Sikorski’s extension theorem

Theorem 1 (Sikorski’s ).

Let A be a Boolean subalgebra of a Boolean algebraMathworldPlanetmath B, and f:AC a Boolean algebra homomorphism from A to a complete Boolean algebra C. Then f can be extended to a Boolean algebra homomorphism g:BC.

Remark. In the categoryMathworldPlanetmath of Boolean algebras and Boolean algebra homomorphisms, this theorem says that every complete Boolean algebra is an injective object.


We prove this using Zorn’s lemma. Let M be the set of all pairs (h,D) such that D is a subalgebraMathworldPlanetmathPlanetmath of B containing A, and h:DC is an algebra homomorphism extending f. Note that M is not empty because (f,A)M. Also, if we define (h1,D1)(h2,D2) by requiring that D1D2 and that h2 extending h1, then (M,) becomes a poset. Notice that for every chain 𝒞 in M,


is an upper bound of 𝒞 (in fact, the least upper bound). So M has a maximal elementMathworldPlanetmath, say (g,E), by Zorn’s lemma. We want to show that E=B.

If EB, pick aB-E. Let r be the join of all elements of the form g(x) where xE and xa, and t the meet of all elements of the form g(y) where yE and ay. r and t exist because C is completePlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath. Since g preserves order, it is evident that rt. Pick an element sC such that rst.

Let F=E,a. Every element in F has the form (e1a)(e2a), with e1,e2E. Define h:FC by setting h(b)=(g(e1)s)(g(e2)s), where b=(e1a)(e2a). We now want to show that h is a Boolean algebra homomorphism extending g. There are three steps to showing this:

  1. 1.

    h is a function. Suppose (e1a)(e2a)=(e3a)(e4a). Then, by the last remark of this entry (, e2Δe4ae1e3, so that g(e2)Δg(e4)=g(e2Δe4)sg(e1e3)=g(e1)g(e3), which in turn implies that (g(e1)s)(g(e2)s)=(g(e3)s)(g(e4)s). Hence h is well-defined.

  2. 2.

    h is a Boolean homomorphism. All we need to show is that h respects and . Let x=(e1a)(e2a) and y=(e3a)(e4a). Then xy=(e5a)(e6a), where e5=e1e3 and e6=e2e4. So

    h(xy) = (g(e5)s)(g(e6)s)
    = ((g(e1)g(e3))s)((g(e2)g(e4))s)
    = (g(e1)s)(g(e2)s)(g(e3)s)(g(e4)s)
    = h(x)h(y),

    so h respects . In additionPlanetmathPlanetmath, h respects , as x=(e2a)(e1a), so that

    h(x) = h((e2a)(e1a))=(g(e2)s)(g(e1)s)
    = (g(e2)s)(g(e1)s)=((g(e1)s)(g(e2)s))
    = h(x).
  3. 3.

    h extends g. If xE, write x=(xa)(xa). Then


This implies that (g,E)<(h,F), and with this, we have a contradictionMathworldPlanetmathPlanetmath that (g,E) is maximal. This completes the proof. ∎

One of the consequences of this theorem is the following variant of the Boolean prime ideal theorem:

Corollary 1.

Every Boolean ideal of a Boolean algebra is contained in a maximal idealPlanetmathPlanetmath.


Let I be an ideal of a Boolean algebra A. Let B=I, the Boolean subalgebra generated by I. The function f:B{0,1} given by f(a)=0 iff aI is a Boolean homomorphism. First, notice that f(a)=0 iff aI iff aI iff f(a)=1. Next, if at least one of a,b is in I, abI, so that f(ab)=0=f(a)f(b). If neither are in I, then a,bI, so (ab)=abI, or abI. This means that f(ab)=1=f(a)f(b).

Now, by Sikorski’s extension theorem, f can be extended to a homomorphismMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath g:A{0,1}. The kernel of g clearly contains I, and is in addition maximal (either a or a is in the kernel of g). ∎


  • As the proof of the theorem shows, ZF+AC (the axiom of choiceMathworldPlanetmath) implies Sikorski’s extension theorem (SET). It is still an open question whether the ZF+SET implies AC.

  • Next, comparing with the Boolean prime ideal theorem (BPI), the proof of the corollary above shows that ZF+SET implies BPI. However, it was proven by John Bell in 1983 that SET is independent from ZF+BPI: there is a model satisfying all axioms of ZF, as well as BPI (considered as an axiom, not as a consequence of AC), such that SET fails.


  • 1 R. Sikorski, Boolean Algebras, 2nd Edition, Springer-Verlag, New York (1964).
  • 2 J. L. Bell, Axiom of Choice, Stanford Encyclopedia of Philosophy (2008).
Title Sikorski’s extension theorem
Canonical name SikorskisExtensionTheorem
Date of creation 2013-03-22 18:01:31
Last modified on 2013-03-22 18:01:31
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 21
Author CWoo (3771)
Entry type Theorem
Classification msc 06E10
Synonym Sikorski extension theorem