Prove a $T_0$ topological group is $T_1$

Question

How to prove that a $T_0$ topological group is $T_1$.

I am a beginner in topological group. Also I want some good reference.

Answer 1

In a topological group, the group operations are continuous. So if you have two points $x \neq y$, and a neighbourhood $U$ of $x$ that does not contain $y$, then $yU^{-1}x$ is a neighbourhood of $y$ that does not contain $x$, where $U^{-1} = \{u^{-1} : u \in U\}$.

We can see that as follows:

$$x \in yV^{-1}x \iff e \in yV^{-1} \iff y \in V$$

for any set $V \subset G$. Since translations are homeomorphisms, $x^{-1}U$ is a neighbourhood of $e$, hence $(x^{-1}U)^{-1} = U^{-1}x$ is also a neighborhood of $e$, and $yU^{-1}x$ is a neighbourhood of $y$.