Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 3 '13 at 18:44
@Sigur That points are closed. – Arthur Jan 3 '13 at 18:46
@Sigur: All definitions of $T_1$ that I’ve seen are equivalent, so it doesn’t matter. – Brian M. Scott Jan 3 '13 at 18:52
@BrianM.Scott, I know that. I've just asked to suggest him to read about $T_1$ spaces. – Sigur Jan 3 '13 at 18:53
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 3 '13 at 18:56
up vote 14 down vote accepted

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.

share|cite|improve this answer

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?

share|cite|improve this answer

By the above argument, there are two cases:

  1. If {1} is closed then the diagonal is closed (as it is the inverese image of {1} under the continuous map $g_1 \times g_2\rightarrow g_1g_2^{-1}$). So G is Hausdorff.
  2. If {1} is open then G is discrete, so it's Hausdorff.
share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.