Two disjoint compact sets in a topological group

Question

Let $(G, \cdot )$ be a compact (Hausdorff) topological group. If $A$ and $B$ are two disjoint compact subsets of $G$, how can we show that there exists a nonempty open set $V$ such $A\cdot \overline{V}\cdot \overline{V}^{-1}$ and B are disjoint?

Answer 1

We can show the stronger (because we don't need that $G$ is Hausdorf, we don't need that $G$ is compact, and we can show that $V$ can be picked as symmetric neighbourhood of $1$) result

Proposition. Let $G$ be a topological group and $A,B\subseteq G$ compact subsets. Then there exists a nonempty open set $V\subset G$ such that $1\in V$ and $V=V^{-1}$ and $A\overline V\overline V^{-1}\cap B=\emptyset$.

Proof. As $A\times B\subset G\times G$ is compact, so is its image $A^{-1}B$ under the continuous map $(x,y)\mapsto x^{-1}y$. Consequently, $U=G\setminus A^{-1}B$ is open. As $A,B$ are disjoint, we have $1\in U$. So $U$ is an open neighbourhood of $1$ with $U\cap A^{-1}B=\emptyset$, or equivalently $AU\cap B=\emptyset$. By the lemmas below, let $V\ni 1$ be open with $V=V^{-1}$ and $VVVV\subseteq U$. Assume $x\in A\overline V\cap B\overline V$. Then all open neighbourhoods of $x$ intersect both $AV$ and $BV$. As one of these neighbourhoods is $xV$, we find $a\in A$, $v_1,v_2,v_3,v_4\in V$, $b\in B$ with $av_1=xv_2$, $bv_3=xv_4$. But then $av_1v_2^{-1}v_4v_3^{-1}=b$, i.e., $AVV^{-1}VV^{-1}=AVVVV\subseteq AU$ intersects $B$, contradiction. Hence $A\overline V$ and $B\overline V$ are disjoint and so are $A\overline V\overline V^{-1}$ and $B$. $_\square$

Lemma 1. Let $G$ be a topological group, $U\ni 1$ open, $n\in\mathbb N$. Then there exists open $V\ni 1$ such that $\underbrace{V\cdot V\cdot\ldots\cdot V}_n\subseteq U$.

Proof. Since $(1,1)$ is in the preimage of $U$ under the continuous map $\mu\colon G\times G\to G$, $(x,y)\mapsto xy$ there exist open neighbourhoods $V_1,V_2$ with $V_1\times V_2\subseteq \mu^{-1}(U)$. Then $V=V_1\cap V_2$ solves the problem for the case $n=2$. By induction we find that the lemma is true for all powers of $2$, and then by letting some factors $=1$ for arbitrary $n$. $_\square$

Lemma 2. Let $G$ be a topological group, $U\ni 1$ open. Then there exists open $V$ with $1\in V\subseteq U$ and $V=V^{-1}$.

Proof. Simply let $V=U\cap U^{–1}$, where $U^{-1}$ is open as preimage of $U$ under the continuos map $x\mapsto x^{-1}$. $_\square$.