We are finishing up a discussion of the Tychonoff Theorem: an arbitrary product of compact spaces is compact (in the product topology, of course). The genesis of this discussion comes from this David Wright article.

In the first post in this series, we gave an introduction to “compactness”.

In the second post, we gave a proof that the finite product of compact spaces is compact.

In the third post, we gave come equivalent definitions of compactness

In particular, we showed that:

1. A space is compact if and only if the space has the following property: if is an infinite union of open sets with no finite subcover, then is a proper subset of ; that is, and

2. A space is compact if and only if the space has the following property: every infinite subset has a perfect limit point. Note: a perfect limit point for a set is a point such that, for every open , (the intersection of every open neighborhood of a perfect limit point with has the same cardinality as .

Note the following about these two facts: each of these facts promises the existence of a specific point rather than the existence/non-existence of a cover of a particular type. Fact 1 promises the existence of an excluded point, and fact 2 promises the existence of a perfect limit point.

When it comes to a point in an infinite of topological spaces, constructing a point is really like constructing a sequence of points (in the case of countable products) or a net of points (in the case of uncountable products). That is, if one wants to construct a point in an infinite product of spaces, one can assume some well ordering of the index used in the product, then construct the first coordinate of the point from the first factor space, the second coordinate from the second factor space, and so on.

We’ll use fact one: the excluded point property to prove Tychonoff’s Theorem.

Proof. Assume that and that is well ordered. We start out by showing that the product of two compact spaces is compact, and use recursion to get the general result.

Let be an infinite union of open sets in with no finite subcover. First, we show that there is some such that for each open set , no finite subcollection of covers . Now if there is some open where is disjoint from every we are done with this step. So assume not; assume that every is a subset of the first factor of some . If it isn’t the case that there is where has no finite subcover of elements of , for each such there is a finite number of elements of that covers . Now since is compact, a finite number of covers , hence a finite subcover of covers ALL of . Hence some point exists such that no finite subcover of covers for any open .

Similarly, we can find so that for all open , no finite subcollection of covers where is a basic open set in that contains . This shows that because, if it were, this single point would lie in some basic open set which, by definition, is a finite subcover.

Now given an arbitrary product with a well ordered index set we can now assume that there is some collection of open sets that lacks a finite subcover and inductively define so that, if is any basic open set containing then no finite subcollection of covers . The point thus constructed lies in no .

Note: if you are wondering why this “works”, note that we assumed NOTHING about the compactness of the remaining product space factors .

And remember that we are using the product topology: an open set in this topology has the entire space as factors for all but a finite number of indices. So we only exploit the compactness of the leading factors.