# 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.

Thanks

• Using nets you can argue as follows: Suppose $a_ib_i \to g \in G$ with $a_i \in A$ and $b_i \in B$. Since $A$ is compact, there is a convergent subnet $a_j \to a \in A$, now ... Oct 12, 2011 at 12:04
• A completely fleshed out version of martini's argument can be found in Theorem (4.4) of Hewitt-Ross, Abstract Harmonic Analysis, I,, which is a good source for such generalities on topological groups (and many other things).
– t.b.
Oct 12, 2011 at 12:49

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

• How to prove *, I don't think it is obvious. Mar 23, 2015 at 14:56
• @XiangYu: I added a sketch for posterity (as I got stuck on this point myself). It is worth noting that this fact holds even if $G$ is not Hausdorff (in which case $A$ need not be closed). Dec 28, 2016 at 19:01
• What is $e$? Are you talking about $x$? Jul 6, 2017 at 9:26
• I think $e$ is the identity element of $G$.
– Kei
Feb 19, 2021 at 3:03

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 $$W_i$$.

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