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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

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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
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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

1 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 $xUU^{-1} \cap BA=\emptyset$.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. $UA\cap UB=\emptyset.$

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