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A Boolean algebra $B$ is defined (e.g. in Jech) to be $\kappa$-saturated if there is no partition $W$ of $B$ where $|W|=\kappa$. He seems to assume that this implies $|W|<\kappa$ for any partition $W$. But why should this be the case?

For example, say that $B$ is $\aleph_1$-saturated. Why does this imply that $B$ is $\aleph_2$-saturated? It's clearly true in the case where $B$ is complete or even $\aleph_3$-complete, but suppose we're not given that. How would one construct a partition of size $\kappa$ given a partition of size $\lambda>\kappa$ in the absence of completeness?

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  • $\begingroup$ I can't make any sense of the question. No, $|W|=\kappa$ certainly does not imply $|W|<\kappa$, as it obviously implies its negation. So what is it that you actually intend to ask? $\endgroup$
    – Emil Jeřábek
    Commented May 21, 2022 at 8:41
  • $\begingroup$ @EmilJeřábek Edited to clarify. $\endgroup$
    – Chad Groft
    Commented May 21, 2022 at 14:00

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Jech defines a partition of a Boolean algebra $B$ as a maximal antichain. Now the cardinalities of maximal antichains in $B$ and its completion can indeed differ: Take $B$ as the finite, cofinite subsets of $\omega_1$ with the canonical Boolean algebra strucure. $B$ has maximal antichains of every nonzero finite cardinality and of size $\omega_1$, but no countably infinite one, so $B$ is $\omega$-saturated but not $\omega_1$-saturated according to Jech's definition.

It seems that, to make the definition work as intended for all Boolean algebras instead of just complete Boolen algebras, one should drop the requirement of maximality and define $B$ to be $\kappa$-saturated if there is no antichain in $B$ of size $\kappa$. This definition agrees with the old one for complete Boolean algebras, but now any Boolean algebra $B$ is $\kappa$-saturated iff $B$'s completion is $\kappa$-saturated. The reason is that $B$ has the same cardinalities of antichains as its completion (as you noted in your answer).

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  • $\begingroup$ This is a really interesting point, thank you! I guess I'll have to pay extra close attention when this concept comes up later. $\endgroup$
    – Chad Groft
    Commented May 26, 2022 at 15:29
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I just realized it follows from the second answer to this question. If $B$ and its completion have the same cardinalities of antichains, then the statement for general $B$ follows from that for complete $B$.

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    $\begingroup$ While $B$ and its completion have the same cardinalities of antichains, they might not have the same cardinalities of maximal antichains, which is what one would need here as a partition is just a maximal antichain. For example the Boolean algebra given by the finite, cofinite subsets of $\omega_1$ has no maximal antichain of size exactly $\omega$ while its completion, the whole powerset of $\omega_1$, certainly does. $\endgroup$
    – Andreas Lietz
    Commented May 23, 2022 at 10:15
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    $\begingroup$ @AndreasLietz This answers the original question, right? You've produced a Boolean algebra with no partition of size $\omega$ that does have a partition of size $\omega_1$. $\endgroup$
    – Gabe Goldberg
    Commented May 23, 2022 at 20:20
  • $\begingroup$ Yes, you're right Gabe! Maybe I should post this as an answer... $\endgroup$
    – Andreas Lietz
    Commented May 25, 2022 at 12:20

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