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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.

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    $\begingroup$ Just to clarify, some people use a definition of a topological group which does not include $T_1$, so those won't be Hausdorff. $\endgroup$ Jan 3, 2013 at 18:44
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    $\begingroup$ @Sigur That points are closed. $\endgroup$
    – Arthur
    Jan 3, 2013 at 18:46
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    $\begingroup$ @Sigur: All definitions of $T_1$ that I’ve seen are equivalent, so it doesn’t matter. $\endgroup$ Jan 3, 2013 at 18:52
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    $\begingroup$ @BrianM.Scott, I know that. I've just asked to suggest him to read about $T_1$ spaces. $\endgroup$
    – Sigur
    Jan 3, 2013 at 18:53
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    $\begingroup$ 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 $H$. $\endgroup$
    – M.Sina
    Jan 4, 2013 at 19:29

3 Answers 3

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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.

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

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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.
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