Proposition

Show that a subset of a topological space is closed if and only if it contains all of its limit points.

Solution

Suppose $A$ is closed. Let $x \notin A$. $A^c$ is a neighborhood of $x$ that does not contain a point of $A$. Thus $x$ is not a limit point of $A$. By taking the contrapositive, if $x \in X$ is a limit point of $A$, $x \in A$. In other words, $A$ contains all of its limit points.

Suppose that $A$ contains all of its limit points. Let $x \in A^c$. Then $x$ is not a limit point of $A$, so there must exist an open set $U$ such that $x \in U$ and $U$ does not contain a point of $A$ other than $x$. Since $x \notin A$, $U$ does not contain any point of $A$. This implies that $x \in U \subset A^c$, so $A^c$ is open. Therefore, $A$ is closed.