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?
Best Answer
It is easier to show that $X \setminus A^\prime$ is open. If $x \in X \setminus A^\prime$, then $x$ is not a limit point of $A$, so there is an open neighbourhood $U$ of $x$ such that $U \cap A \subseteq \{ x \}$. It suffices to show that no other point of $U$ (i.e., no point of $U \setminus \{ x \}$) is a limit point of $A$. Since $X$ is Hausdorff,1 it follows that $U \setminus \{ x \} = U \cap ( X \setminus \{ x \} )$ is open. By choice of $U$ we know that $( U \setminus \{ x \} ) \cap A = \varnothing$. Thus $U \setminus \{ x \}$ witnesses that none of its points are limit points of $A$. So $x \in U \subseteq X \setminus A^\prime$.
1 The separation axiom T1 suffices.