A question about continuous actions of topological groups. Let $G$ act continuously on $X$. If $K$ is a compact subspace of $G$ and $B$ a closed subspace of $X$, is $KB:=\{k.b|k\in K,b\in B\}$ a closed subspace of $X$?
I know this to be true when $X$ is the same as $G$ with the action being the group operation. I think the proof using nets in this case could be adapted to the general case, but I am not completely sure.
 A: Indeed the usual proof idea as used for subsets of topological groups also
applies here:
Suppose that $k_i \cdot b_i$ (for some $i \in I$, a directed set) is a net in $KB$ converging to $x \in X$.
Then $(k_i)_i$ is a net in $K$ so by compactness we have a convergent subnet $(k'_j)_{j \in J}$ (with $J$ also a directed set, and a corresponding connecting map $f: J \to I$) of $(k_i)_i$ (so that $f[J]$ is cofinal in $I$ and $k'_j = k_{f(j)}$ for all $j$ etc.)) converging to $k \in K$. 
By continuity of the inverse operation, we have that $((k')^{-1}_j)_j \to k^{-1}$ as well.
The same connecting map $f$ allows us to define the corresponding subnet $(b'_j)_j$ of $(b_i)_i$ as well by setting $b'_j = b_{f(j)}$ as usual.
It follows by the group action axioms that for all $j$: $$(k'_j)^{-1} \cdot (k'_j \cdot b'_j) = ((k'_j)^{-1} \ast_G k'_j)\cdot b'_j = 1_G \cdot b'_j = b'_j$$
and the left hand side converges to $k^{-1} \cdot x$ (as subnets of convergent nets converge to the same limit and the action is continuous). As the $b'_j$ are all from the closed set $B$, $k^{-1} \cdot x \in B$ as well. Hence $x = k \cdot (k^{-1} \cdot x) \in KB$ as well and the latter set is closed. 
