Urysohn Metrization Theorem

The Urysohn metrization theorem gives a clean, sufficient condition for metrizability: a regular space with a countable basis is metrizable. This covers most spaces encountered in analysis and provides the first deep connection between separation axioms, countability, and metric structure.

Statement

Theorem7.8Urysohn Metrization Theorem

Every second-countable $T_3$ space (i.e., regular Hausdorff space with a countable basis) is metrizable. In fact, it embeds into the Hilbert cube $[0, 1]^{\mathbb{N}}$ , hence into $\mathbb{R}^{\mathbb{N}}$ .

Proof Outline

Proof

Let $X$ be a second-countable regular Hausdorff space with countable basis $\mathcal{B} = \{B_1, B_2, \ldots\}$ .

Step 1: $X$ is normal.

Since $X$ is second-countable, it is Lindel"of (every open cover has a countable subcover). A regular Lindel"of space is normal (Theorem 7.5). So $X$ is normal.

Step 2: Construct countably many Urysohn functions.

Since $\mathcal{B}$ is countable, the set of pairs $(B_m, B_n) \in \mathcal{B} \times \mathcal{B}$ with $\overline{B_m} \subseteq B_n$ is countable. Enumerate these pairs as $(B_{m_k}, B_{n_k})$ for $k = 1, 2, \ldots$ .

For each $k$ , the sets $\overline{B_{m_k}}$ and $X \setminus B_{n_k}$ are disjoint and closed. By the Urysohn lemma, there exists a continuous function $f_k: X \to [0, 1]$ with: $f_k|_{\overline{B_{m_k}}} \equiv 0 \qquad \text{and} \qquad f_k|_{X \setminus B_{n_k}} \equiv 1.$

Step 3: The functions separate points and closed sets.

For any $x \in X$ and closed $C$ with $x \notin C$ : the open set $X \setminus C$ contains $x$ , so by regularity there exists $B_n \in \mathcal{B}$ with $x \in B_n \subseteq X \setminus C$ . By regularity again, there exists $B_m \in \mathcal{B}$ with $x \in B_m \subseteq \overline{B_m} \subseteq B_n$ . Then $(B_m, B_n)$ is one of our pairs, say $(B_{m_k}, B_{n_k})$ . We have $f_k(x) = 0$ and $f_k|_C = 1$ (since $C \subseteq X \setminus B_n$ ).

Step 4: Embed into the Hilbert cube.

Define $F: X \to [0, 1]^{\mathbb{N}}$ by $F(x) = (f_1(x), f_2(x), \ldots)$ . Using the metric $d(\mathbf{a}, \mathbf{b}) = \sum_{k=1}^{\infty} 2^{-k} |a_k - b_k|$ on $[0, 1]^{\mathbb{N}}$ :

$F$ is injective: If $x \neq y$ , Step 3 gives $f_k(x) \neq f_k(y)$ for some $k$ .
$F$ is continuous: Each $\pi_k \circ F = f_k$ is continuous, so $F$ is continuous by the universal property.
$F$ is an open map onto its image: For open $U \ni x$ , Step 3 gives $f_k$ with $f_k(x) = 0$ and $f_k|_{X \setminus U} = 1$ . Then $F(U) \supseteq F(X) \cap \pi_k^{-1}([0, 1/2))$ , which is open in $F(X)$ .

Therefore $F: X \to F(X) \subseteq [0, 1]^{\mathbb{N}}$ is an embedding. Since $[0, 1]^{\mathbb{N}}$ is metrizable (with the metric $d$ above), the subspace $F(X)$ is metrizable, and hence $X$ is metrizable.

■

Consequences

Theorem7.9Metrization of Common Spaces

The following spaces are metrizable (by the Urysohn metrization theorem):

Every second-countable manifold (Hausdorff by definition, locally Euclidean gives regularity).
Every compact Hausdorff space with a countable basis.
Every separable, regular, Lindel"of space.
$\mathbb{R}^n$ , $S^n$ , all compact surfaces, and countable products of these.

ExampleMetrizing the Cantor Set

The Cantor set $C$ is a closed subspace of $[0, 1]$ , hence second-countable and regular. By the Urysohn metrization theorem, it is metrizable. In fact, $C \cong \{0, 1\}^{\mathbb{N}}$ (product of discrete two-point spaces), and the Urysohn embedding maps it into the Hilbert cube.

Limitations

RemarkSecond Countability is Essential

The Urysohn metrization theorem does not apply to:

The Sorgenfrey line $\mathbb{R}_\ell$ : regular but not second-countable, and indeed not metrizable.
The long line: regular, locally metrizable, but not second-countable.
$\mathbb{R}^I$ for uncountable $I$ : not even first-countable.

For spaces that are regular but not second-countable, the more general Nagata-Smirnov metrization theorem is needed.

RemarkEmbedding Dimension

The Urysohn theorem embeds $X$ into $[0,1]^{\mathbb{N}}$ (the Hilbert cube), which is a universal metrizable compact space: every second-countable metrizable space embeds in it. This is optimal in the sense that the Hilbert cube is the "smallest" such universal space (any compact metrizable space embeds in it).