---
tag: KB11T6389G
kind: statement
role: theorem
name: Nagata–Smirnov metrization theorem
msc: ["54E35"]
uses: [0CQHFPGJ45, 03HMJASQBQ, 0S1FJ8VEX9, DZ297MMGV8, 091GJ3FYD7]
about: [091GJ3FYD7]
sources:
  - url: https://en.wikipedia.org/wiki/Nagata%E2%80%93Smirnov_metrization_theorem
same_as:
  wikidata: Q2621667
  wikipedia: Nagata–Smirnov metrization theorem
---

## Statement

Let $X$ be a [[topological space|0CQHFPGJ45]].
A family of subsets of $X$ is *$\sigma$-locally finite* if it is a union of countably many [[locally finite families|DZ297MMGV8]].
Then $X$ is [[metrizable|091GJ3FYD7]] if and only if $X$ is [[regular|03HMJASQBQ]], [[Hausdorff|0S1FJ8VEX9]] and has a $\sigma$-locally finite base.

## Remarks

For a metric space, [[Stone's theorem|7QFKD5JXH6]] gives, for each $n$, a locally finite open refinement $\mathcal{B}_n$ of the cover by balls of radius $1/n$.
The union of the $\mathcal{B}_n$ is a $\sigma$-locally finite base.
For the converse one shows that $X$ is normal and builds countably many locally finite families of continuous functions to $[0,1]$.
These functions embed $X$ into a metrizable product space.
A countable base is $\sigma$-locally finite, since each one-element family is locally finite.
So the [[Urysohn metrization theorem|HBFHDH4EJX]] is a special case.

## Explanations

### L1 human

The theorem gives an exact test for when a topology comes from a distance.
Countably many basic open sets are no longer needed.
It is enough that the basic open sets come in countably many layers, and that near each point only finitely many sets of each layer are present.

### L2 human

Together with the [[Bing metrization theorem|RGCZ2P9QRM]], this is the general solution of the metrization problem.
Its proof uses [[Urysohn's lemma|01EY3VR7ZD]] to build the continuous functions of the embedding.
A useful consequence is that a paracompact Hausdorff space that is locally metrizable is metrizable.