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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!

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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 $X$.

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