free Boolean algebra

Let $A$ be a Boolean algebra and $X\subseteq A$ such that $⟨X⟩=A$. In other words, $X$ is a set of generators of $A$. $A$ is said to be freely generated by $X$, or that $X$ is a free set of generators of $A$, if $⟨X⟩=A$, and every function $f$ from $X$ to some Boolean algebra $B$ can be extended to a Boolean algebra homomorphism $g$ from $A$ to $B$, as illustrated by the commutative diagram below:

where $i:X\to A$ is the inclusion map. By extension of $f$ to $g$ we mean that $g(x)=f(x)$ for every $x\in X$. Any subset $X\subseteq A$ containing $0$ (or $1$) can never be a free generating set for any subalgebra of $A$, for any function $f:X\to B$ such that $f(0)\ne 0$ can never be extended to a Boolean homomorphism.

A Boolean algebra is said to be free if it has a free set of generators. If $A$ has $X$ as a free set of generators, $A$ is said to be free on $X$. If $A$ and $B$ are both free on $X$, then $A$ and $B$ are isomorphic. This means that free algebras are uniquely determined by its free generating set, up to isomorphisms.

A simple example of a free Boolean algebra is the one freely generated by one element. Let $X$ be a singleton consisting of $a$. Then the set $A=\{0,a,a^{\prime },1\}$ is a Boolean algebra, with the obvious Boolean operations identified. Every function from $X$ to a Boolean algebra $B$ singles out an element $b\in B$ corresponding to $a$. Then the function $g:A\to B$ given by $g(a)=b$, $g(a^{\prime })=b^{\prime }$, $g(0)=0$, and $g(1)=1$ is clearly Boolean.

The two-element algebra $\{0,1\}$ is also free, its free generating set being $\mathrm{\varnothing }$, the empty set, since the only function on $\mathrm{\varnothing }$ is $\mathrm{\varnothing }$, and thus can be extended to any function.

In general, if $X$ is finite, then the Boolean algebra freely generated by $X$ has cardinality $2^{2^{|X|}}$, where $|X|$ is the cardinality of $X$. If $X$ is infinite, then the cardinality of the Boolean algebra freely generated by $X$ is $|X|$.