Criteria for a limit point in a certain topological space Let $(X,\tau)$ be a topological space such that for each $x$ in $X,$ $\{x\}$ is closed in $X$. Then prove that an element $x$ is a limit point of a set $A$ iff for each open set $U$ containing $x,$ $U \cap A$ is an infinite set.
Is this statement true? I know that if $(X, \tau)$ a Hausdorff space, the statement is true. Should I show that the give topological space is Hausdorff or is there any proof for saying the above statement is true or false? 
 A: If all singletons are closed, then all finite sets $F \subseteq X$ are closed as well, as closed sets are closed under finite unions.
So if there would be some open neighbourhood $U$ of $x$ such that $U \cap A$ were finite, then $F = (U \cap A) \setminus \{x\}$ is also finite, so closed, and so $X \setminus F$ is open and contains $x$. Then $V = U \cap (X \setminus F)$ is an open neighbourhood $x$ and $V \cap A$ can be empty or $\{x\}$ (as we cut away all other intersection points). So in that case $x$ would not be limit point of $A$.
So we proved by contraposition that if $x$ is a limit point, then $A \cap U$ is infinite for all open neighbourhoods of $x$. The reverse is trivial.
A: Let $x$ be a limit point of $A\subset X$ and $U$ a neighborhood of $x$. Suppose $U\cap A$ is finite, then $V:=U\cap (A\setminus\{x\})$ is also finite so $A\setminus\{x\}$ is closed. But this means $U\setminus V$ is a neighborhood of $x$ which contains no points of $A$ distinct from $x$, contradicting the assumption that $x$ is a limit point of $A$.
The condition of singleton sets being closed is equivalent to the $T1$ separation axiom: given $x\ne y$, there exist neighborhoods $U$, $V$ of $x$, $y$ such that $x\notin U$ and $y\notin V$. This is a weaker condition than the Hausdorff separation axiom. For a counterexample, consider $\mathbb Z$ with the cofinite topology $\mathcal T=\{S\subset\mathbb Z: S\text{ finite or } \mathbb Z\setminus S\text{ finite}\}$.
A: $\lbrace x \rbrace$ being closed for each $x$ doesn't necessarily imply that $X$ is Hausdorff, so trying to prove that $X$ is Hausdorff is a no go. Imitate the proof for the Hausdorff case, this should be quite similar.
