Relationship metric space and $\sigma$-discrete base

Question

Hy, I am newbie here. Can you help me to prove this proposition?

If $X$ metric space, then there is a $\sigma$-discrete base $\mathcal{U}$ for the topology of $X$, i.e., $\mathcal{U}=\bigcup\{\mathcal{U}_{n}:n\in\mathbb{N}\}$ where each $\mathcal{U}_{n}$ is a discrete family af non-empty open subsets of $X$.

A family $\mathcal{A}$ of subsets of a topological space $X$ is said to be discrete if each $x\in X$ has a neighborhood in $X$ that intersect at most one member of $\mathcal{A}$.

Thank you.

This is half of the Bing metrization theorem. Its proof is fairly non-trivial unless you’ve already proved some fairly high-powered results about metric spaces. Do you know, for instance, that metric spaces are paracompact? — Brian M. Scott, Jul 15 '13 at 06:41

score 2 · Answer 1 · answered Aug 11 '21 at 09:50

Since the Wikipedia article given by Brian M. Scott does not have the proof, and somebody might find this question using a search engine as I just did, I'll give a reference from a text book as an answer.

You can find a proof of this in General Topology by Ryszard Engelking. The statement appears as Theorem 4.4.3 on p. 281, although the work is done using Theorem 4.4.1, that every open cover of a metrizable space has a $\sigma$-discrete open refinement.

Relationship metric space and $\sigma$-discrete base

1 Answers1