I want to prove the following Lemma. Let $A$ be a locally finite collection of subset of a topological space $X$. Then the collection $B=\{\bar{a}\mid a\in A\}$ of the closure of elements of $A$ is locally finite. I can see that for any $x\in X$, there is an open neighbourhood U which intersects with finitely many $a\in A$'s. However, the closure of $a$ contains the limits points which maybe extra to $a$. So I do not see why the limit points of $a$'s intersection with $U$ must be empty all the time.
[Math] If a collection is locally finite, then the collection of all closures is also locally finite
general-topology
Best Answer
Suppose $a \in A$ such that $\overline a \cap U \ne \varnothing$. Either $U$ contains a point of $a$ and we are done, or $U$ contains a limit point of $a$, but then $U$ is a neighborhood of a limit point of $a$, so it must contain a point of $a$ itself.