Tychonoff theorem

The Tychonoff Theorem


The Tychonoff/Tihonov/Tikhonov theorem is a theorem of topology, which in its classical form is equivalent to the axiom of choice. Spellings vary; the theorem is named after Андрей Николаевич Тихонов, whose name becomes Tihonov in the transliteration used in slavistics or ‘Andrey Nikolayevich Tikhonov’ in modern English (BGN/PCGN) transliteration, but he originally published in German as ‘A.N. Tychonoff’. The topology on the product space is also called Tikhonov’s (Tihonov’s, Tychonoff) topology.

The theorem states that a product of spaces (indexed over an arbitrary set) is compact if each of the spaces is compact. (Compare the axiom of choice: a product of sets is inhabited if each set is inhabited.)

Notice that no choice whatsoever is needed to prove the analogous theorem for locales: even in constructive mathematics, a product of locales is compact if each of the locales is compact. From that perspective, the statement equivalent to the axiom of choice is this: a product of locales is spatial if each of the locales is spatial. Either of these may be considered a Tychonoff theorem for locales.

To prove the Tychonoff theorem for Hausdorff spaces only, the full axiom of choice is not needed; the ultrafilter theorem (and possibly excluded middle) are enough.

Proof via ultrafilter convergence

One method of proof uses ultrafilter convergence. Let X α αA\langle X_\alpha \rangle_{\alpha \in A} be a family of compact spaces.

  1. Every ultrafilter U αU_\alpha on the underlying set of X αX_\alpha converges to some point x αx_\alpha. (Consider the collection of all closed sets belonging to U αU_\alpha. Finite subcollections have nonempty intersection, so by compactness the intersection of the collection is also nonempty. Let x αx_\alpha be a point belonging to that intersection. Every closed set disjoint from x αx_\alpha belongs to the complement of U αU_\alpha, hence every open set containing x αx_\alpha belongs to U αU_\alpha, i.e., U αU_\alpha converges to x αx_\alpha.)

  2. Let UU be an ultrafilter on αX α\prod_\alpha X_\alpha. Since the set-of-ultrafilters construction SβSS \mapsto \beta S is functorial, we have a push-forward ultrafilter U α=β(π α)(U)U_\alpha = \beta(\pi_\alpha)(U) on each X αX_\alpha, where π α\pi_\alpha is a projection map. Choose a point x αx_\alpha to which U αU_\alpha converges. Then it is easily checked that UU converges to the point x α\langle x_\alpha \rangle in the product space.

  3. Since every ultrafilter UU on αX α\prod_\alpha X_\alpha converges to a point, the product is compact. (If it were not, then we could find a collection of nonempty closed sets whose intersection is empty, but closed under finite intersections. This collection generates a filter which is contained in some ultrafilter UU, by the ultrafilter theorem. Since every ultrafilter UU converges to some xx, it cannot contain any closed set in the complement of xx, hence cannot contain some closed set in the original collection, contradiction.)

Toby: Note that we use excluded middle in two places for a proof by contradiction; I'll try to remove these if possible. (Todd: I’ve rearranged the argument in the first step; the first formulation I gave looked unnecessarily “choice-y”.) We use the axiom of choice in the middle step to combine the x αx_\alpha into a single family x α\langle x_\alpha \rangle; if the X αX_\alpha were Hausdorff, then the x αx_\alpha would be unique, and we would not need the axiom of choice here. Finally, we use the ultrafilter theorem in the last step.

Todd: Yes, although I’m not too fussed about using indirect arguments, since in a topos, excluded middle follows from AC. I don’t know whether it also follows from a weaker choice principle equivalent to the ultrafilter theorem, but I guess I wouldn’t be surprised if it did.

Toby: As far as I know, UF and EM are independent (although I'm not certain). So I'm interested in whether the Hausdorff case requires EM as well as UF.

Todd: I wrote John Bell about this, and he kindly responded. Apparently you are right: UF (or BPIT) implies only a weakened version of EM: ¬x¬¬x=1\neg x \vee \neg\neg x = 1. He referred me to his paper Boolean Algebras and Distributive Lattices Treated Constructively, for this (among much else).

Toby: Great, hard results! Although I'm not sure that his ultrafilter theorem is the same as mine, since he talks about ultrafilters in arbitrary distributive lattices, while I would use a constructively stronger definition of ultrafilter in a powerset. But I should be able to figure out what carries over, now that I have a result to go for.

Proof of converses

Now we will prove that the Tychonoff theorem implies the axiom of choice, while the Tychonoff theorem for Hausdorff spaces implies the ultrafilter theorem. This is done by judicious choice of examples.

Tychonoff theorem implies axiom of choice: let {X α} αA\{X_\alpha\}_{\alpha \in A} be a family of nonempty sets. Let Y αY_\alpha be obtained by adjoining a point pp to X αX_\alpha. Topologize Y αY_\alpha by taking the nontrivial open sets to be X αX_\alpha and {p}\{p\}. Then Y αY_\alpha is compact; assuming Tychonoff, Y= αY αY = \prod_\alpha Y_\alpha is compact. For each α\alpha, put

K α={yY:y αX α}K_\alpha = \{y \in Y: y_\alpha \in X_\alpha\}

Then K αK_\alpha is closed, and any finite intersection of the K αK_\alpha is nonempty (use pp in all but finitely many components). Hence

αX α= αK α\prod_\alpha X_\alpha = \bigcap_\alpha K_\alpha

is nonempty as well, by compactness. Thus the axiom of choice follows.

Tychonoff theorem for Hausdorff spaces implies ultrafilter theorem: let FF be a filter on a set XX, i.e., a filter in the Boolean algebra PXP X. An ultrafilter UU containing FF is tantamount to a Boolean algebra map 2 X22^X \to 2 which sends all of FF to 1, or equivalently to a Boolean algebra map ϕ:B=2 X/F2\phi: B = 2^X/F \to 2. Thus it suffices to construct a Boolean algebra map ϕ:B2\phi: B \to 2 for any Boolean algebra BB.

Let B|B| denote the underlying set of BB. The set of all functions f:B2f: |B| \to 2 is a B|B|-indexed product of copies of 2; considering 2 as a compact Hausdorff space, this set is compact Hausdorff, assuming the Tychonoff theorem for Hausdorff spaces. For each finite subset SBS \subseteq |B|, let K SK_S be the set of functions f:B2f: |B| \to 2 for which the equations

f(xy)=f(x)f(y)f(¬x)=¬f(x)f(x \wedge y) = f(x) \wedge f(y) \qquad f(\neg x) = \neg f(x)

hold for all x,ySx, y \in S. Then K SK_S is a closed subset of the topological space 2 B2^{|B|}, and it is easy (and classical) that it is nonempty (the subalgebra B(S)B(S) generated by SS is finite and admits such an ff, and we can take f(b)=1f(b) = 1 for any bb outside B(S)B(S)). By compactness, the intersection of all the K SK_S is nonempty, and this is precisely the set of Boolean algebra maps ϕ:B2\phi: B \to 2. Thus the ultrafilter theorem follows.

  • This proof may be seen as a classical illustration of the compactness theorem for propositional logic. We may think of a Boolean algebra BB as a Lindenbaum algebra for a propositional theory, and a Boolean algebra map ϕ:B2\phi: B \to 2 as a model for that theory. A model exists iff a model exists for every theory generated by a finite subset of propositions, by the compactness theorem. In fact, the proof above suggests the topological provenance of the term “compactness theorem”.

Revised on October 17, 2011 18:10:02 by Dmitri Pavlov (