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?


$(1)\implies(2)$ 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…


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.