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**I-Think** Need help verifying/completing proof

Here's a definition before the question

Separable: A metric space is separable if it has a countable dense subset

Base: A collection $\displaystyle [V_\alpha]$ of open sets of metric space $\displaystyle X$ is called a base for $\displaystyle X$ if the following is true: For every $\displaystyle x\in{X}$ and every open set $\displaystyle G\subset{X}$ such that $\displaystyle x\in{G}$, we have $\displaystyle x\in{V_{\alpha}}\subset{G} $

Question

Prove every separable metric space has a countable base

Proof

Let $\displaystyle E$ be a countable dense subset of metric space $\displaystyle X$ and let $\displaystyle [V_{\alpha}]$ be a base for $\displaystyle X$

Let $\displaystyle G$ be a neighborhood of $\displaystyle e$: $\displaystyle d(x,e)<q$, $\displaystyle q\in{\mathbb{Q}}, e\in{E}$

$\displaystyle E$ dense, so either $\displaystyle \forall{x\in{E}}$ (in which case $\displaystyle X$ countable, so proof follows) or

$\displaystyle \forall{x\in{E}}$ are limit points of $\displaystyle X$

So $\displaystyle x\in{[V_\alpha]}\subset{G}$, for some $\displaystyle \alpha $and $\displaystyle \forall{x}$

There are a countable number of sets $\displaystyle G$, so the base is countable

QED