TITLE: Topological Hausdorff Group: $ A$ closed, $B$ compact $\Longrightarrow AB$ closed
QUESTION [5 upvotes]: I'm dealing with a puzzling problem and hope some of you can help me.
Let be $(G,\cdot ) $ be a Hausdorff-group and let $A,B \subseteq G$. Show that if $A$ is closed and $B$ is compact, then $AB$ is closed.
Due to I'm considering a topological Hausdorff-group, we have two continous maps:
$\psi: G \times G \rightarrow G, (x,y) \mapsto xy$  and  $ \phi: G \times G \rightarrow G, (x,y) \mapsto x{y}^{-1} $
My idea was to show that $(AB)^{c}$ is open instead. 
First we observe, that if $A$ is closed, then $A^{c}$ is open and due to $\phi$ is continous $\phi^{-1}(A^{c})$ is also open.
Now let's choose an arbitrary $x \in (AB)^{c}$. We see that $\forall y \in B: \phi(x,y)=xy^{-1} \in A^{c}$. (because if it wasn't it would be in $A$ we could conclude $\psi(xy^{-1},y)=xy^{-1}y \in AB$, what would be opposed to the chosen x)
So we know that $\phi(x,y)=xy^{-1} \in A^{c} \Rightarrow (x,y) \in \phi^{-1}(A^{c})$, which is open.
Then we find neighbourhoods $U$ from $x$ and $V$ from $y$ such that:
$\forall y \in B \exists U_{y} \exists V : (x,y) \in U_{y} \times V \subseteq \phi^{-1}(A^{c})$
If I could manage to show, that $U:= \cap_{y \in B} U_{y}$ is open (I don't know this, because it might be an infinite intersection) i think my proof is complete.
Because then I can show that $U \subseteq (AB)^{c}$, so that for arbitrary $x$ $(AB)^{c}$ is a neighbourhood of x. This means that $(AB)^{c}$ is open.
Can anybody give me a hint how I can show, that this intersection is finite? I suppose it must follow from the compactness of B, because I haven't used it yet, but I have no clue how?
Thank you!

REPLY [5 votes]: It is basically the same proof as in a sequential space. 
Let $(a_i, b_i) \in A \times B$ be a net with $a_i b_i \to c \in G$. Then, there is a subnet $b_{i_k}$ with $b_{i_k} \to b \in B$. Due to continuity the limit
$$ a := \lim_k a_{i_k} = \lim_k (a_{i_k} b_{i_k}) b_{i_k}^{-1} = cb^{-1} \in A $$
exists and it follows
$$ c = ab \in AB. $$