Boolean algebras of size equal to singular cardinals

Question

In the paper New Proof of a Theorem of Gaifman and Hales by Solovay, he gives a constructive proof of complete Boolean algebras of size larger than $\aleph_\tau$ for any ordinal $\tau$ which are also countably generated.

His construction can be summarized as follows. Let $\mathscr{B}$ be the Boolean algebra of regular open sets of the product space $X = \aleph_\tau^\omega$ where $\aleph_\tau$ is endowed with the discrete topology. In his paper, he easily shows that $|\mathscr{B}| \geq \aleph_\tau$. In fact, I'm fairly certain that $|\mathscr{B}| = 2^{\aleph_\tau}$.

After thinking about this for some time, I'm curious if there are complete Boolean algebras (maybe or maybe not countably generated) of size something like $\aleph_\omega$. If something like GCH is assumed, I'm not sure if Solovay's constructions could give us something of that size. I'm really just looking for references of these constructions if they do exist. Thanks!

Edit: To be concise, I am looking for a construction of a complete Boolean Algebra with size precisely $\aleph_\omega$.

Answer 1

There does not exist any complete Boolean algebra of cardinality $\aleph_\omega$. More generally, if $B$ is an infinite complete Boolean algebra then $|B|^{\aleph_0}=|B|$.

To prove this, if $a\in B$, we will write $|a|$ for the cardinality of the set $\{b\in B:b\leq a\}$. Note that for any antichain $A\subseteq B$ (i.e., any set of pairwise disjoint elements), we have $|\bigvee A|=\prod_{a\in A}|a|$. Now let $S=\{a\in B:|a|<|B|\}$ and let $A$ be a maximal subset of $S$ that is an antichain. By maximality of $A$, every nonzero $a\leq\neg\bigvee A$ must satisfy $|a|=|B|$. If $\bigvee A\neq 1$, we can then find a countably infinite family $(a_n)$ of disjoint nonzero elements of $B$ below $\neg\bigvee A$, and so $|B|\geq \prod_n|a_n|=|B|^{\aleph_0}$.

The only remaining case is that $\bigvee A=1$. In that case, we have $|B|=\prod_{a\in A}|a|$, so we have written $B$ as a product of strictly smaller cardinals. The proof is now completed by the following lemma.

Lemma: Let $(\kappa_i)_{i\in I}$ be a family of cardinals and $\kappa=\prod_i\kappa_i$, and suppose that $\kappa$ is infinite and $\kappa>\kappa_i$ for all $i$. Then $\kappa^{\aleph_0}=\kappa$.

Proof: Pick a well-ordering on $I$ such that $i<j$ implies $\kappa_i\leq\kappa_j$ (this just amounts to picking a well-ordering of $\{i\in I:\kappa_i=\mu\}$ for each cardinal $\mu$). We can then write the order type of $I$ as $\omega\cdot\alpha+n$ where $\omega\cdot\alpha$ is a limit ordinal and $n<\omega$. Note that the last $n$ terms of the product $\prod_i\kappa_i$ cannot change the value of the product since we assume $\kappa_i<\kappa$ for each $i$. So, we may assume $n=0$ and the order type of $I$ is a limit ordinal.

Now we can partition $I$ into intervals $(J_\beta)_{\beta<\alpha}$ of length $\omega$. We can then partition each $J_\beta$ into infinitely many cofinal subsets, and each of these subsets will have the same product. We can thus partition $I$ into infinitely many subsets $I_n$ (where each $I_n$ contains one of the cofinal subsets of each $J_\beta$) such that $\prod_{i\in I_n}\kappa_i=\lambda$ is the same for each value of $n$. We then have $\kappa=\lambda^{\aleph_0}$ so $\kappa=\kappa^{\aleph_0}$.

Conversely, if $\kappa$ is any infinite cardinal such that $\kappa^{\aleph_0}=\kappa$, then there exists a complete Boolean algebra of cardinality $\kappa$. Indeed, let $B_0$ be the free Boolean algebra on $\kappa$ generators and let $B$ be its completion. Since $B_0$ is ccc, each element of $B$ is a join of countably many elements of $B$, so $|B|\leq \kappa^{\aleph_0}=\kappa$.

Answer 2

To your edit, that is impossible. Here are two papers which prove similar results:

1. Pierce, R. S., A note on complete Boolean algebras, Proc. Am. Math. Soc. 9, 892-896 (1959). ZBL0096.25601.

2. Comfort, W. W.; Hager, A. W., Cardinality of k-complete Boolean algebras, Pac. J. Math. 40, 541-545 (1972). ZBL0217.31302.

Every complete Boolean algebra, and indeed every $\sigma$-algebra, must satisfy $|\mathcal B|^{\aleph_0}=|\mathcal B|$. So in particular, the cardinality of a $\sigma$-algebra, let alone a complete Boolean algebra, must have an uncountable cofinality.

In the first paper, however, Pierce shows that this condition is a complete characterisation. Namely, if $\kappa^{\aleph_0}=\kappa$, then there exists a complete Boolean algebra of cardinality $\kappa$. So you can still have complete Boolean algebras which have singular cardinality. Just not one with a countable cofinality.