$G$ is a hausdorff topological group $\iff \{e \}$ is closed.

Question

I did the right implication:

$U_{y},V_{y}$ open disjoint sets such that $x \in U_{y}, y \in V_{y}$

Then $\cup_{y \in G} V_{y}$ is open such that $y$ is in and $x$ not. Then $\cup_{y \in G} V_{y} = G - \{x\}$ is open, therefore $\{e\}$ is closed.

But how to do the left? I know that for every neighborhood $L$ of $e$ there is a neighborhood $J^{-1}J$ of $e$ such that $ J^{-1}J \subseteq L$, because this was the previous item of this exercise. Perhaps I could use this, but can't see how?

Thanks.

Answer 1

HINT: Assume that $\{e\}$ is closed. Let $\mathscr{U}$ be the family of all open nbhds of $e$.

• Show that $\bigcap\mathscr{U}=\{e\}$. (This is in fact equivalent to the assertion that $\{e\}$ is closed.)

• Show that if $\bigcap_{U\in\mathscr{U}}\operatorname{cl}U=\{e\}$, then $G$ is Hausdorff.

This reduces the problem to showing that $\bigcap_{U\in\mathscr{U}}\operatorname{cl}U=\{e\}$. Let $U\in\mathscr{U}$.

• Show that there is a $V\in\mathscr{U}$ such that $VV\subseteq U$.

Let $x\in\operatorname{cl}V$ be arbitrary. Then $xV^{-1}$ is an open nbhd of $x$ (why?), so $V\cap xV^{-1}\ne\varnothing$, and there is a $y\in V$ such that $xy^{-1}\in V$.

• Use this to show that $x\in VV\subseteq U$, and conclude that $\operatorname{cl}V\subseteq U$.

• Conclude further that $\bigcap_{U\in\mathscr{U}}\operatorname{cl}U\subseteq\bigcap\mathscr{U}$ and hence that $\bigcap_{U\in\mathscr{U}}\operatorname{cl}U=\bigcap\mathscr{U}=\{e\}$.

Added: This is an outline of an argument along the lines of what I think you’ve been doing up to this point. DonAntonio has provided an outline of a nice slick proof along very different lines.

Answer 2

Hints:

(1) A topological space $\;X\;$ is Hausdorff iff the diagonal $\;\Delta:=\{ (x,x)\;/\;x\in X\}\;$ is closed in $\;X\times X\;$ (with the product topology) .

(2) Now look at the continuous (why?) map

$$F:X\times X\to X\;,\;\;f(x,y):=xy^{-1}\;,\;\;\text{and check}\;\;F^{-1}(\{e\})$$