# 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

• The statement you are trying to prove is false as you can see by taking $X$ to be any finite set with the discrete topology. – Rob Arthan Jun 17 '17 at 21:46
• What is $A'$ here? It is not a standard notation for anything in particular. – DanielWainfleet Jun 18 '17 at 0:18
• @RobArthan That's not true. In that case any subset $A \subset X$ won't have any limit points at all. This because $\{p\}$ itself is open, so the statement in OP's question becomes vacuously true. – Demophilus Jun 18 '17 at 1:43
• @DanielWainfleet it is the set of limit points of A, called the derived of A. it is standard notation as far as im concerned – Grassy LittleRoot Jun 18 '17 at 2:12

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.

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.

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 $p\not \in A'.$