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Second countability implies the existence of a countable topological base.

Does it also imply that any open set in a second countable space itself is a countable union of basic open sets?

I mean, in a second countable space together with a (not necessarily countable) topological base, can every open set be written as a countable union of basic open sets?

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Given any second-countable space $X$ and any base $\mathcal B$ for $X$, every open $U \subseteq X$ is the union of a countable subfamily of $\mathcal B$.

First fix a countable base $\mathcal D$ for $X$, and let $U \subseteq X$ be open.

Note that there is a $\mathcal B_U \subseteq \mathcal B$ such that $U = \bigcup \mathcal B_U$. Now for each $V \in \mathcal B_U$ and $x \in V$ there is a $W_{x,V} \in \mathcal D$ such that $x \in W_{x,V} \subseteq V$. It follows that $\mathcal D_U = \{ W_{x,V} : V \in \mathcal B_U, x \in V \} \subseteq \mathcal D$, and is therefore countable. Also, $$ U = \bigcup_{V \in \mathcal B_U} V = \bigcup_{V \in \mathcal B_U} \bigcup_{x \in V} \{ x \} \subseteq \bigcup_{V \in \mathcal B_U} \bigcup_{x \in V} W_{x,V} \subseteq U,$$ so $\bigcup \mathcal D_U = U$.

For each $W \in \mathcal D_U$ pick $V_W \in \mathcal B$ such that $W \subseteq V_W \subseteq U$ (e.g., pick any $V$ for which $W = W_{x,V}$ for some $x \in V$). Then $\mathcal B'_U = \{ V_W : W \in \mathcal D_U \}$ is a countable subfamily of $\mathcal B$. Note, too, that $$ U = \bigcup_{W \in \mathcal D_U} W \subseteq \bigcup_{W \in \mathcal D_U} V_W \subseteq U,$$ and so $\bigcup \mathcal B'_U = \bigcup_{W \in \mathcal D_U} V_W = U$.

(In fact, some countable subfamily of $\mathcal B$ is even a base for $X$: just take $\mathcal B' = \bigcup \{ \mathcal B'_U : U \in \mathcal D \}$ where $\mathcal B'_U \subseteq \mathcal B$ is countable with $\bigcup \mathcal B'_U = U$ for each $U \in \mathcal D$.)

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