In the book Munkres: Topology §30 I met the following problem:
Show that if $X$ has a countable basis $\{B_n\}_{n \in \mathbb{Z}_+}$, then every basis $\mathscr{C}$ for $X$ contains a countable basis for $X$. [Hint: For every pair of indeces $n$, $m$ for which it is possible, choose $C_{n, m} \in \mathscr{C}$ such that $B_n \subset C_{n, m} \subset B_m$].
I understand that $\{C_{n, m}\}_{n, m \in \mathbb{Z}_+}$ is countable collection of sets, but construction of this collection is not obvious for me. Can you give me some other hint, please?
Best Answer
The suggestion is this: whenever the ordered pair $\langle n,m\rangle\in\Bbb Z_+\times\Bbb Z_+$ is such that $B_n\subseteq B_m$ and there is a $C\in\mathscr{C}$ such that $B_n\subseteq C\subseteq B_m$, let $C_{n,m}$ be that set $C$. (If there’s more than one $C\in\mathscr{C}$ such that $B_n\subseteq C\subseteq B_m$, just pick one of them to be $C_{n,m}$.) If there is no such $C\in\mathscr{C}$, let $C_{n,m}=\varnothing$. Then let $\mathscr{C}_0=\{C_{n,m}:\langle n,m\rangle\in\Bbb Z_+\times\Bbb Z_+\}$. Of course some members of $\mathscr{C}_0$ may be empty, but at any rate $\mathscr{C}_0$ is definitely just a countable subset of $\mathscr{C}$. Let $\mathscr{C}_1=\{C\in\mathscr{C}_0:C\ne\varnothing\}$; $\mathscr{C}_1$ is of course still a countable subset of $\mathscr{C}$, and the problem is to prove that $\mathscr{C}_1$ is a base for the topology.
To prove this, let $x\in X$ be arbitrary, and let $U$ be any open nbhd of $x$; we must prove that there is a $C\in\mathscr{C}_1$ such that $x\in C\subseteq U$. Since $\mathscr{B}$ is a base for the topology, there is some $B_m\in\mathscr{B}$ such that $x\in B_m\subseteq U$. Since $\mathscr{C}$ is a base for the topology, there is a $C\in\mathscr{C}$ such that $x\in C\subseteq B_m$. And since (again) $\mathscr{B}$ is a base for the topology, there is a $B_n\in\mathscr{B}$ such that $x\in B_n\subseteq C$. Can you finish it from here?