Show that if $X$ is a separable metric space, every basis for the topology on $X$ contains a finite or countable subset that's also a basis.
I know how to show that every separable metric space has a countable basis, but here the problem is showing that every basis for the topology on $X$ has a finite or countable subset that's also a basis.
Let $S$ be a basis for the topology on $X$. Assume $S$ is uncountably infinite; if it's finite or countable, we can take $S$ itself to be the finite or countable subset that's a basis. Let $B=\{b_1,b_2,\cdots\}$ be a countable dense subset of $X$. Then every open ball in $X$ has a point in $B$. Also, I know how to show that the set of open balls centred at points in $B, $ defined by $S' :=\{B(b_i, \frac{1}{a}) : i,a\in\mathbb{Z}^+\}$ is finite or countable and is a basis for $X$ (i.e. a set $O$ is open in $X$ iff for every $o\in O, \exists U\in S'$ so that $a\in U\subseteq O$). I need to find a subset $S_0$ of $S$ so that $S_0$ is a basis and it is finite or countable. I think I should intuitively find some way to "filter out" unnecessary basis elements so that we only have countably many basis elements and this "filtering" will likely involve using the countable dense subset $B$.
Best Answer
You already know that your separable metric space $X$ has a countable base $\mathcal D$ so let's start from there. (In fact we show that any topological space which has a countable base has the property that every base contains a countable base. No need to say "finite or countable" because finite sets are countable.)
Let $\mathcal B$ be any base for the topology of $X$. Define $$\mathcal P=\{(D,E)\in\mathcal D\times\mathcal D:D\subseteq B\subseteq E\text{ for some }B\in\mathcal B\}.$$ For each pair $(D,E)\in\mathcal P$ choose a set $B_{D,E}\in\mathcal B$ such that $D\subseteq B_{D,E}\subseteq E$, and let $$\mathcal B'=\{B_{D,E}:(D,E)\in\mathcal P\}.$$ $\mathcal B'$ is countable because $\mathcal P$ is countable, because $\mathcal P\subseteq\mathcal D\times\mathcal D$ and $\mathcal D$ is countable. We have to show that $\mathcal B'$ is a base for the topology of $X$. Let $U$ be an open set and let $x\in U$; I have to find a set $B'\in\mathcal B'$ such that $x\in B'\subseteq U$.
Since $\mathcal D$ is a base, there is a set $E\in\mathcal D$ such that $x\in E\subseteq U$. Since $\mathcal B$ is a base, there is a set $B\in\mathcal B$ such that $x\in B\subseteq E$. Since $\mathcal D$ is a base, there is a set $D\in\mathcal D$ such that $x\in D\subseteq B$. Since $D\subseteq B\subseteq E$, and since $D,E\in\mathcal D$ and $B\in\mathcal B$, we have $(D,E)\in\mathcal P$. Let $B'=B_{D,E}\in\mathcal B'$. Then $x\in D\subseteq B'\subseteq E\subseteq U$, so $x\in B'\subseteq U$.