# Set of limit points of a subset of a Hausdorff space is closed.

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?

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Minor typo: every neighborhood of $x$ in $A'$ contains $y$ in $A$ (not $A'$) distinct from $x$. – 1015 Mar 18 '13 at 10:52

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.

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I don't understand why it follows that there exists an open neighbourhood $U$ of $x$ such that $U \cap A \subseteq \{x\}$ because $x$ is not a limit point of $A$. If $x$ is not a limit point of $A$, then either $U \cap A = \varnothing$ or $U \cap A = \varnothing$ AND there exists a neighbourhood $N$ of $x$ such that $U \cap A =$ some set with elements of both $A$ and $N$. So I don't see why $U \cap A \subseteq \{x\}$. – morphic May 18 at 14:59
@ᴇʏᴇs Recall that "$x$ is a limit point of $A$" means that "every (open) neighbourhood $U$ of $x$ contains a point of $A$ different from $x$"; in more symbols "$U \cap ( A \setminus \{x\} ) \neq \varnothing$ for every (open) neighbourhood $U$ of $x$." Then "$x$ is not a limit point of $A$" means that "$U \cap ( A \setminus \{ x \} ) = \varnothing$ for some (open) neighbourhood $U$ of $x$." If $U \cap ( A \setminus \{ x \} ) = \varnothing$, then either $U \cap A = \varnothing$ (if $x \notin A$) or $U \cap A = \{ x \}$ (if $x \in A$); so in any case $U \cap A \subseteq \{ x \}$. – Arthur Fischer May 18 at 15:10

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 $U \subseteq X \setminus A'$.

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If $x\in\overline{A'}$ then every open neighborhood $U$ of $x$ contains a point $y\in A'$. Since $U$ is also a neighborhood of $y$, the T$_1$ property ensures that $U$ contains infinitely many points of $A-\{y\}$. This proves that $x$ is an accumulation point of $A$, i.e. $x\in A'$. Note that Hausdorffness is not necessary for this to work.

Regarding your question about sequences: In general you can not assume that there exists a sequence in $S$ which converges to $x$ if $x\in\overline S$. In first-countable spaces, however, this is true.

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