Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.


share|cite|improve this question
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 ... – martini Oct 12 '11 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 '11 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 $xUU^{-1} \cap BA=\emptyset$.Hence, $xUU^{-1}$ beinh a neighborhood of $x$ not meeting $BA$, it follows that $BA$ is closed, since $x$ is arbitary.

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. $UA\cap UB=\emptyset.$

share|cite|improve this answer
How to prove *, I don't think it is obvious. – Xiang Yu Mar 23 '15 at 14:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.