Bases having countable subfamilies which are bases in second countable space

Question

I don't understand a proof right at the beginning of this document found here.

This proof is on why any base for the open sets in a second countable space has a countable subfamily that is a base.

My trouble comes in the second paragraph of the first proof. It says $\{B_{i_x}\mid x\in C_k\}$ is a countable set whose union is $C_i$. First, why is the set countable? If you take a set $B_{i_x}$ for any $x\in C_k$, what if $C_k$ is not countable itself? And why is the union $C_i$? Should it instead be $C_k$? I don't see $C_i$ mentioned anywhere in the second paragraph.

Answer 1

Another proof: let $\mathcal{B} = \{B_i \mid i \in I \}$ be any base for $X$ and $\mathcal{C} = \{C_n \mid n \in \mathbb{N} \}$ be a countable base for $X$. Consider the set of pairs $$\mathcal{P} = \{ (n,m) \in \mathbb{N} \times \mathbb{N} \mid \exists i \in I: C_n \subset B_i \subset C_m \, \}$$

which is countable, as a subset of all pairs from a countable set. For each $(n,m) \in \mathcal{P}$ fix $i = i(n,m)$ as the $i$ in the definition of $\mathcal{P}$. The set $\mathcal{B}' = \{ B_{i(n,m)} \mid (n,m) \in \mathcal{P} \,\}$ is thus a countable subfamily of $\mathcal{B}$ and is the required base:

Let $O$ be any open subset of $X$ and let $x \in O$. As $\mathcal{C}$ is a base, there exists some $m \in \mathbb{N}$ such that $x \in C_m \subset O$, and applying $\mathcal{B}$ is a base we find $j \in I$ such that $x \in B_j \subset C_m$, and again applying that $\mathcal{C}$ is a base, we find $C_n$ so that $x \in C_n \subset B_j$.

Now note that $(n,m)$ is in $\mathcal{P}$ as witnessed by $j$. We don't know that $j = i(n,m)$ but we don't care, because for $i = i(n,m)$ we also have $x \in B_i \subset O$, as required. So we have found a member of $\mathcal{B}'$ between every open $O$ and its elements, showing $\mathcal{B}'$ is a base.

Answer 2

You start with a base $\mathcal{B}=\{B_{\alpha}\mid \alpha\in \Gamma\}$, and a countable base $\mathcal{C}=\{C_i\mid i\in\mathbb{N}\}$, the latter guaranteed by second countability.

Now you fix $C_k\in\mathcal{C}$. You can find a subset $I_k\subseteq\Gamma$ such that $\cup_{i\in I_k}B_i = C_k$, because $\mathcal{B}$ is a basis. And for each $x\in C_k$, you can pick a single $i_x\in I_k$ such that $x\in B_{i_x}$. And since the base $\mathcal{C}$, you can in turn pick $j_x\in\mathbb{N}$ such that $x\in C_{j_x}\subseteq B_{i_x}$.

Since $\mathcal{C}$ is countable, the set $\{ C_{j_x}\mid x\in C_k\}$ is in fact countable: there are probably many different $x$s that yield the same index $j_x$, but there are only countably many distinct indices. So let $J=\{j_x\mid x\in C_k\}\subseteq \mathbb{N}$. Then for each $j\in J$ there is an $x_j\in C_k$, and we take $B_{i_{x_j}}$. We only need to consider the set $\{B_{i_{x_j}}\mid j\in J\}$: for every $x\in C_k$, $i_x=i_{x_j}$ for some $j$, so $x\in B_{i_{x_j}}$. Thus, the union of these $B_{i_{x_j}}$ is all of $C_k$, and there are only countably many of them because $J$ is countable and we have one per $j\in J$ only.

Yes, the $C_i$ at the end should be $C_k$.

Answer 3

A shorter proof with reduced amount of indices, hopefully no errors.

Let $ \mathcal{C} = \{C_n : n \in \mathbb{N} \}$ be a countable basis. Let $\mathcal{B} = \{B_i : i \in \Gamma \}$ be any basis.

Let $C \in \mathcal{C}$. Then there exists a set $I \subset \Gamma$ such that $C = \bigcup_{i \in I} B_i$. We define the countable set $S := \{n \in \mathbb{N} : \exists i \in I \text{ such that } C_n \subset B_i \}$. For all $n \in S$ we choose a set $B_n$ such that $C_n \subset B_n$. We claim that $C = \bigcup_{n \in S} B_n$. Indeed, let $x \in C$. Then, there exists an $i \in I$ such that $x \in B_i$. Since $\mathcal{C}$ is a basis there exists $n \in \mathbb{N}$ such that $ x \in C_n \subset B_i$. This shows $n \in S$ and $x \in C_n \subset B_n$.