# How to show that topological groups are automatically Hausdorff?

On page 146, James Munkres' textbook Topology(2ed),

Show that $$G$$ (a topological group) is Hausdorff. In fact, show that if $$x \neq y$$, there is a neighborhood $$V$$ of $$e$$ such that $$V \cdot x$$ and $$V \cdot y$$ are disjoint.

Noticeably, the definition of topological group in Munkres's textbook differs from that in wikipedia.

A topological group $$G$$ is a group that is also a topological space satisfying the $$T_1$$ axiom, such that the map of $$G \times G$$ into $$G$$ sending $$x \times y$$ into $$x \cdot y$$ and the map of $$G$$ into $$G$$ sending $$x$$ into $$x^{-1}$$, are continuous maps.

• Just to clarify, some people use a definition of a topological group which does not include $T_1$, so those won't be Hausdorff. Commented Jan 3, 2013 at 18:44
• @Sigur That points are closed. Commented Jan 3, 2013 at 18:46
• @Sigur: All definitions of $T_1$ that I’ve seen are equivalent, so it doesn’t matter. Commented Jan 3, 2013 at 18:52
• @BrianM.Scott, I know that. I've just asked to suggest him to read about $T_1$ spaces. Commented Jan 3, 2013 at 18:53
• For Further Reading; If $G$ is a topologial group the following conditions are equivalent. i) $G$ is a $T_{0}$ space. ii) $G$ is a $T_{1}$ space. iii) $G$ is a $T_{2}$ space. iv) If $\beta_{e}$ is a fundamental system of neighborhoods of $e$ then $\cap \beta_{e} =\{e\}$. v) \{e\} is a closed subgroup of $G$. vi) For all $f:H\rightarrow G$ in $\tau g$, $Kerf$ is a closed subgroup of $H$. Commented Jan 4, 2013 at 19:29

A space $X$ is Hausdorff if and only if the diagonal $\Delta_X\subseteq X\times X$ is closed. Consider the map $G\times G\rightarrow G$ given by $(x,y)\mapsto xy^{-1}$. It is continuous by the axioms for a topological group, and the diagonal is the inverse image of the the identity $\{e\}$, which is closed by assumption. So $G$ is Hausdorff if $\{e\}$ is closed, i.e., if $G$ is $T_1$ (by homogeneity, $T_1$ for $G$ is equivalent to $\{e\}$ being closed).

Given $x\neq y$ in a Hausdorff $G$, let $U_x$ and $U_y$ be disjoint opens around $x$ and $y$, respectively. Both $U_xx^{-1}$ and $U_yy^{-1}$ are opens around $e$, so we can find open $V$ with $e\in V\subseteq U_xx^{-1}\cap U_yy^{-1}$. Then $Vx$ and $Vy$ are disjoint neighborhoods of $x$ and $y$, as desired.

• Why are Vx and Vy closed? Commented Sep 1, 2020 at 19:54
• My argument does not assert that they are closed. They are open. They may or may not be closed, but this is irrelevant. Commented Sep 2, 2020 at 1:49

A topological space $G$ is hausdorff iff the diagonal in $G\times G$ is closed. Can you see how the diagonal is the inverse image of a closed set under a continuous map?

By the above argument, there are two cases:

1. If $$\{1\}$$ is closed then the diagonal is closed (as it is the inverse image of $$\{1\}$$ under the continuous map $$g_1 \times g_2\mapsto g_1g_2^{-1}$$). So $$G$$ is Hausdorff.
2. If $$\{1\}$$ is open then $$G$$ is discrete, so it's Hausdorff.