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Let us say that a complete Boolean algebra $B$ is:

  • countably distributive when for any sequence $(I_n)_{n\in\mathbb{N}}$ of sets and any elements $(u_{n,i})_{n\in\mathbb{N},i\in I_n}$ of $B$ we have $$\bigwedge _ {n\in\mathbb{N}}\, \bigvee _ {i\in I_n} u_{n,i} = \bigvee _ {f\in\prod_{n\in\mathbb{N}} I_n} \, \bigwedge_{n\in\mathbb{N}} u_{n,f(n)}$$

  • countably compact when for any sequence $(v_n)_{n\in\mathbb{N}}$ of elements of $B$, if $\bigwedge_{n\in\mathbb{N}} v_n = 0$ then there is $F \subseteq \mathbb{N}$ finite such that $\bigwedge_{n\in F} v_n = 0$.

Questions: Is here a relation between these two properties (does one imply the other)? If not, might I have examples of Boolean algebras with one but not the other?

(The apparent lack of any mention of the second property everywhere I searched in the literature leaves me to suspect that there is something deeply wrong with this property which I failed to see, so my question is perhaps stupid: if so, I apologize. But, generally speaking, I am interested in what can be said about this “countably compact” property.)

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There are many countably distributive complete Boolean algebras, and this is an important concept in forcing. For example, the canonical forcing to add a Cohen subset (or any number of Cohen subsets) to $\omega_1$ is countably closed and hence also countably distributive. A forcing notion is countably distributive if and only if the forcing extension adds no new countable sequences of ordinals over the ground model.

Meanwhile, there are no infinite complete countably compact Boolean algebras. Every infinite complete Boolean algebra has a countable maximal antichain $a_n$, and by taking the supremum of tails of this antichain $v_n=\bigvee_{k\geq n}a_k$, we get a strictly descending sequence $v_0>v_1>v_2>\cdots$ with meet zero $\bigwedge_n v_n=0$, but no finitely many of the $v_n$ will have meet zero. (You can drop completeness in this argument, provided that in fact you have a countably infinite maximal antichain, since the supremum on the tails are complements of finite joins from the front. Having a countably infinite maximal antichain is equivalent to having any nontrivial countably infinite join $b=\bigvee_n b_n$, since you can disjointify and take complements, etc.)

In general, if a complete Boolean algebra has a countably-closed dense subset (countably closed = all countable descending sequences are nontrivially bounded below), then it will be countably distributive. This is an extremely useful property in forcing, since many natural partial orders are countably closed, and so their Boolean completions are countably distributive.

But the lesson of the second paragraph is that infinite complete Boolean algebras themselves are never countably closed. You have to look at dense subsets as orders to get the countable closure property.

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  • $\begingroup$ Just to be sure, “the canonical forcing which adds a Cohen subset to $ω_1$” is the complete Boolean algebra of regular open subsets of the topological space $\{0,1\}^{ω_1}$ with the product topology — correct? $\endgroup$
    – Gro-Tsen
    Commented Aug 8 at 22:00
  • $\begingroup$ No, you have to use countable support, not finite support. So it isn't the product topology, but the countable-support topology. Countable support is what makes it countably closed forcing and hence countably distributive. Partial order = $2^{<\omega_1}$, ordered by extension. Then take Boolean completion. $\endgroup$
    – Joel David Hamkins
    Commented Aug 8 at 22:06
  • $\begingroup$ Ah, so we start with the product topology on $\{0,1\}^{ω_1}$, then we consider the $G_δ$ subsets of that, they form the basis for another topology (the “$G_δ$ modification” of the product topology, or, I guess we can say, a “countably restricted box product”) on $\{0,1\}^{ω_1}$, and the regular open sets of that topology form the complete Boolean algebra which you speak of (and which is countably distributive). Correct, this time? (Probably not the simplest way to put it, but I want to make sure I got it right.) $\endgroup$
    – Gro-Tsen
    Commented Aug 9 at 9:33
  • $\begingroup$ I think of it as: take the tree $2^{<\omega_1}$ of all countable well ordered binary sequences as an order under extension, and consider the lower-cone topology (basic open = all extensions of a fixed element), and then take the regular open subsets of this space. The tree is the partial order approach to this forcing, the completion is the Boolean algebra approach. The tree is obviously countably closed, and it is dense in the Boolean algebra, which is therefore countably distributive. $\endgroup$
    – Joel David Hamkins
    Commented Aug 9 at 11:08
  • $\begingroup$ Tree grows downward, since the empty sequence corresponds to 1 in the Boolean algebra. The partial order has no 0, but the branches stretch down towards 0 in the BA. $\endgroup$
    – Joel David Hamkins
    Commented Aug 9 at 11:34

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