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Recently it was explained here that in a second countable topological space $X$, any base admits a countable subfamily which is also a base. I know a base $\mathcal{B}$ covers the space $X$, so $\bigcup\mathcal{B}=X$. I also know that any open cover has a countable subcover.

Just curious about a more general statement now. If we have some $(X,\mathcal{T})$ a second countable space, and $\mathcal{U}\subseteq\mathcal{T}$ is any collection of open sets, is it true that $\mathcal{U}$ admits some countable subfamily $\mathcal{V}$ where $\bigcup\mathcal{V}=\bigcup\mathcal{U}$?

That would be nice if it did, because then you could get as a nice corollary that any base contains a countable base.

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    I'd recommend calling $\mathcal{U}$ a "collection of open sets" rather than a "subset of open sets"; the latter sounds a bit confusing...2011-09-18
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    ok, I have changed it.2011-09-18
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    It is common to denote the union of a set with `\bigcup`, and this is really nothing about set theory.2011-09-18

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