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I've been going over previous exams that I've had in topology to study for my comprehensive exam, and I noticed a problem that I missed. I was wondering if anyone could help me out with this problem:

Let $X$ be a Lindelöf topological space and let $A \subset X$. Suppose that every $x \in X$ has a neighborhood $V$ such that $V \cap A$ is countable. Prove that $A$ is countable.

Thanks in advance for any help.

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up vote 2 down vote accepted

For each $x\in X$, take neighborhood $V_x$ such that $V_x\cap A$ is countable. Because $X$ is Lindelöf, a countable subset of the $V_x$ cover $X$, and hence a countable number of the $V_x\cap A$ cover $A$. Thus $A$ is a countable union of countable sets, and therefore countable.

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Sure: Take the union of all the given $V$s: $$S=\cup_{x\in X} V$$ This is an open cover of $X$, so we apply Lindelöf to pass to a countable subcover $$S'=\cup_{i\in \mathbb{N}} V_i$$ Now for $V\in S'$, $V \cap A$ is countable by assumption, and meanwhile $\cup_i V_i \cap A = S \cap A=A$. So we get $A$ as a countable union of countable sets, which is countable by a standard argument-do you know this fact?

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