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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.

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  • $\begingroup$ This is not true in general. Take $X$ a finite set with the discrete topology or more generally a space $X$ with an isolated point $x$. $\endgroup$ – yamete kudasai Dec 16 '18 at 1:09
  • $\begingroup$ This result is from a Wikipedia article: en.wikipedia.org/wiki/T1_space , at the point where it talks about the equivalent formulation of $T_1$ condition. $\endgroup$ – William Sun Dec 16 '18 at 1:12
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    $\begingroup$ @HeroKenzan A limit point can't be isolated ... $\endgroup$ – Noah Schweber Dec 16 '18 at 1:12
  • $\begingroup$ Suppose $x\ne y$. Are $x$ and $y$ limit points of the set $A=\{x,y\}$? $\endgroup$ – bof Dec 16 '18 at 1:23
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    $\begingroup$ Note that, if $A=\{x,y\}$, the point $x$ certainly does not satisfy the condition "every neighborhood of $x$ contains infinitely many points of $A$". Therefore, if the topology is such that "$x$ is a limit point of $A$ if and only if every neighborhood of $x$ contains infinitely many points of $A$," then we can conclude that $x$ is not a limit point of the set $\{x,y\}$. What does that tell you about the topology? $\endgroup$ – bof Dec 16 '18 at 1:31
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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 $T_{1}$.

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