Tychonoff's theorem

Statement

Here is the statement of the theorem.

Strength

The above theorem turns out to be equivalent to the axiom of choice. There is a special case of the theorem saying

A product of compact Hausdorff spaces

X

i

{\displaystyle X_{i}}

is compact Hausdorff. If, in addition, all the

X

i

{\displaystyle X_{i}}

are nonempty, then the product

∏

X

i

{\displaystyle \prod X_{i}}

is nonempty.

This Hausdorff version of the theorem is actually strictly weaker than the axiom of choice and is equivalent to the Boolean prime ideal theorem. (Note for an arbitrary product, the axiom of choice ensures the product is nonempty when each factor is nonempty. But in the Hausdorff case, the above version can ensure the nonempty-ness without Choice. In fact, it implies a version of the axiom of choice; see § Application: the Hahn–Banach theorem.)

For a finite product of compact sets, the theorem is rather trivial (see § Tychonoff's theorem for finite products below) and does not depend on the axiom of choice or weaker variants.

Tychonoff's theorem for finite products

For a finite product of compact spaces, the theorem is substantially easier. For example, it would follow from

if

X

{\displaystyle X}

is a compact space and

Y

{\displaystyle Y}

is any topological space, then the projection

p

:

X

×

Y

→

Y

{\displaystyle p:X\times Y\to Y}

is proper; i.e., the pre-image of a compact set is compact.

Alternatively, the tube lemma can be used to give a direct proof; see Tube lemma § Examples and properties.

Further remark

For a metric space, compactness is equivalent to sequential compactness. Using Cantor's diagonal argument, it is straightforward to show a countable product of sequentially compact spaces is sequentially compact space; thus, for metric spaces at least, this gives the theorem for a countable product.

However, for many applications, taking uncountable products is necessary. For example, taking

L

=

C

(

X

,

I

)

{\displaystyle L=C(X,I)}

to be the set of all continuous maps from a topological space

X

{\displaystyle X}

to the unit

I

=

[

0

,

1

]

{\displaystyle I=[0,1]}

, we have the natural map

X

→

∏

L

Applications

Tychonoff's theorem has been used to prove many other mathematical theorems. These include theorems about compactness of certain spaces such as the Banach–Alaoglu theorem on the weak-* compactness of the unit ball of the dual space of a normed vector space, and the Arzelà–Ascoli theorem characterizing the sequences of functions in which every subsequence has a uniformly convergent subsequence. They also include statements less obviously related to compactness, such as the De Bruijn–Erdős theorem stating that every minimal k-chromatic graph is finite, and the Curtis–Hedlund–Lyndon theorem providing a topological characterization of cellular automata.

As a rule of thumb, any sort of construction that takes as input a fairly general object (often of an algebraic, or topological-algebraic nature) and outputs a compact space is likely to use Tychonoff: e.g., the Gelfand space of maximal ideals of a commutative C*-algebra, the Stone space of maximal ideals of a Boolean algebra, and the Berkovich spectrum of a commutative Banach ring.

In logic, specifically model theory, Tychonoff’s theorem implies the compactness theorem, which says a theory (a set of sentences) has a model if and only if each finite subset of it does.

Proofs

1) Tychonoff's 1930 proof used the concept of a complete accumulation point.

2) The theorem is a quick corollary of the Alexander subbase theorem.

3) The theory of convergence via filters, due to Henri Cartan and developed by Bourbaki, leads to the following criterion: assuming the ultrafilter lemma, a space is compact if and only if each ultrafilter on the space converges. With this in hand, the proof becomes easy: the (filter generated by the) image of an ultrafilter on the product space under any projection map is an ultrafilter on the factor space, which therefore converges, to at least one xi. One then shows that the original ultrafilter converges to x = (xi). See Filters in topology § Tychonoff's theorem for the ingredients used here.

In the above proof, the axiom of choice is used two times: (1) to construct ultrafilters to characterize compactness and (2) to pick limit points

x

i

{\displaystyle x_{i}}

. Conversely, Tychonoff's theorem implies the axiom of choice; see § Proof of the axiom of choice from Tychonoff's theorem below. Thus, the use of the axiom of choice in the proof is unavoidable in general. However, if each factor

X

i

{\displaystyle X_{i}}

is Hausdorff, each limit point

x

i

{\displaystyle x_{i}}

is unique and so the second use of Choice is unnecessary. Ultrafilters, on the other hand, can be constructed by the Boolean prime ideal theorem or BPI (in the form of ultrafilter lemma) and, conversely, the Hausdorff version of the theorem implies BPI; see the remark at the end of § Application: the Hahn–Banach theorem. (Studying the strength of Tychonoff's theorem for various restricted classes of spaces is a topic in set-theoretic topology.)

Proof using nets

A proof here, due to P. Chernoff, is based on the following characterization of compactness in terms of nets:

A space is compact if and only if each net on it has a cluster point.

Given a family of compact sets

X

a

{\displaystyle X_{a}}

indexed by a set

A

{\displaystyle A}

, to show

∏

a

∈

A

X

a

{\displaystyle \prod _{a\in A}X_{a}}

is compact, we can therefore show each net

⟨

x

α

∣

α

∈

I

⟩

{\displaystyle \langle x_{\alpha }\mid \alpha \in I\rangle }

on it has a cluster point.

We shall identity the product

Proof of the axiom of choice from Tychonoff's theorem

To prove that Tychonoff's theorem in its general version implies the axiom of choice, we establish that every infinite cartesian product of non-empty sets is nonempty. The trickiest part of the proof is introducing the right topology. The right topology, as it turns out, is the cofinite topology with a small twist. It turns out that every set given this topology automatically becomes a compact space. Once we have this fact, Tychonoff's theorem can be applied; we then use the finite intersection property (FIP) definition of compactness. The proof itself (due to J. L. Kelley) follows:

Let {Ai} be an indexed family of nonempty sets, for i ranging in I (where I is an arbitrary indexing set). We wish to show that the cartesian product of these sets is nonempty. Now, for each i, take Xi to be Ai with the index i itself tacked on (renaming the indices using the disjoint union if necessary, we may assume that i is not a member of Ai, so simply take Xi = Ai ∪ {i}).

Now define the cartesian product

X

=

∏

i

∈

I

X

i

{\displaystyle X=\prod _{i\in I}X_{i}}

along with the natural projection maps πi which take a member of X to its ith term.

We give each Xj the topology whose open sets are: the empty set, the singleton {i}, the set Xi. This makes Xi compact, and by Tychonoff's theorem, X is also compact (in the product topology). The projection maps are continuous; all the Ai's are closed, being complements of the singleton open set {i} in Xi. So the inverse images πi−1(Ai) are closed subsets of X. We note that

Application: the Hahn–Banach theorem

A standard proof of the Hahn-Banach theorem in functional analysis uses the axiom of choice in the form of Zorn's lemma. Tychonoff's theorem for compact Hausdorff spaces, which is strictly weaker than the axiom of choice and is equivalent to the Boolean prime ideal theorem, can be used to give an alternative proof.

Specifically, Tychonoff's theorem in the Hausdorff case is used to deduce the following variant of the axiom of choice. Given a family of nonempty compact Hausdorff spaces

K

x

,

x

∈

X

{\displaystyle K_{x},\,x\in X}

parametrized by a set

X

{\displaystyle X}

, by a choice function for it, we shall mean a function

c

:

X

→

U

:=

∪

x

∈

X

K

x

{\displaystyle c:X\to U:=\cup _{x\in X}K_{x}}