Given a Boolean algebra $(B, \land, \lor)$ can I find a set $X$ such that $(B, \land, \lor) = (\mathcal{P}(X), \cap, \cup)$?

Question

Suppose I have a set $B$ and operations $\land$ and $\lor$ and $\neg$ such that : $$ (B, \land, \lor, \neg) $$ is a Boolean algebra.

Can I always find a set $X$ such that this Boolean algebra is isomorphic to the canonical Boolean algebra induced on its powerset $\mathcal{P}(X)$ : $$(\mathcal{P}(X), \cap, \cup, \bar{\cdot}) \quad ?$$

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If not, I'd be happy to commit to the two following simplifications :

• it is the case for a subset $\mathcal{F} \subseteq \mathcal{P}(X)$ of the powerset ?
• is it the case when $B$ and $X$ are finite ?

Answer 1

Stone's Theorem says in part that any Boolean algebra is isomorphic to a subalgebra of some power set.

Not every Boolean algebra is isomorphic to a power set; for example a power set cannot be countably infinite.

Answer 2

No. For example, the Boolean Algebra of finite/cofinite subsets of an infinite set is not isomorphic to the powerset of some set.
It is true that any complete and atomic Boolean Algebra is isomorphic to the powerset of some set; in particular, any finite one.
And it is also true that any Boolean Algebra is embeddable into a complete atomic Boolean Algebra, and thus, a powerset Boolean Algebra.