18.901 Problem Set 4

Chris Laas

May 5, 2000

Problem 1

32.4

This proof follows the lines of Theorem 32.1 very closely; only the first paragraph changes, to this:

Let X be a regular Lindelöf space. Let A and B be disjoint closed subsets of X. Each point x of A has a neighborhood U not intersection B. Using regularity, choose a neighborhood V of x whose closure lies in U. By choosing such a neighborhood for each x in A, we constructa covering of A by open sets whose closures do not intersect B. Since X is Lindelöf, there exists a countable subcovering, which we denote by {U_{n}}.

The rest of the proof is identical to pages 200-201; we have then proven that all regular Lindelöf spaces are normal.

34.8

Let X be a regular Lindelöf space that is locally metrizable. For every point x of X there is a neighborhood U of x which is metrizable in the subspace topology. Since X is regular, there is a neighborhood V of x such that V̄ ⊂ U. Since closed subspaces of Lindelöf spaces are Lindelöf, V̄ is Lindelöf, and since subspaces of metric spaces are metrizable, V̄ is metrizable. Thus, by the result of problem 30.5b, V̄ has a countable basis; thus, V has a countable basis, since V is a subspace of V̄.

Thus, for every point x of X, we can construct a neighborhood V of x with a countable basis. The collection of all these neighborhoods is an open covering of X, and since X is Lindelöf, there exists a countable subcovering of X, {V_{n}}. Thus, since X = ⋃ V_{n}, X is a countable union of open subspaces, each of which has a countable basis; thus, X has a countable basis.

Since X is regular and has a countable basis, X is metrizable, by the Urysohn metrization theorem.

Problem 2: 33.4

Let X be normal and let A be a subset of X.

Existence of f implies A is closed G_{δ}

If there exists a continuous function f : X → [0, 1] such that f(x) = 0 for x ∈ A, and f(x) > 0 for x ∉ A, then A = f^{−1}({0}). Since f is continuous and \{0\} is closed in [0, 1], A is closed in X. Let U_{n} = f^{−1} ([0, \frac{1}{n})) for all n ∈ ℤ_{+}; since [0, \frac{1}{n}) is open in [0, 1], U_{n} is open in X. If x ∈ A, then f(x) = 0 ∈ [0, \frac{1}{n}) for all n; thus, x ∈ ⋂ U_{n}; and thus, A ⊂ ⋂ U_{n}. Conversely, if x ⊂ ⋂ U_{n}, then f(x) ∈ [0, \frac{1}{n}) for all n, and thus f(x) = 0. Thus, x ∈ A; therefore, ⋂ U_{n} = A. Thus, A = ⋂ U_{n}, i.e. A is the intersection of a countable collection of open sets of X, and thus A is a closed G_{δ} set in X.

Lemma

A is closed G_{δ} implies existence of f

Conversely, let us assume that A is a closed G_{δ} set in X. Then A = ⋂ U_{n} for a countable collection {U_{n}} of open sets of X. Then, for each n, A and X − U_{n} are disjoint closed subsets of X. Since X is normal, there exists a continuous map f_{n} : X → [0, \frac{1}{n}] such that f_{n}(x) = 0 for every x in A, and f_{n}(x) = \frac{1}{n} for every x in X − U_{n}, by the Urysohn lemma.

Then, we can construct the function f : X → [0, 1] by

f(x) = sup{f_{n}(x) such that n ∈ ℤ_{+}}

It is trivial to show that f(x) = 0 for all x ∈ A: if x ∈ A, then f_{n}(x) = 0 for all n, and thus f(x) = sup{0} = 0. Likewise, to show that f(x) > 0 for all x ∉ A: if x ∉ A, then x ∉ U_{n} for some n; thus, x ∈ X − U_{n}, and thus f(x) ≥ \frac{1}{n}.

The difficult part is to prove that f is continuous. I’m not sure how to do this.

Problem 3

Lemma: Let X be a compact Hausdorff space. X is metrizable if and only if X has a countable basis.

Since X is compact Hausdorff, it is normal and Lindelöf. If X has a countable basis, then X is metrizable, by the Urysohn metrization theorem. If X is metrizable, then, since X is Lindelöf, X has a countable basis, by the result of problem 30.5b.

Lemma: If X is locally compact Hausdorff, X is regular.

Take any point x of X and a neighborhood U of x. By theorem 29.2, there is a neighborhood V of x such that V̄ is compact and V̄ ⊂ U. Thus, by lemma 31.1, X is regular.

34.4a

Since X is locally compact Hausdorff, X is regular. Thus, since X is regular and has a countable basis, X is metrizable, by the Urysohn metrization theorem.

34.4b and 34.5a

I couldn’t prove these, nor find counterexamples. Lots of scribbled notes and facts though...

34.5b

Since Y is compact, Y is Lindelöf. Thus, if Y is metrizable, Y has a countable basis, by the results of problem 30.5b. Thus X has a countable basis, as a subspace of Y.