0
$\begingroup$

Let $G$ be a topological group with identity $e$ such that $\left\{e\right\}$ is closed. Then $G$ is $T_0$ and regular.

$G$ is trivially $T_0$.

Let $C\subset G$ be closed and $C\not\ni g\in G$. For every $h\in C$, I was able to find an open neighborhood $U_h\not\ni g$ of $h$. I am stuck trying to find an open neighborhood $U$ of $g$ disjoint from $\bigcup_{h\in C}U_h$.

$\endgroup$
5
  • $\begingroup$ I'd rather say $G$ is trivially regular, since any topological group is. $\endgroup$ Feb 11, 2018 at 21:32
  • $\begingroup$ trivially $T_0$ why? It needs some argument. $\endgroup$ Feb 11, 2018 at 21:46
  • $\begingroup$ @HennoBrandsma $\{g\}^c$ is an open neighborhood of $h$ $\endgroup$
    – wjmolina
    Feb 11, 2018 at 21:55
  • 1
    $\begingroup$ You still need to explain that by homogeneity $\{e\}$ closed implies that all singletons are closed. We even get Hausdorff for free because $\Delta_X = \mu^{-1}[\{e\}]$ where $\mu(x,y) = xy^{-1}$ is continuous. $\endgroup$ Feb 11, 2018 at 22:03
  • $\begingroup$ That's a good point about Hausdorff. $\endgroup$
    – wjmolina
    Feb 11, 2018 at 22:40

0