# Equivalent formulation of $T_1$ condition.

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.

• 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 $x$. – yamete kudasai Dec 16 '18 at 1:09
• 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. – William Sun Dec 16 '18 at 1:12
• @HeroKenzan A limit point can't be isolated ... – Noah Schweber Dec 16 '18 at 1:12
• Suppose $x\ne y$. Are $x$ and $y$ limit points of the set $A=\{x,y\}$? – bof Dec 16 '18 at 1:23
• 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? – bof Dec 16 '18 at 1:31

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 $$T_{1}$$.