Timeline for A strictly descending chain of subalgebras of $P(\omega)/_{\mathrm{fin}}$
Current License: CC BY-SA 4.0
17 events
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| Apr 18, 2023 at 8:25 | history | edited | KP Hart | CC BY-SA 4.0 |
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| Mar 16, 2023 at 7:27 | history | edited | KP Hart | CC BY-SA 4.0 |
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| Mar 16, 2023 at 7:22 | vote | accept | Rafał Gruszczyński | ||
| Mar 16, 2023 at 7:21 | history | edited | Rafał Gruszczyński | CC BY-SA 4.0 |
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| Mar 8, 2023 at 19:52 | comment | added | KP Hart | @WillBrian An old VU method: one simply kills off all potential embeddings | |
| Mar 8, 2023 at 10:05 | answer | added | KP Hart | timeline score: 9 | |
| Mar 6, 2023 at 15:10 | comment | added | Joel David Hamkins | Ah, I had thought that was what you had had in mind with "some cardinal invariant". The Tarski invariants would seem to be the natural choice for Boolean algebras, but they are more about iterated quotients than embeddings, and so it isn't clear to me exactly what they have to say about embeddings. It seems that subalgebras can sometimes have larger invariants, since I can find an atomic subalgebra inside an atomless algebra, but perhaps in such cases the quotient iteration must get shorter, which could make the Tarski invariants descend in a suitable lexical order or something. | |
| Mar 6, 2023 at 14:48 | comment | added | Will Brian | @JoelDavidHamkins: That's an interesting idea. But I didn't know about the Tarski invariants before today, and at a first glance I'm not sure how they interact with embeddings. | |
| Mar 6, 2023 at 14:06 | comment | added | Joel David Hamkins | @WillBrian, can you say something definite about how the Tarski invariants interact with embeddings? They characterize the theory of the model, and I think for countable models they characterize the model up to isomorphism. But how do they affect embeddability? Perhaps we might hope to prove a negative answer by considering them. | |
| Mar 6, 2023 at 11:55 | comment | added | YCor | Since it hasn't been said explicitly: it's easy to check there is no such chain consisting of countable subalgebras (of any Boolean algebra). In ZFC+CH (the most interesting case to my taste, although not for many set theorists, I expect) the question is equivalent to finding such a chain consisting of BAs of continuum cardinal. | |
| Mar 6, 2023 at 10:20 | comment | added | Will Brian | I like this question. I think it's likely to be tricky because often one shows a non-embedding of $A$ in $C$ by showing some cardinal invariant of the algebra to be smaller in $C$. But of course, we can't decrease a cardinal invariant infinitely often to solve this question. | |
| Mar 6, 2023 at 8:31 | comment | added | Rafał Gruszczyński | @JoelDavidHamkins I cannot come up with any example of Boolean algebra with this property, so maybe I should have started with the more elementary question. The closest I got is an example of Heyting algebras. Start with [0,1] interval treated as a Heyting algebra, and follow the Cantor set construction stages, but delete only the middle third sets of irrationals, instead of the whole intervals. Every $H_{i+1}$ is embeddable in $H_i$, but is not isomorphic. | |
| Mar 6, 2023 at 0:25 | comment | added | Joel David Hamkins | Would it be possible for you to provide an instance of such a descending chain of subalgebras perhaps in another Boolean algebra? | |
| Mar 5, 2023 at 14:46 | comment | added | Rafał Gruszczyński | @YCor Corrected. Thanks for spotting it. | |
| Mar 5, 2023 at 14:45 | history | edited | Rafał Gruszczyński | CC BY-SA 4.0 |
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| Mar 5, 2023 at 12:29 | comment | added | YCor | Your definition of "proper subalgebra" is incorrect. $B\subset C$ proper subalgebra means that $C-B$ is nonempty. Either you're asking the question as asked and then the answer is a trivial yes. Or you're asking that there is no Boolean algebra embedding of $A_i$ into $A_{i+1}$ but then the question should be rephrased. | |
| Mar 5, 2023 at 12:01 | history | asked | Rafał Gruszczyński | CC BY-SA 4.0 |