complete Boolean algebra

A Boolean algebra $A$ is a complete Boolean algebra if for every subset $C$ of $A$, the arbitrary join and arbitrary meet of $C$ exist.

By de Morgan’s laws, it is easy to see that a Boolean algebra is complete iff the arbitrary join of any subset exists iff the arbitrary meet of any subset exists. For a proof of this, see this link (http://planetmath.org/PropertiesOfArbitraryJoinsAndMeets).

For an example of a complete Boolean algebra, let $S$ be any set. Then the powerset $P(S)$ with the usual set theoretic operations is a complete Boolean algebra.

In a complete Boolean algebra, the infinite distributive and infinite deMorgan’s laws hold:

• $x\wedge \bigvee A=\bigvee (x\wedge A)$ and $x\vee \bigwedge A=\bigwedge (x\vee A)$

• ${(\bigvee A)}^{*}=\bigwedge A^{*}$ and ${(\bigwedge A)}^{*}=\bigvee A^{*}$, where $A^{*}:=\{a^{*}\mid a\in A\}$.

In the category of complete Boolean algebras, a morphism between two objects is a Boolean algebra homomorphism that preserves arbitrary joins (equivalently, arbitrary meets), and is called a complete Boolean algebra homomorphism.

Remark There are infinitely many algebras between Boolean algebras and complete Boolean algebras. Let $\kappa$ be a cardinal. A Boolean algebra $A$ is said to be $\kappa$-complete if for every subset $C$ of $A$ with $|C|\le \kappa$, $\bigvee C$ (and equivalently $\bigwedge C$) exists. A $\kappa$-complete Boolean algebra is usually called a $\kappa$-algebra. If $\kappa ={\mathrm{\aleph }}_0$, the first aleph number, then it is called a countably complete Boolean algebra.

Any complete Boolean algebra is $\kappa$-complete, and any $\kappa$-complete is $\lambda$-complete for any $\lambda \le \kappa$. An example of a $\kappa$-complete algebra that is not complete, take a set $S$ with , then the collection $A\subseteq P(S)$ consisting of any subset $T$ such that either $|T|\le \kappa$ or $|S-T|\le \kappa$ is $\kappa$-complete but not complete.

A Boolean algebra homomorphism $f$ between two $\kappa$-algebras $A,B$ is said to be $\kappa$-complete if

$$f(\bigvee \{a\mid a\in C\})=\bigvee \{f(a)\mid a\in C\}$$

for any $C\subseteq A$ with $|C|\le \kappa$.