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.

$\square$