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Math 731 Homework 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 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 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 2 σ. It must be that O = N \ G for some G 2 τ. Since M is S generated by ρ, we know that G = x2G Bρ(x; x), where x > 0 may depend 2 on x. Now, O = N \ G [ = N \ Bρ(x; x) x2G [ = N \ Bρ(x; x) x2G [ = BρN (x; x): x2N\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 = fBα j α 2 Γg be any base for X, and C = fCi j i 2 Ng be the countable base guaranteed by the second countability of X. Our aim is to show that, for each i 2 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 fA j A 2 BCi for some ig will be a countable subfamily of B that is a base for X. To finish the proof, let Ck 2 C. Since B is a base for X, we can write S Ck = i2I Bi for some subset I of the indexing set Γ. Now, for each x 2 Ck, pick a set Bix 2 B with x 2 Bix and ix 2 I. Since C is also a base for X, we can find some Cix with x 2 Cix ⊂ Bix . It follows that fBix j x 2 Ckg 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 2 A, there is n 2 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 2 A such that, for all n 2 N, b; b + n \ A 6= ;. It follows that the set A n fc 2 A j c ≤ bg 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 [ fpg with neighborhoods y 2 Y unchanged in Y ∗ while neighborhoods of p have the form fpg [ (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 2 Y , a set Ux (called the neighborhood system at x) satisfying 1. If U 2 Ux, then x 2 U. 2. If U; V 2 Ux, then U \ V 2 Ux. 3. If U 2 Ux, then there is a set V 2 Ux such that U 2 Uy for each y 2 V . 4. If U 2 Ux and U ⊂ V , then V 2 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 neigh- borhood system satisfying these requirements for each x 2 Y . Let Up contain all sets of the form fpg [ (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 2 Up, then p 2 U. Proof. By definition, U = fpg [ (Y − L) (some L). Hence, p 2 U. Claim 6.3. If U; V 2 Up, then U \ V 2 Up. Proof. Let U = fpg [ (Y − L) and V = fpg [ (Y − K), where L and K are subsets of Y with compact closure. It follows that U \ V = [fpg [ (Y − L)] \ [fpg [ (Y − K)] = fpg [ (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 2 Up. Claim 6.4. If U 2 Up, then there is a set V 2 Up such that U 2 Uy for each y 2 V . Proof. Let U 2 Up and choose any open V 2 Up with V ⊂ U. Since V is open, there exists, for all y 2 V , an open subset G of V with y 2 G and p2 = G. Hence, G 2 Uy. Since y 2 Y , Uy satisfies the property that any superset of G is also contained in Uy. In particular, U 2 Uy, as desired. Claim 6.5. If U 2 Up and U ⊂ V , then V 2 Up. Proof. Let U = fpg [ (Y − L) for some subset L of Y with compact closure. It follows that V = fpg [ (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 2 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 = fG j G open in Y g and τ = fG \ Y j G open in Y ∗g. Evidently, σ ⊂ τ. We also have that, for any open G in Y ∗, G \ Y = [fpg [ (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.
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