Prove $M$ has a countable open base if, and only if, $M$ is separable

general-topologymetric-spaces

I'm in my first analysis class and I'm pretty confused about what direction to take this question.
The definitions I'm working with are:

Open Base: $(U_{\alpha})$ is an open base of M if every open subset of $M$ can be written as the union of $U_{\alpha}$.

Dense: A set $A$ is dense in $M$ if if $\bar{A}=M$ $\iff$ $\forall x\in M, \forall\varepsilon>0,$ $B_{\varepsilon}(x)\cap A\not=\phi$

Separable: A metric space $M$ is separable if it contains a countable dense subset.

I got a hit that we should consider $\{x_n\}$ a countable dense subset in $M$ and the collection of balls with rational radii centered at $x_n$, and I how we could potentially use this to construct an open base… but not how to go from an arbitrary open base to separable.

Thanks for any help!

Best Answer

Going from a countable base to separability is actually the easier direction. Let $\mathscr{B}=\{B_n:n\in\Bbb N\}$ be a base for $M$, and for each $n\in\Bbb N$ let $x_n\in B_n$. Let $D=\{x_n:n\in\Bbb N\}$; now use the fact that $\mathscr{B}$ is a base to show that $D$ is dense in $M$.