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Let $X$ be a metric space. I want to show that: $X$ separable $\Rightarrow X$ second countable.

What I have been able to show thus far:

Let $Y$ be the countable dense subset in $X$.

I have shown for any $x \in X$ and corresponding neighbourhood $N_{x}$ that there exists $y \in Y$ so that $B_{\frac{1}{n}}(y)\subseteq N_{x}$

My question: I do not know whether I am on the right track... I still need to show that some kind of collection $\bigcup_{U \in C}U=N_{x}$ but I am unsure on how to construct it. I do not see how the fact that $B_{\frac{1}{n}}(y)\subseteq N_{x}$ can help.

Any help is greatly appreciated.

SABOY
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1 Answers1

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Let $\{x_i\}_{i\in\mathbb N}$ a dense subset. Then $$\mathcal B=\{B_r(x_i)\mid r\in\mathbb Q, i\in\mathbb N\},$$ is a countable basis of topology. Indeed, let $O$ an open set. Then, for all $x\in O$ there is $r_x>0$ s.t. $B_{r_x}(x)\subset O$. Let $x_i\in B_{r_x}(x)$. You can easily find $\delta _x\in \mathbb Q$ s.t. $B_{\delta _x}(x_i)\subset B_{r_x}(x)\subset O$ and conclude the proof.

Surb
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