reasonable way of defining∏i∈I mi, even

if we make the product dependent on the specific index set I. For example, it is consistentto have a countable collection of sets of size two whose product is not in bijection with theproduct of countably many copies of {0, 1}.

Definition 5 κλ = |λκ|, where AB denotes the set of functions from A to B, so∏i∈I κ =

κ|I|. Also, κ · λ = κ× λ = κλ denotes the product of κ and λ.

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During the proof of Lemma 3 we showed:

Lemma 6 κλ = max{κ, λ} if at least one of κ, λ is infinite, and neither is zero. �

From choice it follows that a product of nonempty sets is nonempty. In fact, it tends to bea rather large object. Before we present quantitative evidence of this statement, one moredefinition is in order. This notion is central to set theoretic combinatorics.

Definition 7 If (X,<) is a linearly ordered set, a subset Y ⊆ X is cofinal in X iff

∀x ∈ X ∃y ∈ Y (x ≤ y).

Obviously, this notion coincides with the notion of an unbounded subset if X has no largestelement. If X has a largest element, a subset of X is cofinal iff it contains this largestelement.

Definition 8 A map into a linear order (X,<) is cofinal iff its range is cofinal. Thecofinality cf(X) of (X,<) is the smallest ordinal β such that there is a cofinal map f : β →(X,<).

Lemma 9 1. cf(0) = 0, cf(α + 1) = 1, and cf(α) is an infinite cardinal for any limitordinal α.

2. cf(cf(α)) = cf(α).

3. Any cofinal subset of a linearly ordered set (X,<) contains a cofinal subset of sizecf(X).

Proof: In item 1, only that cf(α) is an infinite cardinal for α limit needs to be argued.But if there is a cofinal map f : β → α and β ∼ γ, then there is a cofinal map g : γ → α.It follows that cf(α) is a cardinal. It is obviously infinite, since α is a limit ordinal andtherefore no map from a finite set into α will be cofinal.

Item 2 is shown similarly: If β < cf(α) and there is a cofinal map f : β → cf(α), then thereis a cofinal map g : β → α, contradicting the definition of cf(α).

To prove item 3, let Y be a cofinal subset of the linearly ordered set X and let Z bea cofinal subset of Y of smallest cardinality. By transfinite recursion, define a strictlyincreasing sequence ~a = (aα : α < θ) of elements of Z. The construction stops once therange of the sequence is cofinal in Z. Let µ = cf(θ), and select a cofinal subsequence of ~aof length µ. Then this sequence witnesses that cf(X) ≤ µ. By minimality of |Z| we mustnecessarily have |Z| = µ.

Now we argue that cf(X) = µ. Letting Y ′ be the range of a cofinal map f : cf(X) → X,the argument above with Y ′ in place of Y shows that we can assume that f is is strictly

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increasing. The situation is now this: We have two maps g : µ → X and f : cf(X) → X,both strictly increasing and cofinal. Also, µ = cf(µ), from item 2. By transfinite inductiondefine subsequences of f and g as follows: Given (pα : α < β), subsequence of f, and(qα : α < β), subsequence of g, if neither is cofinal, define pβ to be the least element inthe range of f that is an upper bound for all the pα and qα. Now define qβ as the leastmember of the range of g with qβ ≥ pβ. The construction stops when the sequences so builtare cofinal in X.

If X has a largest element, then Z is a singleton, µ = 1, and also cf(X) = 1, so we mayassume that this is not the case. It follows that the construction must stop at a limit ordinalβ. But then (pα : α < β) witnesses that β ≥ cf(X) (by definition of cf(X)), so β = cf(X).Similarly, (qα : α < β) is cofinal in Z and β ≤ µ so β = µ by definition of Z (or directly,from the sequence one easily sees that β ≤ cf(µ) = µ and there is a cofinal map h : β → µ,so β = µ).

We have shown that cf(X) = β = µ. This completes the proof. �

Corollary 10 There is a strictly increasing cofinal map f : cf(α)→ α. �

Definition 11 An infinite cardinal κ is regular iff cf(κ) = κ. An infinite limit ordinal α issingular iff cf(α) < α.

The following is immediate from Lemma 3:

Corollary 12 cf(κ+) = κ+ for all κ.

Proof: If f : α → κ+ and α < κ+, then α and all the f(β) for β ∈ α have size at mostκ. But then sup(ran(f)) =

⋃β f(β) ≤

∑β f(β), and |

∑β f(β)| ≤ κ, by Fact 1. It follows

that f is bounded in κ+. �

That ω is regular does not require choice. Similarly, ℵω has cofinality ω, and choice is notneeded to establish this. On the other hand, Gitik showed that it is consistent with ZF thatevery uncountable initial ordinal has cofinality ω.

If cf(ω1) = ω, then ω1 is the countable union of countable sets, since if f : ω → ω1 is cofinal,then ω1 =

⋃n f(n) and each f(n) is a countable ordinal. On the other hand, if X is a

countable set of ordinals, then its order type is a countable ordinal, and it follows that anycountable union of countable sets of ordinals has size at most ω1. This is a curious situation,in that it is consistent that cf(ω1) = cf(ω2) = · · · = ω, and therefore there are countableunions of countable unions of countable sets that are not countable unions of countable sets,etc.

Homework problem 7. (ZF).

1. Assume that R is the countable union of countable sets. Show that cf(ω1) = ω.

2. Assume that every countable union of countable sets is also a countable union of finitesets. Show that ω1 is regular.

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It is consistent with ZF that every countable union of countable sets is also a countableunion of finite sets, and yet not every countable union of countable unions of finite sets is acountable union of finite sets so, in particular, not every countable union of countable setsis countable. This appears in De la Cruz-Hall-Howard-Keremedis-Rubin [6], from 2008.

Lemma 13 A cardinal κ is regular if and only if it cannot be written as a union of fewerthan κ many sets, each of size less than κ.

Proof: If the right hand side holds, no map f : β → κ with β < κ can be cofinal, so κ isregular.

Assume now that κ =⋃i∈I Xi where |I| < κ and |Xi| < κ for each i ∈ I. Without

loss, I = |I| is an infinite cardinal, and it is least for which such a decomposition of X ispossible. Let κi = |Xi|, and define a function f : |I| → κ by f(α) =

∑β<α κβ, where sums

denote ordinal addition. That the range of f is indeed in κ follows from our assumptionon the minimality of |I|. Notice that the ordinal sup(ran(f)) =

∑α<|I| κα has size κ, as

the decomposition X =⋃α∈|I|Xα provides us with a straightforward bijection between κ

and sup(ran(f)). It follows that in fact sup(ran(f)) = κ, i.e., f is cofinal and therefore κ issingular. �

We are ready to prove the first significant result.

Theorem 14 (Konig) Assume that κi < λi for all i ∈ I. Then∑

i κi <∏i λi.

Notice that∑

i κi =∑

i λi is possible. For example, let I = ω and let each κi be 1 and eachλi be 2. Then both sums equal ω.

Proof: First we show that the inequality holds, and then we argue that it is strict. Forthe first part, we simply need to exhibit an injection from ti∈Iκi =

⋃i∈I κi × {i} into the

collection of functions∏i∈I λi. Simply assign to each (α, i) the function that is zero in each

coordinate except the i-th one, where it takes the value α+ 1. Note that since α < κi < λi,then indeed α+ 1 < λi, and the function is well defined. Clearly, from the function we canrecover (α, i), so the assignment is injective.

Now consider an arbitrary map ϕ :⋃i κi × {i} →

∏i λi. We need to exhibit a function

not in its range. For this, notice that the assignment α 7→ ϕ(α, i)(i) maps κi into λi, andtherefore there is some ordinal g(i) ∈ λi not in its range. The function g so defined is notin the range of ϕ, since g(i) 6= ϕ(α, i)(i) for any i ∈ I and any α ∈ κi. �

Note that at the heart of the proof just given there is again a diagonal argument.

Corollary 15 1. If β is a limit ordinal and (κi : i < β) is a strictly increasing sequenceof nonzero cardinals, then

∑α<β κα <

∏α<β κα.

2. If (κi : i ∈ I) is an I-indexed sequence of nonzero cardinals and κi <∑

j∈I κj for all

i ∈ I, then∑

i κi < (∑

i κi)|I| .

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3. (Cantor) κ < 2κ.

4. For any infinite κ, one has κ < κcf(κ).

5. cf(2κ) > κ.

Proof: 1. Notice that κi < κi+1 for all i ∈ β, so∑

i κi <∏i κi+1 by Konig’s lemma. But,

by an obvious injection,∏i κi+1 ≤

∏i κi.

2. Let λ =∑

a∈I κa. If κi < λ for all i ∈ I, then Konig’s lemma gives us that∑

i κi <∏i λ = λ|I| = (

∑i κi)

|I| .

3. Use Konig’s lemma with I = κ, κα = 1 and λα = 2 for all α < κ. Then κ =∑

α∈κ 1 <∏α∈κ 2 = 2κ.

4. Let f : cf(κ)→ κ be a cofinal function. Let κα = 1+ |f(α)| and λα = κ for all α < cf(κ).Note that κ =

⋃α f(α) ≤

∑α |f(α)| ≤

∑α κα = κ · cf(κ) = κ and

∏α λα = κcf(κ).

5. (2κ)κ = 2κ·κ = 2κ but, by 4., (2κ)cf(2κ) > 2κ. �

Note that once one decodes the argument, the proof of Cantor’s theorem above is exactlythe same as the usual proof. On the other hand, Corollary 15.5 gives us more informationand begins to show that the notion of cofinality is quite relevant in the study of cardinalarithmetic; much more dramatic illustrations of this claim are shown below.

Konig’s lemma is the first genuinely new result in cardinal arithmetic (with choice) pastCantor’s theorem. For example, Specker’s Lemma 18 of Chapter I, and the Halbeisen-ShelahTheorem 19 of Chapter I, are both trivial once AC is assumed.

Definition 16 A cardinal κ is strong limit iff 2ρ < κ for all cardinals ρ < κ.

Notice that the name is justified, since any strong limit (other than ℵ0) is in particular alimit cardinal, i.e., an ℵλ with λ a limit ordinal.

Of course, ℵ0 is a strong limit cardinal and, if GCH holds below ℵω, then ℵω is also stronglimit. Without this assumption, examples of strong limit cardinals can be found by iteratingthe power set operation. More precisely, define the sequence of beth cardinals by:

• i0 = ω,

• iα+1 = 2iα , and

• iλ = supα<λ iα for λ a limit ordinal.

Then the strong limit cardinals are precisely the iλ with λ = 0 or a limit ordinal.

Theorem 17 (Bukovsky-Hechler) 1. If κ is infinite then 2κ = (supρ<κ 2|ρ|)cf(κ).

2. In particular, if κ is strong limit, then 2κ = κcf(κ).

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3. Let κ be singular, and assume that the exponential ρ 7→ 2ρ is eventually constant belowκ; i.e., there is some τ and some ρ < κ such that for all cardinals λ with ρ ≤ λ < κ,one has 2ρ = 2λ = τ. Then also 2κ = τ.

The theorem illustrates that the exponential map must satisfy certain restrictions, at leaston its values on the singular cardinals. It is well known since the beginnings of forcing, fromthe work of Easton, that the exponential is ‘essentially’ arbitrary on the regular cardinalsexcept that, of course, it is monotonic: If λ < τ then 2λ ≤ 2τ , and must obey Konig’slemma, cf(2κ) > κ. However, once large cardinals are considered, these are not the onlyrestrictions. We will show this in Section 4.

Proof: 1. Let κ be infinite and let τ = supρ<κ 2|ρ|. Let f : cf(κ)→ κ be strictly increasing,and let κα = |f(α) \

⋃β<α f(β)| for all α < cf(κ), so κ =

∑α κα. Then

2κ = 2Pα κα =

∏α

2κα ,

as the obvious bijection verifies. But∏α 2κα ≤

∏α τ = τ cf(κ). On the other hand, clearly

2κα ≤ 2κ for all α < cf(κ), so τ ≤ 2κ and therefore τ cf(κ) ≤ (2κ)cf(κ) = 2κ. The result followsfrom these two inequalities.

2. If κ is strong limit and τ is as above, then τ ≤ κ. On the other hand, for any λ < κ,λ < 2λ ≤ τ, so τ = κ, and 2. follows from 1.

3. Assume κ is singular. With τ as above, if λ < κ is sufficiently large, 2λ = τ . But thenτ cf(κ) = τ , since we can take λ > cf(κ). The result follows from 1. �

For example, if 2ℵ0 = 2ℵn = ℵω+1 for all n < ω (this situation is easily achieved by forcing)then also 2ℵω = ℵω+1.

The next result provides an easy ‘algorithm’ to compute any power, provided we know thevalues of the exponential function, the cofinality map, and the gimel function (κ)ג = κcf(κ).I’ll expand on this remark after the proof of the theorem.

Theorem 18 Let κ and λ be infinite cardinals. Let τ = supρ<κ |ρ|λ. Then

κλ =

2λ if κ ≤ 2λ,κ · τ if λ < cf(κ),

τif cf(κ) ≤ λ, 2λ < κ, andρ 7→ |ρ|λ is eventually constant below κ,

κcf(κ) otherwise.

Proof: 1. If κ ≤ 2λ, then 2λ ≤ κλ ≤ (2λ)λ = 2λ.

2. If λ < cf(κ) then any function f : λ → κ is bounded, so λκ =⋃α<κ

λα, and κλ ≤∑α |α|λ = κ · ρ ≤ κλ.

3. Suppose that cf(κ) ≤ λ, 2λ < κ, and ρ 7→ |ρ|λ is eventually constant (and equal to τ) asρ approaches κ.

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Since κ > cf(κ), κ is a singular cardinal and we can choose a strictly increasing sequenceof cardinals (κα : α < cf(κ)) cofinal in κ such that κλα = τ for all α < cf(κ). In particular,τλ = τ .

Note that for each α < cf(κ), κα <∏i∈cf(κ) κi, so also κ = supα κα ≤

∏α κα. [Of course, it

follows from Konig’s lemma that in fact we have a strict inequality, but we only need thisweaker estimate.]

We now have κλ ≤ (∏α κα)λ =

∏α κ

λα =

∏α τ = τ cf(κ) = τ. The other inequality is clear.

4. Finally, suppose that cf(κ) ≤ λ, 2λ < κ, and ρ 7→ |ρ|λ is not eventually constant as ρapproaches κ.

Notice that if ρ < κ then |ρ|λ < κ. Otherwise, µλ = (µλ)λ ≥ κλ ≥ µλ for any ρ ≤ µ < κ,and the map ρ 7→ |ρ|λ would be eventually constant below κ after all. Hence, τ = κ.

Choose an increasing sequence of cardinals (κα : α < cf(κ)) cofinal in κ. Then

κλ ≤ (∏α

κα)λ =∏α

κλα ≤∏α

κ = κcf(κ).

The other inequality is clear. �

By induction, it follows from Theorem 18 that the computation of the function (κ, λ) 7→ κλ

‘reduces’ to computations of the functions λ 7→ 2λ and λ 7→ (λ)ג for λ a cardinal, and thefunction α 7→ cf(α) for α an ordinal. More precisely, if M and N are two models of settheory (with choice) with the same ordinals, and the values of any of these three functionsare the same whether they are computed in M or in N , then also all the powers κλ are thesame, whether they are computed in M or in N.

Forcing has shown that there is not much one can say about the exponential functionλ 7→ 2λ when restricted to successor cardinals and small large cardinals. This indicatesthat to understand cardinal arithmetic, one’s efforts must concentrate on the exponentialfunction on singular and large cardinals, and on the gimel function.

In fact, the gimel function suffices to compute the exponential λ 7→ 2λ, by Theorem 17:First, 2κ = κκ = (κ)ג for κ regular. Assume now that κ is singular. Let τ = supρ<κ 2|ρ|. Ifρ 7→ 2|ρ| is not eventually constant below κ, then clearly cf(τ) = cf(κ), and 2κ = ,(τ)ג byTheorem 17.1, while if ρ 7→ 2|ρ| is eventually constant below κ, then 2κ = τ, by Theorem17.3.

This observation explains why the bulk of research in cardinal arithmetic concentrates onunderstanding the gimel function. As we will see below, this naturally leads to the studyof certain infinite products. This latter study has proved quite fruitful. We will studytwo outcomes: Silver’s theorem in Section 2, and the Galvin-Hajnal results in Section 3.After considering the influence of large cardinals on cardinal arithmetic in Section 4, wewill conclude the Chapter by briefly mentioning in Section 5 how Shelah has extended andgeneralized these theorems with his development of pcf theory.

The following easy consequence of Theorem 18 is particularly useful:

Corollary 19 (Hausdorff) (κ+)λ = κ+ · κλ.

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Proof: The result is immediate from Theorem 18 if λ < κ+ = cf(κ+). If, on the otherhand, λ ≥ κ+ then certainly κ+ ≤ 2λ, and it is easy to see that both sides of the equationwe want to prove equal 2λ. �

For example, ℵℵ0n = ℵn · 2ℵ0 for any n ∈ ω.The following more general result is most easily established by induction on γ :

Homework problem 8. (Tarski) If |γ| ≤ ℵβ then ℵℵβα+γ = ℵ|γ|α+γ · ℵℵβα .

For example, ℵℵ1ω = ℵℵ0ω · ℵℵ10 = ℵℵ0ω · 2ℵ1 . Hence if 2ℵ1 ≤ ℵω, then ℵℵ0ω = ℵℵ1ω .

Now we begin to explore the behavior of infinitary products, mainly as they relate to thegimel function (κ)ג = κcf(κ), but the results to follow are more general.

First, an easy observation.

Lemma 20 Suppose κi ≥ 2 for all i ∈ I. Then (∑

i κi)|I| = (

∏i κi)

|I|.

Proof: There is an obvious injection from the product set∏i κi into I(

⊔i κi). It follows

that∑

i κi ≤∏i κi ≤ (

∑i κi)

|I|, from which the result is immediate. �

Theorem 21 Suppose that µ is a cardinal and (κξ : ξ < µ) is a strictly increasing sequenceof nonzero cardinals with supremum κ. Then κµ =

∏ξ κξ.

Note that the weaker claim that (∏ξ κξ)

µ = κµ is an obvious consequence of Lemmas 3 and20.

Proof: Write µ as a disjoint union of µ sets Aα for α < µ, each of size µ. This is possiblesince µ × µ = µ. Since each Aα has size µ, it is necessarily cofinal in µ. We then haveκµ ≥

∏i∈µ κi =

∏α∈µ

∏i∈Aα κi ≥

∏α κ = κµ. �

For example,∏α<λ ℵα = ℵλλ for any infinite cardinal λ. We will prove a more general result

below, in Theorem 23, but notice for now that, in particular, ℵℵ0ω =∏n ℵn. One of the most

celebrated results of Shelah’s pcf theory (established by a careful analysis of this product)is the fact that ℵℵ0ω ≤ 2ℵ0 + ℵω4 . In particular, if ℵω is strong limit so ℵℵ0ω = 2ℵω by theBukovsky-Hechler Theorem 17.2, then 2ℵω ≤ ℵω4 . In fact, the inequality is strict, by Konig’slemma, since cf(ℵω4) = ω4 < ℵω.The following is another consequence of Theorem 17:

Corollary 22 If there is an ordinal β such that for all α we have that 2ℵα = ℵα+β, thenβ < ω.

Proof: Suppose that β ≥ ω and for all α, 2ℵα = ℵα+β. Let α be least such that β < α+β;α exists since β < β + β. On the other hand, α is a limit ordinal, since 1 + ω = ω andω ≤ β, so (λ+ n) + β = λ+ (n+ β) = λ+ β for any λ and any n < ω, and we would havea contradiction to the minimality of α.

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Let γ < α. Then γ+ β = β and 2ℵα+γ = ℵα+β. By the Bukovsky-Hechler theorem, 2ℵα+α =ℵα+β since ℵα+α is singular (it has cofinality cf(α) ≤ α < α + α ≤ ℵα+α). This is acontradiction since, by assumption, 2ℵα+α = ℵα+α+β > ℵα+β. �

On the other hand, for any nonzero n ∈ ω, Woodin proved that it is consistent with ZFCthat 2ℵα = ℵα+n. The case n = 1 is of course the GCH. (For n = 2 this is a joint result ofForeman and Woodin.) However, for n > 1, this situation is incompatible with the existenceof strongly compact cardinals, as we will see in Section 4.

Theorem 23 (Tarski) 1. For any limit ordinal β,∏ξ<β ℵξ = ℵ|β|β .

2. If β is zero or a limit ordinal then, for and any n ∈ ω,∏ξ≤β+n ℵξ = ℵ|β+n|

β+n .

Theorem 23 does not follow from the proof of Theorem 21. The same argument of Theorem21 actually shows that whenever (κξ : ξ < β) is a strictly increasing sequence of nonzerocardinals and β is a countable limit ordinal, then

∏ξ κξ = κℵ0 = κ|β| where κ = supξ κξ.

However, the argument used to prove Theorem 21 does not apply if β = ω1 +ω, since thereis no way of splitting ω1 + ω into ℵ1 many cofinal sets.

Proof: Notice that item 2. is trivial if β = 0, and an immediate consequence of 1. andHausdorff’s formula, Corollary 19, if β is an infinite limit ordinal.

Now we prove 1.: Write β = |β|+α where α = 0 or a limit ordinal, and proceed by inductionon α, noticing that the case α = 0 follows from Theorem 21.

Notice that any limit ordinal α can be written as α = ω · δ for some nonzero ordinal δ; thisis easily established by induction. If δ is a limit ordinal, then any sequence converging toδ gives rise in a natural way to a sequence of limit ordinals converging to α. On the otherhand, if δ is a successor, then α = λ + ω for some λ = 0 or limit. In summary: Any limitordinal is either a limit of limit ordinals, or else it has the form λ+ ω for λ = 0 or limit.

We divide our induction on α in two cases. Suppose first α = λ + ω, as above. Then∏ξ<β ℵξ = ℵ|β||β|+λ

∏n ℵ|β|+λ+n, by induction. Since |β| = |β|ℵ0, we have

ℵ|β||β|+λ∏n

ℵ|β|+λ+n =∏n

ℵ|β||β|+λℵ|β|+λ+n =∏n

ℵ|β||β|+λ+n,

where the last equality follows from Hausdorff’s formula. Finally,∏n

ℵ|β||β|+λ+n = (∏n

ℵ|β|+λ+n)|β| = (ℵβ)ℵ0·|β| = ℵ|β|β

as wanted, where the previous to last equality is by Theorem 21.

Now suppose that α is a limit of limit ordinals. Choose a strictly increasing sequence(γν : ν < cf(α)) of limit ordinals cofinal in α. We argue according to which of the fourpossibilities described in Theorem 18 holds. If ℵβ ≤ 2|β| then ℵ|β|β = 2|β| ≤

∏ξ<β ℵξ ≤ ℵ

|β|β .

Notice that cf(α) = cf(β) = cf(ℵβ) and cf(β) ≤ |β|, so the second possibility does notoccur.

10

Suppose now that we are in the third possibility, i.e., 2|β| < ℵβ and ρ 7→ ρ|β| is eventuallyconstant as ρ approaches ℵβ. Call τ this eventual value. Then

ℵ|β|β = τ = supν<cf(α)

ℵ|β||β|+γν = supν

∏ξ<|β|+γν

ℵξ,

by induction (it is here that we use the assumption that α is a limit of limit ordinals). Andsupν

∏ξ<|β|+γν ℵξ ≤

∏ξ<β ℵξ ≤ ℵ

|β|β .

Finally, if 2|β| < ℵβ and ρ 7→ ρ|β| is not eventually constant as ρ approaches ℵβ, then ℵ|β|β =

ℵcf(α)β and ℵcf(α)

β =∏ν<cf(α) ℵ|β|+γν , by Theorem 21, but

∏ν<cf(α) ℵ|β|+γν ≤

∏ξ<β ℵξ ≤ ℵ

|β|β ,

and we are done. �

Tarski conjectured that Theorem 21 also holds in this generality. This is independent ofZFC. It follows, for example, from GCH. Its failure is a strong refutation of the SingularCardinals Hypothesis, SCH, see Definition 26 below. Indeed , there are counterexamplesin models of Magidor where SCH fails. Notice that the proof of Theorem 23 shows that thefirst time we run into trouble trying to generalize Theorem 21 has to be at some β of theform |β|+λ+ω for some λ = 0 or limit. Indeed, the inductive proof above in the case thatα is a limit of limit ordinals generalizes without difficulty. However, the proof in the caseα = λ + ω for λ = 0 or limit uses Hausdorff’s formula, and does not apply in general. InJech-Shelah [15], it is shown that there is a counterexample if and only if there is one ofshortest possible length, β = ω1 + ω.

2 Silver’s theorem

From the results from the previous Section, we know that any power κλ can be computedfrom the cofinality and gimel functions (see the comments following the proof of Theorem18). What we can say about the numbers (λ)ג varies greatly depending on whether λ isregular or not. If λ is regular, then (λ)ג = 2λ. As mentioned after the statement of Theorem17, forcing provides us with a great deal of freedom to manipulate the exponential functionκ 7→ 2κ, at least for κ regular. In fact, the following holds:

Theorem 24 (Easton) If GCH holds, then for any definable function F from the class ofinfinite cardinals to itself such that:

1. F (κ) ≤ F (λ) whenever κ ≤ λ, and

2. κ < cf(F (κ)) for all κ,

there is a class forcing P that preserves cofinalities and such that in the extension by P itholds that 2κ = F V (κ) for all regular cardinals κ; here, F V is the function F as computedprior to the forcing extension. �

11

For example, it is consistent that 2κ = κ++ for all regular cardinals κ (as mentioned afterCorollary 22, the same result is consistent for all cardinals, as shown by Foreman andWoodin, although their argument is significantly more elaborate that Easton’s).

There is almost no limit to the combinations that the theorem allows: We could have2κ = κ+16 whenever κ = ℵτ is regular and τ is an even ordinal, and 2κ = κ+17 wheneverκ = ℵτ for some odd ordinal τ. Or, if there is a proper class of weakly inaccessiblecardinals (regular cardinals κ such that κ = ℵκ) then we could have 2κ = the third weaklyinaccessible strictly larger than κ, for all regular cardinals κ, etc.

Morally, Easton’s theorem says that there is nothing else to say about the gimel functionon regular cardinals, and all that is left to be explored is the behavior of (λ)ג for singularλ. In this Section we begin this exploration. However, it is perhaps sobering to point outthat there are several weaknesses in Easton’s result.

First, there is the assumption of GCH in the ground model. Trying to remove this assump-tion leads one to the following question:

Question. Assume that F is as above and also satisfies:

3. F (κ) ≥ 2κ for all regular cardinals κ.

Does it follow that there is some cofinality-preserving forcing extension where 2κ = F V (κ)for all regular cardinals κ?

In the 1980s, Foreman, Magidor and Shelah asked whether the following is consistent:“Every nontrivial (set) forcing either adds a real or collapses a cardinal.”

This “maximality principle,” as they called it, implies that GCH fails everywhere and thereare no inaccessible cardinals (regular strong limit cardinals). Assuming that the principle isconsistent gives us that nothing like Easton’s original approach would work to produce thecofinality preserving extensions we require. This is because Easton’s forcing is a productof class many nontrivial set forcings as factors. Of course, mutual genericity gives us thatonly set many of these factors can add reals, so at least one cardinal (probably a properclass) will be collapsed. This does not rule out that a completely different approach willgive Easton’s result, but it makes the problem potentially more difficult.

The work of Foreman, Magidor and Shelah on this “maximality principle” led to theirdiscovery of Martin’s Maximum, MM, arguably one of the most important events inmodern set theory. A local version of the principle already follows from BPFA, the boundedproper forcing axiom:

Theorem 25 (Todorcevic) BPFA implies that every forcing that adds a subset of ω1 musteither add a real or collapse ω2. �

For the proof, see Todorcevic [24]. All that is needed is that 2ℵ0 = ℵ2 and that every treeT of height and size ω1 is sealed, i.e., there is a function f : T → ω such that wheneverf(x) = f(y) = f(z) and x ≤T y, z, then in fact y ≤T z or z ≤T y. This notion is due toBaumgartner, who called these trees special. That these statements are consequences of

12

BPFA follows from work of Baumgartner and of Justin Moore. Todorcevic requires that theforcing notion is a set, but one can dispense with this assumption.

An examination of the proof of Todorcevic’s theorem actually shows that it answers neg-atively the Question posed above. Very briefly, the argument is this: Assume BPFA andsuppose that a forcing adds a subset of ω1 but no reals. Given a name for this set, we canthen build a tree of height ω1, essentially the tree of attempts to decide the name. Theassumption of BPFA gives that the tree has size at most ω2 and from it a cofinal map fromω1 to ω2 can be constructed.

Assume that P is a forcing answering the Question affirmatively, where the function Fsatisfies F (ℵ0) = ℵ2 and F (ℵ1) = ℵ3. We then have ℵ3 many distinct names for subsets ofω1 to which the above procedure can be applied. It follows that the tree construction mustfail, which only happens because some proper initial segment of the new subset is not inthe ground model, so a new real is added. A mutual genericity argument ensures that theremust be ℵ3 many new reals added this way.

Second, Easton’s forcing preserves cofinalities and therefore cardinals. However, it doesnot preserve in general any significant large cardinals. For example, as we will see in Section4, if κ is a measurable cardinal, then κ is not the first counterexample to GCH so, althoughEaston forcing preserves the fact that κ is weakly inaccessible, it may very easily destroyits measurability. I do not know of any general Easton-like result for definable maps F thattakes into account a significant amount of the large cardinal structure of the universe.

Third, of course, Easton’s result only applies to the regular cardinals. In fact, in themodels obtained by Easton’s method we have no freedom whatsoever to manipulate thegimel function for singular cardinals. This can be made precise as follows:

Definition 26 The singular cardinals hypothesis, SCH, is the statement that for allcardinals λ, we have that λcf(λ) = λ+ + 2cf(λ).

Note that the equality holds if λ is regular, so SCH only has content for singular cardinals,which explains its name. For such λ, notice that the left hand side is always at least as largeas the right hand side, so what SCH says is that (λ)ג is as small as possible for λ singular.

SCH is a trivial consequence of GCH, and it is easy to see that it holds in the models obtainedby Easton’s method. One may then wonder whether it is actually provable. Magidorshowed that this is not the case; for example, beginning with significantly large cardinals,Magidor obtained a model in which κ = ℵω1+ω1 is a strong limit cardinal, (κ)ג = κ+ω+2

and (κ+ω)ג = κ+ω+1.

In this model SCH fails since 2cf(κ)+κ+ = 2ℵ1 +κ+ = κ+ < .(κ)ג There are simpler models ofthe failure of SCH; I mention this one because it also witnesses a failure of Tarski’s conjecture(see the remarks at the end of Section 1), the claim that

∏α<β λα = λ|β| whenever β is a

limit ordinal and (λα : α < β) is a strictly increasing sequence of cardinals cofinal in λ.

In effect, Tarski’s conjecture fails in Magidor model for β = ω1 + ω, as witnessed by thesequence

λα ={ℵα if α < ω1,κ+n if α = ω1 + n.

13

This is because∏α<β

λα =∏α<ω1

ℵα∏n<ω

κ+n = (κ+ω)ג(ℵω1)ג < κג(κ) = (κ)ג = (κ+ω)|ω1+ω|,

where the last equality follows from Tarski’s formula, Homework problem 8 (or from The-orem 18).

Magidor’s counterexample occurs at a singular cardinal of cofinality ω. Silver’s remarkableresult is that this must always be the case:

Theorem 27 (Silver) If SCH holds for all singular cardinals of countable cofinality, thenit holds everywhere. �

Following Galvin-Hajnal [11], I state below Silver’s theorem in great generality, see Theorem43, and prove a particular case that illustrates the combinatorial ideas that guide the proof ofthe result and its many generalizations. It follows from Silver’s theorem that in an Easton-like result that also considers singular cardinals the function F must satisfy additionalrestrictions. I do not know of any general results of this kind, although many particularexamples of possible behaviors and a significant list of restrictions have been identified.

After proving the particular case of Silver’s theorem, I will present in Section 3 a general-ization that indicates the wide scope of the techniques involved in arguments of this kind.These techniques emphasize the role played by the combinatorics of infinitary products. Ina sense, it is natural that the study of the gimel function requires of us some understandingof products. For example, if τ is a regular cardinal and (κα : α < τ) is a strictly increas-ing sequence of cardinals cofinal in κ, then it follows from the results from Section 1 that(κ)ג =

∏α κα.

A key feature of Silver’s result is the use of the notion of stationarity. In a sense, stationarysets are the first nontrivial manifestation of the axiom of choice at the uncountable level.Their definition requires an additional, very important notion.

Definition 28 Let α be an ordinal. A set C ⊆ α is club in α if and only if it is closed inthe order topology of α, and it is cofinal in α. (Club stands for “closed and unbounded.”)

The definition above is unnecessarily general. The empty set is club in α = 0. A closedsubset C of α + 1 is club if and only if α ∈ C. If α has cofinality ω then, for any strictlyincreasing and cofinal function f : ω → α, its range is club in α, and if C ⊆ α is closed inα, then C is club if and only if C contains the range of such an f.

This means that the concept is not interesting if α is a successor; for example, automatically,any intersection of clubs is club. It also means that the concept is not interesting if α hascofinality ω, as we can have disjoint club sets.

We want club sets to capture enough of the structure of α that they can be used as surrogatesfor α. A good working intuition is that club sets provide a measure of “size” of subsets of α.A set is large if it contains a club. In order for this to be a reasonable notion of largeness,

14

we need that no two large sets be disjoint. On the other hand, we do not want that anarbitrary intersection of large sets be large.

This means that the notion of club set is only interesting (or at least, only captures ourworking intuition) for α a limit ordinal of uncountable cofinality (as the following theoremverifies), and from now on when we mention a club subset of α we (perhaps implicitly)assume that α satisfies these non-triviality requirements.

It may be worth pointing out that the topological closure condition of a club set can beeasily restated as follows: A set C ⊆ α is closed in α iff for all β < α, if 0 < β and C ∩ β isunbounded in β, then β ∈ C.Examples of club subsets of ω1 include the set of all countable limit ordinals, the set ofall countable indecomposable ordinals of the form ωδ for δ limit (ordinal exponentiation),and many other, much more elaborate, sets. At first sight, it may look like these sets areactually rather thin, compared with, for example, the set of all countable successor ordinals.However, the closure condition in fact gives us that a club set has many “accumulation”points, and it is this reason what makes them so useful (and large, in some intuitive sense).

It is easy to verify that if C is club in α, then C ′ ( C, the set of limit points of C (in thetopological sense, as a subset of α) is also a club set, and we can iterate this procedure. Infact, we can iterate it for a long while:

Theorem 29 Assume that ω < κ = cf(α). Then for any λ < κ, the intersection of any λmany club subsets of α is club.

Proof: Let (Cγ : γ < λ) be a sequence of club subsets of α, where λ < κ. Let C =⋂γ Cγ ,

so C is closed, and all we need to verify is that it is unbounded.

For this, let β < α be given. By induction on γ, we define a strictly increasing sequence(β0γ : γ < λ) so that β0

i ∈ Ci for all i. Assume that (β0γ : γ < δ) has been defined, and δ < λ.

Since δ < λ < κ, the supremum ε of this sequence is strictly below α. Let β0δ ∈ Cδ be least

such that ε, β < β0δ . Such an ordinal β0

δ exists since Cδ is unbounded in α.

Let β = supγ β0γ . Again, β < α. Repeat the above construction with β in place of β to

obtain a sequence (β1γ : γ < λ). Iterate this procedure ω times. The sequences we have

produced then satisfy that θγ = supn βnγ is actually independent of γ. Call this ordinal θ.Since, by construction, βnγ ∈ Cγ for all n, it follows that θ ∈ Cγ , because Cγ is closed in α(and θ < α since κ > ω).

We have shown that θ ∈⋂γ Cγ = C and, by construction, β < θ. This shows that C is

unbounded. �

When α = κ = cf(κ) is a regular cardinal, we can say even more.

Definition 30 Let κ be a regular cardinal and let (Cα : α < κ) be a sequence of club subsetsof κ. The diagonal intersection of the Cα, in symbols, 4αCα, is the set of all β < κ suchthat β ∈ Cα for all α < β.

15

For example, let Cα = κ \ α. Then 4αCα = κ while if Cα = κ \ (α+ 2), then 4αCα is theset of limit ordinals below κ. What we are doing is “fattening up” the club sets Cα a tinybit before taking their intersection: Given (Cα : α < κ), let Dα = Cα ∪ (α+ 1) for all α+ 1.Then 4αCα =

⋂αDα.

Lemma 31 If κ is regular, the diagonal intersection of any κ many club subsets of κ isclub in κ.

This is as strong a result as we can hope for. Considering again the example of Cα = κ \α,it is clear that we cannot strengthen the result to conclude that the intersection of κ manyclub subsets of κ is club in κ.

Proof: Let (Cα : α < κ) be a sequence of club subsets of α, and let C be their diagonalintersection. The characterization mentioned right before the statement of the lemma showsthat C is closed. Now we show that it is unbounded. For this, let α = α0 < κ be given. SetC0 =

⋂β<α0

Cβ, so C0 is club. Let α1 ∈ C0 be strictly larger than α0. Set C1 =⋂β<α1

Cβ,

so C1 is also club. Let α2 ∈ C1 be strictly larger than α1.

Continuing in this fashion we build a strictly increasing ω-sequence α0 < α1 < . . . ofelements of κ and a sequence of club sets Cn =

⋂β<αn

Cβ such that αn+1 ∈ Cn for all n.Let Cω =

⋂nCn, so Cω is club in κ, and let γ = supn αn, so γ < κ. Note that, for any n,

the ordinals αn+1, αn+2, . . . all belong to Cn, so γ ∈ Cn. Therefore, γ ∈ Cω.If β < γ, then β < αn for some n, so αn+1, αn+2, . . . all belong to Cβ and so does γ. Itfollows that γ ∈ C = 4βCβ. Since α < γ, we have that C is unbounded, as we wanted toshow. �

Definition 32 Let α be a limit ordinal of uncountable cofinality. The set S ⊆ α is station-ary in α iff S ∩ C 6= ∅ for all club sets C ⊆ α.

For example, let λ be a regular cardinal strictly smaller than cf(α). Then

Sαλ := {β < α : cf(β) = λ}

is a stationary set, since it contains the λ-th member of the increasing enumeration of anyclub in α. This shows that whenever cf(α) > ω1, there are disjoint stationary subsets ofα. Below, we show a stronger result. The notion of stationarity is central to most of settheoretic combinatorics.

Lemma 33 Let S be stationary in α.

1. S is unbounded in α.

2. Let C be club in α. Then S ∩ C is stationary.

16

Proof: 1. S must meet κ \ α for all α and is therefore unbounded.

2. Given any club sets C and D, (S ∩C)∩D = S ∩ (C ∩D) 6= ∅, and it follows that S ∩Cis stationary. �

Continuing with the intuition that a club subset of α is large, we should think of stationarysets as being those that are not small. Thinking in terms of measure theory may be helpful:A club set is like a set of full measure, and a stationary set is like a set of positive measure.Clearly, we cannot have disjoint full measure sets, this corresponds to the fact that theintersection of two clubs is club. However, it is possible to have disjoint stationary sets.It is because of this fact that above (see before Definition 28) I made the comment thatstationary sets are a manifestation of the axiom of choice: It is consistent with ZF (forexample, it is a consequence of determinacy) that every stationary subset of ω1 contains aclub. However, under choice any stationary subset of an ordinal α of cofinality κ can besplit into κ many disjoint stationary subsets. This is a result of Solovay that we will provein Chapter III. For ω1 or, in general, for any successor cardinal, this was known prior toSolovay’s theorem; it is a consequence of a nice construction due to Ulam.

Definition 34 Let κ be a cardinal. An Ulam matrix on κ+ is defined as follows: For eachβ < κ+ fix an injection fβ : β → κ. For ι < κ and α < κ+ define

Aια = {β : α < β and fβ(α) = ι}.

For α < κ+ define supp(α) = {ι : Aια is stationary }.

The following observations are immediate from the definition:

1. If ι 6= η then Aια ∩Aηα = ∅ for any α < κ+.

2. If α 6= γ then Aια ∩Aιγ = ∅ for any ι < κ, since the fβ are injective.

3.⋃ιA

ια = κ+ \ (α+ 1) for each α < κ+.

It follows from 3. and Theorem 29 that supp(α) 6= ∅ for each α, in fact,⋃ι∈supp(α)A

ια

contains a club. It follows from the pigeonhole principle that for some ι, the set

{α : ι ∈ supp(α)}

has size κ+.

This shows that there are κ+ many disjoint stationary subsets of κ+. In fact, more is true:Given S stationary, by replacing each Aια with S ∩Aια, the same argument shows:

Theorem 35 (Ulam) For each κ, any stationary subset of κ+ can be split into κ+ manydisjoint stationary sets. �

A little argument with the pigeonhole principle shows that, for every η < κ,

{α < κ+ : supp(α) ⊆ η}

17

has size at most |η| and therefore for all but boundedly many α, supp(α) is unbounded inκ.

Question. Are there κ+ many values of α such that |supp(α)| = κ?

Of course, the answer to the Question is yes if κ is regular.

We are now ready to provide a very useful characterization of stationary sets due to Fodor.

Definition 36 Let S be a set of ordinals. A function f : S → ORD is regressive ifff(α) < α for all 0 < α ∈ S.

Theorem 37 (Fodor) Suppose that S is stationary in the regular cardinal κ, and f : S →κ is a regressive function. Then there is some α such that f−1{α} is stationary.

Proof: Otherwise, for each α < κ there is a club Cα disjoint from f−1{α}. Let C = 4αCα,so C is club by Lemma 31. Let α ∈ C ∩ S. For all β < α, α ∈ Cβ, so f(α) 6= β. Therefore,f(α) ≥ α, and f is not regressive. Contradiction. �

Corollary 38 Let κ be an uncountable regular cardinal and let S ⊆ κ. The following areequivalent:

1. S is stationary.

2. For every regressive function f on S, there is some α such that f−1{α} is unbounded.

3. For every regressive function f on S, there is some α such that f−1{α} is stationary.

4. For every function f on S, there is some stationary subset of S where f is constant,or else there is a stationary subset of S where f is strictly increasing.

Proof: Clearly 3. implies 2., and 1. implies 3. by Theorem 37.

Assume that S is not stationary, and let C be a club set disjoint from S. For every γ ∈ Slarger than min(C), the set C ∩ γ is nonempty and bounded below γ, so it has a largestelement. The map that assigns to γ this largest element is regressive on S \ (min(C) + 1).By mapping every other γ ∈ S (if any) to 0, the map extends to a regressive function on S.However, the preimage of every point in its range is bounded. This proves that 2. implies1.

Obviously, 4. implies 1.

Suppose that S is stationary in κ, that f : S → κ and that T = {α ∈ S : f(α) < α} is notstationary. Let S′ = S \ T, so S′ is stationary and f(α) ≥ α for all α ∈ S′. By induction,define a sequence (aα : α < κ) such that

1. a0 ∈ S,

2. aα+1 ∈ S for all α < κ,

18

3. aα+1 > aα for all α < κ and in fact

4. aα+1 > f(aα) whenever aα ∈ S, and

5. aα = sup{aβ : β < α} whenever α is a limit ordinal.

(Whenever condition 5 holds in a sequence, we say that the sequence is continuous. Ifcondition 3 holds, this indeed corresponds to the enumeration map being continuous in theorder topology.)

That we can satisfy requirements 1–5 follows from the regularity of κ. By construction,C = {aα : α < κ} is a club subset of κ, and therefore C ∩S′ is stationary. By construction,f(aα) ≥ aα > f(aβ) whenever aα, aβ ∈ S′ ∩ C and β < α. This shows that 4. holds. �

If α has uncountable cofinality but is not a regular cardinal, it is not necessarily truethat every regressive function on every stationary subset of α is constant on a stationarysubset. However, a weak version of Fodor’s lemma still holds in this case, namely, for any Sstationary in α and any regressive f : S → α, there is some stationary subset of S in whichf is bounded. The proof is an easy extension of the argument for Theorem 37.

Note that the collection of subsets of κ that contain a club forms a cf(κ)-complete filter(the club filter), and the collection of nonstationary sets forms a cf(κ)-complete ideal (thenonstationary ideal).

Before stating Silver’s theorem, we need one additional result. Recall that SCH was intro-duced in Definition 26, it is the statement that (κ)ג = κ+ + 2cf(κ) for all infinite cardinalsκ. We break the result into two parts, Lemmas 39 and 40, corresponding respectively toTheorems 17 (describing the behavior of the exponential λ 7→ 2λ) and 18 (describing thecomputation of powers κλ).

Lemma 39 Assume SCH. Let κ be singular and let τ = supρ<κ 2ρ. Then

2κ ={

τ if ρ 7→ 2ρ is eventually constant below κ,τ+ otherwise.

Proof: Assume SCH, let κ be singular and let τ be as in the statement of the lemma.By the Bukovsky-Hechler theorem, if ρ 7→ 2ρ is eventually constant below κ, then 2κ = τ,and otherwise, 2κ = .(τ)ג By SCH, we have that (τ)ג = τ+, since we are assuming that2cf(κ) < τ. �

Lemma 40 Assume SCH. Let κ, λ be infinite. Then

κλ =

2λ if κ ≤ 2λ,κ if λ < cf(κ) and 2λ < κ,κ+ if cf(κ) ≤ λ and 2λ < κ.

19

Proof: Now we use Theorem 18. Assume SCH. We argue by induction on κ that theformula holds for all λ. If κ ≤ 2λ, then κλ = 2λ as before.

Let τ = supρ<κ |ρ|λ. If 2λ < κ and λ < cf(κ), then

κλ = κτ = κ sup2λ≤ρ<κ

|ρ|λ ≤ κ sup2λ≤ρ<κ

|ρ|+ = κ,

where the induction hypothesis is used in the last inequality.

Suppose now that 2λ < κ and cf(κ) ≤ λ. If ρ 7→ ρλ is not eventually constant below κ, then

κλ = (κ)ג = κ+ + 2cf(κ) = κ+,

by SCH. Otherwise, ρ 7→ ρλ is eventually constant below κ. If ρ < κ is sufficiently large,then we have that 2λ < ρ, τ = ρλ, and

κλ = τ = ρλ ≤ ρ+ ≤ λ,

where the last inequality is by the induction hypothesis, and we are done. �

Corollary 41 SCH holds iff κλ ≤ 2λ + κ+ for all infinite cardinals κ, λ. �

What the results say is that SCH makes all powers as small as possible modulo the sizeof the exponential function λ 7→ 2λ. Notice that the arguments we gave are local, meaningthat the computations hold at some κ assuming only that SCH holds up to κ.

The following notion proves to be a useful shorthand.

Definition 42 (Shelah) Let κ be a cardinal. A cardinal λ is κ-inaccessible iff µκ < λfor all cardinals µ < λ.

The following is a fairly general statement of Silver’s result (still some additional generalityis possible). I will present some corollaries and explain how they follow from the theorem.Then I will prove in detail a (very) particular case which, however, contains all the requiredingredients for the proof of the general result. The argument I present is due to Baumgartnerand Prikry and is combinatorial in nature. Silver’s original proof used the technique offorcing and depended on a generic ultrapower argument. We will see a similar (forceless)argument in Section 4, in the context of large cardinals. For the original approach, seeSilver [23].

Theorem 43 (Silver) 1. Let λ be a cardinal. Assume that cf(λ) = κ > ω and λ isκ-inaccessible. Let (λα : α < κ) be a strictly increasing and continuous sequence ofcardinals cofinal in λ. Suppose that there is some µ < κ such that

{α < κ :∏β<α

λβ ≤ λ+µα }

is stationary in κ. Then λκ ≤ λ+µ.

20

2. In the situation of item 1., if in addition 2τ ≤ λκ for all τ < λ, then 2λ ≤ λ+µ.

3. If ℵλ is a strong limit singular cardinal of uncountable cofinality, and

{α < λ : (ℵα)ג ≤ ℵα·2}

is stationary in λ, then 2ℵλ < ℵλ·2. �

Corollary 44 (Silver) 1. Let λ be a singular strong limit cardinal of uncountable cofi-nality κ. Let (λα : α < κ) be a strictly increasing and continuous sequence of cardinalscofinal in λ. Suppose that µ < κ and

{α < κ :∏β<α

λβ ≤ λ+µα }

is stationary. Then 2λ ≤ λ+µ.

2. Let λ be a singular cardinal of uncountable cofinality κ, and let µ < κ. If

{δ < λ : 2δ ≤ δ+µ}

is stationary, then 2λ ≤ λ+µ.

3. Suppose that µ < ω1, 2ℵ1 < ℵω1 , and

{α < ω1 : ℵℵ0α ≤ ℵα+µ}

is stationary. Then ℵℵ1ω1≤ ℵω1+µ.

4. Suppose that µ < ω1, ℵω1 is strong limit, and

{α < ω1 : ℵℵ0α ≤ ℵα+µ}

is stationary. Then 2ℵω1 ≤ ℵω1+µ.

5. Suppose that µ < ω1 and{α < ω1 : 2ℵα ≤ ℵα+µ}

is stationary. Then 2ℵω1 ≤ ℵω1+µ.

6. The first counterexample to GCH is not a singular cardinal of uncountable cofinality.

7. The first counterexample to SCH is not a singular cardinal of uncountable cofinality.

Proof: 1. This follows immediately from the theorem, using that 2λ = λκ since λ is stronglimit.

2. Note that λ is strong limit: If δ < λ then δ+µ < λ, since µ < κ. It follows that 2δ < λfor unboundedly many, and therefore for all, δ < λ.

Let (λα : α < κ) be a strictly increasing continuous sequence of cardinals cofinal in λ. Then{α : 2λα ≤ λ+µ

α } is stationary in κ.

21

But∏β<α λβ ≤ 2λα for all α. Now the result follows from 1.

3. For all α < ω1, ℵℵ1α = 2ℵ1ℵℵ0α by Tarski’s formula, Homework problem 8. It follows thatℵℵ1α ≤ ℵα+µ for stationarily many α. But, for any countable limit ordinal α, we have that∏β<α ℵβ = ℵℵ0α . The result now follows from the theorem.

4. By 3., since ℵω1 is strong limit.

5. and 6. By 2.

7. Suppose that SCH holds below λ, that λ is singular, and that κ = cf(λ) is uncountable.Suppose first that 2κ ≥ λ. Then in fact 2κ > λ, since cf(2κ) > κ = cf(λ). We then havethat (λ)ג = 2κ = 2κ + λ+, and SCH also holds at λ.

Suppose now that 2κ < λ. We claim that in fact λ is κ-inaccessible. Assume that this isthe case, and fix a strictly increasing and continuous sequence of cardinals (λα : α < κ)converging to λ. For any limit ordinal α < κ, we have that

∏β<α λβ ≤ λ

|α|α . By SCH below

λ, λ|α|α ≤ λ+

α as long as α is sufficiently large to ensure that 2|α| ≤ 2κ < λα. (We are usingCorollary 41 here, more precisely, the fact that Corollary 41 has a local proof as explainedabove.)

It follows that Theorem 43.1 applies, with µ = 1, and therefore 2κ +λ+ = λ+ ≤ (λ)ג ≤ λ+,and SCH also holds at λ, as wanted.

All that remains is to argue that λ is indeed κ-inaccessible. For this, suppose that δ < λ.Using SCH below λ, by (the local nature of the proof of) Corollary 41 we have that

δκ ≤ δ+ + 2κ < λ,

and we are done. �

Now I state and prove the best known particular case of Silver’s result.

Corollary 45 (Silver) ℵω1 is not the first counterexample to GCH.

Proof: Suppose that 2κ = κ+ for all κ < ℵω1 . In what follows, by∏α<ω1

ℵα+1 we mean theset product (the collection of functions) rather than its cardinality, and similarly for otherproducts. Fix injections πα : P(ℵα)→ ℵα+1 for all α < ω1. For X ⊆ ℵω1 let fX ∈

∏α ℵα+1

be the map fX(α) = πα(X ∩ ℵα). Set F = {fX : X ⊆ ℵω1}.The following properties of F are immediate:

1. Whenever f 6= g ∈ F , there is in fact an α such that f(β) 6= g(β) for all countableβ ≥ α.

2. |F| = 2ℵω1 , since the assignment X 7→ fX is 1-1.

Whenever a collection G of functions f : ω1 → ORD satisfies 1., we will say that G is almostdisjoint.

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Lemma 46 Suppose that G ⊆∏α<ω1

ℵα+1 is almost disjoint, and for all f ∈ G, the set{α : f(α) < ℵα} is stationary in ω1. Then |G| ≤ ℵω1 .

Proof: Given f ∈ G define f : ω1 → ω1 so that ℵ0 + |f(α)| = ℵf(α) for all α. Then f isregressive in a stationary set. By Fodor’s lemma (Theorem 37), there is some γf ∈ ω1 suchthat Sf = {α : f(α) = γf} is stationary.

For γ ∈ ω1 and S ⊆ ω1, let Aγ,S = {f ∈ G : γf = γ and Sf = S}. Then either Aγ,S isempty, or else the map that assigns to f ∈ Aγ,S its restriction f � S is injective, since G isalmost disjoint and S is stationary in ω1 and therefore unbounded.

By definition of γf , if f ∈ Aγ,S then f � S : ω1 → ℵγ+1. But then it follows from GCH that|Aγ,S | ≤ ℵγ+2.

The result follows, since G =⋃γ,S Aγ,S , and there are only ℵ1 many possible values of γ

and ℵ2 many possible values of S. �

Corollary 47 Let g ∈∏α<ω1

ℵα+1 and let G be almost disjoint. Suppose that for all f ∈ G,{α < ω1 : f(α) < g(α)} is stationary. Then |G| ≤ ℵω1 .

Proof: This is immediate from the lemma since∏α g(α) injects into

∏α ℵα. �

To complete the proof of the theorem, we now argue that F can be split into ℵω1+1 manysets to which the corollary applies.

Definition 48 Let Cω1 be the club filter on ω1. Given functions f, g : ω1 → ORD, we saythat f <∗ g iff {α : f(α) < g(α)} contains a club.

Suppose that g ∈ F . Then, by Corollary 47, at most ℵω1 many functions in F are below gon a stationary set, i.e., g <∗ f for all but at most ℵω1 many functions f ∈ F .By induction, we can therefore define a <∗-increasing sequence (fη : η < ℵω+1) of membersof F . But notice that for any f ∈ F , it is the case that f <∗ fη for some η. Otherwise, fcontradicts the previous paragraph. It follows that F =

⋃η Fη, where

Fη = {f ∈ F : f <∗ fη},

and we are done by Corollary 47. �

The arguments above (see the proof of Corollary 44.7, for example) suggest that SCHguarantees that Tarski’s conjecture (see the remarks at the end of Section 1) holds. Thiseasily follows from Lemma 40. We can in fact say a bit more:

Homework problem 9. (Jech-Shelah) Assume that Tarski’s conjecture fails, so forsome limit ordinal β there is some strictly increasing sequence of cardinals (κα : α < β)with limit κ such that

∏α<β κα < κ|β|.

1. Show that cf(β) < |β| < β and there is some cardinal λ < κ such that λ|β| > κ.

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2. Suppose that β is least such that there is a counterexample as above. Then (ℵ0 < |β|and) β = |β|+ ω.

3. If Tarski’s conjecture fails, then there is a cardinal ℵγ of uncountable cofinality τ suchthat γ > τ, ℵτν < ℵγ for all ν < γ, and (ℵγ+ω)ג < .(ℵγ)ג

3 The Galvin-Hajnal theorems

In this Section I want to present two theorems of Galvin and Hajnal that greatly generalizeSilver’s theorem. I begin with a “pointwise” (or everywhere) result, that gives us informationbeyond the pointwise theorems from last Section, like Corollary 45. Then I continue with aresult where the hypotheses, as in Silver’s theorem, are required to hold stationarily ratherthan everywhere. From this result, the full version of Silver’s result can be recovered.

Both results appear in the paper Galvin-Hajnal [11], that I will follow closely. For thenotion of κ-inaccessibility, see Definition 42.

Theorem 49 (Galvin-Hajnal) Let κ, λ be uncountable regular cardinals. Suppose that λis κ-inaccessible. Let (κα : α < κ) be a sequence of cardinals such that

∏α<β κα < ℵλ for

all β < κ. Then also∏α<κ κα < ℵλ.

The second theorem is stated below, see Theorem 75. Theorem 49 is a rather general result;here are some corollaries that illustrate its reach:

Corollary 50 Let κ, λ be uncountable regular cardinals. Suppose that λ is κ-inaccessible.Let τ be a cardinal, and suppose that τσ < ℵλ for all cardinals σ < κ. Then also τκ < ℵλ.

Proof: Apply Theorem 49 with κα = τ for all α < κ. �

Corollary 51 Let κ, λ be uncountable regular cardinals. Suppose that λ is κ-inaccessible.Let τ be a cardinal of cofinality κ, and suppose that 2σ < ℵλ for all cardinals σ < τ. Thenalso 2τ < ℵλ.

Proof: Let (τα : α < κ) be a sequence of cardinals smaller than τ such that τ =∑

α τα, andset κα = 2τα for all α < κ. Then

∏α<β κα = 2

Pα<β τα < ℵλ for all β < κ, by assumption.

By Theorem 49,∏α<κ κα = 2

Pα τα = 2τ < ℵλ as well. �

Corollary 52 Let κ, ρ, τ be cardinals, with ρ ≥ 2 and κ regular and uncountable. Supposethat τσ < ℵ(ρκ)+ for all cardinals σ < κ. Then also τκ < ℵ(ρκ)+ .

Proof: This follows directly from Corollary 50, since λ = (ρκ)+ is both regular and κ-inaccessible. �

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Corollary 53 Let ρ, τ be cardinals, with ρ ≥ 2 and τ of uncountable cofinality κ. Supposethat 2σ < ℵ(ρκ)+ for all cardinals σ < τ. Then also 2τ < ℵ(ρκ)+ .

Proof: This follows directly from Corollary 51 with λ = (ρκ)+. �

Corollary 54 Let ξ be an ordinal of uncountable cofinality. Suppose that 2ℵα < ℵ(|ξ|cf(ξ))+

for all α < ξ. Then also 2ℵξ < ℵ(|ξ|cf(ξ))+ .

Proof: This follows from Corollary 53 with ρ = |ξ|, τ = ℵξ, and κ = cf(ξ). �

Corollary 55 Let ξ be an ordinal of uncountable cofinality. Suppose that ℵσα < ℵ(|ξ|cf(ξ))+

for all cardinals σ < cf(ξ) and all α < ξ. Then also ℵcf(ξ)ξ < ℵ(|ξ|cf(ξ))+ .

Proof: This follows from Corollary 52: If σ < cf(ξ), then ℵσξ = ℵξ supα<ξ ℵσα, by Theorem18. But ξ < (|ξ|cf(ξ))+, so both ℵξ and supα<ξ ℵσα are strictly smaller than ℵ(|ξ|cf(ξ))+ . �

Corollary 56 If 2ℵα < ℵ(2ℵ1 )+ for all α < ω1, then also 2ℵω1 < ℵ(2ℵ1 )+ .

Proof: By Corollary 53. �

Corollary 57 If ℵℵ0α < ℵ(2ℵ1 )+ for all α < ω1, then also ℵℵ1ω1< ℵ(2ℵ1 )+ .

Proof: By Corollary 55. �

Notice that, as general as these results are, they do not provide us with a bound for thesize of 2τ for τ the first cardinal of uncountable cofinality that is a fixed point of the alephsequence, τ = ℵτ , not even under the assumption that τ is a strong limit cardinal.

Now I proceed to the proof of Theorem 49.

Definition 58 Let κ be a cardinal and let A = (Aα : α < κ) be a sequence of nonemptysets. A set F ⊆

∏αAα is an almost disjoint transversal system (a.d.t.) for A iff

|{α : f(α) = g(α)}| < κ whenever f, g ∈ F .

As in the proof of Silver’s theorem, Theorem 49 follows from an analysis of the possiblesizes of a.d.t.s for a particular sequence A.

Theorem 59 Let κ, λ be uncountable regular cardinals, and suppose that λ is κ-inaccessible.Let A = (Aα : α < κ) be a sequence of nonempty sets such that |Aα| < ℵλ for all α < κ.Let F be an a.d.t. for A. Then |F| < ℵλ.

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Before proving Theorem 59, I show how it implies Theorem 49:

Lemma 60 Let κ be a cardinal and let (κα : α < κ) be a sequence of cardinals. For α < κlet Aα be the set product

∏β<α κβ, and set A = (Aα : α < κ). Then there is an a.d.t. F for

A with |F| =∏α<κ κα.

This readily implies Theorem 49:

Proof: Let κ, λ, and (κα : α < κ) be as in the statement of Theorem 49. Let A = (Aα :α < κ) where Aα =

∏β<α κβ for all α < κ. By Lemma 60 there is an a.d.t. F for A of size∏

α<κ κα. By assumption, |Aα| < ℵλ for all α < κ, and it follows from Theorem 59 that|F| < ℵλ as well. �

The argument for Lemma 60 is fairly easy:

Proof: Let τ be the cardinal∏α<κ κα. Consider an injective enumeration (gη : η < τ)

of the set product∏α κα. For η < τ let fη ∈

∏α<κAα be given by fη(α) = gη � α for all

α < κ. Clearly, η < ν < τ implies that {α : fη(α) = fν(α)} is an ordinal below κ, namely,the least β such that gη(β) 6= gν(β). It follows that F = {fη : η < τ} is an a.d.t. for A. �

All that remains is to show Theorem 59.

Proof: The argument is organized as an induction. For this, we need a rank on which toinduct.

Definition 61 Let κ be a cardinal. Let <b,κ be the partial ordering on κORD given by

f <b,κ g iff |{α < κ : f(α) ≥ g(α)}| < κ.

(The subindex b stands for “bounded.”)

Lemma 62 If κ is a regular, uncountable cardinal, then <b,κ is well-founded.

The result is fairly general. For example, instead of the ideal of bounded sets, we couldhave used the ideal of nonstationary sets in Definition 61 (i.e., requiring that f(α) < g(α)for club many values of α), and Lemma 62 would still hold; all we need is a σ-completeideal. Below I follow the usual convention that a τ -complete ideal means one closed underunions of any length less that τ.

Proof: Let C be the filter of cobounded subsets of κ. Notice that C is κ-complete, byregularity of κ. In particular, <b,κ is indeed a partial order: If f <b,κ g and g <b,κ h, thenf <b,κ h because

{α < κ : f(α) < h(α)} ⊇ {α : f(α) < g(α)} ∩ {α : g(α) < h(α)}

contains the intersection of two sets in C, so it is also in C.To prove well-foundedness, consider a <b,κ-decreasing sequence (fn : n < ω), and let An ={α : fn+1(α) < fn(α)}, so An ∈ C for all n. But then A =

⋂nAn ∈ C. In particular, A 6= ∅.

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If α ∈⋂nAn, then ∀n (fn+1(α) < fn(α)), and (fn(α) : n < ω) is a decreasing sequence of

ordinals, contradiction. �

Whenever we are given a well-founded partial order on a set X, we can assign a rank (anordinal) to the elements of X, as follows:

Definition 63 Let (X,<) be well-founded. The rank ‖x‖< of a set x ∈ X is defined bytransfinite recursion as ‖x‖< = sup{‖y‖< + 1 : y < x}.

It follows from standard arguments that, in the situation of Definition 63,

‖ · ‖< : X → ORD

is well-defined. For example, considering some x for which ‖x‖< is not defined (or un-bounded), it follows that for some y < x, either ‖y‖< is not defined or it is also unbounded.This easily gives us a strictly <-decreasing sequence of members of X, contradicting that< is well-founded. It is immediate from the definition that if y < x then ‖y‖< < ‖x‖<.

Definition 64 Given a cardinal κ and a sequence B = (Bα : α < κ) of nonempty sets, let

T (B) = sup{|F| : F is an a.d.t. for B}.

Clearly,

• T (B) = T (C) for C = (|Bα| : α < κ), i.e., T (B) only depends on the cardinalities ofthe terms of the sequence B.

• (Monotonicity). If B = (Bα : α < κ) and C = (Cα : α < κ) are sequences of nonemptysets with |Cα| ≤ |Bα| for all α, then T (C) ≤ T (B).

For f ∈ κλ let T (ℵf ) be T (B) where B = ℵf := (ℵf(α) : α < κ). By monotonicity, Theorem59 will follow if I can show that T (ℵf ) < ℵλ for all f : κ→ λ, since if Aα is finite for someα, one can just replace it with ℵ0. It is this reformulation that is proved by induction onthe <b,κ-rank of f.

We argue by contradiction: Assume the claim above fails, and let f be a counterexampleof least <b,κ rank. It is our goal to show that there must be another counterexample ofstrictly smaller rank. For this, consider the following set:

I = {X ⊆ κ : ∃g ∈ κλ ∀α ∈ X (g(α) < f(α) or g(α) = 0) and T (ℵg) ≥ ℵλ}.

All we need from Lemma 65 below is that I is a nontrivial proper ideal (i.e., it is closedunder finite unions, does not contain κ, and contains the bounded sets), but proving κ-completeness requires about the same amount of work.

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Lemma 65 I is a nontrivial κ-complete proper ideal on κ.

Proof: We prove this in four steps.

• Clearly, I is ⊆-downward closed.

• κ /∈ I.

To see this, suppose κ ∈ I, as witnessed by g, so T (ℵg) ≥ ℵλ and for all α < κ, g(α) < f(α)or g(α) = 0.

By <b,κ-minimality of f, {α < κ : g(α) ≥ f(α)} = {α : f(α) = 0} has size κ.

If F is an a.d.t. for ℵg, then for h 6= j ∈ F , we have that {α : h(α) = j(α)} is bounded, so forκmany values of α, h(α) 6= j(α), and they are finite! Let A be a subset of {α < κ : f(α) = 0}of size κ. The number of functions from A to ℵ0 is |Aω| = ℵκ0 , and there are at most|P(κ)| = 2κ many such sets A, so the number of elements of F is at most 2κ · ℵκ0 = 2κ, and|T (ℵg)| ≤ 2κ < λ, contradiction.

It follows that κ /∈ I, i.e., I is proper.

• If X ⊆ κ is bounded, then X ∈ I.

For X ⊂ κ bounded, let gX : κ → λ be given by gX(α) = f(α) if α /∈ X, and gX(α) = 0otherwise.

I claim that gX witnesses that X ∈ I. Since ∀α ∈ X (g(α) = 0), it suffices to show thatT (ℵgX ) ≥ ℵλ.

Let F be an a.d.t. for ℵf , and define F as {h : h ∈ F} where

h(α) ={h(α) if α /∈ X,

0 otherwise,

so h ∈∏α<κ ℵgX(α) and h 7→ h is injective since X is bounded and h(α) 6= g(α) for

all but boundedly many values of α. Thus, F is an a.d.t. for ℵgX , and it follows thatT (ℵgX ) ≥ T (ℵf ) ≥ λ.

• It remains to argue that I is closed under unions of fewer than κ sets.

For this, let Xβ be sets in I for β < γ, where γ < κ, as witnessed by functions fβ, β < γ.Let X =

⋃β<γ Xβ. Let g : κ→ λ be given by g(α) = minβ<γ fβ(α) for α < κ.

Let µ < ℵλ. We argue that there is an a.d.t. Fµ for ℵg of size at least ℵµ. Since this holdsfor all such µ, it follows that T (ℵg) ≥ ℵλ, and g witnesses that X ∈ I.To build Fµ, fix an a.d.t. Fβ for Xfβ of size µ for each β < γ, and let (fβ,δ : δ < µ) be aninjective enumeration of Fβ.

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Partition X into (possibly empty) sets Yβ, β < γ, such that g(α) = fβ(α) for all α ∈ Yβ.Set

gδ =⋃β<γ

fβ,δ � Yβ

for δ < µ so, if α < κ and α ∈ Yβ, then gδ(α) = fβ,δ(α) < fβ(α) ≤ g(α). Let

Fµ = {gδ : δ < µ}.

Notice that since each Fβ is an a.d.t., if δ1 6= δ2 < µ, then

Zβ = {α ∈ Yβ : fβ,δ1(α) = fβ,δ2(α)}

is bounded in κ, so {α : gδ1(α) = gδ2(α)} =⋃β<γ Zβ is bounded in κ, by regularity of κ,

and therefore Fµ is an a.d.t. for ℵg. �

Split κ as X0 ∪X1 ∪X2 where

X0 = {α < κ : f(α) = 0},

X1 = {α < κ : f(α) is a nonzero limit ordinal},

andX2 = {α < κ : f(α) is a successor}.

As argued in the proof of Lemma 65, |X0| < κ, so the following holds:

Lemma 66 X0 ∈ I. �

We also have:

Lemma 67 X1 ∈ I.

Proof: Since λ is regular and κ < λ, then f : κ → λ is bounded, and we can find ρ < λsuch that f(α) ≤ ρ for all α < κ. Let

G = {g ∈ κλ : ∀α ∈ X1 (g(α) < f(α)) and ∀α /∈ X1 (g(α) = f(α))}.

Then |G| ≤ |ρ|κ < λ.

For each µ < λ fix an a.d.t. Fµ for ℵf with |Fµ| > ℵµ. For µ < λ and g ∈ G, let

Fµ,g = Fµ ∩∏α<κ

ℵg(α),

so Fµ,g is an a.d.t. for ℵg. Moreover, since f(α) is a limit ordinal for all α ∈ X1, we havethat Fµ =

⋃g∈G Fµ,g. For µ < λ sufficiently large so |ρ|κ < µ, we have that |G| < µ and

therefore there must be some gµ ∈ G such that |Fµ,gµ | > ℵµ.

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Since λ is regular, there must be some fixed g ∈ G such that g = gµ for λ many ordinalsµ > |ρ|κ. But then T (ℵg) ≥ λ, and g witnesses that X1 ∈ I, as we wanted to show. �

It follows that X2 /∈ I.For X ⊆ X2 define gX : κ→ λ so gX(α) = f(α) if α /∈ X, and gX(α) + 1 = f(α) otherwise.If X /∈ I, it follows that T (ℵgX ) = ℵρX for some ρX < λ.

Since 2κ < λ, there is some ρ < λ such that 2κ < ℵρ and ρX ≤ ρ for all X ∈ P(X2) \ I. LetF be an a.d.t. for ℵf of size strictly larger than ℵρ+1.

Given g ∈ F and X ∈ P(X2) \ I, let

HX(g) = {h ∈ F : ∀α ∈ X (h(α) < g(α))}.

It follows that HX(g) is an a.d.t. for (Bα : α < κ), where Bα = ℵf(α) if α /∈ X andBα = g(α) otherwise. By definition of gX , we have that |Bα| ≤ ℵgX(α) for all α < κ, and itfollows that there is an a.d.t. H for ℵgX of size |HX(g)|. But then

|HX(g)| = |H| ≤ T (ℵgX ) = ℵρX ≤ ℵρ.

Let H(g) =⋃X∈P(X2)\I HX(g), so

|H(g)| ≤ 2κ · ℵρ = ℵρ < ℵρ+1 < |F|.

If E ⊆ F has size ℵρ+1, there is then some g0 ∈ (F \E)\⋃g∈E H(g) and some g1 ∈ E \H(g0).

We then have that g0, g1 ∈ F , g0 6= g1, g0 /∈ H(g1) and g1 /∈ H(g0). It follows that

{α ∈ X2 : g0(α) = g1(α)} ∈ I,

since it is in fact a bounded subset of κ. Also,

{α ∈ X2 : g0(α) < g1(α)} ∈ I,

since otherwise it is a set X such that HX(g1) is defined and g0 ∈ HX(g1) ⊆ H(g1),contradiction. Similarly,

{α ∈ X2 : g1(α) < g0(α)} ∈ I.

This is a contradiction, because then X2 is the union of three sets in I and is therefore alsoin I. �

3.1 The Galvin-Hajnal rank

We have just shown the Galvin-Hajnal Theorem 49, stating that if κ and λ are uncountableregular cardinals, and λ is κ-inaccessible, then

∏α<κ κα < ℵλ for any sequence (κα : α < λ)

of cardinals such that∏α<β κα < ℵλ for all β < κ.

In particular (see, for example, Corollary 51.7), if cf(ξ) > ω and ℵξ is strong limit, then2ℵξ < ℵ(|ξ|cf(ξ))+ .

30

The argument relied in the notion of an almost disjoint transversal. Assume that κ isregular and uncountable, and recall that if A = (Aα : α < κ) is a sequence of sets, then

T (A) = sup{|F| : F is an a.d.t. for A}.

Here, F is an a.d.t. for A iff F ⊆∏A :=

∏α<κAα and whenever f 6= g ∈ F , then

{α < κ : f(α) = g(α)} is bounded.

With κ, λ as above, Theorem 49 was proved by showing that there is an a.d.t. for

(∏β<α

κβ : α < κ)

of size∏α<κ κα, and then proving that, provided that |Aα| < ℵλ for all α < κ, then

T (A) < ℵλ.In fact, the argument showed a bit more. Recall that if f : κ→ ORD, then

ℵf = (ℵf(α) : α < κ).

Then, for any f : κ→ λ, T (ℵf ) < ℵλ.The proof of this result was inductive, taking advantage of the well-foundedness of thepartial order <b,κ defined on κORD by

f <b,κ g iff {α : f(α) ≥ g(α)} is bounded in κ.

That <b,κ is well-founded allows us to define a rank ‖f‖b for each f : κ → ORD, and wecan argue by considering a counterexample of least possible rank to the statement from theprevious paragraph.

In fact, more precise results are possible. Galvin and Hajnal observed that replacing theideal of bounded sets with the nonstationary ideal (or, really, any normal ideal), results ina quantitative improvement of Theorem 49.

Definition 68 For f, g : κ→ ORD, let

f <NSκ g iff {α < κ : f(α) ≥ g(α)} is nonstationary.

Similarly, define f ≤NSκ g iff {α : f(α) > g(α)} is nonstationary, etc.

Since NSκ, the collection of nonstationary subsets of κ is a κ-complete (in fact, normal)ideal, then <NSκ is well-founded, and

‖f‖ = ‖f‖NSκ

is defined for any f : κ→ ORD. ‖ · ‖ is usually called the Galvin-Hajnal rank of f.

Before stating the result, Theorem 75 below, we prove some preliminary lemmas about thebehavior of the Galvin-Hajnal rank.

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Definition 69 Let κ be a regular uncountable cardinal. The canonical functions for κare defined inductively: A function f : κ→ ORD is an α-th canonical function iff

1. A β-th canonical function fβ : κ→ ORD exists for all β < α,

2. For all β < α, there is a β-th canonical function fβ for κ such that fβ <NSκ f, and

3. For all h : κ → ORD, if for all β < α there is some β-th canonical function fβ for κsuch that fβ <NSκ h, then f ≤NSκ h.

Of course, the idea is that an α-canonical function f witnesses ‖f‖ = α and is canonicaljust as ordinals are canonical among well-ordered sets. We prove this in a series of lemmas.

Lemma 70 If κ is regular and uncountable, and f, g are α-th canonical functions for κ,then f =NSκ g.

Proof: By definition, if f and g are α-th canonical functions for κ, then both f ≤NSκ gand g ≤NSκ f hold, so f =NSκ g. �

By the lemma, we can simply talk about the α-th canonical function, if one exists.

Lemma 71 Let κ be an uncountable regular cardinal, let f : κ → ORD, and let α be suchthat the α-th canonical function fα exists. Then ‖f‖ > α iff fα <NSκ f.

Proof: By induction, ‖fα‖ ≥ α, so if fα <NSκ f then α < ‖f‖.To prove the other implication, argue by induction, and assume that for all β < α and allg : κ→ ORD, if ‖g‖ > β then fβ <NSκ g where fβ is the β-th canonical function.

Suppose that ‖f‖ > α. Then there is some f ′ <NSκ f such that ‖f ′‖ ≥ α. It follows thatfor any β < α, ‖f ′‖ > β. By induction, fβ <NSκ f

′. By definition, fα ≤NSκ f′. But then

fα <NSκ f as wanted. �

Corollary 72 For any regular uncountable cardinal κ, if the α-th canonical function fα forκ exists, then ‖fα‖ = α.

Proof: By induction on α, ‖fα‖ ≥ α. But ‖fα‖ > α is impossible by Lemma 71. �

It is easy to exhibit the first few canonical functions.

Lemma 73 Let κ be a regular uncountable cardinal.

1. The α-th canonical function fα exists for all α < κ+.

2. If β < κ then ‖f‖ ≤ β iff {α : f(α) ≤ β} is stationary.

32

3. ‖f‖ ≤ κ iff {α : f(α) ≤ α} is stationary.

Proof: For α < κ we can set fα to be the function constantly equal to α, and we canalso set fκ to be the identity. It is straightforward from induction and Fodor’s lemma thatDefinition 69 is satisfied. Items 2 and 3 follow now from Lemma 71.

In general, given κ < α < κ+, choose a (nonstrictly) increasing sequence of ordinals

(iξ : ξ < κ)

such that α = supξ(iξ + 1). This is possible, since cf(α) ≤ κ. Now set

fα(ξ) = supξ<α

(fiξ(α) + 1)

for all ξ < κ.

We argue by induction that Definition 69 is satisfied. For this, suppose that fα 6≤NSκ h.Then {ξ < κ : h(ξ) < supρ<ξ(fiρ(ξ) + 1)} is stationary. By Fodor’s lemma, there is somefixed ρ < κ such that {ξ < κ : h(ξ) ≤ fiρ(ξ)} is stationary. But then fiρ 6<NSκ h. �

For κ regular uncountable, it may or may not be the case that fκ+ , the κ+-canonical functionfor κ, exists.

For example, Galvin showed that the ω2-th function f : ω1 → ω1 does not exist in L, whilein a model of Jech, Magidor, Mitchell, and Prikry, the α-th canonical function f : ω1 → ω1

exists for all α. This statement is equiconsistent with the existence of a measurable cardinal(the model is obtained by a construction that gives that NSω1 is precipitous, and it is easyto see that the existence of fα for all α implies this). See Jech-Magidor-Mitchell-Prikry [13].

If one is only interested in the existence of the α-th canonical function for ω1 for all α < θ forsome fixed θ, this does not require any large cardinals, and its consistency was establishedin Jech-Shelah [14].

Given an uncountable regular cardinal κ, forcing allows us to specify in which sense thecanonical functions for κ are, indeed, canonical: These are the functions f : κ→ ORD suchthat the ordinal [f ]G in the internal ultrapower V κ/G is independent of the NSκ-genericfilter G. This characterization is often taken in place of Definition 69.

Lemma 74 Let κ be an uncountable regular cardinal and let f : κ→ ORD.

1. ‖f‖ <

∏f(α)6=0

|f(α)|

+

.

2. If λ is regular uncountable and κ-inaccessible, and f : κ→ λ, then ‖f‖ < λ.

Proof: Suppose g <NSκ f. Let

g′(α) ={g(α) if g(α) < f(α),

0 otherwise.

33

Then g′ =NSκ g so ‖g′‖ = ‖g‖, g′(α) = 0 if f(α) = 0, and if X = {α < κ : f(α) 6= 0}, thenthe restriction of g′ to X is in the set product

∏α∈X f(α). There are at most

∏α∈X |f(α)|

many such functions g′, and since any ordinal below ‖f‖ is represented by one of them, it

follows that ‖f‖ <

∏f(α)6=0

|f(α)|

+

, as wanted.

Item 2 follows immediately, since if f : κ→ λ then f(α) < λ for all α, and

δ = supαf(α) < λ,

by regularity. But then ‖f‖ < (δκ)+ ≤ λ. �

Shelah has shown (by a more elaborate argument) that Lemma 74.2 still holds if one weakensthe κ-inaccessibility assumption on λ to the requirement that τκ < λ for all τ < λ ofcofinality κ. This allows us to weaken the assumption in the same way in the corollariesbelow.

We are now ready to state the second Galvin-Hajnal theorem.

For notational convenience, let’s write T (κ, δ) for T (A) where A = (Aα : α < κ) withAα = δ for all α < κ.

Theorem 75 (Galvin-Hajnal) Let κ be an uncountable regular cardinal, let (δα : α < κ)be a (not necessarily strictly) increasing continuous sequence of cardinals, let ϕ : κ→ ORD

and set ψ(α) = δ+ϕ(α)α for α < κ. If ∆ = 2κ supα<κ T (κ, δα), then T (ψ) ≤ ∆+‖ϕ‖.

The proof of Theorem 75 is by induction on ‖ϕ‖. When the δα are strictly increasing and‖ϕ‖ = 0, this is a result of Erdos, Hajnal, and Milner.

Corollary 76 If κ is uncountable regular and ϕ : κ→ ORD, then T (ℵϕ) ≤ (2κ)+‖ϕ‖.

Proof: In the notation of Theorem 75, take δα = ω for all α < κ, so ψ(α) = ℵϕ(α) andT (ℵϕ) = T (ψ). Note that ∆ = 2κT (κ, ω) = 2κ. �

Notice that Theorem 49 follows immediately:

Corollary 77 Suppose that κ, λ are uncountable regular cardinals and λ is κ-inaccessible.If f : κ→ λ then T (ℵf ) < ℵλ.

Proof: By Corollary 76, T (ℵf ) ≤ (2κ)+‖f‖ and by Lemma 74, ‖f‖ < λ. �

Corollary 78 Let κ, (δα)α<κ,∆, ϕ be as in Theorem 75.

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1. Let (τα : α < κ) be a sequence of cardinals such that

∀β < κ (∏α<β

τα ≤ δ+ϕ(β)β ).

Then∏α<κ τα ≤ ∆+‖ϕ‖.

2. Let (τα : α < κ) be an increasing sequence of cardinals with τ0 ≥ 2, and such that

∀β < κ (∏α<β

τα ≤ δ+ϕ(β)β ).

Then

(∑α<κ

τα

)κ≤ ∆+‖ϕ‖.

3. Let (λα : α < κ) be a strictly increasing sequence of infinite cardinals. If λ =∑

α<κ λα,ρ is a cardinal, and

∀α < κ (ρλα ≤ δ+ϕ(α)α ),

then ρλ ≤ ∆+‖ϕ‖.

Proof: For 1, let Aα =∏β<α τβ and let F be an a.d.t. for (Aα : α < κ) of size

∏β<κ τβ.

By assumption, |Aα| ≤ δ+ϕ(α)α for all α, so |F| ≤ T (ψ) ≤ ∆+‖ϕ‖.

For 2, recall from Lemma 20 that(∑α<κ

τα

)κ=

(∏α<κ

τα

)κ,

and apply item 1.

For 3, let τα = ρλα for α < κ. Then∏β<α τβ = ρ

Pβ<α λβ ≤ ρλα ≤ δ+ϕ(α)

α for all α < κ, andthe result follows from item 1. �

Corollary 79 Let λ be a cardinal of uncountable cofinality κ, and suppose that λ is κ-inaccessible. Let (λα : α < κ) be a strictly increasing and continuous sequence of infinitecardinals cofinal in λ. Let ϕ : κ→ ORD.

1. If ∀α < κ (∏β<α λβ ≤ λ

+ϕ(α)α ), then λκ ≤ λ+‖ϕ‖.

2. If ∀α < κ (2λα ≤ λ+ϕ(α)α ), then 2λ ≤ λ+‖ϕ‖.

3. If 2τ ≤ λκ for all τ < λ and∏β<α λβ ≤ λ

+ϕ(α)α for all α < κ, then 2λ ≤ λ+‖ϕ‖.

Proof: In Corollary 78, take δα = τα = λα and ρ = 2. To prove 1, notice that∏α λα = λκ

and ∆ = 2κ supα T (κ, λα) = supα λκ = λ. Item 2 follows from Corollary 78.3, and item 3follows from the fact that 2λ = λκ. �

Recall the very general statement of Silver’s Theorem 43:

35

Corollary 80 (Silver) Let λ be a κ-inaccessible cardinal of uncountable cofinality κ. Let(λα : α < κ) be a strictly increasing sequence of cardinals cofinal in λ. Suppose that there issome µ < κ such that {α < κ :

∏β<α λβ ≤ λ

+µα } is stationary in κ. Then λκ ≤ λ+µ.

Proof: This follows from Corollary 79.1 and Lemma 73.2. �

I proceed now to sketch the proof of Theorem 75. The argument is complete, modulo twoadditional ingredients (whose proofs I skip) are needed. One is the Erdos-Rado theorem,that we will prove in Chapter III.

Definition 81 Let κ, λ, ρ be cardinals and let n < ω. Given a set X, recall that

[X]n = {Y ⊆ X : |Y | = n}.

The statementκ→ (λ)nρ

means that whenever f : [κ]n → ρ, then there is Y ⊆ κ, |Y | = λ, such that the restriction off to [Y ]n is constant. If this is not the case, we write κ 6→ (λ)nρ .

It is customary in the field to call the functions f as above colorings, and to refer to thesets Y as homogeneous with respect to f.

This “arrow notation” turns to be very useful. For example, the infinitary version of Ram-sey’s theorem is simply ω → (ω)2

2. A well known example due to Sierpinski shows thatc 6→ (ω1)2

2 or, more generally, 2κ 6→ (κ+)22. This will be discussed in Chapter III.

Theorem 82 (Erdos-Rado) For any infinite cardinal κ, (2κ)+ → (κ+)2κ. �

The second ingredient is a result of Hajnal on set-mappings.

Definition 83 Let E be a set. A set-mapping on E is a function f : E → P(E) suchthat ∀x ∈ E (x /∈ f(x)). Given such f,E, we say that H ⊆ E is free with respect to fiff ∀x ∈ H (H ∩ f(x) = ∅).

For example, the function f : κ → P(κ) given by f(α) = {β < κ : β ∈ α} = α is aset-mapping on κ, |f(α)| < κ for all α, and any free H ⊆ κ is a singleton. Ruziewiczconjectured in 1936 that if one strengthens the cardinality restriction on f to |f(α)| < λfor all α < κ, where λ < κ is some fixed cardinal, then there is now a free H of size κ. Thiswas shown by Hajnal in 1961.

Theorem 84 (Hajnal) Let κ and λ be cardinals with λ < κ and κ infinite. Let f be aset-mapping on κ such that |f(α)| < λ for all α < κ. Then there is an H ⊆ κ of size κ thatis free with respect to f. �

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Proof of Theorem 75: Proceed by induction on ‖ϕ‖.Suppose first that ‖ϕ‖ = 0.

Let F be an a.d.t. for ψ. By Lemma 73.2, {α < κ : ϕ(α) = 0} is stationary. Let

X0 = {α : ϕ(α) = 0 and α is limit},

so X0 is also stationary. Note that f(α) < δα for all f ∈ F and α ∈ X0 and thereis therefore a βf (α) < α such that f(α) < δβf (α). (It is here that the continuity of thesequence (δα : α < κ) is used.)

By Fodor’s lemma, for each f ∈ F there is a stationary set Xf ⊆ X0 and an ordinal βf < κsuch that f(α) < δβf for any α ∈ Xf . For fixed sets X,β, let

FX,β = {f ∈ F : Xf = X and βf = β},

and notice that F =⋃X,β FX,β.

Since there are only 2κ such pairs (X,β) and |FX,β| ≤ T (κ, δβ) ≤ ∆, it follows that

|F| ≤ 2κ∆ = ∆,

and we are done.

Suppose now that ‖ϕ‖ = ν > 0 and that the result holds for all ϕ : κ → ORD andcorresponding ψ such that ‖ϕ‖ < ν.

Let F be an a.d.t. for ψ. For f ∈ F and α < κ let ϕf (α) be least such that |f(α)| ≤ δϕf (α)α ,

and let ψf (α) = δ+ϕf (α)α . Notice that ϕf <NSκ ϕ, since ϕf (α) ≥ ϕ(α) only if ϕ(α) = 0, but

the set of these ordinals α is nonstationary since ‖ϕ‖ > 0. It follows that ‖ϕf‖ < ν.

This shows that F =⋃µ<ν Fµ, where

Fµ = {f ∈ F : ‖ϕf‖ = µ},

and therefore |F| ≤ |ν| supµ<ν |Fµ|, and we are done if we show that Fµ ≤ ∆+ν for eachµ < ν.

In fact, |Fµ| ≤ ∆µ+1 for all µ < ν. To see this, fix such a µ and begin by defining aset-mapping H on Fµ by

H(f) = {g ∈ Fµ : g 6= f and g ∈∏α<κ

(f(α) + 1)}.

Clearly, H(f) is an a.d.t. for f + 1. Note that for all α < κ, |f(α) + 1| ≤ δϕf (α)α = ψf (α), so

|H(f)| ≤ T (ψf ). Notice now that the hypotheses of the theorem apply to ϕf and ψf in placeof ϕ and ψ. Since ‖ϕf‖ = µ < ν, by the induction hypothesis we have that T (ψf ) ≤ ∆+µ.

Let λ = ∆+µ+1, and ρ = ∆+µ+2. If, towards a contradiction, |Fµ| ≥ ρ, then by Theorem 84there is a set F ′ ⊆ Fµ of size ρ that is free with respect to H.

Notice that ρ > ∆ ≥ 2κ, and therefore (since H is free) we can inductively find a sequence(fξ : ξ < (2κ)+) with each fξ ∈ F ′ and fξ /∈ H(fη) whenever ξ < η < (2κ)+. By definitionof H, it follows that for all such ξ < η there is an α = α(ξ, η) < κ such that fξ(α) > fη(α).

37

By the Erdos-Rado Theorem 82 there is an infinite set A ⊆ (2κ)+ homogeneous for thecoloring α : [(2κ)+]2 → κ. Let γ < κ be the constant value that the map α takes on [A]2.If ξ < η ∈ A, then fξ(γ) > fη(γ), and therefore there is an infinite decreasing sequence ofordinals, contradiction. �

I conclude with an application.

Definition 85 A structure M = (M,R, . . . ) in a countable language with a distinguishedunary relation (interpreted in M by R ⊆ M) is said to be of type (κ, λ) iff |M | = κ andR = λ.

Chang’s conjecture is the statement that whenever M is of type (ω2, ω1), there is anelementary substructure N ≺M of type (ω1, ω).

Chang’s conjecture is consistent. It holds under MM and can be forced from a Ramseycardinal. A large cardinal of comparable strength is required, since Silver showed thatChang’s conjecture implies the existence of 0].

By arguing about Skolem functions, one can show without too much effort that Chang’sconjecture is equivalent to

ω2 → [ω1]<ωω1,ω,

the statement that whenever f : [ω2]<ω → ω1, there is a subset A of ω2 of order type ω1

such that f ′′[A]<ω is countable.

Theorem 86 (Magidor) Chang’s conjecture implies that if ℵω1 is strong limit, then

2ℵω1 < ℵω2 .

Notice that from Corollary 55, without assuming Chang’s conjecture we have that if ℵω1

is strong limit then 2ℵω1 < ℵ(2ℵ1 )+ . It is open whether Theorem 86 holds without theadditional assumption.

Proof: The following argument is due to Galvin.

By say, Corollary 79, it is enough to show that ‖f‖ < ω2 for all f : ω1 → ω1.

To prove this, let f : ω1 → ORD be such that ‖f‖ ≥ ω2. Using Lemma 73, for β < ω2 let fβbe the β-th canonical function for ω1, so fβ <NSκ fγ <NSκ f for all β < γ < ω2. Let Cβ,γbe a club witnessing this, so for all α < ω1, if α ∈ Cβ,γ then fβ(α) < fγ(α) < f(α).

Define h : [ω2]<ω → ω1 as follows: If x ⊆ ω2 is finite, let

h(x) = min⋂{Cβ,γ : β, γ ∈ x, β < γ},

where we take the minimum of an empty intersection to be zero.

By Chang’s conjecture, there is an A ⊆ ω2 of order type ω1 such that α = suph′′[A]<ω < ω1.

We claim that f(α) ≥ ω1, which concludes the proof.

38

To see the claim, notice first that α ∈ Cβ,γ whenever β, γ ∈ A and β < γ. Suppose otherwise,and fix β1 < γ1 counterexamples. Then

α1 := sup(Cβ1,γ1 ∩ (α+ 1)) = max(Cβ1,γ1 ∩ α) < α.

By definition of α, this means that there must be some finite x ⊂ A such that h(x) > α1

and therefore, by definition of h and α1, it follows that h(x ∪ {β1, γ1}) > α, contradiction.

But the claim concludes the proof, since then fβ(α) < fγ(α) < f(α) for all β < γ in A, andtherefore, since |A| = ω1, we must have f(α) ≥ ω1. �

Besides Galvin-Hajnal [11], a good reference for this Section is the book Erdos-Hajnal-Mae-Rado [9].

4 Large cardinals and cardinal arithmetic

In Section 3 we saw how the powers of singular cardinals (or, at least, of singulars ofuncountable cofinality) satisfy strong restrictions. Here I show that similar restrictionshold at large cardinals. There is much more than one could say about this topic, and theresults I present should be seen as an invitation rather than a full story.

I won’t motivate the large cardinals we will discuss. (In the ideal world, one should probablysay a few words about one’s beliefs in large cardinals, since their existence and even theirconsistency goes beyond what can be done in the standard system ZFC. I will however taketheir existence for granted, and proceed from there.)

4.1 Measurable cardinals

Definition 87 κ is a measurable cardinal iff κ > ω and there is a nonprincipal κ-completeultrafilter over κ.

Recall:

Definition 88 U ⊆ P(X) is an ultrafilter (over X, or on P(X)) iff the following hold:

1. ∀Y ⊆ Z ⊆ X, if y ∈ U , then Z ∈ U .

2. If Y,Z ∈ U , then Y ∩ Z ∈ U .

3. ∅ /∈ U .

4. For all Y ⊆ X, either Y ∈ U or else X \ Y ∈ U .

Conditions 1–3 state that U is a filter. Notice that by 2 and 3, U has the finite intersectionproperty. Conversely, any collection of subsets of X with the finite intersection propertygenerates in a natural way a filter.

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Condition 4 is the ultrafilter condition. Equivalently, one can request that U is maximalunder containment (subject to being a filter). The equivalence is an easy consequence ofZorn’s lemma.

An ultrafilter U over X is nonprincipal iff {x} /∈ U for any x ∈ X. Otherwise, it isprincipal. Clearly, any ultrafilter over a finite set is principal. On the other hand, anyinfinite set admits a nonprincipal ultrafilter, in fact, since the collection of subsets of Xwhose complements have size strictly smaller than X has the finite intersection property,there are uniform ultrafilters over X, i.e., ultrafilters all of whose elements have the samesize as X.

As before, an ultrafilter U is λ-complete iff any intersection of fewer than λ many membersof U is in U . If λ = ω1 we also say that U is σ-complete. Equivalently, by consideringcomplements, U is λ-complete iff whenever a union of fewer than λ many sets is in U , thenat least one of the sets in the union is in U .Clearly, no nonprincipal ultrafilter over κ can be κ+-complete. On the other hand, normalityis a stronger requirement than κ-completeness.

Homework problem 10. Assume that U is a σ-complete nonprincipal ultrafilter over someset X. Let κ be the completeness of U , i.e., κ is smallest such that U is not κ+-complete.Show that κ is measurable.

Ulam introduced measurable cardinals by working on a question on Banach. The problemis whether there could be a nontrivial measure space

(X,P(X), µ).

In detail, µ : P(X) → [0,∞], and µ(⋃n∈ω An) =

∑n µ(An) for any pairwise disjoint

sequence of subsets of X.

The nontriviality condition means several things. First of all, µ(∅) = 0 (otherwise, we musthave µ(E) =∞ for all E ⊆ X.)But this is not enough. For consider any set X, any function f : X → [0,∞) and any(possibly empty, or perhaps nonproper) σ-ideal I of subsets of X. Then we can set

µ(E) ={ ∑

x∈E f(x) if E ∈ I,∞ if E /∈ I,

and certainly (X,P(X), µ) is a measure space.

We then say that a measure space (X,P(X), µ) is nontrivial iff it does not arise in thisfashion.

Assume that (X,P(X), µ) is nontrivial. Let f(x) = µ({x}) if µ({x}) < ∞ and f(x) = 0otherwise. Let I be the σ-ideal generated by the subsets of X of finite measure, and letµ′ be the trivial measure over X generated by f and I. Then µ′(A) ≤ µ(A) for all A ⊆ Xbut µ′ 6= µ, since we are assuming that µ is nontrivial. It follows that there is some A suchthat µ′(A) < µ(A). It must then be the case that A ∈ I, and therefore (by definition ofI) there is an increasing sequence (An : n < ω) of subsets of X such that A =

⋃nAn and

µ(An) <∞ for all n. Necessarily, µ′(An) < µ(An) for some n, and we can define a measureµ′′ : P(An)→ [0,∞] by setting µ′′(E) = µ(E)− µ′(E).

40

Notice that in fact 0 < µ′′(An) < ∞ and µ′′({x}) = 0 for all x ∈ An. By normalizing, wemay as well assume that µ′′(An) = 1.

This shows that the measure problem has a solution iff there is a nontrivial probability space(Y,P(Y ), λ) (so in particular λ(∅) = 0 and λ(Y ) = 1) such that λ({y}) = 0 for all y ∈ Y,and this is the way in which the question is most commonly posed in the literature.

Let add(λ) be the additivity of the measure λ, i.e., the least cardinal κ such that themeasure of any disjoint union of fewer than κ many disjoint subsets of Y is the sum ofthe measures of the sets in the union, understanding that add(λ) = ∞ if there is no suchcardinal.

Let Nλ be the collection of subsets of Y of measure zero. Then Nλ is an ideal. For anyideal I over a set Z we can define add(I) as the least cardinality of a collection of sets inI whose union is not in I, understanding that add(I) =∞ if there is no such cardinal.

If (Y,P(Y ), λ) is a probability space, then either add(λ) = add(Nλ) = ∞ and the space istrivial, or else add(λ) = add(Nλ) is some cardinal κ. In this latter case, there must be asequence (Eα : α < κ) of nonempty pairwise disjoint sets in Nλ whose union E is not inNλ.In this case we can define a measure µ over κ as follows: For x ∈ E let π(x) ∈ κ be theordinal such that x ∈ Eπ(x). Let X ⊆ κ. Then set

µ(X) =λ({x ∈ E : π(x) ∈ X})

λ(E).

Then (κ,P(κ), µ) is a probability space, µ({α}) = 0 for all α ∈ κ, and µ is κ-additive.

Ulam noticed that in this case an interesting dichotomy happens: Either µ is nonatomic,meaning that for for all X ⊆ κ, if µ(X) > 0 then there is some Y ⊆ X such that 0 < µ(Y ) <µ(X), in which case κ ≤ c and we say that κ is (atomlessly) real-valued measurable,or else µ admits an atom X, that necessarily must have size κ. In this case, for all Y ⊆ X,either µ(Y ) = 0 or µ(Y ) = µ(X). By renormalizing and composing with a bijection, thisgives us a probability space (κ,P(κ), ν) where ν : P(κ)→ {0, 1}, ν({α}) = 0 for all α < κ,and ν is κ-complete. Letting U = {Y ⊆ κ : ν(Y ) = 1}, U is a κ-complete nonprincipalultrafilter over κ, i.e., κ is measurable.

Conversely, if κ is measurable and U is a κ-complete nonprincipal ultrafilter over κ, thenχU , the characteristic function of U , is an example of such a probability measure ν.

Theorem 89 (Tarski-Ulam) Suppose that κ is measurable or real-valued measurable.Then κ is weakly inaccessible. If κ is in fact measurable, then it is strong limit and therefore(strongly) inaccessible.

Proof: Fix a witnessing probability ν, so ν is κ-additive, singletons are null, and ν is eithernonatomic or bivalued.

κ is regular. Because, by κ-completeness, every bounded subset of κ, in particular everyordinal below κ, is null, and therefore κ cannot be the union of fewer than κ many ordinalsbelow κ, since any such union has measure zero.

41

κ is limit. For this, proceed by contradiction, assuming that κ = λ+, and consider a λ×λ+

Ulam matrix (Aια : ι < λ, α < κ), see Definition 34. Arguing just as in the proof of Theorem35, notice that

⋃ι<λA

ια = (α, κ) has measure 1 for all α < κ, and by κ-completeness of ν

there must be some ια such that Aιαα has positive measure.

By the pigeonhole principle, there is some fixed ι such that ια = ι for κ many distinct valuesof α. Recall now that Aια∩Aιβ = ∅ whenever α 6= β. By the pigeonhole principle again, sinceκ > ω, there is some nonzero n ∈ ω such that ν(Aια) > 1/n for uncountably many ordinalsα. This is clearly a contradiction.

Assume now that κ is measurable. Then κ is strong limit. Let U be the ultrafiltercorresponding to ν. Let ρ < κ and suppose that {Xf : f ∈ ρ2} ⊆ P(κ) and

⋃f Xf ∈ U .

It is enough to see that Xf ∈ U for some f, since this shows that U is (2ρ)+-complete, so2ρ < κ. Notice that for each α < ρ there is a (least) ε ∈ 2 such that

⋃f(α)=εXf ∈ U . Let

h : ρ→ 2 be the function that to each α < ρ assigns the corresponding ε. Then

Xh =⋂α<ρ

⋃f(α)=h(α)

Xf ∈ U ,

since U is at least ρ+-complete. �

Homework problem 11. Show that if κ is real-valued measurable then κ ≤ c, by showingthat add(ν) ≤ c, where ν is as above. For this, begin by showing that for any X ⊆ κ thereis some Y ⊆ X such that ν(Y ) = ν(X)/2.

Of course, Theorem 89 implies that the existence of measurable cardinals is not provable inZFC, since if κ is strongly inaccessible then Vκ |= ZFC.

Whether all inaccessible cardinals are measurable is a different story. In fact, if κ is mea-surable, then it is the κ-th inaccessible cardinal, and one can show much stronger resultsshowing how vast the increment in strength is between both notions. This we will do bymeans of the ultrapower construction, a key idea in the study of large cardinals.

The reduction indicated above of the measure problem to the question of the existence of aκ-complete probability space (κ,P(κ), ν) where singletons are null is probably classical. Ifollowed the approach of Fremlin [10].

4.2 The ultrapower construction

The study of ultrapowers originates in model theory, although it has found applicationsboth in algebra and in analysis. However, it is accurate to say that it is mainly exploitedin set theory. Here I present the basic idea, showing its close connection to the study ofmeasurable cardinals, see Definition 87.

Suppose first that U is an ultrafilter over a set X. We want to define the ultrapower of theuniverse V of sets by U . The basic idea is to consider the product of X many copies of thestructure (V,∈). We want to “amalgamate” them somehow into a new structure (V , ∈). Forthis, we look for the “typical” properties of the elements {f(x) : x ∈ X} of each “thread”f : X → V, and add an element f to V whose properties in (V , ∈) are precisely these typical

42

properties. We use U to make this precise, by saying that a property ϕ is typical of therange of f iff {x ∈ X : ϕ(f(x))} ∈ U . This leads us to the following definition, due to DanaScott, that adapts the ultrapower construction to the context of proper classes:

Definition 90 Let U be an ultrafilter over a nonempty set X. We define the ultrapower(V X/U , ∈) of V by U as follows:

For f, g : X → V, say that

f =U g iff {x ∈ X : f(x) = g(x)} ∈ U .

This is easily seen to be an equivalence relation. We would like to make the elements ofV X/U to be the equivalence classes of this relation. Unfortunately, these are all properclasses except for the trivial case when X is a singleton, so we cannot within the context ofour formal theory form the collection of all equivalence classes.

Scott’s trick solves this problem by replacing the class of f with

[f ] := {g : X → V : g =U f and rk(g) is least possible}.

Here, as usual, rk(g) = min{α : g ∈ Vα+1} = sup{rk(x) + 1 : x ∈ g}. All the “classes” [f ]are now sets, and we set V X/U = {[f ] : f : X → V }.We define ∈ by saying that for f, g : X → V we have

[f ]∈[g] iff {x ∈ X : f(x) ∈ g(x)} ∈ U .

It is easy to see that ∈ is indeed well defined, i.e., if f =U f ′ and g =U g′ then

{x ∈ X : f(x) ∈ g(x)} ∈ U iff {x ∈ X : f ′(x) ∈ g′(x)} ∈ U .

(The ultrapower construction is more general than as just defined; what I have presentedis the particular case of interest to us.) The remarkable observation, due to Los, is thatthis definition indeed captures the typical properties of each thread in the sense describedabove:

Lemma 91 ( Los) Let ϕ(x1, . . . , xn) be a formula in the language of set theory, and letf1, . . . , fn : X → V. Then V X/U |= ϕ([f1], . . . , [fn]) iff {x ∈ X : ϕ(f1(x), . . . , fn(x)} ∈ U .

Proof: We argue by induction in the complexity of ϕ.

• If ϕ is atomic, the result holds by definition.

• Assume the result for φ, ψ. Suppose that ϕ ≡ φ∧ψ. Then the result also holds for ϕ,since U is both upwards closed and closed under intersections.

• Assume the result holds for φ. Suppose that ϕ ≡ ¬φ. Then the result also holds forϕ, since for any Y ⊆ X, either Y ∈ U or else X \ Y ∈ U .

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• Finally, assume the result holds for φ(x, x1, . . . , xn) and ϕ ≡ ∃xφ(x, x1, . . . , xn). Letf1, . . . , fn : X → V. Then we have, by definition, that V X/U |= ϕ([f1], . . . , [fn]) iffthere is some g : X → V such that V X/U |= φ([g], [f1], . . . , [fn]). This is equivalent,by assumption, to the statement that there is some g : X → V such that

{x ∈ X : φ(g(x), f1(x), . . . , fn(x))} ∈ U .

But this last statement is equivalent to {x ∈ X : ϕ(f1(x), . . . , fn(x))} ∈ U : In onedirection, note that {x ∈ X : φ(g(x), ~f(x))} ⊆ {x ∈ X : ϕ(~f(x))} for any g. Inthe other, for each x such that ϕ(~f(x)) holds pick a witness ιx. Define a functiong : X → V by g(x) = ιx if ϕ(~f(x)) holds, and g(x) = 0 otherwise. Then

{x ∈ X : ϕ(~f(x))} = {x ∈ X : φ(g(x), ~f(x))},

and we are done.

This completes the proof. �

Corollary 92 V X/U ≡ V. In particular, V X/U is a (proper class) model of set theory. �

We arrive at the first connection between the ultrapower construction and measurable car-dinals. By Corollary 92, if U is an ultrafilter over X, then V X/U satisfies the axiom offoundation, and therefore it “believes” that ∈ is a well-founded relation. However, this mayactually be false in V, since a witnessing ∈-decreasing sequence may fail to be an elementof V X/U .Recall that, by Homework problem 10, if a nonprincipal ultrafilter is σ-complete, then thereis a measurable cardinal.

Theorem 93 Let U be an ultrafilter over a set X, and form the ultrapower (V X/U , ∈).Then ∈ is well-founded iff U is σ-complete.

Proof: Suppose first that U is σ-complete. If there is a ∈-decreasing sequence

([fn] : n ∈ ω),

let An = {x ∈ X : fn+1(x) ∈ fn(x)}. Then An ∈ U for all n and therefore⋂nAn 6= ∅

(since it is in fact an element of U . If x is in this intersection, then (fn(x) : n < ω) is an∈-decreasing sequence of sets, contradiction. (This is just the proof of Lemma 62.)

Suppose now that U is not σ-complete. Then there is a sequence (Xn : n ∈ ω) of sets notin U whose union is in U . We may assume that the sets Xn are pairwise disjoint and thattheir union is in fact all of X. Define gn : X → V by gn(x) = max(m − n, 0), where m isthe unique number such that x ∈ Xm. Then

{x ∈ X : gn+1(x) ∈ gn(x)} =⋃m>n

Xm ∈ U ,

since⋃m≤nXm is a finite union of sets not in U and is therefore not in U .

This shows that ([gn] : n ∈ ω) is a ∈-decreasing sequence of elements of V X/U . �

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Lemma 94 Let U be an ultrafilter over a set X, and let f : X → V. Then {[g] : [g]∈[f ]} isa set.

Proof: Let f = {[g] : [g]∈[f ]}. If {x : f(x) = 0} ∈ U , then f = ∅. Otherwise, given [g]∈[f ]let g′ : X → V be given by g′(x) = g(x) if g(x) ∈ f(x) and g′(x) = 0 otherwise. Theng′ =U g and if Y = {x : f(x) 6= ∅} then g′(x) = 0 outside of Y and the restriction of g′ to Yis in

∏x∈Y f(x). Since the map that assigns to g′ its restriction to Y is injective, it follows

that f injects into the set∏x∈Y f(x) and is therefore a set. �

Suppose that U is a σ-complete ultrafilter over a set X (for example, U could be principal).Then (V X/U , ∈) is a well-founded structure by Theorem 93. Lemma 94 says that it isalso set-like. Under these conditions, we can apply Mostowski’s collapsing lemma andreplace (V X/U , ∈) with an isomorphic, transitive structure M :

Lemma 95 (Mostowski) Suppose (A,E) is a (possibly proper class) structure where E ⊆A×A and the following conditions hold:

1. (A,E) |= Extensionality, i.e., for any a, b ∈ A, a = b iff ∀c ∈ A (cEa⇔ cEb).

2. E is set-like, i.e., for all a ∈ A, a = {b ∈ A : bEa} is a set.

3. E is well-founded (in V ).

Then there is a unique (possibly proper) class M that is transitive (i.e., for all a ∈ M,a ⊆M), and a unique isomorphism π : (A,E)→ (M,∈).

The unique M as in Lemma 95 is the Mostowski (or transitive) collapse of A and theunique π is the Mostowski collapsing function.

Proof: Notice first that the E-rank of any a ∈ A is defined. As usual,

‖a‖E = sup{‖b‖E + 1 : bEa};

that E is set-like is used to justify that ‖a‖E is an ordinal if ‖b‖E is an ordinal for all bEa.Hence, if ‖a‖E is not defined, then ‖b‖E is not defined for some bEa, and we obtain anE-decreasing sequence, against well-foundedness. This is just as the argument followingDefinition 63.

Now define π : A → V by π(a) = {π(b) : bEa}. This is well-defined since otherwise, oneimmediately arrives at a contradiction by considering an a of least E-rank for which π(a)is undefined.

Let M = π[A]. Notice that for any a, b ∈ A, if aEb then π(a) ∈ π(b). Also, M is transitive,by definition and, by induction on ‖a‖E , rk(π(a)) = ‖a‖E for any a ∈ A.It follows that π is injective. Otherwise, π(a) = π(b) for some a 6= b. By extensionality,there is some c such that exactly one of cEa and cEb holds. Without loss, say that cEa.Then there is some d such that π(c) = π(d) and dEb. We thus arrive at a contradiction byconsidering the least possible rank of an x ∈M with more than one preimage under π.

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This implies that π is an isomorphism, since if π(a) ∈ π(b) then there is some cEb such thatπ(a) = π(c). But then a = c so aEb.

Finally, suppose that M ′ is some transitive class and π′ is a map such that π′ : (A,E) →(M ′,∈) is an isomorphism. By considering again a possible counterexample of least E-rank,it follows that π′(a) = π(a) for all a ∈ A. This shows that M and π are unique, and we aredone. �

Hence, if U is σ-complete, there is a transitive structure isomorphic to V X/U . We denotethis structure by Ult(V,U). (Sometimes, V X/U is denoted Ult∗(V,U).)

Since Ult(V,U) is transitive and a model of set theory, there is certain resemblance betweenV and Ult(V,U). In the background, we are using that ∆1 formulas are absolute betweentransitive structures that satisfy a modicum of set theory. For example, if x ∈ Ult(V,U)and Ult(V,U) |= “x is a function”, then x is indeed a function. Similarly, since each elementof Vω is (easily) definable, then Vω ⊆ Ult(V,U).

It would take us too long to give a detailed presentation of absoluteness, and it is a standardpart of any introductory course, so I will take it for granted. See for example Chapter IVof Kunen [19].

Consider now an arbitrary set t. Continuing with the motivation indicated at the beginningof the lecture, it should be clear that if ct : X → V denotes the function constantly equalto t, then the typical properties of the elements in the range of ct are just the properties oft, and it is natural to consider the following map:

Definition 96 Let U be an ultrafilter over a set X and, for each t ∈ V, denote by ct : X → Vthe map constantly equal to t. The map

iU : V → V X/U

is defined by iU (t) = [ct].

iU is an example of a class of maps very important in the study of large cardinals.

Definition 97 An elementary embedding j : M → N between two structures (in thesame language L) is a function j from the universe of M to the universe of N such thatfor any formula ϕ(x1, . . . , xn) in language L and any a1, . . . , an ∈M,

M |= ϕ(a1, . . . , an) iff N |= ϕ(j(a1), . . . , j(an)).

Lemma 98 For any ultrafilter U over a set X, the map iU : V → V X/U is an elementaryembedding.

Proof: This is an immediate corollary of Los’s Lemma 91: Given any formula ϕ(~x) andelements ~a ∈ V, we have that ϕ(~a) holds iff {x ∈ X : ϕ(~a)} ∈ U since this set is either X ifϕ(~a) holds, or ∅ otherwise.

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But {x : ϕ(a1, . . . , an)} = {x : ϕ(ca1(x), . . . , can(x))}, and Los’s lemma concludes the proof.�

Let U be an ultrafilter over a set X, and let κ be its additivity, i.e., κ is the least cardinalsuch that U is not closed under intersections of length κ.

We can then define, by means of a projection, a nonprincipal ultrafilterW over κ as follows:There is a sequence (Xα : α < κ) of sets not in U whose union is in U . We can modify thissequence slightly, if necessary, so they are pairwise disjoint and their union is X; this usesthat U is κ-complete.

Now set, for A ⊆ κ,A ∈ W iff

⋃α∈A

Xα ∈ U .

This is clearly an ultrafilter, and it is nonprincipal since otherwise one of the sets Xα wouldbe in U . (This is just like the projection used to define µ from λ in the paragraphs preceedingTheorem 89.)

Define k : V κ/W → V X/U by setting k([f ]W) = [gf ]U where gf : X → V is the functiongiven by gf (x) = f(α) for all x ∈ Xα and all α ∈ κ. Here, I have added subscripts to theclasses [·] to clarify with respect to which ultrafilter they are being formed.

It follows from the definition of W that k is well defined.

V

iW $$IIIIIIIIIIiU // Ult∗(V,U)

Ult∗(V,W)

k77ooooooooooo

It is also immediate from the definition that whenever we have Ult∗(V,W) |= [f1]∈[f2], thenUlt∗(V,U) |= k([f1])∈k([f2]), so k is an embedding.

Lemma 99 With notation as above, k is elementary and the diagram commutes, i.e., k ◦iW = iU .

Proof: Given any x and A, let cAX denote the constant function with domain A and valuex. Then k ◦ iW(x) = k(iW(x)) = k([cκx]W) = [cXx ]U = iU (x), by definition of k. This provescommutativity.

Elementarity is also straightforward: Using notation as above, suppose that

Ult∗(V,U) |= ∃xϕ(x, [gf1 ], . . . , [gfn ])

for some formula ϕ(x, x1, . . . , xn). Let f : X → V be a witness, in the sense that

Yf = {a ∈ X : ∀α < κ (a ∈ Xα → ϕ(f(a), f1(α), . . . , fn(α)))} ∈ U .

Pick an aα ∈ Xα∩Yf for each α < κ such that Xα∩Yf 6= ∅, and define a function g : X → Vby g(a) = f(aα) where α is such that a ∈ Xα and Xα ∩ Yf 6= ∅. If aα is not defined, setg(a) = 0.

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Then g = gh for some h : κ→ V, and Yg ∈ U , where Yg is defined using g exactly as Yf wasdefined from f.

By definition of W,Ult∗(V,W) |= ϕ([h], [f1], . . . , [fn]),

and the elementarity of k follows from the Tarski-Vaught criterion. �

Consider again the case where U is σ-complete, so Ult(V,U) is defined. Let j = π ◦ iU , soj : V → Ult(V,U) is an elementary embedding.

Suppose that U is principal, say U = {Y ⊆ X : x0 ∈ Y } for some fixed x0 ∈ X. Then, forany function f : X → V, we have [f ] = [cf(x0)]. Also, by induction on the ∈-rank of f, wehave that π([f ]) = f(x0), since π([f ]) = {π[g] : gEf} = {g(x0) : g(x0) ∈ f(x0)} = f(x0).But then, with j as above, j(t) = π ◦ iU (t) = π([ct]) = t and we have that j is the identity.

If, on the other hand, (there is a measurable cardinal and) U is nonprincipal, then j is notthe identity. We say that j is non-trivial. Now we can deduce information about V by usingthat there are two sources of resemblance between V and Ult(V,U): The latter is transitive,and j is an elementary embedding.

Note that by elementarity, j(α) is an ordinal for all ordinals α. Hence, j : ORD → ORD isorder preserving, and it follows that j(α) ≥ α for all α.

Corollary 100 (Scott) If there is a measurable cardinal, then there is a nontrivial ele-mentary embedding j : V →M from V into some transitive class M.

Proof: Let κ be measurable and let U be a κ-complete nonprincipal ultrafilter over κ. Letj : V → Ult(V,U) be as above. I will show that j is not the identity by arguing that, infact, j(κ) > κ.

For this, notice that in V κ/U we have [cα]∈[id]∈[cκ] for all α < κ, where id is the identitymap from κ to itself. This uses that bounded subsets of κ are not in U , a consequence ofκ-completeness and U being nonprincipal.

Then, in Ult(V,U), α ≤ j(α) < π([id]) < j(κ) for all α < κ. Hence, π([id]) ≥ κ and j(κ) > κ.�

The fact that some ordinal is moved is not an artifact of the proof but a general fact aboutelementary embeddings.

Lemma 101 Suppose M is a transitive class and j : V → M is elementary. If j = id isthe identity map from V to itself, then M = V. If j is non-trivial, then there is an ordinalα such that j(α) 6= α.

Note that it is not automatic that the fact that j is the identity implies that M = V ; wealso need to assume that the ordinals of M and V coincide. For example, let κ = ωL1 andconsider an elementary substructure M of Lκ with M countable in L. We claim that Mis transitive. To see this, suppose that x ∈ M. Then x ∈ Lκ, so Lκ |= “x is countable”.

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But then the same holds in M. Since ω + 1 ⊆ M, as every member of ω + 1 is definable,it follows that ran(f) ⊂ M where f : ω → x is a surjection in M. But ran(f) = x. By thecondensation lemma, it follows that M = Lα for some ordinal α that is countable in L, andwe have Lα ≺ Lκ.Proof: j(V ) = j(

⋃α Vα) ⊇

⋃α j(Vα). If j is the identity, then j(Vα) = Vα for all α, and

M ⊇ V.Suppose now that j is not the identity, so there is some x such that j(x) 6= x. Pick such aset x of least possible rank. Since y ∈ x implies that rk(y) < rk(x), it follows that j(y) = y,so x ⊆ j(x), and there must be some y ∈ j(x) \ x. Assume now that j is the identity on theordinals. Since rk(t) is an ordinal for all t, it follows that rk(j(x)) = j(rk(x)) = rk(x). Thusrk(y) < rk(x) and y = j(y). Since y ∈ j(x) then j(y) ∈ j(x) and, by elementarity, y ∈ x,contradiction. �

Definition 102 Let j : V → M be a non-trivial elementary embedding from V into sometransitive class M. (We will simply say that j : V → M is a nontrivial elementary embed-ding, or even, that j : V →M is elementary.)

The smallest α such that j(α) > α is called the critical point of j, and denoted cp(j) orcrit(j).

Lemma 103 Let j : V → M be elementary. Then κ = cp(j) is a strongly inaccessiblecardinal.

It follows from the lemma that the existence of elementary embeddings is not provable inZFC. The argument to follow should probably remind the reader of the proof of Theorem89.

Proof: κ is a regular cardinal. Suppose that f : α → κ where α < κ. Then, byelementarity, j(f) : j(α) → j(κ). But j(α) = α since α < κ, and j(f)(β) = j(f)(j(β)) =j(f(β)) = f(β) for all β < α since both β and f(β) are smaller than κ. This shows thatj(f) = f. In particular, sup f ′′α = sup j(f)′′α, so it is an ordinal fixed by j, and f is notcofinal in κ.

κ is uncountable. This is because ω and each natural number are definable, so they arefixed by j.

κ is limit. Otherwise, κ = λ+ for some λ, but j(κ) = (j(λ)+)M , the ordinal that, from thepoint of view of M, is the successor of j(λ). Since M ⊆ V, (j(λ)+)M ≤ j(λ)+. Since λ < κ,j(λ) = λ, and it follows that j(κ) ≤ λ+ = κ, contradicting that j(κ) > κ.

In fact, κ is strong limit. Suppose otherwise, and let ρ < κ be such that 2ρ ≥ κ. So wecan find A ⊆ P(ρ) and a bijection f : A→ κ.

Note that j(A) ⊆ PM (j(ρ)) = PM (ρ) ⊆ P(ρ). Suppose that x ∈ j(A). Then x ⊆ ρ andj(x) ⊆ ρ and for any β < ρ, β ∈ x iff j(β) = β ∈ j(x), so x = j(x). It follows that x ∈ Aand j(A) ⊆ A. On the other hand, if y ∈ A, then y = j(y) ∈ j(A), and we have shown thatj(A) = A.

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Also, j(f)(x) = j(f)(j(x)) = j(f(x)) = f(x) for any x ∈ A, since f(x) < κ. It follows thatj(f) = f, and we have shown that κ = ran(f) is fixed by j, contradiction. �

This is how things stood up for a while: We have shown that the critical point of anyembedding is inaccessible, and that any measurable is inaccessible.

Now the key idea of reflection shows up, and we can prove significantly more.

Theorem 104 Let j : V →M be elementary and let κ = cp(j). Then κ is the κ-th stronglyinaccessible cardinal.

Proof: κ is strongly inaccessible in V by Lemma 103 and therefore in M since M ⊆ V. Itfollows that, for any α < κ, there is in M some inaccessible larger than α and below j(κ),namely, κ. By elementarity, the inaccessibles below κ are unbounded in κ. �

The argument gives much more. For example, κ is inaccessible and limit of inaccessibles,so the same proof shows that κ is limit of inaccessibles that are limit of inaccessibles thatare limit of inaccessibles . . . Even more is true, i.e., we can diagonalize and continue. Thisrequires a new key idea.

Theorem 105 (Keisler) If j : V →M is elementary, then κ = cp(j) is measurable.

Proof: LetU = {X ⊆ κ : κ ∈ j(X)}.

Then U is a nonprincipal ultrafilter over κ:

1. If X ⊆ Y ⊆ κ and X ∈ U then Y ∈ U since j(X) ⊆ j(Y ).

2. U is closed under intersections since j(A ∩B) = j(A) ∩ j(B).

3. ∅ = j(∅), so ∅ /∈ U .

4. If A ⊆ κ then j(κ) = j(A) ∪ j(κ \A), so either A or κ \A is in U .

5. If α < κ, then κ /∈ {α} = j({α}) so U is nonprincipal.

In fact, U is κ-closed: Let γ < κ and let f : γ → U . Then j(⋂α<γ f(α)) =

⋂α<j(γ) j(f)(α).

But γ = j(γ) and therefore j(f)(α) = j(f)(j(α) = j(f(α)) for all such α. It follows thatκ ∈ j(

⋂α<γ f(α)). �

The ultrafilter found by Keisler is particularly nice, since it is normal: Using the samenotation as above, suppose that (Xα : α < κ) is a sequence of elements of U . Let f : κ→ Ube the map f(α) = Xα. Recall that 4αXα = {β < κ : ∀α < β (β ∈ Xα)}. Then κ ∈j(4αXα) iff for all α < κ, κ ∈ j(f)(α). But j(f)(α) = j(f(α)) = j(Xα), and we are done.

Notice that U extends the club filter. This follows either from general arguments aboutnormal filters, or simply by noticing that if C ⊆ κ is club, then κ ∈ j(C) because j(C)∩κ =C is unbounded in κ and j(C) is closed.

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(To see that if F is a normal filter on κ, that contains the cobounded sets then it is κ-complete, given γ < κ and a sequence (Xα : α < γ) extend it to a κ-sequence by settingXα = κ for all α ≥ γ, and notice that 4α<κXα = A ∪

⋂α<γ Xα, where A ⊆ γ. Since

κ \A ∈ F , it follows that⋂α<γ Xα ∈ F as claimed.)

Corollary 106 If κ is measurable, then there is a nonprincipal normal ultrafilter on κ(extending the filter of cobounded sets). �

Recall:

Definition 107 A cardinal κ of uncountable cofinality is Mahlo iff the set of inaccessiblecardinals below κ is stationary.

It follows from the definition that if κ is Mahlo then it is regular: Suppose that X ⊆ κ isunbounded and |X| < κ. Let C be the club of limit points of X \ |X|. Then C contains noinaccessibles, in fact, it contains no regular cardinals, contradiction. It also follows that κis strong limit, since it is a limit of strong limit (in fact, inaccessible) cardinals. This, κ isthe κ-th inaccessible, and an easy diagonal argument shows it is limit of inaccessibles thatare limits of inaccessibles, etc.

Corollary 108 If κ is measurable then it is the κ-th Mahlo cardinal.

Proof: If j : V → M is elementary and cp(j) = κ, then κ ∈ j(C) for any club C ⊆ κ. Itfollows by elementarity that C contains an inaccessible. Hence, κ is Mahlo. By elementarity,C contains a Mahlo, etc. �

Stronger conclusions can be derived by taking j : V → Ult(V,U), where U is normal. Thisis because of the following observation:

Lemma 109 Let U be a nonprincipal κ-complete ultrafilter over the measurable cardinal κ.Then the following are equivalent:

1. U is normal.

2. Whenever X ∈ U and f : X → κ is regressive, then there is ξ < κ such that

{α < κ : f(α) = ξ} ∈ U .

3. π([id]) = κ where [·] denotes (Scott) classes in V κ/U .

Proof: From normality of U we get a version of Fodor’s lemma just as before, so 1 implies2.

2 implies 3, since 2 says that if [f ]∈[id] then [f ] = [cξ] for some ξ < κ.

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3 implies 1, because given any X ⊆ κ, we have that κ ∈ iU (X) iff [id]∈[cX ] iff

{α < κ : id(α) ∈ cX(α)} ∈ U

iff X ∈ U , so U is the ultrafilter derived from the embedding j : V → Ult(V,U), which isnormal by the remark following Keisler’s Theorem 105. �

This is a good place to list a few structural facts about (well-founded) ultrapower embed-dings; I concentrate on the particular case of ultrafilters over cardinals, although essentiallythis suffices by Lemma 99:

Lemma 110 Let U be a κ-complete nonpricipal ultrafilter over the measurable cardinal κ,and let j : V → Ult(V,U) be the corresponding ultrapower embedding. To ease notation,write M instead of Ult(V,U). Then:

1. cp(j) = κ.

2. κM ⊆M, Vκ+1 ⊂M, and (κ+)M = κ+.

3. κ+M 6⊆M.

4. U /∈M so Vκ+2 6⊂M.

5. 2κ ≤ (2κ)M < j(κ) < (2κ)+.

Scott’s result that the existence of measurable cardinals implies that V 6= L is an immediatecorollary, as any elementary j : L→M must have M = L.

This result indicates that there are limits to the embeddings that can be represented bymeans of ultrapowers. For example, even though there is a great deal of resemblancebetween V and M = Ult(V,U), M is not closed under κ+-sequences and does not containVκ+2, and j(κ) is not really a cardinal. Any embedding with these additional propertieswould necessarily capture stronger large cardinal properties and would produce strongerreflection arguments than those once can do in our setting. For example, if Vκ+2 ⊂ M forsome elementary j : V →M with cp(j) = κ, then κ is measurable in M and therefore thereis a normal ultrafilter V over κ that concentrates on measurable cardinals, i.e.,

{α < κ : α is measurable} ∈ V.

The modern template for defining large cardinals stipulates the existence of embeddingsj : V →M with large resemblance between V and M and stronger “correctness” of j. It isnatural to wonder how much resemblance can be asked. Kunen showed, for example, thatM = V is impossible.

Proof: We already know from Corollary 100 that cp(j) ≤ κ. If α = cp(j) < κ, letf : κ → V be such that π([f ]) = α. Then {β < κ : f(β) < cα(β)} = {β : f(β) < α)} ∈ U .By κ-completeness, f =U cβ for some β < α, so j(β) = π([f ]) = α, and cp(j) ≤ β,contradiction. This proves 1.

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If A ∈ Vκ, then an easy argument as in the proof of Lemma 103 gives that A = j(A), soVκ = VM

κ . If A ⊆ Vκ, then it follows that A = j(A) ∩ Vκ and therefore Vκ+1 = VMκ+1 ∈M.

Note that any well-ordering of κ is a subset of κ× κ and can be identified with a subset ofκ. Since Vκ+1 ∈M, all these subsets are in M, and M computes κ+ correctly.

To show that κM ⊂ M, suppose more generally that j′′x = {j(a) : a ∈ x} ∈ M for somex and that y ⊆ M, where |y| ≤ |x|, and conclude that y ∈ M. This gives the result, sincej′′κ = κ.

To see the claim, say y = {π([fz]) : z ∈ x} and let h : j′′x→ y be the map h(j(z)) = π([fz]).It is easy to see that h is well defined. If h ∈M, then y ∈M also, since y = ran(h). To seethat h ∈ M, we find some F such that h = π([F ]). To do this, note that since j′′x ∈ M,there is some f such that π([f ]) = j′′x, and set F : κ→ V to be the function such that, forall α < κ, F (α) : f(α) → V is the map F (α)(z) = fz(α), whenever this makes sense, andF (α) = ∅ otherwise. It is easy to check that, indeed, π([F ]) = h. This proves 2.

On the other hand, j′′κ+ /∈M, which proves 3. To see this, note that if π([f ]) < j(κ+), thenf =U f ′ for some f ′ : κ→ κ+. By regularity of κ+, f ′ is bounded, so there is some α < κ+

such that π([f ]) < j(α). This shows that j′′κ+ is cofinal in j(κ+), so cf(j(κ+)) = κ+.

But j(κ+) > j(κ) and j(κ) > κ is a limit cardinal of M, so in fact j(κ) > κ+, sinceκ+ = (κ+)M . Hence, j(κ+) > κ+. We are done, since j(κ+) is regular from the point ofview of M, by elementarity.

Now we prove 5: First, P(κ) = PM (κ). Since M ⊆ V, it follows that 2κ ≤ (2κ)M . Sincej(κ) is, from the point of view of M, a strong limit cardinal larger than κ, we must have(2κ)M < j(κ). On the other hand, if π([f ]) < j(κ), then f =U f ′ for some f ′ : κ → κ,and there are only |κκ| = 2κ many such functions f ′, so only 2κ many ordinals below j(κ).Therefore, j(κ) < (2κ)+, and 5 follows.

This gives us 4: Since κM ⊂ M, we have κκ = (κκ)M . If U ∈ M, then we could correctlycompute, in M, the value of j(κ), which would prove that (2κ)M < j(κ) < ((2κ)+)M ,contradicting the fact that j(κ) is a cardinal in M. �

The results above are all classical. A good reference is Kanamori [16]. For the few modeltheoretic results we need, any standard reference would work, for example, Chang-Keisler[4].

Before we continue, I want to show a different proof of Lemma 110.2; I think it is cleanerthan the argument above; on the other hand, that argument will be useful below, in theproof of Kunen’s theorem.

Lemma 111 If κ is measurable, U is a κ-complete nonprincipal ultrafilter over κ, andjU : V →M is the corresponding ultrapower embedding, then κM ⊂M.

Proof: Recall that if π is Mostowski’s collapsing function and [·] denotes classes in V κ/U ,then M = {π([f ]) : f ∈ κV }. To ease notation, write 〈f〉 for π([f ]).

Let h : κ→M. Pick f : κ→ V such that for all α < κ, h(α) = 〈f(α)〉.

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Lemma 112 With notation as above, 〈f〉 = jU (f)(〈id〉) for any f : κ→ V.

Proof: For a set X let cX : κ→ V denote the function constantly equal to X. Since π isan isomorphism, Los’s lemma gives us that the required equality holds iff

{α < κ : f(α) = ((cf )(α))(id(α))} ∈ U ,

but this last set is just {α < κ : f(α) = f(α)} = κ. �

From the nice representation just showed, we conclude that 〈f(α)〉 = jU (f(α))(〈id〉) for allα < κ. But for any such α, jU (f(α)) = jU (f)(α) because cp(jU ) = κ by Lemma 110. Hence,h = (jU (f)(α)(〈id〉) : α < κ), which is obviously in M, being definable from jU (f), 〈id〉, andκ. �

The following was shown in the proof of Lemma 109, but it deserves to be isolated.

Lemma 113 If U is a normal nonprincipal κ-complete ultrafilter over the measurable car-dinal κ, then U = {X ⊆ κ : κ ∈ iU (X)}, i.e., we get back U when we compute the normalmeasure derived from the embedding induced by U . �

Finally, the construction in Lemma 99 and preceeding remarks is a particular case of amuch more general result.

Definition 114 Given f : I → J and an ultrafilter D over I, the projection f∗(D) of Dover J is the set of X ⊆ J such that f−1(X) ∈ D.

Clearly, f∗(D) is an ultrafilter over J.

Notice that if κ = add(D), (Xα : α < κ) is a partition of I into sets not in D, and f : I → κis given by f(x) = the unique α such that x ∈ Xα, then f∗(D) is a κ-complete nonprincipalultrafilter over κ. (Of course, κ = ω is possible.)

For a different example, let U be a κ-complete nonprincipal ultrafilter over the measurablecardinal κ, and let f : κ→ κ represent the identity in the ultrapower by U , 〈f〉 = κ. Thenf∗(U) is the normal ultrafilter over κ derived from the embedding induced by U .

Definition 115 Given ultrafilters U and V (not necessarily over the same set), say that Uis Rudin-Keisler below V, in symbols, U ≤RK V, iff there are sets S ∈ U , T ∈ V, and afunction f : T → S such that U � S = f∗(V � T ).

Theorem 116 Let U be an ultrafilter over a set X and V an ultrafilter over a set Y.Suppose that U ≤RK V. Then there is an elementary embedding j : V X/U → V Y /V suchthat j ◦ iU = iV .

Proof: Fix T ∈ U and S ∈ V for which there is a map f : S → T such that

U � T = f∗(V � S).

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Clearly, V X/U ∼= V T /(U � T ) as witnessed by the map [f ]U 7→ [f � T ]U�T , and similarlyV Y /V ∼= V S/(V � S), so it suffices to assume that S = Y and T = X.

Given h : X → V, let h∗ : Y → V be given by h∗ = h ◦ f. Then j([h]U ) = [h∗]V iswell-defined, elementary, and j ◦ iU = iV .

In effect, h =U h′ iff {x ∈ X : h(x) = h′(x)} ∈ U iff {y ∈ Y : h ◦ f(y) = h′ ◦ f(y)} ∈ Viff h∗ =V h′∗, where the second equivalence holds by assumption, and it follows that j iswell-defined.

If cAB denotes the function with domain A and constantly equal to B, then for any x,

j ◦ iU (x) = j([cXx ]U ) = [(cXx )∗]V = [cYx ]V = iV(x)

since (cXx )∗ = cYx by definition of the map h 7→ h∗. This shows that j ◦ iU = iV .

Elementarity is a straightforward modification of the proof of Lemma 99. �

One can show that Theorem 116 “very nearly” characterizes the Rudin-Keisler ordering,see for example Proposition 0.3.2 in Ketonen [17].

4.3 Large cardinals and GCH

With the preliminary technical machinery established, we are ready to show how the pres-ence of large cardinals in the universe may influence its cardinal arithmetic.

Theorem 117 (Scott) Let κ be a measurable cardinal, and let U be a normal nonprincipal,κ-complete measure on κ. If α < κ and

{ρ < κ : 2ρ ≤ ρ+α} ∈ U ,

then 2κ ≤ κ+α. In particular, a measurable cardinal cannot be the first counterexample toGCH. Similarly, if

{ρ < κ : 2ρ ≤ ρ+ρ} ∈ U ,

then 2κ ≤ κ+κ and if{ρ < κ : 2ρ < ρ+ρ} ∈ U ,

then 2κ < κ+κ.

Proof: Let j : V → M be the ultrapower embedding derived from U . If α < κ and{ρ < κ : 2ρ ≤ ρ+α} ∈ U , then M |= 2π([id]) ≤ π([id])+j(α), by Los’s lemma. But j(α) = αsince cp(j) = κ, and π([id]) = κ by Lemma 109, since U is normal.

This shows that, in M, 2κ ≤ κ+α. By Lemma 110, we know that P(κ) ∈M, so 2κ ≤ (2κ)M ,and (κ+α)M ≤ κ+α since M ⊆ V. It follows that 2κ ≤ κ+α in V.

The other two statements are shown similarly. �

The assumption that U is normal cannot be eliminated in the previous theorem. Forexample, Gitik showed that it is consistent to have a measurable cardinal κ and a stationary

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set S ⊂ κ such that 2ν = ν+ for all ν ∈ S but 2κ = κ++. On the other hand, if 2ν = ν+

holds in a club below κ, then it also holds at κ, since any normal ultrafilter extends theclub filter.

Definition 118 Denote by (ℵ′α : α ∈ ORD) the increasing enumeration of the fixed pointsof the ℵ function.

The following is the first result we obtain about fixed points of the aleph sequence:

Theorem 119 Suppose that κ is measurable and that λ = ℵ′κ+κ is strong limit. Then2λ < ℵ′(2κ)+ .

Proof: Let U be a κ-complete nonprincipal ultrafilter over κ, and let j : V → M =Ult(V,U) be the corresponding elementary embedding. The result follows from generalfacts about continuity points of j.

Lemma 120 With notation as above, assume that τ is a limit ordinal.

1. If cf(τ) = κ, then j(τ) > supα<τ j(α) while if cf(τ) 6= κ, then j(τ) = supα<τ j(α).

2. If τ is strong limit and cf(τ) 6= κ, then j(τ) = τ, while if cf(τ) = κ, then 2τ < j(τ).

Proof: To prove 1, we need to determine whether there is some function f : κ → λ suchthat supα<τ j(α) ≤ 〈f〉 < j(τ).

If cf(τ) = κ, let (τα : α < κ) be increasing and cofinal in τ, and take f(α) = τα for allα < κ.

If cf(τ) > κ, then any f : κ→ τ is bounded.

If cf(τ) < κ, the κ-completeness of U implies that there is some β < τ such that {α < κ :f(α) < β} ∈ U .To prove 2, notice first that τ is closed under j: If α < τ, then j(α) < |κα|+ < τ, where thefirst inequality follows from the fact that any ordinal below j(α) is 〈f〉 for some f : κ→ α.

The result now follows from item 1: If cf(τ) 6= κ, then j(τ) = supα<τ j(α) ≤ τ, but τ ≤ j(τ)always holds.

If cf(τ) = κ, note that τ ≤ supα<τ j(α) < j(τ) so, in particular, τ < j(τ). Also, 2τ = τκ,since τ is strong limit. Since κM ⊂ M, then τκ ≤ (τκ)M ≤ (τ j(κ))M . If τ = κ, that2κ < j(κ) was shown in Lemma 110. If κ < τ, then j(κ) < τ, and (τ j(κ))M < j(τ), sincej(τ) is strong limit in M. �

Let now λ = ℵ′κ+κ and suppose that λ is strong limit. Then, by Lemma 120.2, 2λ < j(λ).But j(λ) = (ℵ′j(κ)+j(κ))

M ≤ ℵ′j(κ)+j(κ) < ℵ′(2κ)+ since j(κ) < (2κ)+ by Lemma 110. �

Before we continue, let me remark that the continuity results above depend in an essentialway on the fact that the embedding j being considered is an ultrapower embedding, andthey do not need to hold for other elementary maps.

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As mentioned above, the embeddings j : V → M that measurability provides are ratherrestrictive since, for example, Vcp(j)+2 6⊂ M. The results above show that measurabilityinfluences cardinal arithmetic, so it is reasonable to expect that stronger large cardinalassumptions will provide us with stronger results. This is indeed the case, as the followingexample illustrates. In the next subsection, I will show another result along these lines,deriving a significantly stronger conclusion from the existence of a significantly strongercardinal.

Definition 121 Let α be an ordinal. A cardinal κ is α-strong iff there is an elementaryembedding j : V →M with cp(j) = κ and Vκ+α ⊂M.

κ is strong iff it is α-strong for all α.

Clearly, κ is measurable iff it is 0-strong iff it is 1-strong, but already 2-strength of κ ismore significant than measurability as it implies that U ∈M for any ultrafilter on κ, so inparticular the measurable cardinals below κ form a stationary set.

The standard presentation of strength adds the extra requirement that α < j(κ). This isnot necessary; in fact, if κ is (α+ 2)-strong as defined above, then it is (α+ 2)-strong underthe stronger requirement of this paragraph. This is a consequence of the following famousresult of Kunen:

Theorem 122 (Kunen) Suppose that j : V →M is elementary. Then V 6= M.

Proof: There are several nice proofs of this result. Kunen’s argument uses that any infiniteset x admits an ω-Jonsson function, that is, a map f : [x]ℵ0 → x such that whenever A ⊆ xhas the same size as x, then f ′′[A]ℵ0 = x. Here, as usual, [t]τ = {y ⊆ t : |y| = τ}. This is aresult of Erdos and Hajnal.

Zapletal has found a proof that uses a modicum of pcf theory, namely, Shelah’s result thatany singular cardinal λ admits a scale, that is, there is an increasing sequence

(λα : α < cf(λ))

of regular cardinals cofinal in λ, and a sequence

(fβ : β < λ+)

of functions such that fβ ∈∏α λα for all β < λ+; whenever β < γ < λ+, then fβ <b,cf(λ) fγ ;

and whenever f ∈∏α λα, then f <b,cf(λ) fβ for some β. Here, <b,cf(λ+) is the partial order

induced by the ideal of bounded subsets of cf(λ+), see Definition 61.

There are a few other arguments as well. I present here a proof due to Woodin. Allarguments begin by noticing that if

λ = supn<ω

jn(κ)

then λ = j(λ). Here, j0(κ) = κ and jn+1(κ) = j(jn(κ)). To see this, apply j to both sidesof the displayed equation:

j(λ) = supn<ω

j(jn(κ)) = λ.

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It follows that j(λ+) = λ+, because λ+ ≤ j(λ+) = (λ+)M ≤ λ+. Therefore, j′′λ+ is cofinalin λ+.

LetC = {ξ < λ+ : ξ = j(ξ) and cf(ξ) = ω}.

The same argument that shows that λ is fixed shows that C is unbounded: For any α < λ+

consider the sequence α+ κ, j(α+ κ), j2(α+ κ), . . . Its supremum is in C.

Also, C is ω-closed. (A set A ⊆ λ+ is ω-closed iff whenever α∩A is unbounded in α < λ+

and cf(α) = ω, then α ∈ A.) This is simply because the supremum of an ω-sequence of fixedpoints of j is necessarily fixed by j.

But any ω-closed unbounded subset S of an ordinal γ of uncountable cofinality contains therestriction to Sγω of a club, namely, the club of limit points of elements of S. In particular,this holds of C. It follows that if T ⊆ Sλ+

ω is stationary, then C ∩ T 6= ∅.Woodin’s argument now invokes Ulam’s result that every stationary subset of a successorcardinal τ can be split into τ many disjoint stationary sets, this is Theorem 35. Conse-quently, we can partition Sλ

+

ω = {α < λ+ : cf(α) = ω} into κ many stationary pieces. Letf : κ → P(λ+) be such a partition. By elementarity, j(f) : j(κ) → PM (λ+) is (in M)a partition of Sλ

+

ω into j(κ) many stationary pieces. The result will follow by deriving acontradiction from the assumption that j(f)(κ) is stationary in V.

For suppose that it is. It follows that j(f)(κ) ∩ f(α) is stationary for some α < κ, sincethe club filter on λ+ is λ+-complete and λ+ > κ. But then j(f)(κ) ∩ f(α) must meetC. Let ξ be in this intersection. Then ξ = j(ξ), ξ ∈ f(α) and ξ ∈ j(f)(κ). But thenξ = j(ξ) ∈ j(f(α)) = j(f)(α), so j(f)(α) ∩ j(f)(κ) 6= ∅, contradicting that j(f) is apartition. �

The argument can actually be tightened to prove the following two stronger results:

Theorem 123 (Kunen) There is no elementary j : V → M such that Vλ+2 ⊂ M, whereλ is the first fixed point of j past its critical point.

Proof: It is easy to code (in an absolute way) subsets of λ+ by elements of Vλ+2, soWoodin’s proof of Kunen’s Theorem 122 goes through for j : V → M, since the sets itmakes reference to are subsets of λ+, and we have a contradiction.

To see that there is such a coding, simply notice (as in the appendix to Chapter I) thatthere is a definable bijection between λ and λ × λ, that the map from a well-ordering ofλ to its order type gives via this bijection a (definable) surjection from P(λ) onto λ+, andthat the inverse of this surjection is a (definable) injection of P(λ+) into P(P(λ)) ⊂ Vλ+2.�

Theorem 124 (Kunen) If j : V → M is elementary and λ is the first fixed point of jpast κ = cp(j), then j[λ] /∈M.

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Proof: Suppose otherwise, and define an ultrafilter U over Z = P(λ) by

U = {A ⊆ P(λ) : j[λ] ∈ j(A)}.

Then U is a σ-complete ultrafilter, and we can form the ultrapower embedding

i = iU : V → N = Ult(V,U).

Lemma 125 cp(i) = κ, i(λ) = λ, i[λ] ∈ N and i[P(λ)] ∈ N.

Proof: Begin by noticing that j[λ] = 〈id〉. In effect: Let f : P(λ) → V be arbitrary suchthat 〈f〉 ∈ 〈id〉. This occurs iff {B ⊆ λ : f(B) ∈ B} ∈ U , which (by definition) holds iffj(f)(j[λ]) = j(γ) for some γ < λ. But then {B ⊆ λ : f(B) = γ} ∈ U , and 〈f〉 = j(γ). Weeasily conclude that j[λ] = 〈id〉, and 〈f〉 = j(f)(j[λ]) for all f.

Letting k : N →M be given by k(〈f〉) = j(f)(j[λ]) is now easily checked to be well-definedand elementary, and k ◦ iU = j.

Note now that if α ≤ λ, and f : P(λ)→ V is given by f(B) = ot(B ∩ α), then

〈f〉 = j(f)(j[λ]) = ot(j[λ] ∩ j(α)) = α.

It follows that k � (λ + 1) = id by the definition of k and the same argument. Therefore,i � (λ+ 1) = j � (λ+ 1). In particular, cp(i) = κ, i(λ) = λ, and i[λ] = j[λ] ∈ N.Finally, let X ⊆ λ. Then i(X) = i(

⋃α<λX ∩α) =

⋃α<λ i(X ∩α), since i(λ) = λ. Note that

i[λ] ∈ N implies that λN ⊂ N, by the proof of Lemma 99.2, so P(λ) ∈ N and i � Vλ ∈ N,since |Vλ| = λ. But then

i[P(λ)] = {⋃α<λ

i(X ∩ α) : X ∈ P(λ)} ∈ N,

as required. �

It follows from the lemma that 2λN ⊂ N, again by the proof of Lemma 99.2. But thenVλ+2 ⊂ N, since any element of Vλ+2 has size at most 2λ, and we have a contradiction byTheorem 123. �

Kunen’s result shows that the definition of strong cardinals given above coincides with thestandard one.

Corollary 126 A cardinal κ is strong iff for all α there is an elementary embedding j :V →M with cp(j) = κ, Vα ⊂M, and α < j(κ). In fact, if κ is (α+ 2)-strong, then there isan elementary j : V →M with critical point κ such that Vκ+α+2 ⊂M and κ+α+2 < j(κ).

Proof: Suppose that κ is (α + 2)-strong as witnessed by j : V → M. By Theorem 123,κ+α < λ, where λ is the first fixed point of j past κ, so κ+α (and therefore also κ+α+2)is strictly below jn(κ) for some n < ω.

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Extend j to proper classes by setting j(N) =⋃β j(N ∩ Vβ) for all classes N. Define Mk

for k ≤ n by M0 = V and Ml+1 = j(Ml), so M1 = M. Define jk for 1 ≤ k ≤ n by j1 = jand jl+1 = j(jl) ◦ j, so each jl : V → Ml is elementary, Vκ+α+2 ⊂ Mn, cp(jn) = κ andjn(κ) > κ+ α+ 2. �

Corollary 127 If κ is β+ 2-strong for all β < α, and GCH holds below κ, then it holds forall cardinals below κ+α. Thus if κ is strong and GCH holds below κ then it holds everywhere.

Proof: Let τ be a cardinal below κ+α. Then τ < κ+ρ+2 for some ρ < α. By Corollary 126,there is an embedding j : V →M such that cp(j) = κ, Vκ+ρ+2 ⊂M and κ+ ρ+ 2 < j(κ).Hence, GCH holds in M below j(κ); in particular it holds at τ. But P(τ) ∈ Vκ+ρ+2, so infact 2τ = τ+ in V. �

As mentioned above, it is consistent that GCH fails at a measurable. This is closely relatedto the failure of SCH : Prikry forcing allows us to change the cofinality of a measurable κto ω while preserving all cardinals and without adding any bounded subsets of κ. Hence,if 2κ > κ+, then after Prikry forcing we obtain a singular strong limit of cofinality ω with2κ > κ+, i.e., we have violated SCH.

Moreover, the consistency strength of both assumptions (the existence of a measurablewhere GCH fails and the failure of SCH) is the same: Both statements are equiconsistentover ZFC to the existence of a cardinal κ of Mitchell order κ++. This is a notion significantlystronger than measurability, but weaker than 2-strength.

Definition 128 The Mitchell order of a measurable cardinal κ is defined as follows:First, for normal ultrafilters U and V over κ, set U C V iff U ∈ Ult(V,V). It is easy to seethat this implies that iU (κ) < iV(κ), and therefore C is well-founded.

We define the Mitchell order of κ by

o(κ) = sup{‖U‖C + 1 : U is a normal nonprincipal κ-complete ultrafilter over κ}.

It follows that o(κ) ≥ 1 iff κ is measurable. One can easily check that C is transitive, so it isindeed a partial ordering. The assumption o(κ) = 2 implies that the measurable cardinalsbelow κ form a stationary set, but is stronger than this.

Homework problem 12. Assume that κ is 2-strong and determine o(κ).

One final remark is in order: In the arguments above, I have used proper classes freely,without worrying about whether this can be formalized in ZFC. This is indeed the case, sincejust as the existence of embeddings corresponds to the existence of σ-complete nonprincipalultrafilters, the embeddings discussed here can be coded via sequences of ultrafilters (and so,any mention of proper classes can be eliminated by referring to the (definable) ultrapowersfrom which they arise). Nevertheless, the right setting for Kunen’s theorem is a set theorythat allows classes, so it is perhaps more reasonable to think that we are arguing in either

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GBC or (even better) MKC. My personal opinion is that these matters are not mathematicalin nature and should therefore not worry us.

A good reference for the results of this subsection is Kanamori [16]. For an introduction tothe Rudin-Keisler order, see Comfort-Negrepontis [5].

4.4 Strongly compact cardinals and SCH

Definition 129 A cardinal κ is strongly compact iff it is uncountable, and any κ-complete filter (over any set I) can be extended to a κ-complete ultrafilter over I.

The notion of strong compactness has its origin in infinitary logic, and was formulated byTarski as a natural generalization of the compactness of first order logic. Many distinctcharacterizations have been found.

Let’s begin by defining the (first order) infinitary logics Lκλ.

Definition 130 Let κ and λ be nonzero cardinals or ∞. Given a set of relation, func-tion, and constant symbols (a language), the Lκ,λ formulas of the language are definedinductively by closing under the following clauses:

1. Atomic formulas are Lκ,λ formulas.

2. If φ is an Lκ,λ formula, so is ¬φ.

3. If X is a set of fewer than κ many Lκ,λ formulas, then∧X is also an Lκ,λ formula.

4. If X is a set of fewer than λ many variables, and φ is an Lκ,λ formula, then so is∃X φ.

One can formalize the above even further by, for example, starting with variables xα = (0, α)for each ordinal α, n-ary relation symbols Rα,n = (1, α, n) for all n < ω and all ordinals α,and n-ary function symbols fα,n = (2, α, n) for all n < ω and all ordinals α, and assigninga (well-founded) tree decomposition to each formula.

One then defines the corresponding semantic naturally. The well-foundedness of the treeassociated to each formula is used to guarantee that the obvious inductive definition takesinto account all formulas. For example, given a model M for the language under consid-eration, an assignment π : xα 7→ tα of elements of M to variables, a set X of fewer thanλ many variables, and an Lκ,λ formula φ, we say that M |= ∃X φ[π] holds iff there is afunction τ : X →M such that if π is the map that to xα assigns tα unless xα ∈ X, in whichcase π(xα) = τ(xα), then M |= φ[π].

One says that M |= ψ iff M |= ψ[π] for all assignments π.

One defines free and bound variables as usual, and the standard notions carry through:A sentence is a formula without free variables, a theory is a collection of sentences, anda theory Σ is satisfiable iff it has a model, i.e., a structure M in the language of Σ suchthat M |= φ for all φ ∈ Σ.

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Definition 131 An Lκ,λ theory Σ is ρ-satisfiable iff whenever T ⊆ Σ and |T | < ρ, thenT is satisfiable.

Hence the compactness theorem for first order logic (i.e., for Lω,ω) is the claim that if Σ isan Lω,ω theory that is finitely (i.e., ω-) satisfiable, then Σ itself is satisfiable.

A result of Keisler and Tarski showed that the corresponding notion of compactness forLκ,κ with κ > ω is of large cardinal character in that it at least implies measurability of κ.The argument naturally leads to the notion of a fine measure.

Definition 132 Given a cardinal κ and a set X, denote by Pκ(X) = [X]<κ the collectionof subsets of X of size strictly smaller than κ.

An ultrafilter U over Pκ(X) is fine iff it is κ-complete, nonprincipal, and for any x ∈ X,{σ ∈ Pκ(X) : x ∈ σ} ∈ U .If γ ≥ κ say that κ is γ-compact iff it is uncountable and there is a fine ultrafilter overPκ(γ).

Actually, the notion originally considered was not that of a fine measure, but a closelyrelated one:

Definition 133 Let κ ≤ λ be cardinals. An ultrafilter U over a set X is (κ, λ)-regulariff there is a family (Xα : α < λ) of elements of U such that for every S ⊆ λ of size κ,⋂α∈S Xα = ∅.

Lemma 134 Suppose that κ ≤ λ. Then there is a fine ultrafilter over Pκ(λ) iff there is aκ-complete (κ, λ)-regular ultrafilter.

Proof: Let U be a fine ultrafilter over Pκ(λ) and let Xα = {σ ∈ Pκ(λ) : α ∈ σ} for allα < λ. Then (Xα : α < λ) witnesses that U is (κ, λ)-regular.

Conversely, let D be a κ-complete (κ, λ)-regular ultrafilter as witnessed by the sequence(Xα : α < λ). Consider the projection f :

⋃αXα → Pκ(λ) given by

f(x) = {α < λ : x ∈ Xα}.

Then f∗(D �⋃αXα) is a fine ultrafilter over Pκ(λ). �

Lemma 135 If κ is strongly compact, then it is regular, and therefore measurable.

Proof: Recall that for any nonprincipal ultrafilter U , its additivity add(U), the first τsuch that U is not τ -complete, is either ω or a measurable cardinal. Let F be the filter ofcobounded subsets of κ+. This is a κ+-complete filter so, by assumption, it can be extendedto a κ-complete nonprincipal ultrafilter over κ+. If κ is singular, the additivity of thisultrafilter cannot be κ, but it is not smaller than κ, and it is not larger than κ+, so it wouldhave to be κ+. But, being a successor, κ+ is not measurable, contradiction. �

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Theorem 136 The following are equivalent about the cardinal κ :

1. κ is strongly compact.

2. κ is λ-compact for all λ ≥ κ.

3. κ is uncountable, and for all regular λ ≥ κ there is a uniform κ-complete ultrafilterover λ.

4. For all λ there is an elementary embedding j : V → M with cp(j) = κ such thatwhenever X ⊆ M and |X| ≤ λ then there is some Y ∈ M such that X ⊆ Y andM |= |Y | < j(κ).

5. For all λ there is an elementary embedding j : V → M with cp(j) = κ such thatκM ⊂M and there is a set X ∈M such that j′′λ ⊆ X and M |= |X| < j(κ).

6. κ is uncountable, and the compactness theorem holds for Lκκ : Any κ-satisfiable Lκ,κ-theory is satisfiable.

7. κ is uncountable, and for every λ ≥ κ there is a κ-complete, (κ, λ)-regular ultrafilter.

Proof: 1 implies 2. Let λ ≥ κ. The collection {{x ∈ Pκ(λ) : σ ⊆ x} : σ ∈ Pκ(λ)}generates a κ-complete filter by closing under supersets (because κ is regular, by Lemma135). By 1, there is a κ-complete ultrafilter over Pκ(λ) that extends it. This ultrafilter isclearly fine.

2 implies 5. We may assume that λ ≥ κ. Let U be a fine ultrafilter over Pκ(λ) and formthe ultrapower embedding j : V → M = Ult(V,U). Obviously (by considering the sets{x ∈ Pκ(λ) : α ∈ x} for α ∈ λ), U is not κ+-complete. It follows that cp(j) = κ. From, say,Theorem 116 and preceeding remarks, it follows that κORD ⊂ M. By choice, this impliesthat κM ⊂M.

The result follows, since we can take X = 〈id〉 : That j′′λ ⊆ X simply means that for allα < λ, {x ∈ Pκ(λ) : α ∈ x} ∈ U . That M |= |X| < j(κ) simply means that

Pκ(λ) = {x ∈ Pκ(λ) : |x| < κ} ∈ U .

5 implies 2. If j : V → M and X are as in 5, without loss, X ∈ PMj(κ)(j(λ)); otherwise,replace X with X ∩ j(λ).

DefineU = {A ⊆ Pκ(λ) : X ∈ j(A)}.

Then U is a fine ultrafilter over Pκ(λ).

2 implies 4. Again, take j : V → M = Ult(V,U) where U is a fine measure over Pκ(λ).Given X ⊆ M with |X| ≤ λ, say X = {〈fα〉 : α < λ}, let F : Pκ(λ) → V be given byF (σ) = {fα(σ) : α ∈ σ}. Then Y = 〈F 〉 is as required.

4 implies 1. Let F be a κ-complete filter over a set I and let λ = |F|. Take j : V → Mwitnessing item 4 for λ. Let Y ∈ M extend j[F ] and have size smaller than j(κ) in M.

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Then⋂

(j(F) ∩ Y ) 6= ∅, since in M it is an intersection of fewer than j(κ) members of thej(κ)-complete filter j(F). Let c be in this intersection, and define an ultrafilter by

U = {X ⊆ I : c ∈ j(X)}.

Then U is a κ-complete ultrafilter extending F .(Notice we have shown that 4 and 5 are equivalent, although neither property seems toimply the other directly.)

2 implies 6. Let Σ = {φα : α < λ} be a κ-satisfiable Lκ,κ theory. We may assume thatλ ≥ κ, or there is nothing to show. For any σ ∈ Pκ(λ) let Mσ |=

∧{φα : α ∈ σ}. Let U be

a fine ultrafilter over Pκ(λ), and consider the ultraproduct∏σ∈Pκ(λ)Mσ/U .

This is defined by a straightforward generalization of Definition 90: For f, g ∈∏σMσ set

f =U g iff {σ : f(σ) = g(σ)} ∈ U .

Let [f ] be the equivalence class of f under this equivalence relation. Now we do not needScott’s trick to make sense of [f ], since there are only set-many such functions f to beginwith. Let ∏

σ∈Pκ(λ)

Mσ/U

be the collection of these equivalence classes. Define RU , for R an n-ary relation symbol inthe language of Σ, by

([f1], . . . , [fn]) ∈ RU iff {σ : Mσ |= (f1(σ), . . . , fn(σ)) ∈ R} ∈ U .

Define fU , for f an n-ary function symbol in the language of Σ, by

fU ([g1], . . . , [gn]) = [g] iff {σ : Mσ |= f(g1(σ), . . . , gn(σ)) = g(σ)} ∈ U .

Finally, define cU , for c a constant symbol in the language of Σ, by c = [f ], where

Mσ |= f(σ) = c

for all σ.

It is easy to check that these definitions make sense (i.e., they do not depend on the rep-resentatives of the equivalence classes under consideration), and they turn

∏σ∈Pκ(λ)Mσ/U

into a structure in the language of Σ.

The proof of Los’s lemma generalizes to this setting, since U is κ-complete and, since U isfine, this implies that

∏σ∈Pκ(λ)Mσ/U |= φα for all α < λ.

(Notice that the same argument gives a proof of the compactness of Lω,ω and shows that,over ZF, the existence of nonprincipal ultrafilters over any infinite set guarantees the com-pactness of Lω,ω.)6 implies 1. Let F be a κ-complete filter over a set I. Consider the language that hasa constant symbol X for all X ⊆ I, a different constant symbol c, and a binary relationsymbol ∈. Let Σ be the theory in this language consisting of the union of the Lκ,κ-theory

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of (I ∪ P(I),∈, X)X⊆I and {c∈X : X ∈ F}. The κ-completeness of F ensures that Σ isκ-satisfiable, and 6 implies that it is satisfiable. Let M be a model of Σ and define anultrafilter U over I by

U = {X ⊆ I : M |= c∈X}.It is straightforward to check that U is κ-complete and extends F .(Notice that, over ZF, the same argument shows that compactness for Lω,ω implies (and istherefore equivalent to) the existence of nonprincipal ultrafilters over any infinite set.)

1 implies 3. Use 1 to extend the filter of cobounded subsets of λ to a κ-complete ultrafilterover κ, and notice that it is necessarily uniform.

2 is equivalent to 7. By Lemma 134.

3 implies 7. This is a theorem of Ketonen. Notice first that κ is regular. By induction onλ, one shows that if λ ≥ κ is regular then there is a κ-complete (κ, λ)-regular ultrafilter Dλover λ.

Notice that the base of the induction (λ = κ) is clear.

Definition 137 Let λ be regular. An ultrafilter D over λ is weakly normal iff wheneverf : λ \ {0} → λ is regressive, there is some α < λ such that {β < λ : f(β) < α} ∈ D.

First we check that we may strengthen the assumption on the ultrafilters given by 3:

Lemma 138 Let κ ≤ λ be uncountable regular cardinals. If D is a uniform κ-completeultrafilter over λ then there is a uniform κ-complete weakly normal ultrafilter U over λ suchthat U ≤RK D.

The following notion will prove useful:

Definition 139 Given a σ-complete ultrafilter V and an ordinal α, let

δVα = sup jV [α]

where, as usual, jV : V → Ult(V,V) is the ultrapower embedding.

If V is a σ-complete ultrafilter over λ and 〈f〉 = δVλ , we say that f is the first function ofV.

Proof: Let f be the first function of D. Then U = f∗(D) works. �

It follows that for all regular λ ≥ κ there is a weakly normal uniform κ-complete ultrafilterUλ over λ.

Definition 140 Let U and V be ultrafilters over sets X and Y, respectively. Then U ×V isthe ultrafilter over X × Y given by

A ∈ U × V iff {x ∈ X : {y ∈ Y : (x, y) ∈ A} ∈ V} ∈ U .

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It is straightforward to check that U × V in Definition 140 is indeed an ultrafilter.

Assuming that µ ≥ κ is regular and Dµ exists, we claim that we can take Dµ+ = Uµ+ ×Dµ(more carefully, this is an ultrafilter over µ+ × µ, but we obtain the desired ultrafilter overµ+ via a bijection.)

In effect, it is easy to check that this product is κ-complete and uniform. We need to verifythat it is (κ, µ+)-regular. This follows from a general result.

Lemma 141 Suppose λ is regular and D is a uniform σ-complete ultrafilter over λ. Supposethat Ult(V,D) |= cf(δDλ ) < j(κ). Then D is (κ, λ)-regular.

Proof: Let f be the first function of D, so Y = {α < λ : cf(f(α)) < κ} ∈ D. LetM = Ult(V,D).

For each α ∈ Y let Aα ⊆ f(α) be cofinal and of size smaller than κ, and let F : λ→ V bea function such that F (α) = Aα for all α ∈ Y, so 〈F 〉 is a cofinal subset of δDλ that, in M,has size smaller than j(κ).

By induction define disjoint intervals Iη = (ξη, µη) for η < λ as follows: Given {Iη : η < η′}for some η′ < λ, let ξη′ = (supη<η′ µη) + 1, and let µη′ be the least ordinal µ < λ such that〈F 〉 ∩ (j(ξη′), j(µ)) 6= ∅.Note that 〈F 〉 ∩ j(Iη) 6= ∅ for all η < λ, by construction, and therefore the sets

Xη = {α ∈ Y : Aα ∩ Iη 6= ∅}

are in D for all η < λ. However, if S ⊆ λ has size κ, then⋂η∈S

Xη = {α ∈ Y : ∀η ∈ S (Aα ∩ Iη 6= ∅)} = ∅,

because all the Aα have size smaller than κ and there are κ disjoint intervals Iη for η ∈ S.Hence, (Xη : η < λ) witnesses that D is (κ, λ)-regular. �

Notice that Lemma 141 is in fact an equivalence. In order to make use of this result, weneed to identify the first function for Uµ+ ×Dµ.

Definition 142 Let κ ≤ λ be regular uncountable cardinals. Let U be a weakly normalκ-complete uniform ultrafilter over λ. For each α < λ let Vα be a κ-complete ultrafilter oversome set Xα. The U-sum of the ultrafilters Vα is the ultrafilter D over

Γ = {(α, β) : α < λ, β ∈ Xα}

given byX ∈ D iff {α : {β : (α, β) ∈ X} ∈ Vα} ∈ U .

In particular, if Vα = V for all α, the above reduces to the product U × V.

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Definition 143 In the setting of Definition 142, let 〈gα〉Vα = δVαα for all α < λ, and letfλ : Γ→ V be the function

fλ(α, β) = gα(β).

Lemma 144 In the setting of Definition 143, 〈fλ〉D = δDλ .

In particular, if U = Uµ+ and Vα = Dµ for all α < µ+, then D = Dµ+ (up to a bijection),and fµ+ is the first function of Dµ+ .

Proof: By uniformity of U , 〈fλ〉D ≥ δDλ . Suppose that 〈g〉D < 〈fλ〉D, so

S = {α : {β : g(α, β) < gα(β)} ∈ Vα} ∈ U .

By definition of gα, for each α ∈ S there is tα < α such that

{β : g(α, β) < tα} ∈ Vα.

By weak normality of U , there is a fixed ξ < λ such that {α ∈ S : tα ≤ ξ} ∈ U . But then〈g〉D ≤ j(ξ) < δDλ . �

By a straightforward generalization of Lemma 141, we get:

Corollary 145 For D as in Definition 142, {α : Vα is (κ, cf(α))-regular} ∈ U iff {(α, β) :cf(gα(β)) < κ} ∈ D iff D is (κ, λ)-regular. �

In particular, Dµ+ is indeed (κ, µ+)-regular.

More generally, if the inductive assumption holds for all α < λ and λ is regular, we canchoose Vα to be a κ-complete (κ, cf(α))-regular ultrafilter over some set Xα, and define Dλto be the Uλ-sum of the ultrafilters Vα. Then Dλ is (κ, λ)-regular, and we are done. �

Corollary 146 Let κ be γ-compact for all γ ∈ [κ, τ ], where τ is strongly compact. Then κis strongly compact.

Proof: By the local nature of the proof of the equivalence of conditions 2 and 3 in Theorem136. �

In terms of cardinal arithmetic, our goal in this subsection is to prove the following theoremof Solovay:

Theorem 147 (Solovay) Assume that κ is strongly compact. Then SCH holds above κ,i.e., for all λ ≥ κ, λcf(λ) = 2cf(λ) + λ+.

The proof below is due to Matteo Viale and uses the notion of a covering matrix, that hasproved very useful as well in the presence of forcing axioms like PFA or MM. I have hadoccasion of using both the theory developed in the course of the proof of Theorem 136, andViale’s machinery, in my own research.

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Definition 148 Let κ be a cardinal. A precovering matrix for κ is an array D =(K(n, α) : n < ω, α < κ) of subsets of κ such that

α ⊆⋃n

K(n, α)

for all α < κ.

We say that a precovering matrix is nontrivial iff there is some β < κ such that

ot(K(n, α)) < β

for all n < ω and α < κ. The least such β is denoted βD.

Viale’s notion of a covering matrix satisfies several additional requirements. For our appli-cation, precovering suffices.

Lemma 149 Suppose that κ is a singular cardinal of cofinality ω. Then there is a nontrivialprecovering matrix D for κ+ with βD = κ.

Proof: Let (κn : n < ω) be a strictly increasing sequence of regular cardinals cofinal in κ,and for each nonzero η < κ+ let φη : κ→ η be surjective. Define the entries of D inductivelyby

K(n, α) =⋃{K(n, β) ∪ {β} : β ∈ φα[κn]}.

Clearly this works. In fact, for any n and all α, ot(K(n, α)) < κ+n . �

Definition 150 The covering property holds at κ, CP(κ), iff for any precovering matrixD for κ there is an unbounded set A ⊆ κ such that

[A]ℵ0 ⊆⋃n,α

[K(n, α)]ℵ0 .

Viale’s definition of the condition CP is more restrictive, but for our application, this moregeneral version suffices. Notice that CP(κ) says something about cardinal arithmetic at κ,at least if there are nontrivial precovering matrices for κ.

Corollary 151 If CP(κ) holds then

cf(κ)ℵ0 ≤ κ supρ<βD

ρℵ0

for any precovering matrix D for κ.

Proof: This is immediate from the displayed equation in Definition 150. �

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Theorem 152 (Viale) Suppose that κ is strongly compact. Then CP(λ) holds for all reg-ular λ ≥ κ.

Proof: Let λ ≥ κ be regular, and let D be a precovering matrix for λ. Let U be aκ-complete uniform ultrafilter over λ. For each α < λ and each n < ω let

Aαn = {β > α : α ∈ K(n, β)}.

Since⋃nA

αn = (α, λ) ∈ U , there is some nα such that Aαnα ∈ U for all α.

For each n < ω letAn = {α : Aαn ∈ U}.

Since⋃nAn = λ by the above, there is some n such that An ∈ U .

In particular, An is unbounded in λ, and we claim that An witnesses the covering propertyfor D : Let X ⊂ An be countable. Then (by definition of An) Aαn ∈ U for all α ∈ X, so⋂α∈X A

αn ∈ U . In particular, this set is nonempty. Let β ∈

⋂α∈X A

αn. Then (by definition

of Aαn) β ≥ sup(X) and α ∈ K(n, β) for all α ∈ X. Hence, X ⊆ K(n, α). �

Proof of Theorem 147: Let κ be strongly compact. We show that λcf(λ) = 2cf(λ) + λ+

for all λ ≥ κ by induction on λ. The result is clear if λ is regular, and follows by inductionif cf(λ) > ω, by Silver’s theorem. Assume now that cf(λ) = ω. Let D be a nontrivialprecovering matrix for λ+ with βD = λ.

By CP(λ+) and Corollary 151, λℵ0 ≤ (λ+)ℵ0 ≤ λ+ supτ<λ τℵ0 = λ+ since τℵ0 ≤ τ+ for allτ ∈ [κ, λ), by induction (by the local nature of the proof of Lemma 40). The result followsimmediately. �

Strongly compact cardinals are a bit of an oddity in the large cardinal hierarchy in thatthey are of really high consistency strength (for example, if κ is κ+-compact, then there areinner models of ZF+ADR that contain all the reals; this is significantly above in consistencystrength than, say, the existence of a proper class of strong cardinals). However, in termsof size they need not be too large; in fact, Magidor showed that it is consistent that thefirst strongly compact is also the first measurable cardinal.

In contrast with the situation from strong cardinals shown last lecture, the following is stillopen:

Question 25 (Woodin) Assume that κ is strongly compact and that GCH holds below κ.Does it follow that it holds everywhere?

For more on infinitary logic, see Barwise-Feferman [2]. Of particular interest is Part C(Infinitary languages), 269–441, and even more specifically, the two articles Nadel [20] andDickmann [7].

For the general theory of strongly compact cardinals, see Kanamori [16] and Ketonen [17].For more on the covering property, see Viale [25].

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5 PCF theory

To close the topic of cardinal arithmetic, this Section is a summary introduction to SaharonShelah’s pcf theory. Rather, it is just motivation to go and study other sources; there aremany excellent references available, and I list some below. Here I just want to give youthe barest of ideas of what the theory is about and what kinds of results one can achievewith it. All the results mentioned are due to Shelah unless otherwise noted. All the notionsmentioned are due to Shelah as far as I know.

Here are some useful references, in rough historical order: Burke-Magidor [3], Jech [12],Shelah [21], Shelah [22], Kojman [18], Abraham-Magidor [1], and Eisworth [8].

Pcf came out of Shelah’s efforts to extend and improve the Galvin-Hajnal results on powersof singulars of uncountable cofinality (see Section 3). His first success was an extension tocardinals of countable cofinality. For example:

Theorem 153 1. If ℵω is strong limit, then 2ℵω < ℵ(2ℵ0 )+ .

2. In fact, for any limit ordinal ξ, ℵ|ξ|ξ < ℵ(2|ξ|)+ .

3. If δ is limit and δ = α+ β, β 6= 0, then ℵcf(δ)δ < ℵα+(|β|cf(β))+ . �

Notice that this is an improvement of the Galvin-Hajnal bounds even if ξ has uncountablecofinality. As impressive as this result may be, Shelah felt that it was somewhat lackingin that the bound was not “absolute” since 2|ξ| can be arbitrarily large. The best knownresult of pcf theory remedies this issue:

Theorem 154 If δ is a limit ordinal and 2|δ| < ℵα+δ then ℵ|δ|α+δ < ℵα+|δ|+4 . �

These results are obtained by a careful study of the set products∏

a =∏κ∈a κ for sets of

cardinals a. The basic idea is the following: Let κ ≤ λ be cardinals, and let cov(λ, κ) be thecofinality of ([λ]κ,⊆), i.e., the minimal size of a collection of subsets of λ of size κ such thatany such subset is covered by one of the family:

cov(λ, κ) = min{|F| : F ⊆ [λ]κ,∀A ∈ [λ]κ∃B ∈ F (A ⊆ B)}.

Then clearly λκ = 2κ + cov(λ, κ) : On the one hand, λκ = |[λ]κ|; if F is cofinal in [λ]κ, thenany subset of λ of size κ is in

⋃{[B]κ : B ∈ F}, and this set has size at most κκ|F|. On the

other hand, obviously 2κ + cov(λ, κ) ≤ λκ.Hence, to bound λκ, at least if 2κ < λ, reduces to understanding cov(λ, κ). Note that if2κ ≥ λ, then trivially λκ < ℵ(2κ)+ .

More generally:

Definition 155 Let λ, κ, θ, σ be cardinals such that λ ≥ θ ≥ σ, κ ≥ ℵ0, θ > 1, σ > 1, andeither κ ≥ θ, or else (κ+ = θ and cf(θ) < σ).

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Let cov(λ, κ, θ, σ) denote the least µ such that there is a family P of size µ of subsets of λ,each of size less than κ, such that whenever t ⊆ λ and |t| < θ, there is some P ′ ⊆ P with|P| < σ and t ⊆

⋃A∈P ′ A.

To understand these covering numbers, Shelah switches to the study of cofinalities of(∏

a,≤I) for different ideals I; so his results directly relate to the Galvin-Hajnal approach.The switch is justified in view of the following result (the cov vs pp theorem):

Definition 156 A partially ordered set (P,<) has true cofinality κ iff there is a <-increasing and cofinal sequence of length κ of elements of P. If this is the case, we writetcf(P,<) = κ.

The definition above is nontrivial. For example, cf(ω × ω1) = ω1, where P = ω × ω1 isordered by (n, α) < (m,β) iff n < m and α < β. However, there is no increasing ω1-sequence cofinal in P. Similarly, we can arrange that cf(P,<) is singular, although this isimpossible if P has a true cofinality.

Definition 157 If cf(λ) ≤ κ < λ then the pseudopower ppκ(λ) is the supremum of thetrue cofinalities of (

∏a, <I) where a is a set of at most κ regular cardinals below λ and

unbounded in λ, and I varies among the ideals over a extending the ideal of bounded setssuch that tcf(

∏a, <I) exists.

pp(λ) denotes ppcf(λ)(λ) and if Γ is a class of ideals, then ppΓ(λ) is defined just as pp(λ),but with only ideals in Γ being considered. If J is an ideal (extending the ideal of boundedsets), then ppJ(λ) denotes ppΓ(λ), where Γ is the class of ideals extending J.

Theorem 158 Let Γ = Γ(θ, λ) denote the set of (proper) λ-complete ideals over somecardinal θI < θ.

Suppose that σ is regular, σ > ℵ0, and λ ≥ κ ≥ θ > σ. Then:

1. sup{ppΓ(λ∗) : λ∗ ∈ [κ, λ], σ ≤ cf(λ∗) < θ}+ λ = cov(λ, κ, θ, σ) + λ.

2. Moreover, if µ = cov(λ, κ, θ, σ) is regular and larger than λ, then for some I ∈ Γ(θ, σ)and (λα : α ∈ dom(I)), we have µ ≤ tcf(

∏α λα, <I) and θ < λα ≤ λ.

3. In item 1, the inequality ≤ holds without needing to require that σ > ℵ0.

4. In item 1, if both sides of the equation are larger than λ, then the sup is in fact a max.�

For a proof, see Theorem 5.4 of Shelah [22]. Shelah notes the following conclusions:

Corollary 159 Suppose that λ is a singular cardinal of uncountable cofinality. Then

cov(λ, λ, (cf(λ))+, cf(λ)) = ppΓ((cf(λ))+,cf(λ))(λ). �

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Lemma 160 Suppose that λ ≥ κ ≥ θ > σ, cf(κ) ≥ σ, and either cf(θ) ≥ σ, or 2<θ < λ.Then

λ<θ = cov(λ, κ, θ, σ)<σ +∑α<κ

|α|<θ. �

Corollary 161 Assume that cf(λ) > ℵ0. Then

λθ = cov(λ, λ, (cf(λ))+, cf(λ)) = ppΓ((cf(λ))+,cf(λ))(λ),

unless λ ≤ 2θ, in which case λθ = 2θ, or cf(λ) > θ and λ ≤ ρθ for some ρ < λ, in whichcase λθ = supρ<λ ρθ. �

If I is a prime ideal (i.e., the set of complements of members of an ultrafilter), then <Iis a total quasi-order on

∏a, i.e.,

∏a/I, the collection of equivalence classes under the

relation (f =I g iff {κ : f(κ) 6= g(κ)} ∈ I), is linearly ordered under the relation [f ] <I [g]iff {κ : f(κ) ≥ g(κ)} ∈ I. In effect, in this case

∏a/I is just the ultrapower of a by the

ultrafilter dual to I. By Los’s lemma, this is a linearly ordered set.

This leads to the name pcf, which stands for possible cofinalities of ultraproducts of sets a

of regular cardinals.

5.1 A detour

Before we continue, a few words are in order on bounds for fixed points of the aleph function.For them, pcf theory does not apply directly, so different approaches are required. Forexample, the following extension of the ideas of Galvin-Hajnal that, combined with Theorem158, provides bounds for powers λκ for “many” λ even if they are fixed points.

Theorem 162 Suppose that κ is regular and (λi : i ≤ κ) is increasing and continuous, andlet J be a normal ideal on κ. Suppose that ppJ(λi) ≤ λ+h(i)

i for all i. Then ppJ(λκ) ≤ λ+‖h‖κ .

�

Definition 163 By induction, define the classes Cα of cardinals:

1. C0 is the class of infinite cardinals.

2. Cα+1 = {λ ∈ Cα : λ = ot(λ ∩ Cα)}.

3. Cλ =⋂α<λCα for λ limit.

For example, C1 is the class of fixed points of the aleph sequence.

Definition 164 By induction on i define the cardinals ℵiα(λ) as follows:

1. ℵ0α(λ) = λ+α.

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2. ℵi+1α (λ) is defined by induction on α as follows:

• ℵi+10 (λ) = λ.

• ℵi+1α+1(λ) = ℵiγ(λ), where γ = ℵi+1

α (λ) + 1.

• If δ is limit, ℵi+1δ (λ) = supα<δ ℵi+1

α (λ).

3. If i is limit, define ℵiα(λ) by induction on α :

• ℵi0(λ) = λ.

• ℵiα+1(λ) = supj<i ℵjα+1(ℵiα(λ)).

• If δ is limit, ℵiδ(λ) = supξ<δ ℵiξ(λ).

Note that the function ℵiα(λ) is monotone (nonstrictly) increasing in i, α, and λ. Also, forfixed i,

{ℵi+1δ (λ) : δ limit} = {µ : ℵiµ(λ) = µ}.

If ξ is limit,{ℵξα(λ) : α > 0} =

⋂i<ξ

{µ : ℵiµ(λ) = µ}.

If i > 0, Ci = {ℵiα(ℵ0) : α > 0, i and α is limit}.One then has, for example, the following bounds:

Theorem 165 Let nor denote the class of normal ideals.

1. If ζ < ω1, then ppnor(ℵζω1(i2(ℵ1))) < ℵζi2(ℵ1)+

(i2(ℵ1)).

2. If ζ < ω1, λ is the ω1-st member of Cζ and λ > i2(ℵ1), then ppnor(λ) is strictlysmaller than the i2(ℵ1)-st member of Cζ .

3. If cf(δ) = ω1, ℵδ > 2ℵ1 and ℵδ is not a limit of weakly inaccessible cardinals, thenthere are no weakly inaccessible cardinals in (ℵδ,ppnor(ℵδ)]. �

Shelah found several additional results about bounds for powers of fixed points. However,there are inherent limitations to these results, as illustrated by the following:

Theorem 166 (Gitik) If GCH holds, κ is a strong cardinal, κ+ < µ, and there are noinaccessibles above κ, then there is a forcing extension that preserves cofinalities ≥ κ, andwhere the following hold:

1. κ is the first member of Cω,

2. 2κ = µ+, and

3. GCH holds below κ. �

The understanding of cardinal arithmetic at fixed points is still severely more limited thanat other cardinals.

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5.2 pcf

Definition 167 Let a be a (nonempty) set of cardinals.

1. A cardinal λ is a possible cofinality of∏

a iff there is an ultrafilter U over a suchthat tcf(

∏a/U) = λ.

2. pcf(a) is the set of possible cofinalities of∏

a.

3. J<λ(a) = {b ⊆ a : tcf(∏

a/D) < λ for all ultrafilters D over a that contain b}.

Clearly, J<λ is a (not necessarily proper) ideal. By considering principal ultrafilters, a ⊆pcf(a). One easily verifies that a1 ⊆ a2 implies pcf(a1) ⊆ pcf(a2) and that pcf(a ∪ b) =pcf(a) ∪ pcf(b).

More interesting results are obtained assuming that a is not “too large” relative to itsmembers.

Theorem 168 Assume that a is a set of regular cardinals and that |a|+ < min(a). Then,for any ultrafilter D over a, cf(

∏a/D) < λ iff J<λ(a) ∩ D 6= ∅. �

It follows that (for a as in the theorem) the sequence

(J<λ(a) : λ ∈ ORD)

is increasing and continuous. Note that if κ = |∏

a|, then P(a) = J<κ+(a), so the displayedsequence only requires λ ≤ κ+.

It also follows that in the definition of pp(λ) one can restrict oneself to prime ideals, providedthat λ < ℵλ.

Corollary 169 With a as in Theorem 168:

1. |pcf(a)| ≤ 2a.

2. pcf(a) has a largest element.

The number of ultrafilters over an infinite set I is 22|I| , so the bound on the size of pcf isnontrivial.

Proof: 1. If λ ∈ pcf(a), then J<λ+ ) J<λ. Since the sequence of ideals J<τ is increasingand continuous, and all are contained in P(a), the bound follows.

2. If κ = |∏

a| then J<κ+ = P(a), so there is a least λ such that a ∈ J<λ+ . By continuity ofthe sequence of ideals J<τ , it must be the case that a /∈ J<λ, so λ is regular and the largestmember of pcf(a). �

Stronger results are obtained when a is an interval of regular cardinals:

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Theorem 170 Suppose that a is an interval of regular cardinals, without a largest element,a = [min(a), sup(a))∩REG, and suppose that |a| < min(a). Then |

∏a| is the largest element

of pcf(a) and pcf(a) is also an interval of regular cardinals. �

It follows immediately that ℵ|δ|δ < ℵ(2|δ|)+ : If 2|δ| ≥ ℵδ this is clear. Otherwise, one cantake a = [(2|δ|)+,ℵδ) ∩ REG in Theorem 170.

It also follows that, for example, if c < ℵω, then ℵℵ0ω is regular.

By Theorem 170, pp(ℵω) = ℵℵ0ω =∏n>1 ℵn = max(pcf({ℵn : n > 1})), and therefore the

bound c < ℵω ⇒ ℵℵ0ω < ℵω4 follows from the following theorem:

Theorem 171 If δ is limit and δ < ℵα+δ, then pp(ℵα+δ) < ℵα+|δ|+4 . �

In general, for a a “short” interval of regular cardinals, |pcf(a)| < |a|+4. The main openproblem in the area is whether |pcf(a)| > |a| is even possible. It has been recently shownby Velickovic and Er-rhaimini that Shelah’s proof cannot be improved to give the bound

|pcf(a)| < |a|+2.

More specifically, if a is an interval of regular cardinals such that |a| < min(a), then the pcfoperator has the following properties for any X,Y ⊆ pcf(a) :

1. X ⊆ pcf(X), pcf(X) ∪ pcf(Y ) = pcf(X ∪ Y ), pcf(pcf(X)) = pcf(X).

2. If γ ∈ pcf(X), then there exists X ′ ⊆ X with |X ′| = |a|, such that γ ∈ pcf(X ′).

3. pcf(X) has a maximal element.

4. If ν < max pcf(a) is a singular cardinal of uncountable cofinality then there exists aclub C ⊆ ν such that max pcf({λ+ : λ ∈ C}) = ν+.

Abstractly, properties 1–4 suffice to show that |pcf(a)| < |a|+4. Essentially, the propertiesallow one to identify pcf with a topological closure operator, and one can then argue directlyabout the corresponding topology. A height can be attached to these structures, and theEr-rhaimini-Velickovic theorem is that for a = {ℵn : n > 1} it is consistent that there aretopological structures satisfying properties 1–4 of height arbitrarily large below ℵ3.

5.3 Kunen’s theorem revisited

Closely related to property 4 above is the following result about the existence of scales:

Theorem 172 For any singular cardinal λ there is an increasing sequence (λi : i ∈ cf(λ))of regular cardinals cofinal in λ such that tcf(

∏i λi, <b,cf(λ)) = λ+. �

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Zapletal used this result to obtain a nice proof of Kunen’s theorem that there are noembeddings j : V → V. His argument is as follows:

Suppose towards a contradiction that j : V → M is elementary and that if λ is the firstfixed point of j past its critical point, then j[λ] ∈ M. Fix (λi : i < ω) as in Theorem 172with λ0 > κ. Let F = (fα : α < λ+) be strictly <b,ω-increasing and cofinal in

∏i λi. Then

j(F ) is <b,ω-increasing and cofinal in j(∏i λi) =

∏i j(λi) in the sense of M. Since j[λ] is

cofinal in λ, then in M, j[F ] is also <b,ω-increasing and cofinal in∏i j(λi).

Now define a function g with domain ω by g(i) = sup(j[λ] ∩ j(λi)) = sup(j[λi]). Note thatsince j(λi) is regular in M and strictly larger than λi, then g(i) ∈ j(λi), so g ∈

∏i j(λi).

However, j(fα)(i) = j(fα(i)) < sup(j[λi]) = g(i) for all i < ω and all α < λ, so j[F ] is notcofinal in

∏i j(λi) after all, since g bounds it. Contradiction.

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