Product Topology and Compactness

  • Thread starter Bleys
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Please if someone could help me understand something I saw in a proof. It's about proving that if X,Y is compact then their product (with product topology) is compact.

Suppose that X and Y are compact. Let F be an open cover for XxY. Then, for y in Y, F is an open cover for Xx{y}, which is compact. Hence F has a finite subcover
Fy = [tex]\{ U^{y}_{1}\times V^{y}_{1},...,U^{y}_{n}\times V^{y}_{n} \} [/itex] where y is in all [tex]V^{y}_{i}[/itex]

This is the step I don't get. Why is y in all those sets? I understand y must be in at least one (to be covered) but why all? The proof uses this fact to construct a non-empty set so it's pretty crucial, but for the life of me I don't understand how to deduce it :(
I was thinking maybe it's because of the axiom of choice, but I don't know much about that to even be sure it involves it.
 

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  • #2
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There is a subtle point here. Not every element of a subcover of F needs to contain y, but the point is that we can pick a subcover of F such that the property holds.

Indeed, F is a cover of the compact space [tex]X\times\{y\}[/tex], thus it has a compact subcover [tex]\{U_1\times V_1,...,U_n\times V_n\}[/tex]. This subcover does not need to satisfy our assumptions, indeed it is possible that y is not in Vi. But if this is the case, then [tex](U_i\times V_i)\cap (X\times \{y\})=\empty[/tex]. Thus [tex]U_i\times V_i[/tex] is not really important in our cover, and it can be safely removed.

After removing all such sets, we end up with a finite subcover of F, which does satsify the condition. We call this cover Fy.
 
  • #3
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aah, thanks a bunch!
I had thought about "picking" certain elements of the subcover such that y was in [itex]V_{i}[/itex], but then I was wondering if the remaining [itex]U_{i}[/itex] would still cover X, so I was getting confused (forgetting slightly that I was dealing with Cartesian products!).

Thanks for your swift reply
 

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