Every boolean algebra is the algebra of clopen sets of some topological space, by Stone's representation theorem.

Not every boolean algebra is isomorphic to the powerset algebra of some set, though. Although, every finite boolean algebra is isomorphic to a powerset algebra. I think the powerset algebras are the algebras of definable sets when we restrict ourselves to structures with an empty signature.

Is there a nice way to characterize which boolean algebras you can hit by considering the algebras of definable sets in a given structure? I'm curious about the class consisting of the boolean algebra for each $(M, A)$ where $M$ is a first-order structure is $A \subset M$ is a set of parameters.

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    $\begingroup$ There is no point mentioning a parameter set $A$ if it is fixed: definable subsets of $M$ with parameters in $A$ are the same as parameter-free definable subsets of $M$ where you enlarge the language with a constant for each element of $A$. $\endgroup$ Commented Jun 2, 2022 at 22:28
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    $\begingroup$ It's not true that power set algebras are the algebras of definable sets with an empty signature (and a fixed set of parameters). The definable subsets of $M$ with parameters in $A$ with respect to the empty signature are the sets that are either a finite subset of $A$ or the complement of such a set. Up to isomorphism, this gives the algebras that consist of all the finite or cofinite subset of some set, not the full power set. $\endgroup$ Commented Jun 2, 2022 at 22:42

1 Answer 1


Every Boolean algebra is an algebra of definable sets. More strongly, given any set $X$ and any subalgebra $B\subseteq\mathcal{P}(X)$, there is a first-order structure on $X$ whose algebra of definable subsets (without parameters) is $B$. To do this, just make a unary relation for each element of $B$ which is true on that subset of $X$. Clearly then each element of $B$ is definable. For the converse, any formula can involve only finitely many of the relations, and so it suffices to consider the case where $B$ is finitely generated and hence finite. In that case, $B$ just consists of the unions of elements of some finite partition of $X$. Any two elements of the same block of the partition are indistinguishable with respect to all the unary relations, and so any definable subset must be a union of blocks, and thus an element of $B$.


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