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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!

William Elliot
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pcalc
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    See https://math.stackexchange.com/questions/71983/product-of-compact-and-closed-in-topological-group-is-closed – peter a g Nov 07 '17 at 20:46
  • Hi! Thanks! I already know this thread, but I don't fully understand the way given there, so I tried to find my own solution. There are a couple of steps, which need to be made clearer, but I didn't overcome it. Maybe you have some further explainations for me? Or maybe another hint how this intersection can be finite? Thank you! – pcalc Nov 07 '17 at 22:14

2 Answers2

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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 $$ c = ab \in AB. $$

user251257
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  • Hi! Thanks four your respond! I don't really see how this solves my problem with the finite intersection. Maybe I've overseen something? – pcalc Nov 07 '17 at 22:15
  • @pcalc in your proof: the system ${V_y \mid y\in B}$ is an open cover of $B$. Now use compactness to select a finite sub cover. – user251257 Nov 07 '17 at 23:25
  • Yes, that's true! Due to the compactness of $B$ one can follow that there exists a finite subcover made up of some $V_{y}$. I'm not sure how I can follow that there also exists a finite number of $U_{y}$. Does the line $\forall y \in B \exists U_{y} \exists V_{y}: (x,y) \in U_{y} \times V_{y} \subseteq \phi^{-1}(A^{c})$ still hold, if I take a finite number of $U_{y}$ ? – pcalc Nov 08 '17 at 08:31
  • @pcalc the $U_y$ and $V_y$ come in pairs... if you only need finitely many $V_y$'s, you only need finitely many $U_y$'s. – user251257 Nov 08 '17 at 16:19
  • I think i got it! Thank you! – pcalc Nov 10 '17 at 17:10
  • How is the net defined? What is the directed set involved? How is the partial order ≼ defined? EDIT: I see. By net, you basically meant "sequence" rather than the more general "net". – Petra Axolotl Aug 28 '23 at 21:15
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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 $WB^{-1}\subset A^c$.

William Leynoid
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