# Second countable spaces and countable subcovers

I have an issue with the proof of the following proposition:

Let $$X$$ be a second countable topological space. Then every open cover of $$X$$ has a countable subcover.

Proof:

As $$X$$ is second countable, its topology admits a countable basis that is an open cover for $$X$$. Let $$U$$ be an open cover for $$X$$. Define $$B'$$ to be the subset of the countable basis such that $$B \in B'$$ $$\iff$$ $$B$$ is contained in some element of $$U$$. This is possible because $$\bigcup_{B\in \mathbb{B}}B$$ $$=$$ $$\bigcup_{A \in U}A$$. ($$\mathbb{B}$$ is the countable basis) and so every subset of the countable basis is contained in some set in $$U$$.

Therefore for each element $$B \in \mathbb{B}$$ $$\exists$$ $$U_B \in U$$ such that $$B \subseteq U_B$$

The question I have is, why is the following set countable $$\{$$ $$U_B:$$ B $$\in \mathbb{B}$$ $$\}$$.

Is everything else so far with the proof correct?

A family of sets indexed by a countable set is countable

The idea is correct, but is not properly developped. You define $$\mathbb B'$$ to be the set of all $$B \in \mathbb B$$ which are contained in some $$U \in \mathbb U$$ ( I changed the notation a little bit - $$\mathbb U$$ is the given open cover of $$X$$). This definition has nothing to do with $$\bigcup_{B\in \mathbb{B}}B = \bigcup_{A \in \mathbb U}A$$ - both sides are trivially $$= X$$. However, as a subset of the countable set $$\mathbb B$$ also $$\mathbb B'$$ is countable.
We now show that $$\mathbb B'$$ is a cover of $$X$$. In fact, each $$x \in X$$ is contained in some $$U \in \mathbb U$$. Since $$\mathbb B$$ is a basis, we find $$B \in \mathbb B$$ with $$x \in B \subset U$$. Hence $$B \in \mathbb B'$$ and $$x \in B$$.
Next, for each $$B \in \mathbb B'$$ choose $$U_B \in \mathbb U$$ such that $$B \subset U$$. The set of these $$U_B$$ is a countable subcover of $$\mathbb U$$.
• Since $\mathbb B'$ is countable and for each $B \in \mathbb B'$ we have chosen exactly one $U_B$, the set $\mathbb U' = \{ U_B \mid B \in \mathbb B' \}$ is a countable subset of $\mathbb U$. It is also a cover because each $x \in X$ is contained in some $B \in \mathbb B'$, hence in some $U_B \in \mathbb U'$. Aug 3 '19 at 8:12
$$\Bbb B$$ countable $$\implies \{U_B: B\in\Bbb B\}$$ countable.