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 subspace of X is normal. Proposition 1.2. X is completely normal if and only if, whenever A and B are subsets of X with A ∩ ClX (B) = ClX (A) ∩ B = ∅, then there are disjoint open 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 are contained 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 = ∅.

1 For closedness, note that the closure of a set in a topological space is always closed in that space. By the complete normality of X, we have that Y is normal. Hence, there exist 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 that ClY (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 subsets of 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 ⊃ A and V ⊃ B. It follows that the sets U ∩ Y and V ∩ Y are disjoint open subsets of Y with (U ∩ Y ) ⊃ A and (V ∩ Y ) ⊃ B. Hence, Y is normal, and so X is completely normal.

2 Problem 15B3

Lemma 2.1. If M is metrizable and N ⊂ M, then the subspace N is metrizable 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 the relative topology on N and let ρN be the restriction of ρ to N. We show that σ is generated by ρN . Let O ∈ σ. It must be that O = N ∩ G for some G ∈ τ. Since M is S generated by ρ, we know that G = x∈G Bρ(x, x), where x > 0 may depend

2 on x. Now,

O = N ∩ G [ = N ∩ Bρ(x, x) x∈G [ = N ∩ Bρ(x, x) x∈G [ = BρN (x, x). x∈N∩G

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 a metric 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 has a countable subfamily that is a base.

Proof. Let X be a second countable space, B = {Bα | α ∈ Γ} be any base for X, and C = {Ci | i ∈ N} be the countable base guaranteed by the 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 (call this countable subcollection BCi ). Since the union of countably-many sets each 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 write S Ck = 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, we can ﬁnd some Cix with x ∈ Cix ⊂ Bix . It follows that {Bix | x ∈ Ck} is a countable set (we chose one element of B for each element of the countable set C) whose union is Ci (by construction, every element of Ci is present in the union and every Bix is contained in Ci).

3 4 Problem 16B3

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

Proof. Let A be a set of real numbers that is well-ordered by the usual order. i We claim that, for each a ∈ A, there is n ∈ such that a, a + 1 ∩A = ∅. a N na Observe that the truth of this claim implies the countability of A, as each interval will contain a distinct rational number. Suppose now, for the purpose of contradiction, that the claim does not hold. That is, there exists some 1 b ∈ A such that, for all n ∈ N, b, b + n ∩ A 6= ∅. It follows that the set A \{c ∈ A | c ≤ b} has no least element, which is a contradiction with the fact 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 a locally compact, non-compact Hausdorff space X can be applied to any space Y . That is, Y ∗ = Y ∪ {p} with neighborhoods y ∈ Y unchanged in Y ∗ while neighborhoods of p have the form {p} ∪ (Y − L), where L is a subset of Y with compact closure. Y ∗ is called the Alexandroff extension of Y .

6 Problem 19A1

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

Proof. Recall that one can deﬁne a topology on a set Y by specifying, for each 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.

4 Using this approach, we call a set open whenever it contains a neighborhood of each of its points. Observe that, since Y is a topological space, there already exists a neighborhood system satisfying these requirements for each x ∈ Y . Let Up contain all sets of the form {p} ∪ (Y − L), where L is a subset of Y with compact closure. 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 are subsets 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 that L ∪ 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 each y ∈ V .

Proof. Let U ∈ Up and choose any open V ∈ Up with V ⊂ U. Since V is open, there exists, for all y ∈ V , an open subset G of V with y ∈ G and p∈ / G. Hence, G ∈ Uy. Since y ∈ Y , Uy satisﬁes the property that any superset 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 that ClY (K) ⊂ ClY (L). As a closed subset of compact space is compact, we have that K has compact closure. Hence, V ∈ Up.

5 Therefore, the neighborhood systems as deﬁned indeed give a topology on Y ∗. We conclude by showing that its relative topology on Y is the original topology. 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 merely having compact closure? For example, (0, 1) has compact closure in R, yet R − (0, 1) is not open.) Pending the resolution of the above parenthetical remark, we will have shown that σ = τ, and so the relative topology on Y is precisely the original topology.

7 Problem 19A4

Proposition 7.1. Y ∗ is Hausdorff if and only if Y is locally compact and Hausdorff. Proof. (⇒) Let x, y ∈ Y ⊂ Y ∗. Since Y ∗ is Hausdorff, we can find disjoint open neighborhoods U and V in Y ∗ with x ∈ U and y ∈ V . If neither U nor V 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 violate 0 0 disjointness). Define the set U = Y − ClY (L). We have that x ∈ U , y ∈ V with U 0 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 compact neighborhood, it follows that Y is locally compact. (⇐) Let x, y ∈ Y ∗. If neither x nor y is p, then the fact that Y is Hausdorff implies that x and y can be put into disjoint open neighborhoods. Suppose now that y = p. Choose, by the local compactness of Y , some compact neighborhood V ⊂ Y of x. It follows that IntY (V ) is an open set ∗ containing x. We also have that IntY (Y − V ) = IntY ({p} ∪ (Y − V )), which is an open neighborhood (as it is a member of the base) containing p. Thus,

6 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 Lindel¨of.

Proof. Let C be a basic open (in E) cover of R. That is, C = {[xα, yα) | 0 S 0 xα, yα ∈ R, α ∈ Γ}. Let C = {(xα, yα) | [xα, yα) ∈ C}. Finally, let A = C . Evidently, C0 covers A. Now, observe that C0 contains open sets from the usual topology on R. Since R with the usual topology is second-countable, it is Lindel¨of. Hence, there is a countable subcollection of C0 (and so of C) covering A. Next, we claim that A misses only countably-many points of R. To see this, let x and y be distinct real numbers that are not elements of A. Without loss of generality, let x < y. Since C covers R, it follows that x and y are left endponts of distinct intervals of C. Denote these intervals by [x, x0) and [y, y0), respectively. Evidently, [x, x0) and [y, y0) are disjoint (if not, then y ∈ (x, x0), and so y ∈ A). Hence, we can identify with each of x and y a distinct rational number belonging to (x, x0) and (y, y0), respectively. Thus, A misses only countably-many elements of R, and so there is a countable subcollection 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 Lindel¨of,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 Lindel¨ofspace contains an accumulation point.

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

7 finite. Let C denote the collection {Ux | x ∈ X}. Since C is an open cover of the Lindelöfspace X, C admits a countable subcover C0 of X (and so of A). Since |A ∩ U| is finite for all U belonging to the countable collection C0, 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 for some filter on X if and only if it has the finite intersection property. Proof. (⇒) We establish the contrapositive. That is, suppose A is a nonempty collection of subsets of X that fails to satisfy the finite intersection property. Tk It follows that there are sets A1,...,Ak ∈ A such that i=1 Ai = ∅. Consider Tn now the collection C = { i=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 empty set). Hence, A is not a subbase for any filter on X, thus establishing the contrapositive. (⇐) Let S be a nonempty collection of subsets of X having the finite intersection property. Define B to be the collection \ {F ⊂ X | S0 ⊂ F for some finite subcollection S0 of S}.

Observe first that B is a nonempty collection of nonempty subsets of X, as S satisfies the finite intersection property. We further claim that B is in fact a base for a filter on X, and so S will be a subbase for this filter. To that end, 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 ﬁlter on X.

8 Proof. Let F be the intersection of filters Fα for α ∈ Γ. We claim that F is itself a filter. First, observe that ∅ ∈/ F and X ∈ F, since this is the case for each of the Fα. Next, let F1 and F2 be elements of F. It follows that F1 and F2 belong to 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 for all 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 a filter, as desired.

12 Problem 12B3

Proposition 12.1. Every filter is the intersection of the ultrafilters containing it. T 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 of contradiction, that this inclusion is strict. It follows that there is some set A with A/∈ F and A ∈ U. Consider the filter G generated by F ∪ {Ac} (this is indeed a filter, since A/∈ F and so ∅ ∈/ G). We have that F ⊂ G ⊂ Uβ, c for some β ∈ Γ. Now, since A ∈ U, A ∈ Uβ, but also A ∈ Uβ, which is a contradiction. 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 have caliber ℵ if, whenever U is a family of open subsets of X with |U| = ℵ, then there exists a subfamily V of U with |V| = ℵ and T V= 6 ∅.

Proposition 13.2. Every separable space has caliber ℵ1. Proof. Let X be a separable space having countable dense subset D. Let U be 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 ∈ D belonging to uncountably-many open sets of U. (Were this not the case, each of the elements of the countable set D appears in only countably-many open

9 sets of U, which would imply that U is countable.) Let V = {U ∈ U | x ∈ U}. T 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 provided every family of disjoint open subsets of X is countable.

Proposition 14.2. If X has caliber ℵ1, then X satisﬁes the countable chain condition. Proof. We establish the contrapositive. To that end, suppose that X does not satisfy the countable chain condition. It follows that there is an uncountable family U of disjoint open sets of X. By the Axiom of Choice, we may assume that |U| = ℵ1 (if |U| > ℵ1, we use the Axiom of Choice to reduce it to a collection of size ℵ1). Now, since U is a collection of disjoint sets, it follows T that U = ∅, and so X does not have caliber ℵ1.

15 Problem 16D3

Proposition 15.1. A regular space is Lindel¨ofif and only if each open cover has a countable subcollection whose closures cover (i.e. has a countable dense subsystem). Proof. (⇒) Let X be a Lindel¨ofspace. Any open cover of X admits a countable 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 each x ∈ X, let Ux be an open set in U containing x. Since X is regular, there is an open set Vx containing x with Vx ⊂ Ux. Let V be the collection of sets Vx for all x ∈ X. Evidently, V is also a cover of X. By hypothesis, V admits a 0 countable subcollection V = {Vi | i ∈ N} such that [ Vi = X. i∈N

By construction, each Vi is contained in some open set (call it Ui) belonging to U. It follows that the collection {Ui | i ∈ N} is a countable subcollection of U covering X. Therefore, X is Lindel¨of.

10 16 Problem 17C1

Definition 16.1. A compact space X is maximal compact provided every strictly finer topology on X is noncompact. Proposition 16.2. A compact space X is maximal compact if and only if every compact subset is closed. Proof. (⇒) We establish the contrapositive. To that end, let (X, τ) be a compact 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 0 be the topology generated by τ ∪ {X − K}. Evidently, τ 0 is a strictly finer topology than τ. We claim next that (X, τ 0) is compact. To that end, let C be an open cover (using sets from τ 0) of X. If X −K/∈ C, then only open sets of τ appear in the cover. Since X is compact under τ, we can find a finite subcover of X. Suppose then that X − K ∈ C. Since K is compact under τ and K is disjoint from X − K, we can find a finite subcover of K using elements of C − {X − K}. Taking this cover of K together with X − K gives the desired cover of X. In either case, we conclude that (X, τ) is not maximal, thus establishing the contrapositive. (⇐) We establish the contrapositive. To that end, suppose that (X, τ) is not maximal compact. Let τ 0 denote a strictly finer compact topology than τ. Since τ 0 is strictly finer than τ, we can find a closed set F belonging to τ 0 but not belonging to τ. Since F is a closed (under τ 0) subset of a compact space, F is compact under τ 0. Hence, every open cover of F by open sets of τ 0 admits a finite subcover. In particular, any open cover of F using only open 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 the contrapositive.

17 Problem 17F4 Q 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. Q Proof. (⇒) Let Xn be countably compact. That is, every countable open Q cover of Xn admits a finite subcover. For a fixed k, let C be a countable

11 Q 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, D Q 0 is a countable open cover of Xn, which admits a finite subcover D by the Q countable compactness of Xn. It follows that the collection of all Uk from the open sets belonging to D0 (that is, the collection containing the factor 0 corresponding to Xk from each open set in D ) is a finite subcollection of C covering Xk, and so Xk is countably compact. As k was arbitrary, it follows that each of the product spaces is countably compact. (⇐) As every countably compact space is sequentially compact, it suffices Q to show that Xn is sequentially compact. To that end, let hxni be a sequence in the product space. For each i, hxn(i)i is a sequence in Xi. Since Xi is first countable and countably compact, it is sequentially compact. Hence, hxn(i)i has a convergent subsequence hxnk (i)i. (This is not itself enough to conclude that the sequence in the product has a convergent subsequence, as, in general, the product of sequentially compact spaces need not be sequentially 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 countably compact.

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

Proposition 18.2. Closed subspaces of countably compact spaces are countably compact.

Proof. Let X be countably compact, F a closed subspace of X, and C a countable open cover of F . For each C ∈ C, there is an open set UC ∈ X such 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 countable compactness of X admits a ﬁnite subcover {X − F } ∪ {U1,...,Un | Ui ∈ X}.

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

19 Problem 22F1

Lemma 19.1. Let x and y be real numbers both strictly greater than −1. If x y x ≤ y, then 1+x ≤ 1+y . Proof. We have that

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

Remark 19.2. One can think of the previous lemma as follows: Suppose we x mL of salt and y mL of salt and we add them each to separate 1 mL x containers of water. The ratio 1+x measures the concentration of salt in y the ﬁrst container and 1+y measures the concentration of salt in the second container. Since x is a smaller amount of salt than y, then surely the ﬁrst x y container is less concentrated than the second. That is, 1+x ≤ 1+y . Proposition 19.3. If ρ is a metric on X, then both

ρ1(x, y) = min{1, ρ(x, y)} and ρ(x, y) ρ (x, y) = 2 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 what follows, 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),

13 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 ρ(x, y) ρ (x, y) = 2 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 ρ(x, y) ρ (x, y) = 2 1 + ρ(x, y) ρ(y, x) = (since ρ is symmetric) 1 + ρ(y, x)

= 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) ρ(x, z) + ρ(z, y) ≤ (by lemma). 1 + ρ(x, y) 1 + ρ(x, z) + ρ(z, y)

14 The lefthand side of the inequality is precisely ρ2(x, y), and so we have that ρ(x, z) + ρ(z, y) ρ (x, y) ≤ 2 1 + ρ(x, z) + ρ(z, y) ρ(x, z) ρ(z, y) = + 1 + ρ(x, z) + ρ(z, y) 1 + ρ(x, z) + ρ(z, y) ρ(x, z) ρ(z, y) ≤ + (since ρ is positive deﬁnite) 1 + ρ(x, z) 1 + ρ(z, y)

= 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, ) = Bρ(y, y),

y∈Bρ(x,) 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, ) = Bρ1 (y, y),

y∈Bρ(x,) and so (X, ρ) ⊂ (X, ρ1).

Next, consider the basic open set Bρ1 (x, ) of (X, ρ1). As before, [ Bρ1 (x, ) = Bρ1 (y, y)

y∈Bρ1 (x,) where each y depends on y. In particular, we may insist that 0 < y ≤ 1 for all y. With this restriction, we have, for each y,

Bρ1 (y, y) = Bρ(y, y). Taken together, we see that [ Bρ1 (x, ) = Bρ(y, y),

y∈Bρ1 (x,)

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 as in 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, ) = Bρ2 (y, y),

y∈Bρ2 (x,) where each y depends on y. Observe now that, for all y, ρ(y, z) ρ (y, z) = ⇒ = 2 y 1 + ρ(y, z) y

⇒ ρ(y, z) = y(1 + ρ(y, z)) ⇒ ρ(y, z) = y , 1 − y where we insist that 0 < y < 1 for all y. Taken together, we see that

[ y Bρ2 (x, ) = Bρ(y, ), 1 − y y∈Bρ2 (x,) and so (X, ρ2) ⊂ (X, ρ).

20 Problem 22F2

Proposition 20.1. Every metric generating the topology of a compact metrizable space is bounded. Proof. We proceed by establishing the contrapositive. To that end, let ρ be an unbounded metric generating a topology on a set X. Define, for all n ∈ 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 be S any 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 does not 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 closed ball around a ∈ X of radius > 0 (i.e. the set {y ∈ X | ρ(a, y) ≤ }). Let b ∈ 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 the reverse inclusion. Therefore, B(a, ) = B(b, ), and so both a and b are centers. 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 all closed balls in X. Evidently, F covers X. Observe next that, for any two balls of F, either they are disjoint or one is contained in the other. To see this, let F1,F2 ∈ F with x ∈ F1 ∩ F2 (such an x exists if F1 and F2 are not 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) and F2 = B(a, 2), and so F1 ⊂ F2. From this fact, we conclude that F is a base for X (any F1 and F2 with nontrivial intersection has one containing the other). Now, closed balls are certainly closed in this topology, as F = F for all 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 ( 0 if x = y d(x, y) = 1 2n otherwise, 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, B N N covers N . To see that it is, in fact, a base for N , let B1,B2 ∈ B and let 1 1 x ∈ B1 ∩ B2. Now, B1 is of the form B(a, 2m ) and B2 is of the form B(b, 2n ). Without loss of generality, let m ≤ n. It follows that x(i) = a(i) = b(i) for 1 N all i ≤ m. Hence, x ∈ B(x, 2m ) ⊂ B1 ∩ B2, and so B is a base for N . Now, the sets of B can be expressed as

n ∞ 1 Y Y B x, = {x(i)} × . 2n N i=1 i=n+1

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

23 Problem 17F1

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

Proof. (⇒) We proceed by establishing the contrapositive. To that end, suppose there is a sequence hxni 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 S hxni. Denote by N the open set {Nx | x is not an element hxni}. By construction, N contains no element of the sequence hxni. Finally, let C denote the collection of open sets {Nx | x is an element of hxni} ∪ {N}. We see 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 not countably compact, thus establishing the contrapositive. (⇐) Let X be a space with the property that every sequence has a cluster point. Take any family of closed sets Cn having the finite intersection Tn property and let xn be any point belonging to i=1 Ci. Let x be a cluster point of the sequence hxni and let O be any open set containing x. Since x is a cluster point, O contains infinitely-many xn, and hence O intersects every Cn. As O was arbitrary, we conclude that x belongs to every Cn. Hence, T∞ x ∈ n=1 Cn, 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) = St(x, U). x∈U

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

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

Proof. Let U∆V∆W. Our goal is to show that U ∗ < W. To that end, let U ∈ U. Now, [ St(U, 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). x∈U S (My problem now is that there is no guarantee that Wx ∈ W, nor can S x∈U I force x∈U St(x, V) ⊂ St(x0, V) for any single x0.) 1 Proposition 24.3. If Un is the cover of a metric space (X, d) by 3n -balls ∗ about each of its points, then Un+1 < Un.

1 Proof. Let U ∈ Un+1. By deﬁnition, U = B(x, 3n+1 ) for some x ∈ X. Con- sider now any point y ∈ St(U, Un+1). Now, y ∈ V for some V ∈ Un+1 with 3 1 U ∩ V 6= ∅. It follows that d(x, y) < 3n+1 = 3n , as x is the center of a 1 2 1 3n+1 -ball and V has diameter 3n+1 . Hence, St(U, Un+1) ⊂ B(x, 3n ) ∈ Un, as desired.

Proposition 24.4. If U is an open cover of X, V is an open barycentric reﬁnement 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, for some U ∈ U. We claim that x ∈ FU for this U. Suppose, for the purpose of contradiction, that this is not the case. This means that x ∈ St(X −U, V). In particular, there exists some V ∈ V with x ∈ V and V ∩(X −U) 6= ∅. On the other hand, V is an open set containing x, and so V ⊂ St(x, V) ⊂ U. Having arrived at a contradiction (V cannot be both a subset of U and also intersect X − 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 in some connected subset of X. Deﬁne ≈ in X by x ≈ y if and only if there is no decomposition X = U ∪ V into disjoint open sets with one containing x and the other containing y.

25 Problem 26B1

Proposition 25.1. The relation ∼ is an equivalence on X. The equivalence class [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 X containing it (that is, x ∼ x). The relation ∼ is symmetric, as “x and y lie together in some connected subset” has the same meaning as “y and x lie together in some connected subset”. To see that ∼ is transitive, suppose that x ∼ y and y ∼ z for some x, 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 that U ∩ V 6= ∅ (as y ∈ U ∩ V ), and so U ∪ V is connected. Hence, x ∼ z, as they lie 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, we see that y ∈ Cx. Next, let y ∈ Cx. Since x and y both lie in the connected component 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 the equivalence class of x the quasicomponent of x in X. The quasicomponent of x 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 separating x from itself (that is, x ∼ x). The relation ≈ is symmetric, as U and V are indistinguishable in the deﬁnition. That is, the phrase “one containing x and the other containing y” 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 and V 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 V represent 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 so x 6≈ y. Hence, we have that x 6≈ z ⇒ (x 6≈ y) ∨ (y 6≈ z), thus establishing the 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 containing x. In particular, y ∈ Cx. Since Cx is connected, we see that x cannot be separated 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] ⊂ F by establishing the contrapositive. To that end, suppose that y∈ / F . This means there is some clopen subset Fy of X such that x ∈ Fy but y∈ / Fy. c c It follows that Fy and Fy are disjoint open sets with x ∈ Fy, y ∈ Fy , and c X = Fy ∪Fy . 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 space where the components and quasicomponents may disagree.)

27 Extra Problem

1 + Proposition 27.1. Let A denote the set { n | 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 to the same quasicomponent.

Proof. Evidently, C(0,0) = {(0, 0)}, since {(0, 0)} is the only connected subset of 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 V such 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 number 1 1 1 such that both U ∩ ( k × [0, 1]) 6= ∅ and V ∩ ( k × [0, 1]) 6= ∅. Now, as k × [0, 1] is a connected subset of X, it must belong entirely to U or V , contradicting our 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 same quasicomponent.