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Aug 23, 2023 at 7:32 vote accept Dominic van der Zypen
Aug 22, 2023 at 22:04 answer added KP Hart timeline score: 7
Aug 10, 2023 at 21:13 comment added Ali Enayat @An elaboration of Noah Schweber's comment: By a classical 1963 theorem of Parovicenko every Boolean algebra of cardinality $\aleph_1$ can be embedded into P(ω) mod FIN.
Aug 9, 2023 at 12:21 comment added Will Brian @DominicvanderZypen: I think my answer to this question of yours answers this one as well: mathoverflow.net/questions/441234/…
Aug 9, 2023 at 7:51 comment added Dominic van der Zypen @AndreasBlass thanks for your comment! I suspect it is not known whether the statement "${\frak c}$ cannot be embedded in ${\cal P}(\omega)/\text{(fin)}$" implies ${\sf (CH)}$?
Aug 8, 2023 at 22:25 comment added Joel David Hamkins It might be nice to have a definitive summary account of the universality of $P(\omega)/\text{fin}$ posted as an answer.
Aug 8, 2023 at 21:58 comment added KP Hart And Ken Kunen showed (implicitly) that after adding as many Cohen reals as you wish the ordinal $\omega_2+1$ is not a continuous image of $\omega^*$, hence its clopen algebra does not embed into $\mathcal{P}(\omega)/\mathrm{fin}$.
Aug 8, 2023 at 21:47 comment added Andreas Blass @NoahSchweber's suspicion is correct. A negative answer is consistent with ZFC. In fact, the unembeddable poset $P$ can be just the ordinal $\mathfrak c$. This was proved by Peter Dordal in "A model in which the base matrix tree cannot have cofinal branches" (Journal of Symbolic Logic 52 (1987) pp. 651--664).
Aug 8, 2023 at 20:27 comment added Noah Schweber This is true if $\mathsf{CH}$ holds; I suspect it's independent in general, but I'm not sure.
Aug 8, 2023 at 20:23 history asked Dominic van der Zypen CC BY-SA 4.0