Proposition

Let $X$ be a topological space and let $\mathscr{U}$ be an open cover of $X$.

Suppose we are given a basis for each $U \in \mathscr{U}$ (when considered as a topological space in its own right). Show that the union of all those bases is a basis for $X$.
Show that if $\mathscr{U}$ is countable and each $U \in \mathscr{U}$ is second countable, then $X$ is second countable.

Solution

1

For each $U \in \mathscr{U}$, let $\mathcal{B}_U$ denote a basis for $U$. Let $\mathcal{B} = \bigcup_{U \in \mathscr{U}} \mathcal{B}_U$. We will show that $\mathcal{B}$ is a basis for $X$.

Every element in $\mathcal{B}$ is an open subset of some $U \in \mathscr{U}$. $\mathscr{U}$ is a collection of open subsets of $X$. Thus each element in $\mathcal{B}$ is an open subset of open subset of $X$. We have shown that this implies that each element in $\mathcal{B}$ is open in $X$
Let $V \subset X$ be open and $x \in V$. Then $x \in U$ for some $U \in \mathscr{U}$. Then $x \in U \cap V$. Since $U \cap V$ is an open subset of $U$, $x \in B \subset U \cap V$ for some $B \in \mathcal{B}_U$. Moreover, such a $B$ is in $\mathcal{B}$.

Therefore, $\mathcal{B}$ is a basis for $X$. (See Lemma 13.2 of Munkres)

2

For each $U \in \mathscr{U}$, let $\mathcal{B}_U$ denote a countable basis. The existence of such a basis is guaranteed because each $U \in \mathscr{U}$ is second countable. As shown above, the union of $\mathcal{B}_U$’s forms a basis for $X$. Since $\mathscr{U}$ is countable, it is the countable union of countable sets, so it is countable.