Why this intersection must be infinite if X is Hausdorff? Let $X$ be a hausdorff space with a topology and $A \subset X$, if $p \in A^{'}$ then why for any open set $G$ we must have $(G-\{p\})\cap A$ infinite? i try to construct a open set that must lie in this insetersection and by the fact that this space is Hausdorff all finite sets are closed so by contradiction i would prove what i want but i get to nowhere
 A: This is not so hard if you try to prove this by contradiction. 
So assume $A \subset X$ and $p$ is a limit point of $A$ (to be clear, $p$ is a limit point of $A$ if and only if for every neighbourhood $O$ of $p$ we have that $(O-\{p\})\cap A$ is non-empty). Let $O$ be any open neighbourhood of $p$. Assume $(O-\{p\}) \cap A$ is finite. Since $X$ is Hausdorff, we have that $(O-\{p\}) \cap A$ is closed. In other words $O':=A^c \cup O^c \cup \{p\}$ is open. Clearly $O'$ is an open neighbourhood of $p$, and thus $O' \cap O$ is also an open neighbourhood of $p$. Then we must have that $(O' \cap O - \{p\}) \cap A$ is non-empty, which is clearly a contradiction. 
A: As p is a limit point, there is some p1 /= p in U1 = (G - {p}) cap A.
Since p in open V1 = G - {p1}, there is some p2 /= p in (V1 - {p}) cap A.  
Continue in this way to construct an infinite sequence of distinct points
within G.  Note that instead of Hausdorf, only T1 is needed.
A: It suffices that $X$ is a $T_1$ space.
For $p\in x$ and $A\subset X,$ suppose there exists open $G\subset X$ such that $p\in G$ and  $B=(G$ \  $\{p\})\cap A$ is finite. Then $B$ is closed .(Because finite subsets of a  $T_1$ space are closed).
So $G'=G$ \ $B$ is open with $p\in G'$ and $(G'$ \ $\{p\})\cap A=\phi,$ so $p\not \in A'.$ 
