5
$\begingroup$

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 $B$.

$\endgroup$
1
  • $\begingroup$ It is preferred that "countable" means "finite or countably infinite" so that "uncountable" and "not countable" mean the same thing. $\endgroup$ Commented Jun 16, 2021 at 4:50

2 Answers 2

3
+50
$\begingroup$

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 $x\in B'\subseteq U$.

$\endgroup$
2
$\begingroup$

A topological space is called Lindelof if every open cover has a countable sub-cover.

  • Prove that every separable/2nd-countable metric space is Lindelof.
  • Prove that every open subset of a separable metric space is separable (so Lindelof.)

Let let $\{U_n\}_{n\in \mathbb{N}}$ be a countable basis, and let $\mathscr{V}=\{V_\alpha\}_{\alpha\in A}$ be a basis of any cardinality. Then every $U_i$ is a union of some sub-collection of $V_\alpha$. Also, $U_i$ is Lindelof. So $U_i$ is a union of some countable subfamily $\mathscr{F}_i$ of $V_\alpha$-s. Stated precisely, for each $U_i$

  • $\mathscr{F}_i \subseteq \mathscr{V}$
  • $\mathscr{F}_i$ is countable
  • $U_i=\cup \{x: x\in \mathscr{F}_i\}$

Now let $\mathscr{B} = \cup_i \mathscr{F}_i$. Then $\mathscr{B}$ is countable. Each open set $O$ is a union of $U_{i_k}$ for some subsequence $i_k$, and each of these $U_{i_k}$ is a union of members of $\mathscr{B}$. So $\mathscr{B}$ is a basis.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .