Topological Groups are Hausdorff

A Topological Group is a Group Object in the Category $\mathbf{Top}$ of Topological Spaces and Continuous Functions.

Concretely, this is a topological space $G$, endowed with two continuous functions:

$$\begin{array}{rl}\bullet & :G\times G\to G\\ ({)}^{-1}& :G\to G\end{array}$$

and a distinguished element $e\in G$, satisfying the usual group axioms:

$$\begin{array}{r}e\bullet x=x=x\bullet e\\ x^{-1}\bullet x=e=x\bullet x^{-1}\\ a\bullet (b\bullet c)=(a\bullet b)\bullet c\end{array}$$

An extra assumption, made at least by Munkres, is that $G$ is a $T_1$ space. This means that for any points $x,y\in G,x\ne y$, we have neighborhoods $x\in U_x,y\in U_y$ that don’t contain the other point, i.e. $x\notin U_y,y\notin U_x$.

This extra assumption is enough to show that $G$ is a Hausdorff space. When $x\ne y$, not only can we find neighborhoods $U_x,U_y$ that don’t contain the other point, but that are distinct, with $U_x\cap U_y=\varnothing$.

Hausdorff $=$ closed diagonal

Another characterization of Hausdorff is:

A space $X$ is Hausdorff, if and only if

The set:

$$\Delta :=\{(x,x)|x\in X\}$$

is closed in $X^2$.

Proof:

$⟹$

We show that there exists an open neighborhood $N(x,y)$ around any point $(x,y)$ with $x\ne y$, such that $N(x,y)\cap \Delta =\varnothing$. We then have that:

$$\bigcup _{x\ne y}N(x,y)=X^2-\Delta$$

is open, meaning $\Delta$ is closed.

Since $X$ is Hausdorff, and $x\ne y$, there exists neighborhoods $U_x,U_y$ of $x$ and $y$, respectively, such that $U_x\cap U_y=\varnothing$.

$U_x\times U_y$ is evidently a neighborhood of $(x,y)$. Furthermore, assume $(x^{\prime },y^{\prime })\in U_x\times U_y$. Because $U_x$ and $U_y$ are distinct, we must have $x^{\prime }\ne y^{\prime }$. This means that $U_x\times U_y\cap \Delta =\varnothing$.

This is our neighborhood $N(x,y)$.

$⟸$

If $\Delta$ is closed, then the set $X^2-\Delta$ is open. This means that if $x\ne y$, then we can find a neighborhood $U_x\times U_y\subseteq X^2-\Delta$ of $(x,y)$, with $U_x$, $U_y$ open in $X$, by definition of the product topology.

Since $U_x\times U_y\subseteq X^2-\Delta$, this means that

$$x^{\prime }\in U_x,y^{\prime }\in U_y⟹(x^{\prime },y^{\prime })\notin \Delta$$

That is to say:

$$x^{\prime }\in U_x,y^{\prime }\in U_y⟹x^{\prime }\ne y^{\prime }$$

Which means that $U_x\cap U_y=\varnothing$. Our space is thus Hausdorff.

$\square$

$T_1⟹$ Hausdorff, for a Group

First, note that the function $f=(x,y)\mapsto x{y}^{-1}$ is continuous. This is because it is the composition of two continuous maps:

$$\bullet \circ (id\times ({)}^{-1})$$

For $G$ to be Hausdorff, we need $\Delta$ to be closed. Note that $\Delta =f^{-1}(\{e\})$, since:

$$f^{-1}(\{e\})=\{(x,y)|x{y}^{-1}=e\}=\{(x,y)|x=y\}=\Delta$$

It then suffices to show that $\{e\}$ is closed in $G$, since its preimage under $f$ will also be closed, because $f$ is continuous.

This is easy to show, provided we assume that $G$ is a $T_1$ space. This assumption implies that for each $x\ne e$, we can find a neighborhood $U_x$ of $x$ with $e\notin U_x$.

If take the union of all of these neighborhoods, we have:

$$\bigcup _{x\ne e}{U}_x=G-\{e\}$$

Which means that $\{e\}$ must be closed.

$\square$