Sikorski’s extension theorem
Theorem 1 (Sikorski’s ).
Let A be a Boolean subalgebra of a Boolean algebra B, and f:A→C a Boolean algebra homomorphism from A to a complete Boolean algebra C. Then f can be extended to a Boolean algebra homomorphism g:B→C.
Remark. In the category of Boolean algebras and Boolean algebra homomorphisms, this theorem says that every complete Boolean algebra is an injective object.
Proof.
We prove this using Zorn’s lemma. Let M be the set of all pairs (h,D) such that D is a subalgebra of B containing A, and h:D→C 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 D1⊆D2 and that h2 extending h1, then (M,≤) becomes a poset. Notice that for every chain 𝒞 in M,
(⋃{h∣(h,D)∈𝒞},⋃{D∣(h,D)∈𝒞}) |
is an upper bound of 𝒞 (in fact, the least upper bound). So M has a maximal element, say (g,E), by Zorn’s lemma. We want to show that E=B.
If E≠B, pick a∈B-E. Let r be the join of all elements of the form g(x) where x∈E and x≤a, and t the meet of all elements of the form g(y) where y∈E and a≤y. r and t exist because C is complete. Since g preserves order, it is evident that r≤t. Pick an element s∈C such that r≤s≤t.
Let F=⟨E,a⟩. Every element in F has the form (e1∧a)∨(e2∧a′), with e1,e2∈E. Define h:F→C by setting h(b)=(g(e1)∧s)∨(g(e2)∧s′), where b=(e1∧a)∨(e2∧a′). We now want to show that h is a Boolean algebra homomorphism extending g. There are three steps to showing this:
-
1.
h is a function. Suppose (e1∧a)∨(e2∧a′)=(e3∧a)∨(e4∧a′). Then, by the last remark of this entry (http://planetmath.org/BooleanSubalgebra), e2Δe4≤a≤e1↔e3, so that g(e2)Δg(e4)=g(e2Δe4)≤s≤g(e1↔e3)=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.
h is a Boolean homomorphism. All we need to show is that h respects ∨ and ′. Let x=(e1∧a)∨(e2∧a′) and y=(e3∧a)∨(e4∧a′). Then x∨y=(e5∧a)∨(e6∧a′), where e5=e1∨e3 and e6=e2∨e4. So
h(x∨y) = (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 addition, h respects ′, as x′=(e′2∧a)∨(e1∧a′), so that
h(x′) = h((e′2∧a)∨(e1∧a′))=(g(e′2)∧s)∨(g(e1)∧s′) = (g(e2)′∧s)∨(g(e1)∧s′)=((g(e1)∧s)∨(g(e2)∧s′))′ = h(x)′. -
3.
h extends g. If x∈E, write x=(x∧a)∨(x∧a′). Then
h(x)=(g(x)∧s)∨(g(x)∧s′)=g(x).
This implies that (g,E)<(h,F), and with this, we have a contradiction 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 ideal.
Proof.
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 a∈I is a Boolean homomorphism. First, notice that f(a)=0 iff a∈I iff a′∉I iff f(a′)=1. Next, if at least one of a,b is in I, a∧b∈I, so that f(a∧b)=0=f(a)∧f(b). If neither are in I, then a′,b′∈I, so (a∧b)′=a′∨b′∈I, or a∧b∉I. This means that f(a∧b)=1=f(a)∧f(b).
Now, by Sikorski’s extension theorem, f can be extended to a homomorphism 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). ∎
Remarks.
-
•
As the proof of the theorem shows, ZF+AC (the axiom of choice) 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.
References
- 1 R. Sikorski, Boolean Algebras, 2nd Edition, Springer-Verlag, New York (1964).
- 2 J. L. Bell, http://plato.stanford.edu/entries/axiom-choice/The 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 |