# Is every Boolean algebra a subalgebra of a free one?

It is well-known - and indeed follows from usual algebraic facts - that every Boolean algebra is a quotient of a free one. Is every Boolean algebra moreover a subalgebra of a free one?

I believe this should be known, but I have not been able to locate this in the obvious places (notably the Handbook of Boolean Algebras). I am especially interested in whether one could show this result without appealing to any choice principle.

• why is this tagged with category theory and model theory? Commented Jul 25 at 10:03
• @preferred_anon, I was not sure whether that was appropriate, but my rationale is that the answer could come from any of these communities: people with strong connections to category theory, such as point-free topologists, work with Stone spaces, and it seemed plausible to me (at the time of writing) that there would be some sort of construction there that would explain it; model theorists also work with Boolean algebras, and some people with very deep knowledge of the structure and combinatorial properties of Boolean algebras would call themselves model theorists. Commented Jul 25 at 14:40

Is every Boolean algebra a subalgebra of a free one?

No. Free Boolean algebras do not contain uncountable chains. This is the main result of

Alfred Horn
A property of free Boolean algebras
Proc. Amer. Math. Soc. 19 (1968), 142-143.

A slightly simplified proof appears as Theorem 1 of

Alexander Abian
Two properties of free Boolean algebras
Colloquium Mathematicum 29 (1974), 51-53.

Therefore, any Boolean algebra containing an uncountable chain (like $${\mathcal{P}}(X)$$, $$X$$ infinite) cannot be embedded into a free one.

• Don't you intend $X$ to be uncountable? Commented Jul 26 at 6:38
• @tomasz: I allow $X$ to be countably infinite. For example, ${\mathcal{P}}(\mathbb Q)$ contains an uncountable chain of sets of the form $\mathbb Q\cap (-\infty,r)$, $r\in \mathbb R$. Commented Jul 26 at 6:58
• Right. I was thinking of well-ordered chains. I think the claim is probably easier to show for those (that such chains don't exist in free Boolean algebras). Commented Jul 26 at 7:13

I now realize this has to be false: every free Boolean algebra satisfies the countable chain condition. Hence any Boolean algebra not satisfying this condition will be a counterexample.

• Hi Rodrigo, here are some further observations in case they might be useful. Firstly, the answer is yes for countable BAs: every countable BA is known to be projective. Secondly, subalgebras of free BAs need not be projective, but they inherit a weaker form of this property, namely they contain a dense projective subalgebra. Two relevant keywords are "weak projectivity" (not that kind of weak projectivity) and "Cohen algebras". A superficial look at the literature gives me the impression that this is not a consequence of the countable chain condition, but I haven't seen an explicit example. Commented Jul 26 at 9:55