Separable metric spaces have countable bases.

Question

Suppose $X$ is a separable metric space, and $E$ is a countable dense subset of $X$. I want to show that the set $$B=\{B_r(x)\mid (r,x)\in\Bbb Q^+\times E\}$$ is a basis for $X$. We want to show $$X=\bigcup_{(r,x)}B_r (x),$$ The backward inclusion is easy. For the forward inclusion, suppose $y\in X$. Fix $x\in E$, and take $r\in\Bbb Q^+$ so that $d(x,y)<r/2$. Then $y\in B_r (x)$. Fix $p,q\in E$ and $r,s\in\Bbb Q^+$, and consider $x\in B_r(p)\cap B_s(q)$. Since $B_r(p)\cap B_s(q)$ is open, there is a neighbourhood $B_h(x)\subset B_r(p)\cap B_s(q)$ for some $h\in \mathbb{R}^+$. Fix $y\in E$ so that $y\in B_h(x)$. Choose a $h'\in\mathbb{Q}$ so that $d(x,y)<h'<h$. Then $B_{h'}(y)\subset B_r(p)\cap B_s(q)$ and $x\in B_{h'}(y)$. Does this work?

Answer 1

No, it does not work. By that argument, if $x\in X$, then $\{B_r(x)\mid r\in\Bbb Q^+\}$ is a basis of the topolgy of $X$, since $X=\bigcup_{r\in\Bbb Q^+}B_r(x)$. What you are supposed to prove is that, if $A$ is an open subset of $X$, then $A$ can be written as an union of balls of the type $B_r(x)$, with $x\in E$ and $r\in\Bbb Q^+$.

That's not hard. For each $a\in A$, take $r\in\Bbb Q^+$ such that $B_r(a)\subset A$. Since $E$ is dense, there is some $x\in A\cap B_{r/2}(a)$. So, $a\in B_{r/2}(x)$ and $B_{r/2}(x)\subset A$. Can you take it from here?