In

mathematicsMathematics is the study of quantity, space, structure, and change. Mathematicians seek out patterns and formulate new conjectures. Mathematicians resolve the truth or falsity of conjectures by mathematical proofs, which are arguments sufficient to convince other mathematicians of their validity...

, Tychonoff's theorem states that the

productIn topology and related areas of mathematics, a product space is the cartesian product of a family of topological spaces equipped with a natural topology called the product topology...

of any collection of

compactIn mathematics, specifically general topology and metric topology, a compact space is an abstract mathematical space whose topology has the compactness property, which has many important implications not valid in general spaces...

topological spaceTopological spaces are mathematical structures that allow the formal definition of concepts such as convergence, connectedness, and continuity. They appear in virtually every branch of modern mathematics and are a central unifying notion...

s is compact. The theorem is named after

Andrey Nikolayevich TychonoffAndrey Nikolayevich Tikhonov was a Soviet and Russian mathematician known for important contributions to topology, functional analysis, mathematical physics, and ill-posed problems. He was also inventor of magnetotellurics method in geology. Tikhonov originally published in German, whence the...

, who proved it first in 1930 for powers of the closed

unit intervalIn mathematics, the unit interval is the closed interval , that is, the set of all real numbers that are greater than or equal to 0 and less than or equal to 1...

and in 1935 stated the full theorem along with the remark that its proof was the same as for the special case. The earliest known published proof is contained in a 1937 paper of

Eduard ČechEduard Čech was a Czech mathematician born in Stračov, Bohemia . His research interests included projective differential geometry and topology. In 1921–1922 he collaborated with Guido Fubini in Turin...

.

Several texts identify Tychonoff's theorem as the single most important result in general topology [e.g. Willard, p. 120]; others allow it to share this honor with

Urysohn's lemmaIn topology, Urysohn's lemma is a lemma that states that a topological space is normal if and only if any two disjoint closed subsets can be separated by a function....

.

## Definition

The theorem crucially depends upon the precise definitions of

compactnessIn mathematics, specifically general topology and metric topology, a compact space is an abstract mathematical space whose topology has the compactness property, which has many important implications not valid in general spaces...

and of the

product topologyIn topology and related areas of mathematics, a product space is the cartesian product of a family of topological spaces equipped with a natural topology called the product topology...

; in fact, Tychonoff's 1935 paper defines the product topology for the first time. Conversely, part

of its importance is to give confidence that these particular definitions are the correct (i.e., most useful)

ones.

Indeed, the Heine–Borel definition of compactness — that every covering of a space by open sets admits a finite subcovering — is relatively recent. More popular in the 19th and early 20th centuries was the Bolzano–Weierstrass criterion that every sequence admits a convergent subsequence, now called sequential compactness. These conditions are equivalent for metrizable spaces, but neither implies the other on the class of all topological spaces.

It is almost trivial to prove that the product of two sequentially compact spaces is sequentially compact — one passes to a subsequence for the first component and then a subsubsequence for the second component. An only slightly more elaborate

"diagonalization" argumentCantor's diagonal argument, also called the diagonalisation argument, the diagonal slash argument or the diagonal method, was published in 1891 by Georg Cantor as a mathematical proof that there are infinite sets which cannot be put into one-to-one correspondence with the infinite set of natural...

establishes the sequential compactness of a countable product of sequentially compact spaces. However, the product of

continuum many copies of the closed unit interval fails to be sequentially compact.

This is a critical failure: if X is a completely regular

Hausdorff spaceIn topology and related branches of mathematics, a Hausdorff space, separated space or T2 space is a topological space in which distinct points have disjoint neighbourhoods. Of the many separation axioms that can be imposed on a topological space, the "Hausdorff condition" is the most frequently...

, there is a natural embedding from X into [0,1]

^{C(X,[0,1])}, where C(X,[0,1]) is the set of continuous maps from X to [0,1]. The compactness of [0,1]

^{C(X,[0,1])} thus shows that every completely regular Hausdorff space embeds in a compact Hausdorff space (or, can be "compactified".) This construction is the

Stone–Čech compactificationIn the mathematical discipline of general topology, Stone–Čech compactification is a technique for constructing a universal map from a topological space X to a compact Hausdorff space βX...

. Conversely, all subspaces of compact Hausdorff spaces are completely regular Hausdorff, so this characterizes the completely regular Hausdorff spaces as those that can be compactified. Such spaces are now called

Tychonoff spaceIn topology and related branches of mathematics, Tychonoff spaces and completely regular spaces are kinds of topological spaces.These conditions are examples of separation axioms....

s.

## 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 compactness of the unit ball of the

dual spaceIn mathematics, any vector space, V, has a corresponding dual vector space consisting of all linear functionals on V. Dual vector spaces defined on finite-dimensional vector spaces can be used for defining tensors which are studied in tensor algebra...

of a

normed vector spaceIn mathematics, with 2- or 3-dimensional vectors with real-valued entries, the idea of the "length" of a vector is intuitive and can easily be extended to any real vector space Rn. The following properties of "vector length" are crucial....

, 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 graphIn general the notion of criticality can refer to any measure.But in graph theory, when the term is used without any qualification, it almost always refers to the chromatic number of a graph....

is finite, and the

Curtis–Hedlund–Lyndon theoremThe Curtis–Hedlund–Lyndon theorem is a mathematical characterization of cellular automata in terms of their symbolic dynamics. It is named after Morton L. Curtis, Gustav A...

providing a topological characterization of

cellular automataA cellular automaton is a discrete model studied in computability theory, mathematics, physics, complexity science, theoretical biology and microstructure modeling. It consists of a regular grid of cells, each in one of a finite number of states, such as "On" and "Off"...

.

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 spaceIn mathematics, the Gelfand representation in functional analysis has two related meanings:* a way of representing commutative Banach algebras as algebras of continuous functions;...

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.

## Proofs of Tychonoff's theorem

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.

More modern proofs have been motivated by the following considerations: the approach to compactness via convergence of subsequences leads to a simple and transparent proof in the case of countable index sets. However, the approach to convergence in a topological space using sequences is sufficient when the space satisfies the first axiom of countability (as metrizable spaces do), but generally not otherwise. However, the product of uncountably many metrizable spaces, each with at least two points, fails to be first countable. So it is natural to hope that a suitable notion of convergence in arbitrary spaces will lead to a compactness criterion generalizing sequential compactness in metrizable spaces that will be as easily applied to deduce the compactness of products. This has turned out to be the case.

3) The theory of convergence via filters, due to

Henri CartanHenri Paul Cartan was a French mathematician with substantial contributions in algebraic topology. He was the son of the French mathematician Élie Cartan.-Life:...

and developed by Bourbaki in 1937, leads to the following criterion: assuming the ultrafilter lemma, a space is compact if and only if each

ultrafilterIn the mathematical field of set theory, an ultrafilter on a set X is a collection of subsets of X that is a filter, that cannot be enlarged . An ultrafilter may be considered as a finitely additive measure. Then every subset of X is either considered "almost everything" or "almost nothing"...

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 x

_{i}. One then shows that the original ultrafilter converges to x = (x

_{i}). In his textbook,

MunkresJames Raymond Munkres is a Professor Emeritus of mathematics at MIT and the author of several texts in the area of topology, including Topology , Analysis on Manifolds, Elements of Algebraic Topology, and Elementary Differential Topology...

gives a reworking of the Cartan–Bourbaki proof that does not explicitly use any filter-theoretic language or preliminaries.

4) Similarly, the Moore–Smith theory of convergence via nets, as supplemented by Kelley's notion of a

universal netIn mathematics, more specifically in general topology and related branches, a net or Moore–Smith sequence is a generalization of the notion of a sequence. In essence, a sequence is a function with domain the natural numbers, and in the context of topology, the range of this function is...

, leads to the criterion that a space is compact if and only if each universal net on the space converges. This criterion leads to a proof (Kelley, 1950) of Tychonoff's theorem, which is, word for word, identical to the Cartan/Bourbaki proof using filters, save for the repeated substitution of "universal net" for "ultrafilter base".

5) A proof using nets but not universal nets was given in 1992 by Paul Chernoff.

## Tychonoff's theorem and the axiom of choice

All of the above proofs use the

axiom of choice (AC) in some way. For instance, the third proof uses that every filter is contained in an ultrafilter (i.e., a maximal filter), and this is seen by invoking

Zorn's lemmaZorn's lemma, also known as the Kuratowski–Zorn lemma, is a proposition of set theory that states:Suppose a partially ordered set P has the property that every chain has an upper bound in P...

. Zorn's lemma is also used to prove Kelley's theorem, that every net has a universal subnet. In fact these uses of AC are essential: in 1950 Kelley proved that Tychonoff's theorem implies the axiom of choice. Note that one formulation of AC is that the Cartesian product of a family of nonempty sets is nonempty; but since the empty set is most certainly compact, the proof cannot proceed along such straightforward lines. Thus Tychonoff's theorem joins several other basic theorems (e.g. that every nonzero vector space has a basis) in being equivalent to AC.

On the other hand, the statement that every filter is contained in an ultrafilter does not imply AC. Indeed, it is not hard to see that it is equivalent to the

Boolean prime ideal theoremIn mathematics, a prime ideal theorem guarantees the existence of certain types of subsets in a given abstract algebra. A common example is the Boolean prime ideal theorem, which states that ideals in a Boolean algebra can be extended to prime ideals. A variation of this statement for filters on...

(BPIT), a well-known intermediate point between the axioms of Zermelo-Fraenkel set theory (ZF) and the ZF theory augmented by the axiom of choice (ZFC). A first glance at the second proof of Tychnoff may suggest that the proof uses no more than (BPIT), in contradiction to the above. However, the spaces in which every convergent filter has a unique limit are precisely the Hausdorff spaces. In general we must select, for each element of the index set, an element of the nonempty set of limits of the projected ultrafilter base, and of course this uses AC. However, it also shows that the compactness of the product of compact Hausdorff spaces can be proved using (BPIT), and in fact the converse also holds. Studying the strength of Tychonoff's theorem for various restricted classes of spaces is an active area in

set-theoretic topologyIn mathematics, set-theoretic topology is a subject that combines set theory and general topology. It focuses on topological questions that are independent of ZFC. A famous problem is the normal Moore space question, a question in general topology that was the subject of intense research. The...

.

The analogue of Tychonoff's theorem in

pointless topologyIn mathematics, pointless topology is an approach to topology that avoids mentioning points. The name 'pointless topology' is due to John von Neumann...

does not require any form of the axiom of choice.

### 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 productIn mathematics, a Cartesian product is a construction to build a new set out of a number of given sets. Each member of the Cartesian product corresponds to the selection of one element each in every one of those sets...

of non-empty sets is nonempty. It is actually a more comprehensible proof than the above (probably because it does not involve Zorn's Lemma, which is quite opaque to most mathematicians as far as intuition is concerned!). 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 FIP definition of compactness (the FIP is sure convenient!). Anyway, to get to the proof itself (due to J.L. Kelley):

Let {A

_{i}} 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 X

_{i} to be A

_{i} with the index i itself tacked on (renaming the indices using the

disjoint unionIn mathematics, the term disjoint union may refer to one of two different concepts:* In set theory, a disjoint union is a modified union operation that indexes the elements according to which set they originated in; disjoint sets have no element in common.* In probability theory , a disjoint union...

if necessary, we may assume that i is not a member of A

_{i}, so simply take X

_{i} = A

_{i} ∪ {i}).

Now the define cartesian product

along with the natural projection maps π

_{i} which take a member of X to its ith term.

Now here's the trick: we give each X

_{i} the topology whose open sets are the cofinite subsets of X

_{i}, plus the empty set (the cofinite topology) and the singleton {i}.

This makes X

_{i} compact, and by Tychonoff's theorem, X is also compact (in the product topology). The projection maps are continuous; all the A

_{i}s are closed, being complements of the

singleton open set {i} in X

_{i}. So the inverse images π

_{i}^{−1}(A

_{i}) are closed subsets of X. We note that

and prove that these inverse images are nonempty and have the FIP. Let i

_{1}, ..., i

_{N} be a finite collection of indices in I. Then the finite product A

_{i1} × ... × A

_{iN}
is nonempty (only finitely many choices here, so no AC needed!); it merely consists of N-tuples. Let a = (a

_{1}, ..., a

_{N}) be such an N-tuple. We "extend" a to the whole index set: take a to the function f defined by f(j) = a

_{k} if j = i

_{k}, and f(j) = j otherwise. This step is where the addition of the extra point to each space is crucial (we didn't go through all that trouble for nothing!), for it allows us to define f for everything outside of the N-tuple in a precise way without choices (we can already "choose," by construction, j from X

_{j} ). π

_{ik}(f) = a

_{k} is obviously an element of each A

_{ik} so that f is in each inverse image; thus we have

By the FIP definition of compactness, the entire intersection over I must be nonempty, and we are done.