Tychonoff’s Theorem
One of the biggest results in point-set topology is Tychonoff’s Theorem: the fact that the product of any family of compact spaces is again compact. Unsurprisingly, the really tough bit comes in when we look at an infinite product. Our approach will use the dual definition of compactness.
Let’s say that a collection of closed sets has the finite intersection hypothesis if all finite intersections of members of the collection are nonempty, so compactness says that any collection satisfying the finite intersection hypothesis has nonempty intersection. We can then form the collection
of all collections of sets satisfying the finite intersection hypothesis. This can be partially ordered by containment —
if every set in
is also in
.
Given any particular collection we can find a maximal collection containing it by finding the longest increasing chain in
starting at
. Then we simply take the union of all these collections to find the collection at its top. This is almost exactly the same thing as we did back when we showed that every vector space is a free module! And just like then, we need Zorn’s lemma to tell us that we can manage the trick in general, but if we look closely at how we’re going to use it we’ll see that we can get away without Zorn’s lemma for finite products.
Anyhow, this maximal collection has two nice properties: it contains all of its own finite intersections, and it contains any set which intersects each set in
. These are both true because if
didn’t contain one of these sets we could throw it in, make
strictly larger, and still satisfy the finite intersection hypothesis.
Now let’s assume that is a collection of closed subsets of
satisfying the finite intersection hypothesis. We can then get a maximal collection
containing
. Then given an index
we can consider the collection
of closed subsets of
and see that it, too, satisfies the finite intersection hypothesis. Thus by compactness of
the intersection of this collection is nonempty. Letting
be a closed set containing one of these intersection points
, we see that the preimage
meets every
, and so must itself be in
.
Okay, so let’s take the point for each index and consider the point
in
with
-th coordinate
. Then pick some set
containing
from the base for the product topology. For all but a finite number of the
,
. For those finite number where it’s smaller, the closure of
contains the point
, and so
is in
. So their finite intersection must be nonempty, and so is
itself!
Now, since is in
, it must intersect each of the closed sets in the original collection
. Since the only constraint on
is that it contain
, this point must be a limit point of each of the sets in
. And because they’re closed, they must contain all of their limit points. Thus the intersection of all the sets in
is nonempty, and the product space is compact!