Revisiting Metrizability

We are now ready to answer the question: when is a topological space metrizable? While our answer isn’t the most general answer, it covers many useful kinds of spaces that we are interested in.

Theorem 1 (Urysohn Metrizability). Let $K$ be a Hausdorff space that is regular and second-countable. Then $K$ is metrizable.

We will follow the proof in this post.

Lemma 1. $K$ is normal.

Proof. Since $K$ is second-countable, let $\mathcal B$ be a countable basis for $K$ . Let $C, D$ be disjoint closed subsets of $K$ .

For each $x \in C$ , there exist a basis element $B_x \ni x$ that is contained in the open set $K \backslash D$ . Since $K$ is regular, we can assume $\bar B_x \subseteq K \backslash D$ without loss of generality. Then $\{ B_x : x \in C \} \subseteq \mathcal B$ yields a countable sub-cover of $C$ . Re-index this collection into $\{ B_i : i \in \mathbb N\}$ .

In a similar manner, obtain a countable sub-cover $\{B_j' : j \in \mathbb N\}$ of $D$ . For each $m,n$ , define the open sets

$\displaystyle U_m := B_m \backslash \bigcup_{i=1}^m \bar B_i,\quad V_n := B_n' \backslash \bigcup_{j=1}^n \bar B_j'.$

It is clear that $\{U_m : m \in \mathbb N\}$ and $\{ V_n : n \in \mathbb N\}$ form open covers of $C, D$ respectively. Defining the open sets

$\displaystyle U := \bigcup_{m=1}^\infty U_m,\quad V := \bigcup_{n=1}^\infty V_n,$

it is not hard to verify that $U \cap V = \emptyset$ , so that $U, V$ form disjoint neighborhoods of $C,D$ respectively, as required.

Lemma 2. There exists a countable collection $\mathcal F \subseteq \mathcal C(K, [0,1])$ of continuous functions such that for any $a \in X$ and any neighbourhood $U_a \ni a$ , there is some $n \in \mathbb N$ such that $f_n(a) > 0$ and $f_n(X \backslash U_a) = \{ 0 \}$ .

Proof. Since $K$ is regular and second-countable, it is normal, and we can apply Urysohn’s lemma eventually.

Since $K$ is second-countable, it has a countable basis $\mathcal B = \{B_i : i \in \mathbb N\}$ . Let $\mathcal I \subseteq \mathbb N \times \mathbb N$ denote the collection of indices $(m, n)$ such that $\bar B_m \subseteq B_n$ . For any $x \in K$ , find a basis element $B_n \ni x$ . Since $K$ is regular, find a basis element $B_m \ni x$ such that $\bar B_m \subseteq B_n$ . Thus, $\mathcal I$ is nonempty.

By this setup, for any $(m, n) \in \mathcal I$ , $\bar B_m$ and $K \backslash B_n$ are disjoint closed subsets of $K$ . Apply Urysohn’s lemma to find a continuous map $f_{m, n} : K \to [0, 1]$ such that $f_{m, n}(\bar B_m) = \{1\}$ and $f_{m, n}(K \backslash B_n) = \{0\}$ . The required countable collection therefore is given by $\mathcal F := \{ f_{m,n} : (m, n) \in \mathcal I \}$ .

Lemma 3. Let $f : K \to L$ be a map that is continuous and bijective. Suppose for any open set $U \subseteq K$ , $f(U) \subseteq L$ is open. Then $f^{-1}$ is continuous. In this case, we say that $f$ is an open map.

Proof. Fix any open set $U \subseteq K$ . Since $f$ is bijective, $(f^{-1})^{-1}(U) = f(U)$ is open, as required.

Now we can prove Theorem 1.

Proof of Theorem 1. By Lemma 2, let $\mathcal F = \{ f_n : n \in \mathbb N\}$ be a countable collection of continuous functions that satisfies the properties detailed in Lemma 2. Define the map $f : K \to \mathbb R^\omega$ by

$f(x) := (f_1(x), f_2(x), f_3(x), \dots).$

Since each $f_i$ is continuous, $f$ is continuous with respect to the product topology endowed on $\mathbb R^\omega$ . It is also injective because $\mathbb R$ is Hausdorff. Since $\mathbb R^\omega$ is metrizable, the subspace $f(K) \subseteq \mathbb R^\omega$ is metrizable. All that remains is to prove that $f^{-1}$ is continuous. By Lemma 3, it suffices to check that $f$ maps open sets in $K$ to open sets in $L$ .

Fix any open set $U \subseteq K$ . To prove that $f(U)$ is open, fix $f(x) \in f(U)$ so that $x \in U$ . We seek a neighbourhood $V$ of $f(x)$ such that $V \subseteq f(U)$ . Since $K$ is regular and second-countable, by Lemma 2, there exists $n \in \mathbb N$ such that $f_n(x) > 0$ and $f_n(K \backslash U) = \{0\}$ . Define the neighbourhood $V := f(K) \cap \pi_n^{-1}((0, \infty))$ of $f(x)$ .

It remains to prove that $V \subseteq f(U)$ . To that end, fix $f(y) \in V \equiv f(K) \cap \pi_n^{-1}((0, \infty))$ . By definition, $f_n(y) = (\pi_n \circ f)(y) > 0$ . By construction, if $y \notin U$ , then $f_n(y) = 0$ . By contrapositive, $y \in U$ , so that $f(y) \in f(U)$ , as required.

We have finally answered the metrizability question, at least partly: when can we obtain a metrizable space? One answer would be: when the space is Hausdorff and regular and second-countable, by Theorem 1. We can ask another useful question: when can we extend continuous functions on a subset $C \subseteq K$ to the whole space? This is the content of the Tietze extension theorem, which we will explore in an exercise.

The next topic of interest brings us back into compactness. When can we obtain spaces that are compact? How can we create compact spaces from non-compact ones? These questions launch us into the idea of compactification, and eventually helps us identify compact collections of functions.

—Joel Kindiak, 30 Apr 25, 1530H

KindiakMath

Revisiting Metrizability

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Revisiting Metrizability

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