Product Topology and Compactness

[Bleys]

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 = \{ U^{y}_{1}\times V^{y}_{1},...,U^{y}_{n}\times V^{y}_{n} \} [/itex] where y is in all V^{y}_{i}[/itex]<br /> <br /> 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 :(<br /> 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.

 

[micromass]

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 X\times\{y\}, thus it has a compact subcover \{U_1\times V_1,...,U_n\times V_n\}. This subcover does not need to satisfy our assumptions, indeed it is possible that y is not in V_i. But if this is the case, then (U_i\times V_i)\cap (X\times \{y\})=\empty. Thus U_i\times V_i 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 F_y.

 

[Bleys]

aah, thanks a bunch!
I had thought about "picking" certain elements of the subcover such that y was in V_{i}, but then I was wondering if the remaining U_{i} would still cover X, so I was getting confused (forgetting slightly that I was dealing with Cartesian products!).

Thanks for your swift reply