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I already know the following definition for $T_1$ axiom:

$(1)$ Let $X$ a topological space. We say $X$ satisties $T_{1}$ axiom if all the finite subset of $X$ are closed.

Now, I want to prove that definition is equivalent to the following definition:

$(2)$ Let $X$ a topological space. We say $X$ satisfies $T_{1}$ axiom if for all distincts $x,y\in X$, there are neighborhoods $x\in U_{x}, y\in U_{y}$ in $X$ such that $x\notin U_{y}$ and $y\notin U_{x}$.

To prove $(1)\Rightarrow(2)$, we know that $A:=\{x,y\}$ is closed in $X$ for all distincts $x,y\in X$. Suppose that for all neighborhood $U_{x}\ni x,U_{y}\ni y$ we have $x\in U_{y}$ and $y\in U_{x}$. How can I get a contradiction with the fact that $A^{c}$ is open?

To prove $(2)\Rightarrow (1)$, we can show that $\{x,y\}$ is closed in $X$ for all distincts $x,y\in X$, and that is enough, but I don't know how to do that.

Can someone give me some hints?

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$(1)\implies(2)$ Take $x,y\in X$, with $x\neq y$. Since $\{x\}$ is closed, $X\setminus\{x\}$ is open and, since $y\neq x$, $y\in X\setminus\{x\}$. There you have it: $X\setminus\{x\}$ is a neighborhhod of $y$ no which $x$ doesn't belong.

$(2)\implies(1)$ If $x\in X$, then, for each $y\in X\setminus\{x\}$, lete $A_y$ be an open set such that $x\notin A_y$. Then $\bigcup_{y\in X\setminus\{x\}}A_y=X\setminus\{x\}$. This proves that $X\setminus\{x\}$ is open. So, $\{x\}$ is closed. Since this occurs for every singleton…

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