TITLE: Set of limit points of a subset of a Hausdorff space is closed.
QUESTION [13 upvotes]: Let $X$ be a Hausdorff space and $A\subset X$.  Define $A'=\{x\in X\mid x\text{ is a limit point of }A\}$.  Prove that $A'$ is closed in $X$.
Relevant information: 
(1.) Every neighborhood of a point $x\in A'$ contains a point $y\in A'$ distinct from $x$ (in fact, in a Hausdorff space, every neighborhood of $x$ contains $\infty$-many points of $A$ distinct from $x$)
(2.) In a Hausdorff space, every sequence has a unique limit.
On first glance, it should be an easy proof, but I've made little progress.  I was planning on showing $\overline{A'}=A'$.  Firstly, $A'\subset \overline{A'}$ trivially.  To show $\overline{A'}\subset A'$, proceed by contradiction.  Assume there exists $x\in\overline{A'}$ such that $x\not\in A'$.  This should yield an easy contradiction but I don't see it.  In particular, I'm unsure if the fact that $x\in\overline{A'}$ implies that there actually exists a sequence in $A'$ converging to $x$.  By (2) we know all sequences have unique limits, but do we know that elements in the closure are limits of sequences?  If this is true, it should yield an easy contradiction.  Any help?

REPLY [2 votes]: Show that the complement is open.  Let $x \in X \setminus A'$.  Then there is an open neighborhood $U$ of $x$ such that $U \cap A$ is either empty or is $\{x\}$. If $U$ contained any members of $A'$, then $U$ would intersect $A$ at infinitely many points since $X$ is Hausdorff, so $U \subseteq X \setminus A'$.