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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

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    $\begingroup$ 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. $\endgroup$ – Rob Arthan Jun 17 '17 at 21:46
  • $\begingroup$ What is $A'$ here? It is not a standard notation for anything in particular. $\endgroup$ – DanielWainfleet Jun 18 '17 at 0:18
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    $\begingroup$ @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. $\endgroup$ – Demophilus Jun 18 '17 at 1:43
  • $\begingroup$ @DanielWainfleet it is the set of limit points of A, called the derived of A. it is standard notation as far as im concerned $\endgroup$ – Grassy LittleRoot Jun 18 '17 at 2:12
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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.

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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.

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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 $p\not \in A'.$

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