Timeline for Is every countable model of ZFC a subset of some parameter free definable pointwise-definable model of ZFC?
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11 events
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| Jun 25, 2023 at 11:49 | comment | added | Zuhair Al-Johar | @AliEnayat, True, but that is not really the question. The question pre-suppose the existence of countable models of ZFC. I'll add this to the question. Thanks. | |
| Jun 23, 2023 at 13:55 | comment | added | Ali Enayat | Looking back at the question, and taking it as face value, the answer is trivially yes, since there are models of ZFC in which Con(ZFC) fails, and therefore sentences of the form "every model of ZFC has such-and-such property" are automatically true. | |
| Jun 22, 2023 at 19:10 | comment | added | Zuhair Al-Johar | @NoahSchweber, if $M$ is a countable model of ZFC must there exists some set $N$ such that $M \subset N$ and $N$ is pointwise-definable model of ZFC, and $N$ itself is a parameter free definable element of $V$. | |
| Jun 22, 2023 at 19:02 | comment | added | Noah Schweber | To clarify (and @AliEnayat) I think the point is that we're looking at how $N$ itself sits inside $V$. That is, if $M$ is a countable (transitive?) model of $\mathsf{ZFC}$, must there exist some $N$ such that $M\in N$, $N$ is a p.d. model of $\mathsf{ZFC}$, and $N$ is itself a definable element of the whole set-theoretic universe $V$. Zuhair, is this accurate? | |
| Jun 22, 2023 at 18:51 | comment | added | Zuhair Al-Johar | @AliEnayat, I didn't get your argument. What I asked was about if every countable model M of ZFC can be a subset of some pointwise-definable model N of ZFC and such that N itself is a parameter free definable set. That was my question. So, is it the case? | |
| Jun 22, 2023 at 18:27 | comment | added | Ali Enayat | @ZuhairAl-Johar Forcing extensions need not be pointwise definable, the point of specifying them is that "subset" can be strengthened to "submodel with the same class of ordinals" in your question. In other words, given any countable model M of ZFC, there is a pointwise definable model N of ZFC such that (1) M is a submodel of N, and (2) N has no more ordinals than M. | |
| Jun 22, 2023 at 17:19 | comment | added | Zuhair Al-Johar | To complete your proof you need to add the proof that at least one of those forcing extensions is parameter free definable, like proving that it is an element of some point-wise definable model of ZFC, or something similar to that. | |
| Jun 22, 2023 at 16:35 | comment | added | Zuhair Al-Johar | I don't think so, because you have continuum many of them | |
| Jun 22, 2023 at 16:06 | comment | added | Wojowu | @ZuhairAl-Johar I slightly misunderstood your question. The property you ask for is not itself expressible in the language of ZFC, but it will hold in any pointwise definable model of ZFC. | |
| Jun 22, 2023 at 12:21 | comment | added | Zuhair Al-Johar | What is the proof that these forcing extensions are parameter free definable? Or that at least one of them should be? | |
| Jun 22, 2023 at 11:25 | history | answered | Wojowu | CC BY-SA 4.0 |