# $\left\{e\right\}$ Closed Implies $T_0$ and Regular Topological Group [duplicate]

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 $\bigcup_{h\in C}U_h$.

• I'd rather say $G$ is trivially regular, since any topological group is. Feb 11, 2018 at 21:32
• trivially $T_0$ why? It needs some argument. Feb 11, 2018 at 21:46
• @HennoBrandsma $\{g\}^c$ is an open neighborhood of $h$ Feb 11, 2018 at 21:55
• 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. Feb 11, 2018 at 22:03
• That's a good point about Hausdorff. Feb 11, 2018 at 22:40