# Limit points Topology

I'm trying to prove the following:

Th: A subset of a topological space is closed iff it contains all of its limit points.

Defn of a limit point of a subset $A$ is the following: $p \in X$ is a limit point of $A \subseteq X$ if for every neighbourhood of $p$, $U$ there exists an element $a \in A$ such that $a\in U \cap A$ and $a \neq p$.

Is this adequate for the positive statement?

Assuming we have a closed subset $A\subseteq X$, and $p \in X$ is a limit point of A. Since $A$ is open, $X \setminus A$ is open. Now suppose $p \notin A$, then $X\setminus A$ is a neighbourhood of $p$ so by the definition of a limit point there exists an element $a \in A \cap X\setminus A$ which is not possible so we must have $p \in A$.

For the converse statement I am stuck. I could assume $A$ is open, but if I reach a contradiction then it doesn't immediately imply $A$ is closed. How to go about proving the converse statement?

• Instead assume that $A$ is not closed, which means that $A^c$ is not open. In particular, this means there is some point $y \notin A$ such that every open set containing $y$ also contains a point of $A$. Hence... Oct 18 '14 at 17:40

## 1 Answer

Suppose $A$ is closed. Suppose $x$ is a limit point of $A$. Suppose (for a contradiction) that $x \notin A$. Then, as you noted, $U = X \setminus A$ is open, and contains $x$. So as $x$ is a limit point, $U$ should contain a point of $A$ different from $x$. But this is absurd, as $U$ is by definition disjoint from $A$. Contradiction, so $x \in A$, and $A$ contains all its limit points.

Suppose that $A$ is not closed. This means that $U = X \setminus A$ is not open. So there is some point $x \in U$ that is not an interior point of $U$, so for every open set $O$ that contains $x$, $O \nsubseteq U = X \setminus A$. This means that for every open set $O$ that contains $x$, $O \cap A \neq \emptyset$. And this point of intersection is certainly not equal to $x$ (as $x \in U$, so $x \notin A$ and the intersection point is in $A$). So by definition, $x$ is a limit point of $A$ (in your definition), and $x \notin A$. So $A$ does not contain all its limit points.

This shows the equivalence of being closed and containing all your limit points.