Every boolean algebra is a product of the binary boolean algebra.

Question

I am reading 'From Peirce to Skolem: A neglected chapter in the history of logic'. There the author mentions that the Stone representation theorem 'says that every Boolean algebra is a subalgebra of a direct product of two-element Boolean algebras'. The Stone representation theorem that i know of says the classical isomophism between boolean algebras and set algebras (subalgebras of a powerset with intersection, union and empty set, full set and complement). Since the author does not give reference to the statement, i was wondering where can i find the proof of this statement or can someone give a quick argument.

Answer 1

Note that a direct product $\prod\limits_{s \in S} 2$ of Boolean algebras is isomorphic to the Boolean algebra $P(S)$. This gives us isomorphisms between sub-Boolean algebras of $\prod\limits_{s \in S} 2$ and sub-Boolean algebras of $P(S)$. Note that a sub-Boolean algebra of $P(S)$ is exactly an algebra of sets on $S$.

Thus, if we take your known version of the Stone Representation Theorem (SRT), we can start with a Boolean algebra $B$. Then $B$ is isomorphic to an algebra of sets $J$ on set $S$, which is in turn isomorphic to a sub algebra $K$ of $\prod\limits_{s \in S} 2$.

Conversely, we can start with the book’s version of the SRT. Begin with a Boolean algebra $B$, and find an isomorphic subalgebra $K$ of $\prod\limits_{s \in S} 2$. Then $K$ is in turn isomorphic to an algebra $J$ of sets on $S$.

Therefore, the two statements of the SRT are equivalent.