Timeline for Non-homogeneous forcing and HOD
Current License: CC BY-SA 3.0
14 events
when toggle format | what | by | license | comment | |
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Oct 7, 2016 at 7:58 | vote | accept | Mohammad Golshani | ||
Oct 7, 2016 at 7:56 | vote | accept | Mohammad Golshani | ||
Oct 7, 2016 at 7:57 | |||||
Oct 6, 2016 at 14:24 | answer | added | Yair Hayut | timeline score: 8 | |
Oct 1, 2016 at 15:18 | comment | added | Mohammad Golshani | Yes. I mean their regular open Boolean algebras are isomorphic | |
Oct 1, 2016 at 14:37 | comment | added | Joel David Hamkins | By "forcing isomorphic", you mean that the regular open algebras are isomorphic? | |
Oct 1, 2016 at 14:33 | comment | added | Joel David Hamkins | Ah, actually, the need for separativity is obviated by the requirement in 1 that it is not "forcing isomorphic" to any homogeneous forcing notion. | |
Oct 1, 2016 at 14:23 | comment | added | Mohammad Golshani | @JoelDavidHamkins Thanks, I edited the question. | |
Oct 1, 2016 at 14:23 | history | edited | Mohammad Golshani | CC BY-SA 3.0 |
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Oct 1, 2016 at 14:10 | comment | added | Joel David Hamkins | Thanks for the edit. You should also assume that $\mathbb{P}$ is separative, or use Boolean algebras, since every partial order is forcing equivalent to a non-cone-homogeneous partial order, simply by replacing each point with a (different cardinality) linear set, and this would enable some easy instances. | |
Oct 1, 2016 at 14:07 | history | edited | Mohammad Golshani | CC BY-SA 3.0 |
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Oct 1, 2016 at 13:58 | comment | added | Joel David Hamkins | Could you clarify precisely which notion of homogeneity you have in mind? There are some stronger versions, and some weaker versions. It is easy to make some examples with the stronger versions failing, but I don't yet have an example for the weak versions. | |
Oct 1, 2016 at 13:50 | history | edited | Mohammad Golshani | CC BY-SA 3.0 |
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Oct 1, 2016 at 12:41 | history | edited | Mohammad Golshani | CC BY-SA 3.0 |
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Oct 1, 2016 at 12:07 | history | asked | Mohammad Golshani | CC BY-SA 3.0 |