# Separable metric space

#### Ubistvo

##### New member
Let $(M,d)$ be a metric space. Prove that if $M$ has a countable base, then it's separable.

Let $(G_n)_{n\in\mathbb Z^+}$ be a base of $(M,d),$ for each $n\in\mathbb Z^+$ let $G_n\ne\varnothing$ and let $x_n\in G_n.$ Let $D=\{x_n:n\in\mathbb Z^+,\,G_n\ne\varnothing\},$ then $D$ is countable. Let's prove that $\overline D=M.$ In efect, if $x\in M$ and $V\in V(x),$ then since $\{G_n\}_{n\in\mathbb Z^+}$ is a base, then $\exists n\in\mathbb Z^+$ such that $x\in G_n\subset V$ and $V\cap D\ne\varnothing,$ so since $x_n\in D$ then $x_n\in G_n\subset V,$ so $x\in \overline D,$ and then $D$ is dense in $M.$

Is there another way to solve this?

#### Opalg

##### MHB Oldtimer
Staff member
There is nothing wrong with that proof, so why would you want another way?

#### Ubistvo

##### New member
Ah, I just wanted another point of view.