# partitions of Boolean algebras

A partition of a Boolean algebra is a collection of pairwise disjoint nonzero elements with supremum 1. For any infinite Boolean algebra $A$ let $a(A)$ be the least size of an infinite partition of $A$. Is $a(A+B)=\min\{a(A),a(B)\}$, where $A+B$ is the free product of $A$ and $B$?

• In forcing terms, is $A+B$ just the lottery sum of the two Boolean algerbas? – Asaf Karagila Nov 20 '17 at 16:34
• I think he means the analogue of product forcing, but with Boolean algebras that are not necessarily complete (since then there is always a countable partition). Don, could you tell us a little more about what you mean by free product? – Joel David Hamkins Nov 20 '17 at 16:39
• If he means the lottery sum (direct sum), then the answer is easily yes, since a partition of $1$ in $A\oplus B$ gives a partition of $1_A$ in $A$ and similarly in $B$ of the same size. The more interesting question is with products, since there are Ramsey-like issues concerning on which coordinate the incompatibility arises. – Joel David Hamkins Nov 20 '17 at 16:54
• @Joel: With product, I suspect that a product of a Suslin tree with itself will provide a counterexample. – Asaf Karagila Nov 20 '17 at 17:11
• The free product is the coproduct. Given disjoint presentations of Boolean algebras by generators and relations, $\langle G_1|R_1\rangle$ and $\langle G_2|R_2\rangle$, the free product is the BA with presentation $\langle G_1\cup G_2|R_2\cup R_2\rangle$. Alternatively, the free product of $B_1$ and $B_2$ is the BA with Stone space that is the product of the Stone space of $B_1$ and the Stone space of $B_2$. – Keith Kearnes Nov 20 '17 at 17:29

It seems this is an open question. First note that (since the answer to the corresponding question for lottery sums is yes) the following are equivalent:

1. $a(A+B)=\min\{a(A),a(B)\}$ for every $A,B$
2. $a(A+A)=a(A)$ for every $A$.

Regarding 2, one should read the article The minimal size of infinite maximal antichains in direct products of partial orders by Miloš Kurilić (Order, 2017). The article starts (and ends) with the open question:

Is there a partial order $\mathbb{P}$ such that $a(\mathbb{P})<a(\mathbb{P}\times \mathbb{P})$?

There are several partial results which imply that the answer to the OP´s question is yes in at least two cases: When the boolean algebras are atomic (see Theorem 6.1) and when the boolean algebras have size at most $\aleph_1$ (see Corollary 4.2).

Kurilić also mentions that it is not even clear what happens with the algebra $A:=\mathcal{P}(\omega)/\mathrm{Fin}$. It is known by results of Spinas that $a(A+A)=a(A)$ in models of $\mathfrak{a}=\mathfrak{b}$ (so it is true under $CH$ or $MA$). However, it is not known whether $a(A+A)<a(A)$ is consistent.