# Product Topology and Compactness

## Main Question or Discussion Point

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 = $$\{ U^{y}_{1}\times V^{y}_{1},...,U^{y}_{n}\times V^{y}_{n} \} [/itex] where y is in all [tex]V^{y}_{i}[/itex] This is the step I don't get. Why is y in all those sets? I understand y must be in at least one (to be covered) but why all? The proof uses this fact to construct a non-empty set so it's pretty crucial, but for the life of me I don't understand how to deduce it :( I was thinking maybe it's because of the axiom of choice, but I don't know much about that to even be sure it involves it. ## Answers and Replies Related Topology and Analysis News on Phys.org There is a subtle point here. Not every element of a subcover of F needs to contain y, but the point is that we can pick a subcover of F such that the property holds. Indeed, F is a cover of the compact space [tex]X\times\{y\}$$, 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 Vi. 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 Fy.

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!).