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More generally: must complete Boolean algebras of the same cardinality and with the same cardinality of atoms be isomorphic?

Jeremy
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  • The answer to your "more general" question is an easy no. A complete BA is the sum of a complete atomic BA and a complete atomless BA. Two complete BAs may have the same number of atoms, but one may be atomic and the other nonatomic; or they may both be nonatomic, but have different size atomless parts. The answer to the title question is also no but I'll let an expert on BAs answer it. – bof Jan 12 '14 at 02:10
  • Thanks. Yeah, I misformulated the general principle I was interested in -- I intended an additional same-cardinality requirement on atomless elements. But I see that, since complete BA's can be factored into a complete atomic BA and a complete atomless BA, the general question I intended is equivalent to the titular one. – Jeremy Jan 12 '14 at 02:51

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No. The quotient of the Boolean algebra $B$ of Borel subsets of $[0,1]$ by the ideal of meager (= first Baire category) sets and the quotient of the same $B$ by the ideal of sets of Lebesgue measure zero are both atomless, and they both have the cardinality of the continuum, but they are not isomorphic. Specifically, the latter satisfies a weak distributive law that fails in the former.

Andreas Blass
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  • Or in other words: Cohen is not Random. :-) – Asaf Karagila Jan 26 '16 at 11:35
  • @AsafKaragila To finish translating my answer into the language of forcing, the last sentence says that random is $\omega^\omega$-bounding and Cohen isn't. – Andreas Blass Jan 26 '16 at 22:20
  • Yes, I figured that would have been the weak distributive law. But "Cohen is not Random" is a nice statement that will confuse everyone who is not a set theorist. – Asaf Karagila Jan 26 '16 at 22:22