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Separable metric space

Ubistvo

New member
Dec 19, 2012
10
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
Feb 7, 2012
2,725
There is nothing wrong with that proof, so why would you want another way?
 

Ubistvo

New member
Dec 19, 2012
10
Ah, I just wanted another point of view. :D