Topological Hausdorff Group: $ A$ closed, $B$ compact $\Longrightarrow AB$ closed

Question

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!

Answer 1

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 $$ c = ab \in AB. $$

Answer 2

I think we can use the Tube Lemma to solve this: Consider the continuous function $\phi:(x,y)\mapsto xy^{-1}$. Suppose $c\notin AB$, then $c\times B\subset \phi^{-1}(A^c)$. Since $B$ is compact, there is some neighborhood $W$ of $c$ s.t. $W\times B\subset\phi^{-1}(A^c)$, whence $WB^{-1}\subset A^c$.

Answer 3

If you are a primitive ape like me, you can do this without any nets or Tube Lemmas. Here is how:

Take $z\in (AB)^C$. This means that for any $b\in B$, $zb^{-1}\not\in A$ and because $A$ is assumed to be closed, this implies that $zb^{-1}\in U_b$ where $U_b$ is open and does not intersect $A$. Now notice that $e\in z^{-1} U_b b$ and hence, by the usual properties of topological groups (see chapter 9 of Cohn's Measure Theory book- no greater book has seen the light of day), there is a neighborhood $V_b$ of the identity such that it is symmetric (i.e., $V_b=V_b^{-1}$) and:

$$e\in V_b\subseteq V_b V_b\subseteq z^{-1} U_b b$$

Now by compactness of $B$ we may assume that $B\subseteq \cup_{i=1}^{n} b_iV_{b_i}$. I claim that $V=z\cap_{i=1}^n V_{b_i}$ is a neighborhood of $z$ such that it does not intersect $AB$. If it did, there would exist $a\in A$, $b\in B$ and $i\in \cap_i V_{b_i}$ such that:

$$zi=ab\Rightarrow a= zi b^{-1}\in z V_{b_i}(b_iV_{b_i})^{-1}= zV_{b_i}V_{b_i}b_i^{-1}\subseteq z z^{-1}U_{b_i} b_i b^{-1}_i=U_{b_i}$$

Of course this is a contradiction as $a\in A$ and $U_{b_i}\cap A=\emptyset$. Hence, $(AB)^C$ is open and we are done!