# Proof that any metric space has a $\sigma$-locally finite base

A proof proving that the metric topological space $$(X,d)$$ has a $$\sigma$$-locally finite base:

For every $$x\in X$$ and $$n\in \mathbb{N}$$, consider $$\{B(x,\frac{1}{n})\}_{x\in X, n\in \mathbb{N}}$$, which is a base. since $$X$$ is paracompact, then every open cover $$\{B(x,\frac{1}{n})\}_{x\in X}$$ has a locally finite open refinement $$\mathbb{V_n}$$, which completes the proof.

I understand the paracompact part, so I know that $$\mathbb{V_n}$$ should be a locally finite open cover for every $$n$$, but why the countable union of these refinements must be a base? Could someone give me some ideas about this?

Thank you!

Suppose that $$x\in U$$, where $$U$$ is open; then there is an $$n\in\Bbb Z^+$$ such that $$B\left(x,\frac1n\right)\subseteq U$$. For each $$k\in\Bbb Z^+$$ there is a $$V_k\in\Bbb V_k$$ such that $$x\in V_k$$, and since $$V_k\in\Bbb V_k$$, there is a $$y_k\in X$$ such that $$V_k\subseteq B\left(y_k,\frac1k\right)$$. In particular, $$x\in V_{2n}\subseteq B\left(y_{2n},\frac1{2n}\right)$$. If $$z\in B\left(y_{2n},\frac1{2n}\right)$$, then $$d(x,z)\le d(x,y_{2n})+d(y_{2n},z)<\frac1{2n}+\frac1{2n}=\frac1n$$, so $$z\in B\left(x,\frac1n\right)\subseteq U$$. Thus, $$x\in V_{2n}\subseteq U$$. Since this is the case for every $$x\in X$$ and open nbhd $$U$$ of $$x$$, $$\bigcup_{n\in\Bbb Z^+}\Bbb V_n$$ is a base for $$X$$.