4.5Urysohn's Metrization Theorem

Proof.

We show $X$ is metrizable by embedding it into a metrizable space $Y$. If $f$ is the appropriate embedding, note that $(X \cong f(X)) \subseteq Y$. Since $f(X)$ is a subspace of the metric space $Y$, $f(X)$ takes on a metric topology, and by homeomorphism so does $X$. As aforementioned, we let $Y$ be the space $\bR^\omega$ together with its product topology, and we show there is an embedding of $X$ into $[0, 1]^\omega$.

We first prove that there exists a countable collection of continuous functions $f_n: X \to [0, 1]$ such that, for any point $x_0 \in X$ and any neighborhood $U$ of $x_0$, there exists some $n \in \bZ_+$ such that $f_n$ is positive at $x_0$ and vanishes outside of $U$. Since $X$ is regular and second-countable, it is normal. By Urysohn's lemma, for any fixed $x_0 \in X$ and neighborhood $U$ of $x_0$, the disjoint closed sets $\{x_0\}$ and $X \setminus U$ can be separated by a continuous function $f: X \to [0, 1]$ such that $f(x_0) = 1$ and $f(X \setminus U) = \{0\}$. We now go one step further to show that we can extract a single countable collection of such functions that works for every possible $(x_0, U)$ pair simultaneously.

Let $\{B_n\}_{n \in \bZ_+}$ be a countable basis for $X$. For each pair of indices $m, n \in \bZ_+$ for which $\overline{B}_n \subseteq B_m$, we apply Urysohn's lemma to separate the disjoint closed sets $\overline{B}_n$ and $X \setminus B_m$ by a continuous map $g_{m, n}: X \to [0, 1]$ satisfying $g_{m, n}(\overline{B}_n) = \{1\}$ and $g_{m, n}(X \setminus B_m) = \{0\}$. Because this collection $\{g_{m, n}\}$ is indexed by a subset of $\bZ_+ \times \bZ_+$, it is countable and can be reindexed as $\{f_n\}_{n \in \bZ_+}$.

To see that this collection satisfies the requirement, let $x_0 \in X$ and let $U$ be an open neighborhood of $x_0$. By the definition of a basis, there is some $B_m$ such that $x_0 \in B_m \subseteq U$. By regularity of $X$, there is some basis element $B_n$ such that $x_0 \in B_n \subseteq \overline{B}_n \subseteq B_m$. The function $g_{m, n}$ is therefore defined for this pair; it yields $g_{m, n}(x_0) = 1 > 0$ and vanishes on $X \setminus B_m \supseteq X \setminus U$, completing the proof.

Now consider the map $F: X \to \bR^\omega$ defined by $F(x) = (f_k(x))_{k \in \bZ_+}$. At once we note that $F$ is continuous since each component is continuous, a nice property of the product topology. Moreover, it is injective: for any $x \ne y$ in $X$, there exists an open neighborhood $U$ of $x$ that excludes $y$ (by $X$ being $T_1$). Then there is some $n \in \bZ_+$ such that $f_n(x) > 0$ and $f_n(X \setminus U) = 0$ (and so $f(y) = 0$), meaning $F(x) \ne F(y)$. Thus $F$ is a continuous bijection onto its image $Z = F(X)$, and our final task is to show $F$ maps open sets to open sets.

Suppose $O \subseteq X$ is open: we seek to show $F(O)$ is open in $\bR^\omega$. Since $F$ is a bijection, there is some unique $x_0 \in X$ such that $F(x_0) = z_0$. Let $N \in \bZ_+$ be the index for which $f_N(x_0) > 0$ and $f_n(X \setminus O) = \{0\}$. Finally, let

Note that $(0, \infty)$ is open in $\bR$, and by continuity of the projection map $\pi_N$, we have that $V$ is an open subset of $\bR^\omega$. Thus $W$ is an open subset of $Z$ in its subspace topology. We show that

implying that $F(O)$ is open. Since $f_N(x_0) > 0$, we have that $z_0 \in [\pi_N^{-1}((0, \infty)) = V]$, and $z_0 \in Z$ is immediate by $F$ being a bijection—thus $z_0 \in W$. Now let $w \in W$ be arbitrary. Since $Z \subseteq W$, there is some $x \in X$ such that $F(x) = z$, and so $\pi_N(z) \in (0, \infty)$. Then

meaning $f_N(x) > 0$ and so $x \in O$. Thus $w \in F(O)$, meaning $W \subseteq F(O)$, and so $F(O)$ is open. Thus $F$ is a homeomorphism onto an image (an embedding), completing the proof that $X$ is metrizable. ❦