Proposition

Let $X$ be a Hausdorff space, let $A \subset X$, and let $A’$ denote the set of limit points of $A$. Show that $A’$ is closed in $X$.

Solution

Let $x \in X \setminus A’$. Then $x$ is not a limit point of $A$. This implies that $x$ has a neighborhood $U$ such that $(U \setminus \{ x \}) \cap A = \emptyset$. Since $X$ is Hausdorff, $\{ x \}$ is closed. (See Proposition 2.37 on P.32 of Introduction to Topological Manifolds.) Thus $U \setminus \{ x \}$ is open.

Let $y \in U \setminus \{ x \}$. Then $U \setminus \{ x \}$ is $y$’s neighborhood that does not intersect $A$ at all. Therefore, $y$ is not a limit point of $A$.

This implies that $x \in U \subset X \setminus A’$. Therefore, $X \setminus A’$ is open in $X$, so $A’$ is closed in $X$.