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.


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 $\mathbb U$.

  • $\begingroup$ Why is it a countable sub cover of U? $\endgroup$ Aug 2 '19 at 23:24
  • $\begingroup$ 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 $U_B \in \mathbb U'$. $\endgroup$
    – Paul Frost
    Aug 3 '19 at 8:12

$\Bbb B$ countable $\implies \{U_B: B\in\Bbb B\}$ countable.


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