Ok, now lets prove the following: If are compact spaces, then is compact (in the usual product topology). Note: this effectively proves that the finite product of compact spaces is compact. One might call this a “junior” Tychonoff Theorem.
Proof. We will prove this theorem a couple of times; the first proof is the more elementary but less elegant proof. It can NOT be easily extended to show that the arbitrary product of compact spaces is compact (which is the full Tychonoff Theorem).
We will show that an open covering of by basis elements of the form , open in and open in has a finite subcover.
So let be an open cover of . Now fix and consider the subset . This subset is homeomorphic to and is therefore compact; therefore there is a finite subcollection of which overs , say Note that each is an open set in which contains and there are only a finite number of these. Hence is also an open set which contains . Also know that
We can do this for each and so obtain an open cover of by and because is compact, a finite subcollection of these covers . Call these . For each one of these, we have .
So, our finite subcover of is .
Now while this proof is elementary, it doesn’t extend to the arbitrary infinite product case.
So, to set up such an extension, we’ll give some “equivalent” definitions of compactness. Note: at some point, we’ll use some elementary cardinal arithmetic.