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

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 $\mathcal{P}(X)$ : $$(\mathcal{P}(X), \cap, \cup, \bar{\cdot}) \quad ?$$

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 ?