I was asked to prove the following theorem:
A topological space if $T_1$ if and only if the following holds:
For any subset $A$ of $X$, $x$ is a limit point of $A$ if and only if every neighborhood of $x$ contains infinitely many points of $A$.
I know how to prove the $\rightarrow$ direction, but given the equivalence of limit point vs. number of points it intersects with $A$, I cannot think of how we can link this to the $T_1$ condition.
Assume that if $A\subseteq X$ and $x$ is a limit point of $A$ then every open neighbourhood $U$ of $x$ contains infinitely points of $A$.
Recall that one of the many equivalent formulations of $X$ being $T_{1}$ is that all singletons in $X$ are closed. Then let $x\in X$ and consider the singleton set $\{x\}$. If $\{x\}$ is not closed then $\{x\}$ has some limit point $y\neq x$ in $X$. Then every open neighbourhood of $y$ contains infinitely points of $\{x\}$, but thats clearly impossible. Therefore $\{x\}$ is closed and hence $X$ is $T_{1}$.