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Let $G$ be a Hausdorff topological group. Let $H$ be a subgroup of $G$ such that $H$ is a discrete subspace of $G$.

Is $H$ a closed subgroup of $G$?

I thought this is obviously true, but I failed to prove it.

Makoto Kato
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1 Answers1

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Here's a proof:

We fix some $x\not\in H$, and seek some neighborhood of $x$ containing no points of $H$. Choose some $U_0$ a neighborhood of the identity $e$ such that $U_0 \cap H = \{e\}$. Then consider the preimage of $U_0$ under the map $G \times G \to G, (x,y) \mapsto xy^{-1}$. This yields a neighborhood $U_1$ of $x$ with the property that if $y,z\in U_1$ such that $yz^{-1}\in H$, we have $y = z$. In particular, $U_1\cap H$ has at most one element. We are already done if no such element exists, so assume $U_1 \cap H = \{h\}$. Since $G$ is Hausdorff there is a neighborhood $U_2$ of $x$ not containing $h$, so that $U_1 \cap U_2$ is a neighborhood of $x$ not containing any points of $H$, as desired.

Rolf Hoyer
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  • Your argument does not seem to make sense. $U_1$ is a subset of $G \times G$ so cannot really contain the point $x \in G$. – Tuo Jun 12 '18 at 23:10
  • @TuoTuo I have implicitly set $U_1$ to be the intersection of two open sets $A,B$ where $A \times B$ is an open neighborhood of $(e,e)$ in $G \times G$. – Rolf Hoyer Jun 13 '18 at 02:50