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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.

Martin Sleziak
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Metta World Peace
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    Just to clarify, some people use a definition of a topological group which does not include $T_1$, so those won't be Hausdorff. – Keenan Kidwell Jan 03 '13 at 18:44
  • What is $T_1$ for Munkres? – Sigur Jan 03 '13 at 18:45
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    @Sigur That points are closed. – Arthur Jan 03 '13 at 18:46
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    @Sigur: All definitions of $T_1$ that I’ve seen are equivalent, so it doesn’t matter. – Brian M. Scott Jan 03 '13 at 18:52
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    @BrianM.Scott, I know that. I've just asked to suggest him to read about $T_1$ spaces. – Sigur Jan 03 '13 at 18:53
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    Hint: $X$ is Hausdorff iff the diagonal is closed. Use the fact that $e$ is closed and work with some good continuous function related with the diagonal. – Sigur Jan 03 '13 at 18:56
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    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 $H$.

     – M.Sina Jan 04 '13 at 19:29

3 Answers3

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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.

Keenan Kidwell
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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?

Hagen von Eitzen
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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.
cqfd
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