$F$ closed, $K$ compact $\implies FK$ is closed in group $G$

Question

Let $G$ be a Hausdorff topological group, let $F$ be a closed and let $K$ be compact, both subsets of $G$.

Then $FK$ is closed in $G$.

Attempt:

$aF$ is closed in $G$ for each $a \in G$.

All we have to do is show that $FK$ is compact since a compact subset of a Hausdorff space is closed.

$FK \subset \bigcup\limits_i U_i$

Nevermind, found this: https://math.stackexchange.com/questions/71983/product-of-compact-and-closed-in-topological-group-is-closed — Daniel Donnelly, May 25 '18 at 18:41

score 0 · Accepted Answer · answered May 25 '18 at 01:24

0

Consider a sequence $f_nk_n$ such that $f_n\in F, k_n\in K$ which converges towards $y$. You can extract $k_{n_p}$ which converges towards $k$, the sequence $(f_{n_p}k_{n_p})k_{n_p}^{-1}=f_{n_p}$ converges towards $yk^{-1}$, since $F$ is closed, $yk^{-1}\in F$ implies that $y\in FK$.

answered May 25 '18 at 01:24

Tsemo Aristide

87,475

How can you be sure that $k_{n_p} \to k$ exists? – Daniel Donnelly May 25 '18 at 01:32
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You may want to consider rephrasing this in terms of nets, since $G$ is only assumed to be a Hausdorff topological group. – Aloizio Macedo May 25 '18 at 01:33

$F$ closed, $K$ compact $\implies FK$ is closed in group $G$

1 Answers1