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We work in Kelley-Morse class theory and assume global choice. Let B be a complete Boolean algebra that is a proper class (B is complete in the sense that every subclass of it has a minimal upper bound). Let's assume that B is NOT atomic. Do we know that B must have an anti-chain that is also a proper class?

I was trying to show that B has a maximal chain that is a proper class, which would produce an anti-chain that is a proper class. But I'm not clear if a maximal chain in B has to be a proper class.

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  • $\begingroup$ Could you please add your own thoughts on it? I mean what you have thought so far, and please be a little more clear. $\endgroup$
    – Knight wants Loong back
    Commented Aug 3, 2020 at 16:57
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    $\begingroup$ I added the background theory. Is it better now? $\endgroup$
    – user123
    Commented Aug 3, 2020 at 17:03
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    $\begingroup$ It might help to consider simpler versions first. E.g. is there an infinite complete Boolean algebra with only finite antichains? $\endgroup$
    – Noah Schweber
    Commented Aug 3, 2020 at 17:05
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    $\begingroup$ Thinking in ZFC with an inaccessible cardinal $\kappa$, consider the regular open algebra for adding $\kappa$ Cohen reals. That has only countable antichains, but its usual definition would make it a collection of subsets of $\kappa$. Thanks to the ccc, though, it has cardinality only $\kappa$, so there should be a copy of it that's a subset of $\kappa$. That seems to put it into the range of Kelley-Morse class theory, interpreted with sets (classes) being everything of rank $<(\leq)\,\kappa$. $\endgroup$
    – Andreas Blass
    Commented Aug 3, 2020 at 17:10

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