Compact Sets, The Tikhonov Product Theorem

The Tikhonov Product Theorem

Like the well ordering principle, this theorem is equivalent to the axiom of choice. You must assert them both or deny them both.

Compact is usually described in terms of open sets and open covers, but there is a complementary definition that refers to closed sets and the finite intersection property. We're going to use that here. If you're not familiar with that definition you may want to revisit the intro page.

A collection of sets is "larger" than another if it includes more sets. This is a partial ordering on families of sets.

Restrict attention to collections of sets that have the finite intersection property. Once again, collection X is larger than collection Y if X brings in additional sets, and both X and Y have the finite intersection property. Verify that the union of an ascending chain yields an even larger collection with the finite intersection property. We can now use zorn's lemma to select a maximal collection of sets with the finite intersection property.

Let M be such a collection. Let U and V be two sets in this collection. Clearly U and V intersect; let W = U∩V. Let F be any finite subcollection of M, and note that the intersection of F and U and V is the intersection of F with W. The former is nonempty; so is the latter. If we bring in W, M still has the finite intersection property. Since M is maximal, it already contains W. In other words, the finite intersection of any of the sets in M produces another set in M.

Let C be a set that happens to intersect everything in M. Let F be a finite subcollection, and let G be their nonempty intersection. Since G is in M, C and G intersect. This means C intersects all the members of F. We can bring C into M, and it still has the finite intersection property. Since M is maximal, it already contains C. In other words, M contains any set that intersects all the members of M.

Let S_i, for i in some indexing set, be an indexed family of topological spaces, with P as product, using the weak product topology. Let e_i be the i^th projection from P onto S_i.

If P is compact, then each S_i is the continuous image of P under e_i, and is compact.

Do you see the axiom of choice hidden in the above statement? Without choice, P could be empty, whence it does not pproject onto S_i at all. We need choice to make sure P actually exists, as a product space; then the image of P is indeed S_i, which is compact.

Conversely, assume each space S_i is compact. Let T be a collection of closed sets in P with the finite intersection property. Embed T in a collection M of (not necessarily closed) sets, that is maximal with respect to the finite intersection property. Project M down to S_i to give a collection of sets in S_i, which we will call M_i. Show that M_i has the finite intersection property.

It follows that the closures of the sets in M_i, relative to S_i, also have the finite intersection property. Since S_i is compact, there is some q_i in S_i that lies in the closure of each set in M_i. Let the point q in P have projections q_i.

What does this mean? Let W be any set in M and project W down to W_i in S_i. The closure of W_i includes q_i. Any open set in S_i containing q_i intersects W_i.

Select an arbitrary set W from M, and keep it on standby. Let U be a base open set in P that contains q, but not just any base open set. Let U restrict one component, say S_i, to a base open set V, and leave the other components unconstrained. Now V intersects W_i in a point we will call y. Let x be a point in W, whose i^th component is y. There certainly is such a point, since the projection of W is W_i. Since y lies in V, x lies in U. Thus x is in W∩U.

Since U intersects every W in M, U belongs to M. This holds for every U of this form, i.e. containing q and restricting one component. The intersection of two such sets constrains two components. The intersection of finitely many such sets builds a base open set in P. Since M is closed under finite intersection, M includes every base open set in P that contains q.

Let C be a closed set taken from our original collection T, which is part of M. Since C intersects every open set containing q, C contains q. Therefore T has a nonempty intersection, namely q, and P is compact.

I don't have a proof handy for the compact product implying choice. It's just something I heard or read somewhere. If you have a proof, please send it along. Meantime, I'll try to come up with one while jumping on the trampoline.

Avoiding Choice

consider the subspace of S cross 0, which inherits a topology from the product space P. A base set in P is a base set B in S, cross an interval in [0,1]. If that interval happens to include 0, the base set intersects S:0 in B:0. In other words, we get B back again. Thus S:0 is homeomorphic to S.

Let U be an open cover for the product space P. When P is restricted to S:0, and U is restricted to same, U admits a finite subcover, since S is compact.

Let t be the supremum of the set of real numbers v, such that S cross [0,v] has a finite subcover in U. When v = 0 U has a subcover, so t is well defined.

The subspace S:t has a subcover in U, just as S:0 did. Call this subcover U₁.

We need to prove U₁ has a positive thickness across all of S. Express U₁ as a union of base sets in P. At this point our finite subcover might become infinite, but S is compact, so a finite subcover will do. Thus S:t is covered by finitely many base sets. These are base sets in P, consisting of base sets in S cross open intervals in [0,1] about t. Intersect these open intervals to find an open interval about t of radius ε. This is the uniform thickness we are looking for.

Select v between t-ε and t, and S:[0,v] has a subcover under U. Call this U₂, and let U₃ be the union of U₁ and U₂. Now U₃ is a finite subcover for S:[0,v], for any v between t and t+ε. This contradicts the selection of t, unless t = 1. Therefore S:[0,1] has a finite subcover under U, and P is compact.