HOMEWORK 8

(1) Munkres §30 Exercise 5 (a) Every metrizable, separable space has a countable basis.

Proof: Suppose X is metrizable and Q is a countable dense subset of X. Let d be a metric on X that induces the topology + with which X is equipped. Let n ∈ Z , and for each x ∈ Q, 1 let B (x) be the ball of radius centered at x. Since the set n n An = {Bn(x)|x ∈ Q} is indexed by Q, An is countable. Let [ B = An. Then B is countable as the countable union of + n∈Z countable sets. The claim is that B is basis for the topology on X. First, let x ∈ X. Fix n > 0. Since Q is dense in X, there 1 exists a q in Q such that d(x, q) < ; that is x ∈ Bn(q) ∈ B. n This verifies the first basis axiom. Second, suppose Bn(x) and Bm(y) are two members of B that have a nonempty intersection I. Let z ∈ I. Choose k so that 1  1 − d(x, z) 1 − d(y, z) < min n , m , k 2 2

and choose p ∈ Q so that z ∈ Bk(p) (since Q is dense). It will be shown that Bk(p) ⊂ I. Let q ∈ Bk(p). We have d(q, x) < d(q, p) + d(p, x) < d(q, p) + d(p, z) + d(z, x) 1 1 < + + d(z, x) k k 2 < + d(z, x) k  1 1  < min − d(x, z), − d(y, z) + d(x, z) n m 1 ≤ . n

Hence, q ∈ Bn(x). Similar computations show that q ∈ Bm(y). Thus, q ∈ I, and one can conclude that Bk(p) ⊂ I, verifying the second basis axiom. 

(b) Every metrizable Lindelof space has a countable basis.

Proof: Let X be metrizable and Lindelof. Let d be a metric on X that induces the topology with which X is equipped. For + each n ∈ Z , x ∈ X, let Bn(x) be defined as in (a), and observe that these balls cover X, so since X is Lindeloff, there exists a countable subcollection An of these balls that also covers X. [ Let B = An, which is countable as the countable union + n∈Z 1 2 HOMEWORK 8

+ of countable sets. If x ∈ X, then for any n ∈ Z , An covers X, so there is some ball in An, and hence in B, that contains x. Next, suppose x ∈ Bn(x1) ∩ Bm(x2) for two elements x1, x2 in 1 1  1 1  X. Let < min − d(x, x ), − d(x, x ) . Let x ∈ X k 2 n 1 m 2 0 such that x ∈ Bk(x0). Let y ∈ Bk(x0) be arbitrary. Then we have

d(y, x1) ≤ d(y, x0) + d(x, x0) + d(x, x1) 2 ≤ + d(x, x ) k 1 1 ≤ . n This implies that Bk(x0) ⊂ Bn(x1). Similar computations show that Bk(x0) ⊂ Bm(x2), and hence Bk(x0) ⊂ Bn(x1) ∩ Bm(x2), so that B is in fact a basis. 

(2) Prove that if X has unique limits and is first countable, then X is Hausdorff.

Proof: Assume the given hypotheses for X. Let x and y be + + distinct points in X. Let A = {An|n ∈ Z } and B = {Bn|n ∈ Z } be countable bases at x and y respectively, with the proviso that Bn 6= Bm if n 6= m. The latter proviso is admissible, since one can remove redundant elements from a countable basis without changing the countability or the status of being a basis. Let n1 = 1. Having chosen n1, n2, . . . , nj, let nj+1 be the least positive integer

not included in {n1, n2, . . . , nj} such that Anj+1 ⊂ Anj , if such an integer exists; otherwise let nj+1 = nj. This furnishes a decreasing

(with respect to set inclusion) subsequence {Anj } of A. Construct

an analogous subsequence {Bmj } of B. For each positive integer j,

let Ij = Anj ∩ Bmj . Since X is not Hausdorff, each Ij is nonempty, so choose an element xj ∈ Ij. Let U be an open subset of X that contains x. Since A is a countable basis at x, there exists a positive

integer j such that Anj ⊂ X. Then since X ⊃ Anj ⊃ Anj+1 ⊃

Anj+2 ⊃ · · · , it follows that U contains all xk for k ≥ j. Since U was an arbitrary open set, it follows that {xk} coverges to x. Let V be an open subset of X that contains y. Since B is a countable basis at y, there exists a positive integer

l such that Bnl ⊂ X. Then since X ⊃ Bnl ⊃ Bnl+1 ⊃ Bnl+2 ⊃ · · · , it follows that U contains all xk for k ≥ l. Since V was an arbitrary open set, it follows that {xk} coverges to y. This contradicts the uniqueness of limits. Hence, X is Hausdorff. 

(3) Munkres §30 Exercise 4: Every compact metrizable space has a countable basis.

Proof: This follows from 5(b) above: Every compact space is Lindelof, since finite covers are in particular countable covers. But here is an argument carried out independently of 5(b). Let X be a compact metrizable space with a countable basis. Since X is metrizable, let d be a metric on X that induces the + topology with which X is equipped. Let n ∈ Z be fixed. To each 1 1 x ∈ X, let B(x; ) be the ball centered at x with radius . Then n n HOMEWORK 8 3

[ 1 B(x; ) contains X as a subset, since in particular for each n x∈X 1 1 x ∈ X, x ∈ B(x , ). Thus, {B(x; )|x ∈ X} is an open cover of 0 0 0 n n X. Since X is compact, the latter cover admits a finite subcover, so 1 there exists a finite subset A ⊂ X such that A = {B(x; )|x ∈ A} n n + is a cover of X. Since n was arbitrary in Z , one can construct such + [ a collection An for each n ∈ Z . The claim is that B = An is + n∈Z a basis for the topology on X. By proving the claim, one proves the theorem, since B is countable as a countable union of finite sets. + Let x ∈ X, n ∈ Z . Since An is a cover of X, there exists some 1 1 x0 ∈ X such that x ∈ B(x0; ). Since B(x0; ) ∈ B, this verifies n n one of the two basis axioms. For the second one, recycle “n00 and 1 1 suppose B(x; ) and B(y; ) are two members of B that have a n m nonempty intersection I. Let z ∈ I. Choose k so that 1  1 − d(x, z) 1 − d(y, z) < min n , m , k 2 2 1 and choose p ∈ X so that z ∈ B(p; ). It will be shown that k 1 1 B(p; ) ⊂ I. Let q ∈ B(p; ). We have k k d(q, x) < d(q, p) + d(p, x) < d(q, p) + d(p, z) + d(z, x) 1 1 < + + d(z, x) k k 2 < + d(z, x) k  1 1  < min − d(x, z), − d(y, z) + d(z, x) n m 1 ≤ . n 1 1 Hence, q ∈ B(x; ). Similar computations show that q ∈ B(y; ). n m 1 Thus, q ∈ I, and one can conclude that B(p; ) ⊂ I, verifying the k second basis axiom.