Math 731 Homework

Austin Mohr

June 14, 2012

1 Problem 15B1

Definition 1.1. A space X is said to be completely normal if every subspaceof X is normal.

Proposition 1.2. X is completely normal if and only if, whenever A and Bare subsets of X with A∩ClX(B) = ClX(A)∩B = ∅, then there are disjointopen sets U ⊃ A and V ⊃ B.

Proof. (⇒) Let A and B be given with A ∩ ClX(B) = ClX(A) ∩ B = ∅.Denote by Y the subspace X − (ClX(A)∩ClX(B)). Now, both A and B arecontained in Y . To see this, observe that

A ∩ (ClX(A) ∩ ClX(B)) = (A ∩ ClX(B)) ∩ ClX(A)= ∅ ∩ ClX(A)= ∅.

Similarly,

B ∩ (ClX(A) ∩ ClX(B)) = (ClX(A) ∩B) ∩ ClX(B)= ∅ ∩ ClX(B)= ∅.

Next, we claim that ClY (A) and ClY (B) are disjoint, closed sets in Y .For disjointness, obvsere that

ClY (A) ∩ ClY (B) ∩ Y ⊂ ClX(A) ∩ ClX(B) ∩ Y= ∅.

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For closedness, note that the closure of a set in a topological space is alwaysclosed in that space.

By the complete normality of X, we have that Y is normal. Hence, thereexist sets U and V that are open in Y (and so open in X) with U ⊃ ClY (A)and V ⊃ ClY (B). As A and B are both contained in Y , we further have thatClY (A) ⊃ A and ClY (B) ⊃ B. Therefore, U ⊃ A and V ⊃ B, as desired.

(⇐) Let Y be a subspace of X and let A and B be disjoint closed subsetsof Y . Observe that

A ∩ ClX(B) = A ∩B= ∅,

and similarly

ClX(A) ∩B = A ∩B= ∅.

By hypothesis, there are disjoint open subsets U and V of X with U ⊃ Aand V ⊃ B. It follows that the sets U ∩ Y and V ∩ Y are disjoint opensubsets of Y with (U ∩ Y ) ⊃ A and (V ∩ Y ) ⊃ B. Hence, Y is normal, andso X is completely normal.

2 Problem 15B3

Lemma 2.1. If M is metrizable and N ⊂M , then the subspace N is metriz-able with the topology generated by the restriction of any metric which gen-erates the topology on M .

Proof. Let τ be the topology on M generated by a metric ρ. Let σ be therelative topology on N and let ρN be the restriction of ρ to N . We show thatσ is generated by ρN .

Let O ∈ σ. It must be that O = N ∩ G for some G ∈ τ . Since M isgenerated by ρ, we know that G =

⋃x∈GBρ(x, �x), where �x > 0 may depend

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on x. Now,

O = N ∩G

= N ∩⋃x∈G

Bρ(x, �x)

=⋃x∈G

N ∩Bρ(x, �x)

=⋃

x∈N∩G

BρN (x, �x).

Hence, O is the union of open balls with respect to the metric ρN . Therefore,σ is generated by the ρN , as desired.

Proposition 2.2. Every metric space is completely normal.

Proof. By the lemma, any subspace of a metric space is itself a metric space.As every metric space is T4 (hence, normal), it follows that every subspace of ametric space is normal. That is, every metric space is completely normal.

3 Problem 16B2

Proposition 3.1. Any base for the open sets in a second countable space hasa countable subfamily that is a base.

Proof. Let X be a second countable space, B = {Bα | α ∈ Γ} be anybase for X, and C = {Ci | i ∈ N} be the countable base guaranteed bythe second countability of X. Our aim is to show that, for each i ∈ N,Ci can be represented as the union of countably-many members of B (callthis countable subcollection BCi). Since the union of countably-many setseach having countably-many members is again countable, the set {A | A ∈BCi for some i} will be a countable subfamily of B that is a base for X.

To finish the proof, let Ck ∈ C. Since B is a base for X, we can writeCk =

⋃i∈I Bi for some subset I of the indexing set Γ. Now, for each x ∈ Ck,

pick a set Bix ∈ B with x ∈ Bix and ix ∈ I. Since C is also a base for X, wecan find some Cix with x ∈ Cix ⊂ Bix . It follows that {Bix | x ∈ Ck} is acountable set (we chose one element of B for each element of the countableset C) whose union is Ci (by construction, every element of Ci is present inthe union and every Bix is contained in Ci).

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4 Problem 16B3

Proposition 4.1. Any increasing chain of real numbers that is well orderedby the usual order must be countable.

Proof. Let A be a set of real numbers that is well-ordered by the usual order.

We claim that, for each a ∈ A, there is na ∈ N such that(a, a+ 1

na

]∩A = ∅.

Observe that the truth of this claim implies the countability of A, as eachinterval will contain a distinct rational number. Suppose now, for the purposeof contradiction, that the claim does not hold. That is, there exists someb ∈ A such that, for all n ∈ N,

(b, b+ 1

n

]∩ A 6= ∅. It follows that the set

A \ {c ∈ A | c ≤ b} has no least element, which is a contradiction with thefact that A is well-ordered, thus completing the proof.

5 The One-Point Compactification:

Construction

The procedure used to to obtain the one-point compactification X∗ of alocally compact, non-compact Hausdorff space X can be applied to any spaceY . That is, Y ∗ = Y ∪ {p} with neighborhoods y ∈ Y unchanged in Y ∗ whileneighborhoods of p have the form {p} ∪ (Y − L), where L is a subset of Ywith compact closure. Y ∗ is called the Alexandroff extension of Y .

6 Problem 19A1

Proposition 6.1. The assignment of neighborhoods in Y ∗ described above isvalid.

Proof. Recall that one can define a topology on a set Y by specifying, foreach x ∈ Y , a set Ux (called the neighborhood system at x ) satisfying

1. If U ∈ Ux, then x ∈ U .

2. If U, V ∈ Ux, then U ∩ V ∈ Ux.

3. If U ∈ Ux, then there is a set V ∈ Ux such that U ∈ Uy for each y ∈ V .

4. If U ∈ Ux and U ⊂ V , then V ∈ Ux.

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Using this approach, we call a set open whenever it contains a neighborhoodof each of its points.

Observe that, since Y is a topological space, there already exists a neigh-borhood system satisfying these requirements for each x ∈ Y . Let Up containall sets of the form {p} ∪ (Y − L), where L is a subset of Y with compactclosure. We proceed by verifying that Up satisfies the above requirements.

Claim 6.2. If U ∈ Up, then p ∈ U .

Proof. By definition, U = {p} ∪ (Y − L) (some L). Hence, p ∈ U .

Claim 6.3. If U, V ∈ Up, then U ∩ V ∈ Up.

Proof. Let U = {p} ∪ (Y − L) and V = {p} ∪ (Y −K), where L and K aresubsets of Y with compact closure. It follows that

U ∩ V = [{p} ∪ (Y − L)] ∩ [{p} ∪ (Y −K)]= {p} ∪ (Y − (L ∪K)).

Now, ClY (L∪K) = ClY (L)∪ClY (K), each of which is compact by assump-tion. As the union of two compact spaces is again compact, we conclude thatL∪K is indeed a subset of Y with compact closure. Hence, U ∩V ∈ Up.

Claim 6.4. If U ∈ Up, then there is a set V ∈ Up such that U ∈ Uy for eachy ∈ V .

Proof. Let U ∈ Up and choose any open V ∈ Up with V ⊂ U . Since V isopen, there exists, for all y ∈ V , an open subset G of V with y ∈ G andp /∈ G. Hence, G ∈ Uy. Since y ∈ Y , Uy satisfies the property that anysuperset of G is also contained in Uy. In particular, U ∈ Uy, as desired.

Claim 6.5. If U ∈ Up and U ⊂ V , then V ∈ Up.

Proof. Let U = {p} ∪ (Y −L) for some subset L of Y with compact closure.It follows that V = {p} ∪ (Y − K) with K ⊂ L, which in turn gives thatClY (K) ⊂ ClY (L). As a closed subset of compact space is compact, we havethat K has compact closure. Hence, V ∈ Up.

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Therefore, the neighborhood systems as defined indeed give a topology onY ∗. We conclude by showing that its relative topology on Y is the originaltopology. To that end, let σ = {G | G open in Y } and τ = {G ∩ Y |G open in Y ∗}. Evidently, σ ⊂ τ . We also have that, for any open G in Y ∗,

G ∩ Y = [{p} ∪ (Y − L)] ∩ Y (some L with compact closure in Y )= Y − L.

(Do we here require the stronger notion that L be compact rather than merelyhaving compact closure? For example, (0, 1) has compact closure in R, yetR− (0, 1) is not open.)

Pending the resolution of the above parenthetical remark, we will haveshown that σ = τ , and so the relative topology on Y is precisely the originaltopology.

7 Problem 19A4

Proposition 7.1. Y ∗ is Hausdorff if and only if Y is locally compact andHausdorff.

Proof. (⇒) Let x, y ∈ Y ⊂ Y ∗. Since Y ∗ is Hausdorff, we can find disjointopen neighborhoods U and V in Y ∗ with x ∈ U and y ∈ V . If neither U norV contains p, then U and V suffice to show that Y is Hausdorff. Otherwise,let it be that U = {p} ∪ (Y − L) for some L with compact closure in Y(it cannot be that both U and V are of this form, as this would violatedisjointness). Define the set U ′ = Y − ClY (L). We have that x ∈ U ′, y ∈ Vwith U ′ and V disjoint and open.

To see that Y is locally compact, let x ∈ Y be given. Since Y ∗ is Haus-dorff, there is an open set U of Y ∗ such that x /∈ U . Now, U = {p}∪ (Y −L)for some subset L of Y with compact closure. Hence, x ∈ L ⊂ ClY (L),which is compact. Since Y is Hausdorff and every point of Y has a compactneighborhood, it follows that Y is locally compact.

(⇐) Let x, y ∈ Y ∗. If neither x nor y is p, then the fact that Y isHausdorff implies that x and y can be put into disjoint open neighborhoods.Suppose now that y = p. Choose, by the local compactness of Y , somecompact neighborhood V ⊂ Y of x. It follows that IntY (V ) is an open setcontaining x. We also have that IntY (Y

∗−V ) = IntY ({p}∪ (Y −V )), whichis an open neighborhood (as it is a member of the base) containing p. Thus,

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for any x, y ∈ Y ∗, we have produced two disjoint open sets containing them.Hence, Y ∗ is Hausdorff.

8 Problem 16D1

Proposition 8.1. The Sorgenfrey line E is Lindelöf.

Proof. Let C be a basic open (in E) cover of R. That is, C = {[xα, yα) |xα, yα ∈ R, α ∈ Γ}. Let C ′ = {(xα, yα) | [xα, yα) ∈ C}. Finally, let A =

⋃C ′.

Evidently, C ′ covers A. Now, observe that C ′ contains open sets from theusual topology on R. Since R with the usual topology is second-countable,it is Lindelöf. Hence, there is a countable subcollection of C ′ (and so of C)covering A.

Next, we claim that A misses only countably-many points of R. To seethis, let x and y be distinct real numbers that are not elements of A. Withoutloss of generality, let x < y. Since C covers R, it follows that x and y areleft endponts of distinct intervals of C. Denote these intervals by [x, x′) and[y, y′), respectively. Evidently, [x, x′) and [y, y′) are disjoint (if not, theny ∈ (x, x′), and so y ∈ A). Hence, we can identify with each of x and y adistinct rational number belonging to (x, x′) and (y, y′), respectively. Thus,A misses only countably-many elements of R, and so there is a countablesubcollection of C covering A.

We have shown that C admits a countable subcover for both A and R\A.Taking these subcovers together yields the desired subcover of R.

Corollary 8.2. The Sorgenfrey line is a T4-space.

Proof. Since the Soregenfrey line is regular and Lindelöf, it is normal. Fur-thermore, it is T1. Taken together, we have that the Sorgenfrey line is T4.

9 Problem 16D2

Proposition 9.1. Every uncountable subset of a Lindelöf space contains anaccumulation point.

Proof. We establish the contrapositive. To that end, let A be a subset of aLindelöf space X such that A contains no accumulation points. It followsthat, for each x ∈ X, there is an open neighborhood Ux of x with |Ux ∩ A|

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finite. Let C denote the collection {Ux | x ∈ X}. Since C is an open coverof the Lindelöf space X, C admits a countable subcover C ′ of X (and so ofA). Since |A ∩ U | is finite for all U belonging to the countable collection C ′,it follows that A is countable, thus establishing the contrapositive.

10 Extra Problem

Proposition 10.1. A nonempty collection of subsets of X is a subbase forsome filter on X if and only if it has the finite intersection property.

Proof. (⇒) We establish the contrapositive. That is, supposeA is a nonemptycollection of subsets of X that fails to satisfy the finite intersection property.It follows that there are sets A1, . . . , Ak ∈ A such that

⋂ki=1Ai = ∅. Consider

now the collection C = {⋂ni=1 Ai | Ai ∈ A for all i}. Evidently, ∅ ∈ C, and

so it cannot be a base for any filter on X (as no filter contains the emptyset). Hence, A is not a subbase for any filter on X, thus establishing thecontrapositive.

(⇐) Let S be a nonempty collection of subsets of X having the finiteintersection property. Define B to be the collection

{F ⊂ X |⋂S ′ ⊂ F for some finite subcollection S ′ of S}.

Observe first that B is a nonempty collection of nonempty subsets of X, asS satisfies the finite intersection property. We further claim that B is in facta base for a filter on X, and so S will be a subbase for this filter. To thatend, let B1, B2 ∈ B. By definition of B,

B1 ∩B2 =(⋂S1)∩(⋂S2)

(some S1,S2 ⊂ S)

=⋂

(S1 ∪ S2) .

Evidently, S1∪S2 is a finite subcollection of S, as both S1 and S2 are. Hence,B1 ∩B2 ∈ B, and so B is a base for the filter {F ⊂ X | B ⊂ F for some B ∈B}.

11 Problem 12B1

Proposition 11.1. The intersection of any number of filters on X is a filteron X.

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Proof. Let F be the intersection of filters Fα for α ∈ Γ. We claim that F isitself a filter.

First, observe that ∅ /∈ F and X ∈ F , since this is the case for each ofthe Fα.

Next, let F1 and F2 be elements of F . It follows that F1 and F2 belongto Fα for all α ∈ Γ, and so F1 ∩ F2 ∈ Fα for all α ∈ Γ. Hence, F1 ∩ F2 ∈ F .

Finally, let F ∈ F and let G be any set containing F . Since F ∈ Fα forall α ∈ Γ, it follows that G ∈ Fα for all α ∈ Γ. Hence, G ∈ F .

Taken together, the above three observations verify that F is indeed afilter, as desired.

12 Problem 12B3

Proposition 12.1. Every filter is the intersection of the ultrafilters contain-ing it.

Proof. Let F be a filter and let U =⋂α∈Γ Uα, where the Uα constitute all

ultrafilters containing F . Certainly, F ⊂ U . Suppose, for the purpose ofcontradiction, that this inclusion is strict. It follows that there is some set Awith A /∈ F and A ∈ U . Consider the filter G generated by F ∪ {Ac} (thisis indeed a filter, since A /∈ F and so ∅ /∈ G). We have that F ⊂ G ⊂ Uβ,for some β ∈ Γ. Now, since A ∈ U , A ∈ Uβ, but also Ac ∈ Uβ, which is acontradiction. Therefore, it must be that F = U .

13 Problem 16C1

Definition 13.1. Let ℵ be any cardinal number. A space X is said to havecaliber ℵ if, whenever U is a family of open subsets of X with |U| = ℵ, thenthere exists a subfamily V of U with |V| = ℵ and

⋂V 6= ∅.

Proposition 13.2. Every separable space has caliber ℵ1.

Proof. Let X be a separable space having countable dense subset D. Let Ube a collection of open subsets of X with |U| = ℵ1. As D is dense in X,each open set of U contains a point of D. It follows that there exists x ∈ Dbelonging to uncountably-many open sets of U . (Were this not the case, eachof the elements of the countable set D appears in only countably-many open

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sets of U , which would imply that U is countable.) Let V = {U ∈ U | x ∈ U}.Evidently, |V| = ℵ1 and x ∈

⋂V . Therefore, X has caliber ℵ1.

14 Problem 16C3

Definition 14.1. We say X satisfies the countable chain condition providedevery family of disjoint open subsets of X is countable.

Proposition 14.2. If X has caliber ℵ1, then X satisfies the countable chaincondition.

Proof. We establish the contrapositive. To that end, suppose that X does notsatisfy the countable chain condition. It follows that there is an uncountablefamily U of disjoint open sets of X. By the Axiom of Choice, we may assumethat |U| = ℵ1 (if |U| > ℵ1, we use the Axiom of Choice to reduce it to acollection of size ℵ1). Now, since U is a collection of disjoint sets, it followsthat

⋂U = ∅, and so X does not have caliber ℵ1.

15 Problem 16D3

Proposition 15.1. A regular space is Lindelöf if and only if each open coverhas a countable subcollection whose closures cover (i.e. has a countabledense subsystem).

Proof. (⇒) Let X be a Lindelöf space. Any open cover of X admits acountable subcover U . Evidently, U ⊂ U for all U ∈ U , and so the set{U | U ∈ U} is also a countable cover of X.

(⇐) Let X be a regular space and let U be an open cover of X. For eachx ∈ X, let Ux be an open set in U containing x. Since X is regular, there isan open set Vx containing x with Vx ⊂ Ux. Let V be the collection of sets Vxfor all x ∈ X. Evidently, V is also a cover of X. By hypothesis, V admits acountable subcollection V ′ = {Vi | i ∈ N} such that⋃

i∈N

Vi = X.

By construction, each Vi is contained in some open set (call it Ui) belongingto U . It follows that the collection {Ui | i ∈ N} is a countable subcollectionof U covering X. Therefore, X is Lindelöf.

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16 Problem 17C1

Definition 16.1. A compact space X is maximal compact provided everystrictly finer topology on X is noncompact.

Proposition 16.2. A compact space X is maximal compact if and only ifevery compact subset is closed.

Proof. (⇒) We establish the contrapositive. To that end, let (X, τ) be acompact space having a compact subset K that is not closed. Furthermore,let B be a base for the open sets of X.

Now, since K is not closed, X −K is not open. Let τ ′ be the topologygenerated by τ ∪ {X −K}. Evidently, τ ′ is a strictly finer topology than τ .

We claim next that (X, τ ′) is compact. To that end, let C be an opencover (using sets from τ ′) of X. If X−K /∈ C, then only open sets of τ appearin the cover. Since X is compact under τ , we can find a finite subcover ofX. Suppose then that X − K ∈ C. Since K is compact under τ and K isdisjoint from X − K, we can find a finite subcover of K using elements ofC −{X −K}. Taking this cover of K together with X −K gives the desiredcover of X.

In either case, we conclude that (X, τ) is not maximal, thus establishingthe contrapositive.

(⇐) We establish the contrapositive. To that end, suppose that (X, τ) isnot maximal compact. Let τ ′ denote a strictly finer compact topology thanτ . Since τ ′ is strictly finer than τ , we can find a closed set F belonging to τ ′

but not belonging to τ . Since F is a closed (under τ ′) subset of a compactspace, F is compact under τ ′. Hence, every open cover of F by open setsof τ ′ admits a finite subcover. In particular, any open cover of F using onlyopen sets of τ admits a finite subcover, and so F is compact under τ . Hence,under τ , F is a compact subset of X that is not closed, thus establishing thecontrapositive.

17 Problem 17F4

Proposition 17.1. Let X1, X2, . . . all be first countable. The product∏Xn

is countably compact if and only if each Xn is countably compact.

Proof. (⇒) Let∏Xn be countably compact. That is, every countable open

cover of∏Xn admits a finite subcover. For a fixed k, let C be a countable

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open cover of Xk. For each open C ∈ C, let UC =∏Ui. where Uk = C and

Ui = Xi for i 6= k. Denote by D the collection {UC | C ∈ C}. Evidently, Dis a countable open cover of

∏Xn, which admits a finite subcover D′ by the

countable compactness of∏Xn. It follows that the collection of all Uk from

the open sets belonging to D′ (that is, the collection containing the factorcorresponding to Xk from each open set in D′) is a finite subcollection of Ccovering Xk, and so Xk is countably compact. As k was arbitrary, it followsthat each of the product spaces is countably compact.

(⇐) As every countably compact space is sequentially compact, it sufficesto show that

∏Xn is sequentially compact. To that end, let 〈xn〉 be a se-

quence in the product space. For each i, 〈xn(i)〉 is a sequence in Xi. Since Xiis first countable and countably compact, it is sequentially compact. Hence,〈xn(i)〉 has a convergent subsequence 〈xnk(i)〉. (This is not itself enough toconclude that the sequence in the product has a convergent subsequence, as,in general, the product of sequentially compact spaces need not be sequen-tially compact. Hence, I have to make more explicit use of first countability,but I do not see how to do this.)

18 Problem 17F5

Proposition 18.1. Continuous images of countably compact spaces are count-ably compact.

Proof. Let f be a continuous function from the countably compact space Xonto Y and let C be a countable open cover of Y . By the continuity of f , theset {f−1(C) | C ∈ C} is an open cover of X. Since X is countably compact,{f−1(C) | C ∈ C} admits a finite subcover {f−1(C1), . . . , f−1(Cn) | Ci ∈ C}of X. Since f is onto, it follows that the Ci cover Y . Hence, there is a finitesubcover of the countable cover C, and therefore Y is countably compact.

Proposition 18.2. Closed subspaces of countably compact spaces are count-ably compact.

Proof. Let X be countably compact, F a closed subspace of X, and C acountable open cover of F . For each C ∈ C, there is an open set UC ∈ Xsuch that UC ∩ F = C. Now, since F is closed, X − Y is open in X. Hence,{X − F} ∪ {UC | C ∈ C} is an open cover of X, which by the countablecompactness of X admits a finite subcover {X−F}∪{U1, . . . , Un | Ui ∈ X}.

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It follows that {U1 ∩ F, . . . , Un ∩ F} is a finite subcollection of C coveringF .

19 Problem 22F1

Lemma 19.1. Let x and y be real numbers both strictly greater than −1. Ifx ≤ y, then x

1+x≤ y

1+y.

Proof. We have that

x ≤ yx+ xy ≤ y + xy

x(1 + y) ≤ y(1 + x)x

1 + x≤ y

1 + y(since x, y strictly greater than −1).

Remark 19.2. One can think of the previous lemma as follows: Supposewe x mL of salt and y mL of salt and we add them each to separate 1 mLcontainers of water. The ratio x

1+xmeasures the concentration of salt in

the first container and y1+y

measures the concentration of salt in the secondcontainer. Since x is a smaller amount of salt than y, then surely the firstcontainer is less concentrated than the second. That is, x

1+x≤ y

1+y.

Proposition 19.3. If ρ is a metric on X, then both

ρ1(x, y) = min{1, ρ(x, y)}

and

ρ2(x, y) =ρ(x, y)

1 + ρ(x, y)

are metrics equivalent to ρ on X.

Proof. We first establish that ρ1 and ρ2 are indeed metrics on X. In whatfollows, let x, y, z be arbitrary points in X.

For the positive definiteness of ρ1, observe that

ρ1(x, y) = min{1, ρ(x, y)}≥ 0 (since ρ is positive definite),

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with equality if and only if ρ(x, y) = 0, which occurs if and only if x = y(since ρ is positive definite).

For the symmetry of ρ1, observe that

ρ1(x, y) = min{1, ρ(x, y)}= min{1, ρ(y, x)} (since ρ is symmetric)= ρ1(y, x).

For the subadditivity of ρ1, observe that

ρ1(x, y) = min{1, ρ(x, y)}≤ min{1, ρ(x, z) + ρ(z, y)} (since ρ is subadditive)≤ min{1, ρ(x, z)}+ min{1, ρ(z, y)}= ρ1(x, z) + ρ(z, y).

Therefore, ρ1 is a metric on X.For the positive definiteness of ρ2, observe that

ρ2(x, y) =ρ(x, y)

1 + ρ(x, y)

≥ 0 (since ρ is positive definite),

with equality if and only if ρ(x, y) = 0, which occurs if and only if x = y(since ρ is positive definite).

For the symmetry of ρ2, observe that

ρ2(x, y) =ρ(x, y)

1 + ρ(x, y)

=ρ(y, x)

1 + ρ(y, x)(since ρ is symmetric)

= ρ2(y, x).

For the subadditivity of ρ2, observe that

ρ(x, y) ≤ ρ(x, z) + ρ(z, y) (since ρ is subadditive)ρ(x, y)

1 + ρ(x, y)≤ ρ(x, z) + ρ(z, y)

1 + ρ(x, z) + ρ(z, y)(by lemma).

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The lefthand side of the inequality is precisely ρ2(x, y), and so we have that

ρ2(x, y) ≤ρ(x, z) + ρ(z, y)

1 + ρ(x, z) + ρ(z, y)

=ρ(x, z)

1 + ρ(x, z) + ρ(z, y)+

ρ(z, y)

1 + ρ(x, z) + ρ(z, y)

≤ ρ(x, z)1 + ρ(x, z)

+ρ(z, y)

1 + ρ(z, y)(since ρ is positive definite)

= ρ2(x, z) + ρ2(z, y).

Therefore, ρ2 is a metric on X.We show next that ρ and ρ1 are equivalent by showing (X, ρ) and (X, ρ1)

have the same open sets. To that end, consider the basic open set Bρ(x, �)of (X, ρ). We have that

Bρ(x, �) =⋃

y∈Bρ(x,�)

Bρ(y, �y),

where each �y depends on y. By definition of ρ1, we have, for each y,

Bρ1(y, �y) ⊂ Bρ(y, �y).

Taken together, we see that

Bρ(x, �) =⋃

y∈Bρ(x,�)

Bρ1(y, �y),

and so (X, ρ) ⊂ (X, ρ1).Next, consider the basic open set Bρ1(x, �) of (X, ρ1). As before,

Bρ1(x, �) =⋃

y∈Bρ1 (x,�)

Bρ1(y, �y)

where each �y depends on y. In particular, we may insist that 0 < �y ≤ 1 forall y. With this restriction, we have, for each y,

Bρ1(y, �y) = Bρ(y, �y).

Taken together, we see that

Bρ1(x, �) =⋃

y∈Bρ1 (x,�)

Bρ(y, �y),

15

and so (X, ρ1) ⊂ (X, ρ).We show next that ρ and ρ2 are equivalent by showing that (X, ρ) and

(X, ρ2) have the same open sets. That (X, ρ) ⊂ (X, ρ2) follows precisely asin the proof that (X, ρ) ⊂ (X, ρ1), as Bρ2(x, �) ⊂ Bρ(x, �) for all x ∈ X and� > 0.

Next, consider the basic open set Bρ2(x, �) of (X, ρ2). As before,

Bρ2(x, �) =⋃

y∈Bρ2 (x,�)

Bρ2(y, �y),

where each �y depends on y. Observe now that, for all y,

ρ2(y, z) = �y ⇒ρ(y, z)

1 + ρ(y, z)= �y

⇒ ρ(y, z) = �y(1 + ρ(y, z))

⇒ ρ(y, z) = �y1− �y

,

where we insist that 0 < �y < 1 for all y. Taken together, we see that

Bρ2(x, �) =⋃

y∈Bρ2 (x,�)

Bρ(y,�y

1− �y),

and so (X, ρ2) ⊂ (X, ρ).

20 Problem 22F2

Proposition 20.1. Every metric generating the topology of a compact metriz-able space is bounded.

Proof. We proceed by establishing the contrapositive. To that end, let ρbe an unbounded metric generating a topology on a set X. Define, for alln ∈ N, Bn to be the open set B(x, n) for some fixed x ∈ X. The collection{Bn | n ∈ N} covers X but admits no finite subcover. To see this, let C beany finite subcollection of {Bn | n ∈ N}. By construction,

⋃C ⊂ B(x,M),

where M denotes the maximum n such that Bn ∈ C. Since ρ is unbounded,there is a point y ∈ X with ρ(x, y) > M , and so y /∈ B(x,M). Hence, C doesnot cover X, and so X is not compact, thus establishing the contrapositive.

16

21 Extra Problem 1

Lemma 21.1. Every point of a closed ball in an ultrametric space is a center.

Proof. Let (X, ρ) be an ultrametric space, and let B(a, �) denote the closedball around a ∈ X of radius � > 0 (i.e. the set {y ∈ X | ρ(a, y) ≤ �}). Letb ∈ B(a, �) and let x ∈ B(b, �). It follows that

ρ(a, x) ≤ max{ρ(a, b), ρ(b, x)}≤ �,

and so x ∈ B(a, �). Hence, B(b, �) ⊂ B(a, �), and a similar proof gives thereverse inclusion. Therefore, B(a, �) = B(b, �), and so both a and b arecenters.

Proposition 21.2. Every ultrametric space has a base of clopen sets.

Proof. Let X be an ultrametric space and let F denote the collection of allclosed balls in X. Evidently, F covers X. Observe next that, for any twoballs of F , either they are disjoint or one is contained in the other. To seethis, let F1, F2 ∈ F with x ∈ F1 ∩ F2 (such an x exists if F1 and F2 arenot disjoint). Now, F1 has the form B(a, �1) and F2 has the form B(b, �2).Without loss of generality, let �1 ≤ �2. By the lemma, F1 = B(a, �1) andF2 = B(a, �2), and so F1 ⊂ F2. From this fact, we conclude that F is abase for X (any F1 and F2 with nontrivial intersection has one containingthe other).

Now, closed balls are certainly closed in this topology, as F = F forall F ∈ F . To see that they are also open, observe that, by the lemma,F contains a basic neighborhood of each of its points (namely, F itself).Therefore, F is a base of clopen sets for X.

22 Extra Problem 2

Proposition 22.1. The metric given by

d(x, y) =

{0 if x = y1

2notherwise, where n = max{n | x(i) = y(i)∀i ≤ n}

gives the product topology on NN.

17

Proof. Let B denote the collection of all open balls of NN. Evidently, Bcovers NN. To see that it is, in fact, a base for NN, let B1, B2 ∈ B and letx ∈ B1∩B2. Now, B1 is of the form B(a, 12m ) and B2 is of the form B(b,

12n

).Without loss of generality, let m ≤ n. It follows that x(i) = a(i) = b(i) forall i ≤ m. Hence, x ∈ B(x, 1

2m) ⊂ B1 ∩B2, and so B is a base for NN.

Now, the sets of B can be expressed as

B

(x,

1

2n

)=

n∏i=1

{x(i)} ×∞∏

i=n+1

N.

In other words, the basic open sets are products of open sets Un of N whereUn = N for all but finitely-many n. Hence, B induces the product topologyon NN.

23 Problem 17F1

Proposition 23.1. A space is countably compact if and only if each sequencehas a cluster point.

Proof. (⇒) We proceed by establishing the contrapositive. To that end,suppose there is a sequence 〈xn〉 having no cluster point in X. That is,for every x ∈ X, there exists a neighborhood (without loss of generality,an open neighborhood) Nx of x such that Nx \ {x} contains no point of〈xn〉. Denote by N the open set

⋃{Nx | x is not an element 〈xn〉}. By

construction, N contains no element of the sequence 〈xn〉. Finally, let Cdenote the collection of open sets {Nx | x is an element of 〈xn〉} ∪ {N}. Wesee that C is a countable cover of X that admits no finite subcover. Indeed,omitting any Nx results in a subcollection missing x. Therefore, X is notcountably compact, thus establishing the contrapositive.

(⇐) Let X be a space with the property that every sequence has a clus-ter point. Take any family of closed sets Cn having the finite intersectionproperty and let xn be any point belonging to

⋂ni=1 Ci. Let x be a cluster

point of the sequence 〈xn〉 and let O be any open set containing x. Since x isa cluster point, O contains infinitely-many xn, and hence O intersects everyCn. As O was arbitrary, we conclude that x belongs to every Cn. Hence,x ∈

⋂∞n=1Cn, and so X is countably compact.

18

24 Problem 20B

Lemma 24.1. Let U be a collection of open sets. For all U ∈ U ,

St(U,U) =⋃x∈U

St(x,U).

Proof. (To be included if the lemma turns out to be useful)

Proposition 24.2. A barycentric refinement of a barycentric refinement ofa cover W is a star-refinement of W.

Proof. Let U∆V∆W . Our goal is to show that U ∗ < W . To that end, letU ∈ U . Now,

St(U,U) =⋃x∈U

St(x,U) (by lemma)

⊂⋃x∈U

Vx, for some Vx ∈ V (since U∆V)

⊂⋃x∈U

St(x,V)

⊂⋃x∈U

Wx, for some Wx ∈ W (since V∆W).

(My problem now is that there is no guarantee that⋃x∈U Wx ∈ W , nor can

I force⋃x∈U St(x,V) ⊂ St(x0,V) for any single x0.)

Proposition 24.3. If Un is the cover of a metric space (X, d) by(

13n

)-balls

about each of its points, then Un+1 ∗ < Un.

Proof. Let U ∈ Un+1. By definition, U = B(x, 13n+1 ) for some x ∈ X. Con-sider now any point y ∈ St(U,Un+1). Now, y ∈ V for some V ∈ Un+1 withU ∩ V 6= ∅. It follows that d(x, y) < 3

3n+1= 1

3n, as x is the center of a(

13n+1

)-ball and V has diameter 2

3n+1. Hence, St(U,Un+1) ⊂ B(x, 13n ) ∈ Un,

as desired.

Proposition 24.4. If U is an open cover of X, V is an open barycentricrefinement of U , and for each U ∈ U we define FU = X−St(X−U,V), then{FU | U ∈ U} is a closed cover of X.

19

Proof. The fact that each FU is closed is immediate, since it is the comple-ment of a union of open sets. Let now x ∈ X. Since V∆U , St(x,V) ⊂ U , forsome U ∈ U . We claim that x ∈ FU for this U . Suppose, for the purpose ofcontradiction, that this is not the case. This means that x ∈ St(X−U,V). Inparticular, there exists some V ∈ V with x ∈ V and V ∩(X−U) 6= ∅. On theother hand, V is an open set containing x, and so V ⊂ St(x,V) ⊂ U . Havingarrived at a contradiction (V cannot be both a subset of U and also intersectX − U nontrivially), we conclude that x ∈ FU . As x was chosen arbitrarily,we see that {FU | U ∈ U} is indeed a closed cover of X, as desired.

Define ∼ in any space X by x ∼ y if and only if x and y lie together insome connected subset of X. Define ≈ in X by x ≈ y if and only if there isno decomposition X = U ∪ V into disjoint open sets with one containing xand the other containing y.

25 Problem 26B1

Proposition 25.1. The relation ∼ is an equivalence on X. The equivalenceclass [x] of x is just the component Cx of x in X.

Proof. The relation ∼ is reflexive, as x lies within the connected subset of Xcontaining it (that is, x ∼ x).

The relation ∼ is symmetric, as “x and y lie together in some connectedsubset” has the same meaning as “y and x lie together in some connectedsubset”.

To see that ∼ is transitive, suppose that x ∼ y and y ∼ z for somex, y, z ∈ X. Let U denote the connected subset of X containing x and y,and let V denote the connected subset of X containing y and z. We see thatU ∩V 6= ∅ (as y ∈ U ∩V ), and so U ∪V is connected. Hence, x ∼ z, as theylie together in the connected subset U ∪ V of X.

Taken together, we conclude that ∼ is an equivalence relation on X.Now, let y ∈ [x]. This means that x and y lie in some connected subset

of X. Since Cx is the union of all connected subsets of X containing x, wesee that y ∈ Cx. Next, let y ∈ Cx. Since x and y both lie in the connectedcomponent Cx, we see that y ∈ [x]. Hence, [x] = Cx, as desired.

20

26 Problem 26B2

Proposition 26.1. The relation ≈ is an equivalence on X. We call theequivalence class of x the quasicomponent of x in X. The quasicomponent ofx in X is the intersection of all clopen subsets of X that contain x.

Proof. The relation ≈ is reflexive, since there can be no decomposition sep-arating x from itself (that is, x ∼ x).

The relation ≈ is symmetric, as U and V are indistinguishable in thedefinition. That is, the phrase “one containing x and the other containingy” has the same meaning as “one containing y and the other containing x”.

To see that ≈ is transitive, we establish the contrapositive. To that end,suppose that x 6≈ z. This means that there are disjoint open sets U andV such that (without loss of generality) x ∈ U , z ∈ V , and X = U ∪ V .Now, it must be that y belongs to one of U and V . If y ∈ U , then U and Vrepresent a decomposition of X of the appropriate type separating y from z,and so y 6≈ z. Similarly, if y ∈ V , then U and V separate x from y, and sox 6≈ y. Hence, we have that x 6≈ z ⇒ (x 6≈ y) ∨ (y 6≈ z), thus establishingthe contrapositive.

Taken together, we conclude that ≈ is an equivalence relation on X.Let now F denote the intersection of all clopen subsets of X containing x.

Let y ∈ F . This means that y belongs to every clopen subset of X containingx. In particular, y ∈ Cx. Since Cx is connected, we see that x cannot beseparated from y by a pair of disjoint open sets whose union is X (otherwise,these sets would disconnect Cx). Hence, y ∈ [x]. Next, we show that [x] ⊂ Fby establishing the contrapositive. To that end, suppose that y /∈ F . Thismeans there is some clopen subset Fy of X such that x ∈ Fy but y /∈ Fy.It follows that Fy and F

cy are disjoint open sets with x ∈ Fy, y ∈ F cy , and

X = Fy∪F cy . Hence, y /∈ [x], thus establishing the contrapositive. Therefore,[x] = F , as desired.

21

(I could make no sense of Willard’s picture, so I provide a different spacewhere the components and quasicomponents may disagree.)

27 Extra Problem

Proposition 27.1. Let A denote the set { 1n| n ∈ Z+} and let X be the

space (A × [0, 1]) ∪ {(0, 0)} ∪ {(0, 1)} with the relative Euclidean topology.The points {(0, 0)} and {(0, 1)} belong to separate components, but belong tothe same quasicomponent.

Proof. Evidently, C(0,0) = {(0, 0)}, since {(0, 0)} is the only connected subsetof X containing (0, 0). We claim next that {(0, 0), (0, 1)} ⊂ [(0, 0)]. Suppose,for the purpose of contradiction, that there are disjoint open sets U and Vsuch that (without loss of generality) (0, 0) ∈ U , (0, 1) ∈ V , and U ∪V = X.Since U is open, (0, 0) ( U . Similarly, (0, 1) ( V . Let k be a natural numbersuch that both U ∩ ( 1

k× [0, 1]) 6= ∅ and V ∩ ( 1

k× [0, 1]) 6= ∅. Now, as 1

k× [0, 1]

is a connected subset of X, it must belong entirely to U or V , contradictingour previous assertion that both U and V intersect it nontrivially. Therefore,(0, 0) and (0, 1) belong to separate components, yet they belong to the samequasicomponent.

22