Does a closed set in a metric space contain only it's limit points?
My thinking: A set has interior and boundary points. Interior points will be limit points as every neighbourhood around them will contain at least one point of the set. Boundary points of a closed set are its limit points. A closed set contains all its boundary points. Hence all points in a closed set are limit points.
Is this argument incorrect in any way?
Let $\langle X,d\rangle$ be a metric space, and let $C$ be a closed set in $X$. Then $X\setminus C$ is an open set, so for each $x\in X\setminus C$ there is an $\epsilon_x>0$ such that $B(x,\epsilon_x)\cap C=\varnothing$. Thus, no point of $X\setminus C$ is a limit point of $C$. This shows that $C$ contains all of its limit points.
Conversely, suppose that $A\subseteq X$, and $A$ contains all of its limit points. Then if $x\in X\setminus A$, $x$ is not a limit point of $A$, so there is an $\epsilon_x>0$ such that $B(x,\epsilon_x)\cap A=\varnothing$. Let $U=X\setminus A$. What we just showed can be rephrased as follows: if $x\in U$, there is an $\epsilon_x>0$ such that $B(x,\epsilon_x)\subseteq U$. By definition, then, $U$ is an open set, and $A=X\setminus U$ must be closed.
In short, we’re proved that a subset of $X$ is closed if and only if it contains all of its limit points.
However, a subset of $X$ need not have any limit points, so you can’t make the stronger statement that a subset of $X$ is closed if and only if it is the set of its limit points. In $\Bbb R$, for example, every finite set is closed, but a finite set has no limit points. For a more interesting example, the set $\Bbb Z$ of integers is closed and has no limit points.
No this is not correct. The set $\{0\}\subset \mathbb{R}$ does not contain a limit point, but is closed.
If $X$ is equipped with the (metrizable) discrete topology then limitpoints do not exist in $X$.
