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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 ...2011-10-12
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    A completely fleshed out version of martini's argument can be found in [Theorem (4.4)](http://books.google.com/books?id=uf11K1wXEYUC&pg=PA17) of Hewitt-Ross, *[Abstract Harmonic Analysis, I,](http://books.google.com/books?id=uf11K1wXEYUC)*, which is a good source for such generalities on topological groups (and many other things).2011-10-12

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