Product of compact and closed in topological group is closed

Question

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.

Thanks

Answer 1

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 $AU\cap BU=AU\cap BU^{-1}=\emptyset$.

Answer 2

We can use the Tube Lemma (in Munkres's text) to solve this problem: Consider the continuous function $\phi:(x,y)\mapsto xy^{-1}$. Suppose $c\notin AB$, then $c\times B\subset \phi^{-1}(A^c)$. Since $B$ is compact, there is some neighborhood $W$ of $c$ s.t. $W\times B\subset\phi^{-1}(A^c)$, whence $WB^{-1}\subset A^c$.

Remark: to prove the Tube Lemma, just take the neiborhood $W_y\times V_y$ for every $y\in B$, then choose finite subcollection $\{W_i\times V_i\}$ covering $c\times B$ since $B$ is compact, and take the intersection of corresponding $W_i$.

Answer 3

The * point in @M.sina's answer is really useful. However, I had a hard time following his sketch and failed to find it elsewhere, so I decided to write a more detailed proof. Hope this isn't off topic for this question.

*. 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: $$1. AU\cap BU=\emptyset\\2. UA\cap UB=\emptyset$$

Proof: Since $B$ is closed, $G\setminus B$ is open and for every $a_i \in G \setminus B$, there exists a symmetric neighborhood $V_{i1}$ of $e$ such that $a_iV_{i1}^2\cap B=\emptyset$ and a symmetric neighborhood $V_{i2}$ of $e$ such that $V_{i2}^2a_i\cap B=\emptyset$, take $V_i=V_{i1}\cap V_{i2}$. $V_i$ is also open, symmetric, and has the disjoint property.

Since $\bigcup_{\{i|a_i \in G\setminus B\}} (a_iV_i\cap V_ia_i)$ covers $G\setminus B \supseteq A$ and $A$ is compact, there are some $a_1,\dots,a_n$ and neighbourhoods $V_1,\ldots,V_n$ of $e$ such that $\bigcup_{i=1}^n (a_iV_i\cap V_ia_i)\supseteq A$. Then for any $a\in (a_iV_i\cap V_ia_i)$ we have $aV_i \subseteq a_iV_i^2$ and $V_ia \subseteq V_i^2a_i$, so $V=\bigcap_{i=1}^n V_i$ satisfies $AV\cap B=\emptyset$ and $VA\cap B=\emptyset$.

Then find some symmetric neighborhood $U$ of $e$ such that $U^2 \subseteq V$. We have $$\emptyset=AV\cap B\supseteq AU^2\cap B$$ Suppose that $AU\cap BU\neq \emptyset $, then there exist $au_1=bu_2\Rightarrow au_1u_2 ^{-1}=b$, contradicting $AU^2\cap B= \emptyset $, so $AU\cap BU=\emptyset$. Similarly, $\emptyset=VA\cap B\supseteq U^2A\cap B$, so $UA\cap UB=\emptyset$.