I'm trying to prove the following:
Show that a subset of a topological space is closed if and only if it contains all of its limit points.
Is my proof valid?
Definition of limit point:
$p$ is a limit point of a subset, if every neighborhood of $p$ contains a point in the subset other than $p$ (aka accumulation point).
Lets call the subset, $A$.
In this case we will take $A$ is a closed subset as a given. Lets assume that $p$ is a limit point of $A$, and $p \notin A$. Thus $p \in \partial A $ because only at the boundary can a point, not in the set, have every neighborhood with points that ARE in the set.(More specifically, because EVERY neighborhood of $p$ intersects $A$.) However, we are given that $A$ is closed and closed sets contain all their boundary points. Thus, $p$ cannot exist (RAA). So if a subset is closed it must contain all of its limit points.
Now the converse. In this case we will take $A$ contains all of its limit points as a given. Lets assume that $A$ is not closed, and thus does not contain all of its boundary points. Let $b$ be a point such that, $b \in \partial A$ and $b \notin A$. However $b$ must be a limit point, because every neighborhood of a boundary point contains a point in A. Thus $A$ does not contain all of it limit points (RAA). So, if a subset contains all of its limits points it must be closed.