# Countable subfamily with union equal to that of the containing family

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... – Arturo Magidin Sep 18 '11 at 5:21
ok, I have changed it. – fourtops Sep 18 '11 at 5:26
It is common to denote the union of a set with \bigcup, and this is really nothing about set theory. – Asaf Karagila Sep 18 '11 at 6:31

Yes, this is true. The property you want is equivalent to the space being hereditarily Lindelöf (we apply Lindelöfness to the union of the original family), and second countable spaces are hereditarily Lindelöf, as second countable implies Lindelöf and both properties are hereditary.

So what you want reduces to: let $\mathcal{U}$ be an open cover of a second countable space $X$, then it has a countable subcover (this is the definition of $X$ being Lindelöf).

To see this, fix a countable base $\{ B_n \mid n \in \mathbb{N} \}$ for $X$. For every $x \in X$, fix $U_x \in \mathcal{U}$ that contains it, and $n(x) \in \mathbb{N}$ such that $x \in B_n(x) \subset C_x$. The set $M = \{n(x) \mid x \in X\}$ is countable (subset of $\mathbb{N}$); for many $x, y$ we will have $n(x) = n(y)$. So there is a countable subset $X'$ of $X$ such that $M = \{n(x) \mid x \in X' \}$, by picking (Axiom of choice....) some fixed $x$ for every $n(x)$ that occurs in $M$.

Then clearly, $\{ U_x \mid x \in X' \}$ is a countable subcover of $\mathcal{U}$.

It does not help you directly in the quest for a countable subset of a base that is still a base, in the presence of a countable base, I think, but this is a quite useful fact.

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The same argument applies.

Suppose $\mathcal{U}$ is a family of open sets, and let $\mathcal{C}=\{C_i\mid i\in\mathbb{N}\}$ be a countable base for the topology.

For each $x\in\cup\mathcal{U}$, there exists $U_x\in\mathcal{U}$ such that $x\in U_x$, and there exists $i_x\in\mathbb{N}$ such that $x\in C_{i_x}\subseteq U_x$.

For each $j\in J=\{i_x\in\mathbb{N}\mid x\in\cup\mathcal{U}\}$ there exists $y_j\in\cup\mathcal{U}$ such that $i_{y_j} = j$; in particular, $x_{i_{y_j}}\in U_{j}$. Let $\mathcal{V}=\{U_{y_j}\mid j\in J\}$. $J$ is clearly countable, and we claim that $\cup\mathcal{V}=\cup\mathcal{U}$.

That $\cup\mathcal{V}\subseteq\cup\mathcal{U}$ holds is immediate. For the reverse inclusion, let $x\in\cup\mathcal{U}$. Then $i_x\in J$, and so there exists $y_j$ such that $x\in U_{y_j}\in \mathcal{U}$; hence $x\in\cup\mathcal{U}$, as desired.

(It's the same argument as used in the question you asked before, and as the argument that proves that a second countable space is Lindelöf: every open cover has a countable cover. )

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OK, I'm sensing a pattern here. Thanks for answering both my questions, I'll try to pick this idea up. – fourtops Sep 18 '11 at 5:47