Math 525 More notes related to Metrizability

In (X, τ), we define A to be dense in B iff A ⊆ B ⊆ Clτ A. A is dense in X iff Clτ A = X.

A topological space which contains a countable, dense subset is said to be separable.

On R, τs, τu, τcof, τo and τi are separable, while τd and τcoc are not (why not?). Since (R, τd) is metrizable, it follows that metrizable spaces need not be separable.

Prop A30 Let (X, τ) be a topological space.

(a) If (X, τ) is second countable, then (X, τ) is separable.

(b) If (X, τ) is metrizable, then (X, τ) is second countable iff (X, τ) is separable.

Useful Lemma: Let(X, τ) be a topological space, let B be any basis for τ, and let D ⊆ X.

Then D is dense (in X) iff whenever φ =6 B ∈ B, D ∩ B =6 φ.

Proof of Prop A30 :

(a) Assume (X, τ) is 2nd countable. Then τ has a countable basis of non-empty sets

{Un n ∈ N}. By the Axiom of Choice, we can choose an xn from each nonempty Un. By ¯ the previously¯ stated lemma, the set D = {xn n ∈ N} is a countable, dense subset of X. ¯ ¯ (b) (⇒) follows from part (a). To prove (⇐), assume (X, τ) is separable (in addition to being metrizable). Let D = {xn n ∈ N} be a countable dense subset of X. For any k ∈ N, ¯ 1 define the collection Bk = {B(xn¯, k ) n ∈ N}, and let B = Bk. Note that B is just the k∈N ¯ [ countable collection of all balls of radius¯ 1/k (for all k ∈ N) centered at every xn ∈ D. Claim: B is a basis for τ.

To prove this, let x ∈ U ∈ τ. Then ∃ǫ > 0 such that B(x, ǫ) ⊆ U. Choose k ∈ N such

1 ǫ 1 that k < 2 . Since D is dense and B(x, k ) is an element of a known basis for τ, the lemma

1 1 tells us that there must be an xn ∈ D ∩ B(x, k ). Now, B(xn, k ) is a nbhd of x, since

1 1 1 xn ∈ B(x, k ) ⇒ d(xn, x) < 1/k ⇒ x ∈ B(xn, k ). Also, B(xn, k ) is contained by U, since

1 1 1 ǫ ǫ y ∈ B(xn, k ) ⇒ d(x, y) ≤ d(xn, y)+ d(xn, y) < k + k < 2 + 2 = ǫ ⇒ y ∈ B(x, ǫ) ⊆ U. Therefore B is a countable basis for τ, which makes (X, τ) 2nd countable. ¥ Remark: We showed previously that (R, τs) is separable but not 2nd countable. Thus (R, τs) is not metrizable. Recall that a hereditary property is one that is preserved by subspaces.

Prop A31 Metrizability is hereditary, but separability is generally not. However, any sub- space of a separable, metrizable space is separable.

Proof : If(X, τ) is metrizable by some metric d and (A, τA) is a subspace, then the metric d|A×A = dA defined by dA(x, y)= d(x, y) for all x, y ∈ A is a metric on A which induces τA.

To prove the last assertion in Prop A31, let (X, τ) be separable and metrizable. Then by

Prop A30, (X, τ) is second countable. Since second-countability and metrizability are hered- itary, any subspace of (X, τ) is second countable, and hence, by Prop A30(a), separable. ¥

To show that separability is not hereditary in general, consider (R, τs) × (R, τs), for which basic open sets in the product topology have the form [a, b) × [c,d). Q × Q is dense in

2 (R, τs) × (R, τs), so this product space is separable. But if A = {(x, −x) ∈ R x ∈ R}, then ¯ ¥ the subspace (A, τA) is an uncountable discrete space, which is not separable.¯

Prop A32 Let (X, τ) = (Xi, τi). Then each component space (Xj, τj) is homeomorphic i∈I to a subspace of (X, τ). Y

Proof : A subspace of (X, τ) that (Xi0 , τi0 ) is homeomorphic to can be constructed as follows.

Choose an arbitrary, fixed element of X, say (bi). Then define

A = {x =(xi) ∈ X xi = bi for i =6 i0, and xi0 ∈ Xi0 }. ¯ In other words, A consists¯ of elements of X which are unconstrained only in the i0th compo- nent. Let pi0 : X → Xi0 denote the i0th projection map, and let pi0 denote the restriction of p to A. Then p is continuous, and is clearly a bijection. To show that p −1 is continuous, io i0 b i0 we can show that p (U) is open in X whenever U is open in A. Since p is a bijection and b i0 i0 i0b bijections preserve both unions and intersections, we may assume without loss of generality b b −1 that U = V ∩ A, where V is a subbasic open set in τ of the form V = pi (Vi) with Vi ∈ τi for each i. Then Vi0 if i = i0 −1  pi0 (U)= pi0 (pi (Vi) ∩ A)= Xi0 if i =6 i0 and yi ∈ Vi .   φ if i =6 i0 and yi 6∈ Vi b b   −1 In every case, pi0 (U) is open, so pi0 is an open map and pi0 is continuous, and thus pi0 is a homeomorphism between A (as a subspace of the product space) and (X , τ ). ¥ b b b i0 i0 b

Remark If (Y, µ) is homeomorphic to a subspace of (X, τ), then (Y, µ) is called a gener- alized subspace of (X, τ). Prop A32 shows that each component space, as a generalized subspace of the product space, inherits hereditary properties from the product space.

Prop A33 A product space (X, τ) = (Xi, τi) is metrizable iff each (Xi, τi) is metrizable i∈I Y and only countably many of the Xi’s contain two or more points.

Proof : (⇒) Assume (X, τ) is metrizable. Then by Props A31 and A32, each component space (Xi, τi) is metrizable. Also, (X, τ) must be first countable, and so by Prop A22(b), all but countably many component spaces must be indiscrete. Since metric spaces are T2, the only indiscrete metrizable spaces are singleton spaces, so only countably many component spaces are not singleton spaces (i.e. contain 2 or more points).

(⇐) As we saw in the proof of Prop A32, singleton component spaces do not alter a product space up to homeomorphism. So we may assume here that I = N and (X, τ)= (Xn, τn). n∈N Y Since each (Xn, τn) is metrizable by assumption, there is a bounded metric d¯n on Xn which induces τn such that d¯n(xn, yn) ≤ 1 for all xn, yn ∈ Xn. For x =(xn) ∈ X and y =(yn) ∈ X, define d : X × X → [0, ∞) by ∞ d¯ (x , y ) d(x, y)= n n n . 2n n=1 X

Note that d(x, y) ∈ [0, 1] for all x, y ∈ X, and d(x, y) = 0 iff d¯n(xn, yn) = 0 for all n, which is true iff xn = yn for all n, which is true iff x = y. Also, it should be clear that d(x, y) = d(y, x) for all x, y ∈ X. To show that d is a metric, it remains to verify the triangle inequality. If z =(zn) ∈ X, then since each d¯n is a metric, we have d¯ (x , z ) d¯ (x , y ) d¯ (y , z ) n n n ≤ n n n + n n n for all n ∈ N, 2n 2n 2n which implies that

∞ ∞ ∞ d¯ (x , z ) d¯ (x , y ) d¯ (y , z ) d(x, z)= n n n ≤ n n n + n n n = d(x, y)+ d(y, z). 2n 2n 2n n=1 n=1 n=1 X X X Thus d is a metric on X. We still need to show that the topology τ ′ induced by d equals the product topolgy τ. Let x =(xn) ∈ X. A basic τ-nbhd U of x has the form ∞ ¯ ¯ ¯ U = Bd1 (x1, ǫ1) × Bd2 (x2, ǫ2) × · · · × Bdn (xn, ǫn) × Xk . Ãk=n+1 ! Y ǫ1 ǫ2 ǫ d¯ (x , y ) ǫ Let ǫ = min , ,..., n . If d(x, y) < ǫ, then for k ∈ {1, 2,...,n}, k k k < ǫ ≤ k , 2 22 2n 2k 2k n o ′ ¯ so yk ∈ Bdk (xk, yk) and hence y ∈ U. Thus x ∈ Bd(x, ǫ) ⊆ U, and so τ ≤ τ . ∞ 1 ǫ To show τ ′ ≤ τ, let ǫ > 0 be given, and choose m ∈ N such that < . One may 2k 2 k=m+1 verify that X ∞ ǫ ǫ x ∈ B ¯ (x1, ) × · · · × B ¯ (x , ) × X ⊆ B (x, ǫ). d1 2 dm m 2m k d Ãk=m+1 ! Y The given product set is τ-open, so τ ′ ≤ τ, and consequently τ = τ ′. ¥

The metric defined in the proof of Prop A33 is not the only metric which could have been used. In Theorem 20.5, Munkres proves that Rω is metrizable using a metric D which would generalize to the following:

d (x , y ) D(x, y) = sup i i i i i ½ ¾ for all x, y ∈ X = (Xi, τi), where di is the standard bounded metric on Xi. These two i∈I metrics are not theY same, but they induce the same topology on the product space X.

Sequential Convergence

One definition of a sequence s on a set X is that it’s a function s : Z+ → X such that

+ s(n) = xn for all n ∈ Z , in which case we usually write s = (xn). A sequence t on X is

+ called a subsequence of s =(xn) if there is a strictly increasing function σ : N → Z such

+ that t = s ◦ σ. If σ(k)= nk for all k ∈ Z , we write t = s(nk)= xnk . A sequence (xn) in X is eventually in a set A ⊆ X iff ∃m ∈ N such that xn ∈ A for all n ≥ m. A sequence (xn) is said to be frequently in A iff ∀k ∈ N, ∃m > k such that xm ∈ A. If(xn) is eventually in every τ-nbhd of x, then we say (xn) τ-converges to x, or

τ x is a τ-limit of (xn), which is often written as (xn) → x.

If (xn) is frequently in every τ-nbhd of x, then we say x is a τ-cluster point of (xn). If

τ (X, τ) is a topological space and τ is induced by some metric d, then (xn) → x iff ∀ǫ > 0,

∃m ∈ N such that d(xn, x) < ǫ ∀n ≥ m. A sequence (xn) is called almost-constant or eventually constant iff ∃x ∈ X and ∃m ∈ N such that xn = x for all n ≥ m.

Prop A34 Let (X, τ) be a topological space.

(a) Every almost-constant sequence converges to the constant.

τ τ (b) If (xn) → x, then every subsequence (xnk ) → x.

µ τ (c) If µ is another topology on X and τ ≤ µ, then (xn) → x =⇒ (xn) → x.

Proof : Homework, maybe.

Examples:

1. In any discrete space, each singleton {x} is open, and in order for a sequence (xn) to con- verge to x, (xn) must be eventually in {x}, which means (xn) must be eventually constant.

2. In any indiscrete space (X, τi), every sequence τi-converges to every point.

Remark: These first two examples reinforce the idea of Prop A34(c) that finer topologies have fewer convergent sequences and coarser topologies allow more convergent sequences.

3. An uncountable set X with τcoc has the same sequential convergence as (X, τd).

4. In an infinite space X with the cofinite topology, a sequence (xn) which has no constant subsequence converges to every point in X, whereas a sequence (xn) which has a constant subsequence converges to the given constant iff the constant subsequence is unique.

1 1 5. In the Sorgenfrey line (R, τs), n τs-converges to 0, but − n diverges. ¡ ¢ ¡ ¢ Useful Lemma: Let(X, τ) be first countable. For each x ∈ X, there is a countable, nested nbhd basis {Un} of x such that U1 ⊇ U2 ⊇ . . . , and if xn ∈ Un for all n, then (xn) → x.

Proof : For x ∈ X, let {Vn n ∈ N} be a countable nbhd basis at x (which exists by the first ¯ countability assumption), and¯ let U1 = V1, U2 = V1 ∩ V2, . . ., Un = V1 ∩ V2 ∩ . . . ∩ Vn, . . ..

Then B = {Un n ∈ N} is the desired nested nbhd basis at x. If xn ∈ Un for all n, then since ¯ the Un’s are nested¯ and form a nbhd basis, it’s (hopefully) clear that (xn) is eventually in any nbhd U of x, which means (xn) → x. ¥

Prop A35 Let (X, τ) be a topological space.

(a) If (X, τ) is T2, then sequential limits are unique.

(b) A first countable space (X, τ) is T2 iff every convergent sequence has a unique limit.

Proof : Homework, maybe.

Recall that if set X is uncountable, then in (X, τcoc), the only convergent sequences are the almost constant sequences. This shows that Prop A35(b) is false without the assumption of

first countability. Recall also that if X is infinite, then in (X, τcof), any sequence which has no constant subsequence converges to every point in X. This shows that convergent sequences in T1 spaces need not have unique limits. One can, however, prove that a topological space

(X, τ) is T1 iff every almost constant sequence converges only to the given constant.

Prop A36 Let (X, τ) be a topological space, and let A,B be subsets of X.

(a) If (xn) is a sequence in B and (xn) → x, then x ∈ Clτ B.

(b) If (X, τ) is first countable, then Clτ B = {x ∈ X ∃(xn) → x with xn ∈ B for all n}. ¯ (c) If A is closed, xn ∈ A for all n, and (xn) → x, then¯ x ∈ A.

(d) If(X, τ) is first countable, then A is closed iff (xn) → x and xn ∈ A for all n ⇒ x ∈ A. ¡ ¢ Proof : (d) follows from (b) and (c), which are left as homework. To prove (a), sssume

(xn) → x with xn ∈ B for all n. Then for any nbhd U of x, (xn) is eventually in U, and since xn ∈ B for all n, it follows that U ∩ B =6 φ. Since U was arbitrary, x ∈ Clτ B. ¥ In any topological space, the closed sets uniquely determine the topology, and for first countable spaces, it follows from Prop A36(d) that the convergence of sequences uniquely determines the topology. This is not true in general, since (as we’ve seen) the discrete and cocountable topologies on an uncountable set have the same sequential convergence. Note that τd is first-countable while τcoc is not. Prop A36(d) assures us that there is no other first countable topology with the same sequential convergence as τd.

Prop A37 Let f :(X, τ) → (Y, µ) be a function between topological spaces.

τ µ (a) If f is continuous and (xn) → x, then (f(xn)) → f(x).

τ µ (b) If (X, τ) is first countable, then f is continuous iff (xn) → x ⇒ (f(xn)) → f(x).

Proof : Essentially the same as that of Theorem 21.3 in Munkres. Munkres proves many of these ideas with first countability replaced by metrizability. Since metrizability implies first countability, our results imply his. He does remark about this at the top of page 131.

Prop A38 Let (X, τ)= (Xi, τi). If (xn) is a sequence in X, then i∈I Y τ τi (xn) → x iff (pi(xn)) → pi(x) for all i ∈ I.

Proof : (⇒): Follows by Prop A37(a), since all the projection maps pi are continuous.

τi (⇐): Assume (pi(xn)) → pi(x) for all i ∈ I. Let U be a basic nbhd of x. Then U is of the form Ui, where Ui =6 Xi for only finitely many i. Let F = {i1,i2,...,ik} be the finite i∈I Y index set for which Ui =6 Xi. For each i ∈ F , there is (by assumption) an mi such that pi(xn) ∈ Ui for all n ≥ mi. If we let m = sup{mi}, then xn ∈ U for all n ≥ m. So(xn) is i∈F eventually in every nbhd of x, and therefore (xn) → x. ¥

Remarks: Recalling that members in the product space X are functions on the index set

I, Prop A38 justifies calling the product topology τ the topology of pointwise (or com- ponentwise) convergence. Note that the above proof of (⇐) would not work for the box topology, since the given proof requires that only finitely many of the Ui are restricted. n In a real analysis, one typically learns that a sequence (xk)=(x1k, x2k,...,xnk) in R con- verges to x = (x1, x2,...,xn) iff (xjk) → xj in R for all j = 1, 2,...,n. This result is a special case of Prop A38 for finite products. In general, when one talks about pointwise convergence of a sequence of functions in analysis, the underlying topology is the product topology on the appropriate function space.

At the end of §21, Munkres defines uniform convergence of a sequence of functions (fn) which map from a set X to a metric space Y . The theorem that follows (Theorem 21.6) is commonly presented in real analysis courses. Note that the functions fn : X → Y can

X be interpreted as points xn in the product space Y = Y . Munkres remarks about this i∈X after proving Theorem 21.6; he points out that theY condition

(fn) → (f) uniformly is equivalent to the condition

τρ¯ (fn) → (f)

X when fn and f are considered as elements of the metric space (Y , τρ¯).