2020-12-24

Topological Groups are Hausdorff

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

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

\begin{aligned} \bullet &: G \times G \to G \cr ()^{-1} &: G \to G \end{aligned}

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

\begin{aligned} e \bullet x = x = x \bullet e \cr x^{-1} \bullet x = e = x \bullet x^{-1} \cr a \bullet (b \bullet c) = (a \bullet b) \bullet c \end{aligned}

An extra assumption, made at least by Munkres, is that $G$ is a $T_1$ space. This means that for any points $x, y \in G, x \neq y$ , 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 $x \neq y$ , not only can we find neighborhoods $U_x, U_y$ that don't contain the other point, but that are distinct, with $U_x \cap U_y = \empty$ .

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:

$\implies$

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

\bigcup_{x \neq y} N(x, y) = X^2 - \Delta

is open, meaning $\Delta$ is closed.

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

$U_x \times U_y$ is evidently a neighborhood of $(x, y)$ . Furthermore, assume $(x', y') \in U_x \times U_y$ . Because $U_x$ and $U_y$ are distinct, we must have $x' \neq y'$ . This means that $U_x \times U_y \cap \Delta = \empty$ .

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

$\impliedby$

If $\Delta$ is closed, then the set $X^2 - \Delta$ is open. This means that if $x \neq y$ , 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' \in U_x, y' \in U_y \implies (x', y') \notin \Delta

That is to say:

x' \in U_x, y' \in U_y \implies x' \neq y'

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

$\square$

$T_1 \implies$ Hausdorff, for a Group

First, note that the function $f = (x, y) \mapsto xy^{-1}$ 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) | xy^{-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 $x \neq e$ , 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 \neq e} U_x = G - \{e\}

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

$\square$

Hausdorff === closed diagonal

T1⟹T_1 \impliesT1​⟹ Hausdorff, for a Group

Hausdorff $=$ closed diagonal

$T_1 \implies$ Hausdorff, for a Group