# Every separable metric space has a countable base

A collection $\{V_{\alpha}\}$ of open subsets of $X$ is said to be a base for $X$ if the following is true: For every $x\in X$ and every open set $G\subset X$ such that $x\in G$, we have $x \in V_\alpha \subset G$ for some $\alpha$. In other words, every open set in $X$ is the union of а subcollection of $\{V_{\alpha}\}$.

Prove that every separable metric space has a countable base.

Proof: Let $(X,d)$ be a metric space and $D=\{d_j\}_{j\in \mathbb{N}}$ be the countable and dense subset of $X$. Let $U$ is an open set in $X$ then for $x\in U$ $\exists \varepsilon_x>0$ such that $N_{\varepsilon_x}(x)\subset U$. Then $\exists d_j\in D$ such that $d(x, d_j)<{\varepsilon_x}/{4}$ $\Rightarrow$ $d_j\in N_{{\varepsilon_x}/{4}}(x)\subset U$.

It's easy to check that $N_{{\varepsilon_x}/{4}}(d_j)\subset N_{{\varepsilon_x}}(x)$. Indeed, if $z\in N_{{\varepsilon_x}/{4}}(d_j)$ then $d(z,d_j)<\varepsilon_x/4$ and since $d(x, d_j)<{\varepsilon_x}/{4}$ then by triangle inequality: $$d(z,x)\leqslant d(z,d_j)+d(x, d_j)<{\varepsilon_x}/{4}+{\varepsilon_x}/{4}<{\varepsilon_x}.$$ Hence $z\in N_{{\varepsilon_x}}(x)$ and inclusion $N_{{\varepsilon_x}/{4}}(d_j)\subset N_{{\varepsilon_x}}(x)$ holds.

By the way $N_{r}(d_j)\subset N_{{\varepsilon_x}}(x)\subset U$ for $r\in \mathbb{Q}$ and $r<{\varepsilon_x}/{4}$. Thus for every $x\in U$ we associate an open ball with center at some point $d_j$ of $D$ and rational radius $r$ such that $x\in N_{r}(d_j)\subset N_{{\varepsilon_x}}(x)\subset U$.

Thus $$U=\bigcup\limits_{d_j\in U} N_{r_j}(d_j).$$ This set equality can be verified very easy. Inclusion $\subseteq$ I described in above paragraph. Converse inclusion is not difficult: if $z\in \bigcup\limits_{d_j\in U} N_{r_j}(d_j)$ then $z\in N_{r_j}(d_j)$ for some $d_j\in U$. But these balls arranged so that $N_{r_j}(d_j)\subset U$ then $z\in U$.

Hence collection of balls $\{N_{r}(d): r\in \mathbb{Q}, d\in D\}$ is a countable base.

Is my proof correct? Can anyone check it please? Sorry if this topic is repeated but I would like to know is my proof correct.

• Your proof is correct. But you should at some point mention what your candidate for a countable base is and why it's countable. It's pretty clear for somebody who allready knows the result but maybe a little "hidden" to somebody who has no idea about it. – Maik Pickl Jun 3 '16 at 11:55