Tychonoff’s theorem implies AC

In this entry, we prove that TychonoffPlanetmathPlanetmath’s theoremMathworldPlanetmath implies that product of non-empty set of non-empty sets is non-empty, which is equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath to the axiom of choiceMathworldPlanetmath (AC). This fact, together with the fact that AC implies Tychonoff’s theorem, shows that Tychonoff’s theorem is equivalent to AC (under ZF). The proof was first discovered by John Kelley in 1950, and is now an exercise in axiomatic set theory.


Let C be a non-empty collectionMathworldPlanetmath of non-empty sets. Let Y be the generalized cartesian product of all the elements in C. Our objective is show that Y is non-empty.

First, some notations: for each AC, set XA:=A{A}, D:={XAAC}, X the generalized cartesian product of all the XA’s, and pA the projection from X onto XA.

We break down the proof into several steps:

  1. 1.

    Y is equipollentMathworldPlanetmath to Z:={pA-1(A)AC}.

    An element of X is a function f:DD, such that f(XA)XA for each AC. In other words, either f(XA)A, or f(XA)=A. An element of Y is a function g:CC such that g(A)A for each AC. Finally, hpA-1(A) iff h(XA)A.

    Given gY, define g*X by g*(XA):=g(A)A. Since A is arbitrary, g*Z. Conversely, given hZ, define hY by h(A):=h(XA), which is well-defined, since h(XA)A. Now, it is easy to see that the function ϕ:YZ given by ϕ(g)=g* is a bijection, whose inversePlanetmathPlanetmathPlanetmathPlanetmath ϕ-1:ZY is given by ϕ-1(h)=h. This shows that Y and Z are equipollent.

  2. 2.

    Next, we topologize each XA in such a way that XA is compactPlanetmathPlanetmath.

    Let 𝒯A be the coarsest topologyMathworldPlanetmathPlanetmath containing the cofinite topologyMathworldPlanetmath on XA and the singleton {A}. A typical open set of XA is either the empty setMathworldPlanetmath, or has the form S{A}, where S is cofinite in A.

    To show that XA is compact under 𝒯A, let 𝒟 be an open cover for XA. We want to show that there is a finite subset of 𝒟 covering XA. If XA𝒟, then we are done. Otherwise, pick a non-empty element S{A} in 𝒟, so that A-S, and is finite. By assumptionPlanetmathPlanetmath, each element in A-S belongs to some open set in 𝒟. So to cover A-S, only a finite number of open sets in 𝒟 are needed. These open sets, together with S{A}, cover XA. Hence XA is compact.

  3. 3.

    Finally, we prove that Z, and therefore Y, is non-empty.

    Apply Tychonoff’s theorem, X is compact under the product topology. Furthermore, πA is continuousPlanetmathPlanetmath for each AC. Since {A} is open in XA, and A=XA-{A}, A is closed in XA, and thus so is pA-1(A) closed in X.

    To show that Z is non-empty, we employ a characterizationMathworldPlanetmath of compact space: X is compact iff every collection of closed setsPlanetmathPlanetmath in X having FIP has non-empty intersectionMathworldPlanetmath (http://planetmath.org/ASpaceIsCompactIfAndOnlyIfTheSpaceHasTheFiniteIntersectionProperty). Let us look at the collection 𝒮:={pA-1(A)AC}. Given A1,,AnC, pick an element aiAi, since Ai by assumption. Note that this is possible, since there are only a finite number of sets. Define f:DD as follows:

    f(XA):={aiif A=Ai for some i=1,,n,Aotherwise.

    Since f(XAi)=aiAi, fpAi-1(Ai) for each i=1,,n. Therefore,


    Since pA1-1(A1),,pAn-1(An) are arbitrarily picked from 𝒮, the collection 𝒮 has finite intersection property, and since X is compact, Z=S must be non-empty.

This completesPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath the proof. ∎

Remark. In the proof, we see that the trick is to adjoin the set {A} to each set AC. Instead of {A}, we could have picked some arbitrary, but fixed singleton {B}, as long as BA for each AC, and the proof follows essentially the same way.


  • 1 T. J. Jech, The Axiom of Choice. North-Holland Pub. Co., Amsterdam, 1973.
  • 2 J. L. Kelley, The Tychonoff’s productPlanetmathPlanetmath theorem implies the axiom of choice. Fund. Math. 37, pp. 75-76, 1950.
Title Tychonoff’s theorem implies AC
Canonical name TychonoffsTheoremImpliesAC
Date of creation 2013-03-22 18:45:38
Last modified on 2013-03-22 18:45:38
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 7
Author CWoo (3771)
Entry type Proof
Classification msc 54D30
Classification msc 03E25