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?