# Sikorski’s extension theorem

###### Theorem 1 (Sikorski’s ).

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

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 $(h_{1},D_{1})\leq(h_{2},D_{2})$ by requiring that $D_{1}\subseteq D_{2}$ and that $h_{2}$ extending $h_{1}$, then $(M,\leq)$ becomes a poset. Notice that for every chain $\mathcal{C}$ in $M$,

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

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

Let $F=\langle E,a\rangle$. 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. 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), $e_{2}\Delta e_{4}\leq a\leq e_{1}\leftrightarrow e_{3}$, so that $g(e_{2})\Delta g(e_{4})=g(e_{2}\Delta e_{4})\leq s\leq g(e_{1}\leftrightarrow e% _{3})=g(e_{1})\leftrightarrow g(e_{3})$, 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. 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

 $\displaystyle h(x\vee y)$ $\displaystyle=$ $\displaystyle(g(e_{5})\wedge s)\vee(g(e_{6})\wedge s^{\prime})$ $\displaystyle=$ $\displaystyle((g(e_{1})\vee g(e_{3}))\wedge s)\vee((g(e_{2})\vee g(e_{4}))% \wedge s^{\prime})$ $\displaystyle=$ $\displaystyle(g(e_{1})\wedge s)\vee(g(e_{2})\wedge s^{\prime})\vee(g(e_{3})% \wedge s)\vee(g(e_{4})\wedge s^{\prime})$ $\displaystyle=$ $\displaystyle h(x)\vee h(y),$

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

 $\displaystyle h(x^{\prime})$ $\displaystyle=$ $\displaystyle h((e_{2}^{\prime}\wedge a)\vee(e_{1}\wedge a^{\prime}))=(g(e_{2}% ^{\prime})\wedge s)\vee(g(e_{1})\wedge s^{\prime})$ $\displaystyle=$ $\displaystyle(g(e_{2})^{\prime}\wedge s)\vee(g(e_{1})\wedge s^{\prime})=((g(e_% {1})\wedge s)\vee(g(e_{2})\wedge s^{\prime}))^{\prime}$ $\displaystyle=$ $\displaystyle h(x)^{\prime}.$
3. 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 $(g,E)<(h,F)$, 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 $B=\langle I\rangle$, 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.

## 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 SikorskisExtensionTheorem 2013-03-22 18:01:31 2013-03-22 18:01:31 CWoo (3771) CWoo (3771) 21 CWoo (3771) Theorem msc 06E10 Sikorski extension theorem