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Math 525 More Notes Related to Metrizability in (X, Τ), We Define a To 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.
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