Sikorski’s extension theorem

Theorem 1 (Sikorski’s ).

Let $A$ be a Boolean subalgebra of a Boolean algebra $B$, and $f\mathrm{:}A\mathrm{\to }C$ a Boolean algebra homomorphism from $A$ to a complete Boolean algebra $C$. Then $f$ can be extended to a Boolean algebra homomorphism $g\mathrm{:}B\mathrm{\to }C$.

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 $h:D\to C$ is an algebra homomorphism extending $f$. Note that $M$ is not empty because $(f,A)\in M$. Also, if we define $(h_1,D_1)\le (h_2,D_2)$ by requiring that $D_1\subseteq D_2$ and that $h_2$ extending $h_1$, then $(M,\le )$ becomes a poset. Notice that for every chain $𝒞$ in $M$,

$$(\bigcup \{h\mid (h,D)\in 𝒞\},\bigcup \{D\mid (h,D)\in 𝒞\})$$

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 $E\ne B$, pick $a\in B-E$. Let $r$ be the join of all elements of the form $g(x)$ where $x\in E$ and $x\le a$, and $t$ the meet of all elements of the form $g(y)$ where $y\in E$ and $a\le y$. $r$ and $t$ exist because $C$ is complete. Since $g$ preserves order, it is evident that $r\le t$. Pick an element $s\in C$ such that $r\le s\le t$.

Let $F=⟨E,a⟩$. Every element in $F$ has the form $(e_1\wedge a)\vee (e_2\wedge a^{\prime })$, with $e_1,e_2\in E$. Define $h:F\to C$ by setting $h(b)=(g(e_1)\wedge s)\vee (g(e_2)\wedge s^{\prime })$, where $b=(e_1\wedge a)\vee (e_2\wedge a^{\prime })$. 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 $(e_1\wedge a)\vee (e_2\wedge a^{\prime })=(e_3\wedge a)\vee (e_4\wedge a^{\prime })$. Then, by the last remark of this entry (http://planetmath.org/BooleanSubalgebra), $e_2\mathrm{\Delta }{e}_4\le a\le e_1\leftrightarrow e_3$, so that $g(e_2)\mathrm{\Delta }g(e_4)=g(e_2\mathrm{\Delta }{e}_4)\le s\le g(e_1\leftrightarrow e_3)=g(e_1)\leftrightarrow g(e_3)$, which in turn implies that $(g(e_1)\wedge s)\vee (g(e_2)\wedge s^{\prime })=(g(e_3)\wedge s)\vee (g(e_4)\wedge s^{\prime })$. Hence $h$ is well-defined.

2. $h$ is a Boolean homomorphism. All we need to show is that $h$ respects $\vee$ and ${}^{\prime }$. Let $x=(e_1\wedge a)\vee (e_2\wedge a^{\prime })$ and $y=(e_3\wedge a)\vee (e_4\wedge a^{\prime })$. Then $x\vee y=(e_5\wedge a)\vee (e_6\wedge a^{\prime })$, where $e_5=e_1\vee e_3$ and $e_6=e_2\vee e_4$. So

   $h(x\vee y)$ $=$ $(g(e_5)\wedge s)\vee (g(e_6)\wedge s^{\prime })$

   $=$ $((g(e_1)\vee g(e_3))\wedge s)\vee ((g(e_2)\vee g(e_4))\wedge s^{\prime })$

   $=$ $(g(e_1)\wedge s)\vee (g(e_2)\wedge s^{\prime })\vee (g(e_3)\wedge s)\vee (g(e_4)\wedge s^{\prime })$

   $=$ $h(x)\vee h(y),$

   so $h$ respects $\vee$. In addition, $h$ respects ${}^{\prime }$, as $x^{\prime }=(e_2^{\prime }\wedge a)\vee (e_1\wedge a^{\prime })$, so that

   $h(x^{\prime })$ $=$ $h((e_2^{\prime }\wedge a)\vee (e_1\wedge a^{\prime }))=(g(e_2^{\prime })\wedge s)\vee (g(e_1)\wedge s^{\prime })$

   $=$ $(g{(e_2)}^{\prime }\wedge s)\vee (g(e_1)\wedge s^{\prime })={((g(e_1)\wedge s)\vee (g(e_2)\wedge s^{\prime }))}^{\prime }$

   $=$ $h{(x)}^{\prime }.$

3. $h$ extends $g$. If $x\in E$, write $x=(x\wedge a)\vee (x\wedge a^{\prime })$. Then

   $$h(x)=(g(x)\wedge s)\vee (g(x)\wedge s^{\prime })=g(x).$$

This implies that , 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 $f:B\to \{0,1\}$ given by $f(a)=0$ iff $a\in I$ is a Boolean homomorphism. First, notice that $f(a)=0$ iff $a\in I$ iff $a^{\prime }\notin I$ iff $f(a^{\prime })=1$. Next, if at least one of $a,b$ is in $I$, $a\wedge b\in I$, so that $f(a\wedge b)=0=f(a)\wedge f(b)$. If neither are in $I$, then $a^{\prime },b^{\prime }\in I$, so ${(a\wedge b)}^{\prime }=a^{\prime }\vee b^{\prime }\in I$, or $a\wedge b\notin I$. This means that $f(a\wedge b)=1=f(a)\wedge f(b)$.

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

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