Product of compact and closed in topological group is closed

This could be classified as “homework”, but I tried to solve this, made research online, and still failed, so I’ll be glad to get some hints.

Let $G$ be a topological group, let $A$ be a compact subset of $G$, and let $B$ be a closed subset of $G$. Prove that $AB$ is closed.

If both $A$ and $B$ are not compact, but closed, this can fail, for example, if we let $A$ be the set of integers and $B$ the set of integer multiples of $\pi$, then both are closed, but $A+B$ is a proper dense subset of $\mathbb R$, so can’t be closed. Also if $A$ is compact but $B$ is not closed, this easily fails.


Solutions Collecting From Web of "Product of compact and closed in topological group is closed"

Let ‎$ x‎\in G‎\setminus BA‎ $‎. Then $B^{-1} x ‎\cap A =\emptyset$, and $B^{-1} x$ is closed. since $A$ is compact there exists a neighborhood $U$ of $e$ such that $$B^{-1} xU\cap AU=\emptyset. ^{*}$$
But this implies that $xUU^{-1} \cap BA=\emptyset$. Since $xUU^{-1}$ is a neighborhood of $x$ not meeting $BA$, it follows that $BA$ is closed, since $x$ is arbitrary.

Similarly, $AB$ is closed.

*. Let $B$ be a closed subset and $A$ a compact subset of a topological group $G$ such that $A\cap B=\emptyset.$ Then there exists a neighborhood $U$ of $e$ such that:

  1. $AU\cap BU=\emptyset$

  2. $UA\cap UB=\emptyset.$

Sketch of proof: by compactness, there are some $a_1,\ldots,a_n$ and neighbourhoods $V_1,\ldots,V_n$ of $e$ such that $a_iV_i^2\cap B=\emptyset$ and $\bigcup_i a_iV_i\supseteq A$. Then for any $a\in a_iV_i$ you have that $aV_i\subseteq a_iV_i^2$, so $V=\bigcap_i V_i$ satisfies $AV\cap B=\emptyset$. Then find some $U\subseteq V$ symmetric such that $U^2\subseteq V$. Then $\emptyset=AV\cap B\supseteq AU^2\cap B$, so $AU\cap BU=AU\cap BU^{-1}=\emptyset$.