Timeline for Is every class that does not add sets necessarily added by forcing?
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12 events
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| Jan 10, 2013 at 18:10 | comment | added | Joel David Hamkins | Yes, I was thinking that the class should satisfy GBC, and not merely be amenable, since this is what Jonas is talking about. I have in mind some constructions that I have recently been looking at with Gunter Fuchs, where we do with transitive models (given large cardinals) things that you and Roman Kossak have done with $\omega_1$-like models of arithmetic, and I think our methods works for this rather classless situation also, but I'll have to check it through. | |
| Jan 10, 2013 at 17:59 | comment | added | Ali Enayat | @Joel: regarding your parenthetical clause: rather classless models $M$ have the property (by definition) that if $X$ is an amenable subset of $M$, then $X$ is definable in $M$. Now if $M$ is a well-founded model of $ZF$ of uncountable cofinality, then we can use the reflection theorem and an elementary chain argument to show that $Sat_M$ is an amenable subset of $M$. This shows that no well-founded model of $ZF$ can be rather classless. But maybe by weakening the notion of "rather classless" by replacing "amenable" by "satisfies replacement", one can arrange a well-founded such model. | |
| Jan 10, 2013 at 17:32 | comment | added | Joel David Hamkins | Ah, of course, that is exactly what we need! So this means that one cannot prove that every $M$ admits such a non-forcing extension, since a rather classless model has no extensions at all with the same sets. In particular, there will be $\omega_1$-like models with a positive answer to Jonas's query. (And I believe that with sufficient large cardinals, one can even do this with transitive models of height $\omega_1$.) | |
| Jan 10, 2013 at 17:19 | comment | added | Ali Enayat | @Joel: Nice answer; the question you pose in your comment is intriguing. Of course, we want to concentrate on countable $M$ because there are such things as "rather classless models" of $ZF$, all of whose classes are definable. | |
| Jan 10, 2013 at 16:48 | comment | added | Joel David Hamkins | Meanwhile, it seems that the compactness argument shows that every model $M$ can be elementarily embedded into a model that admits a satisfaction class (just add the elementary diagram of $M$ to the theory before applying compactness). Thus, the existence of a satisfaction class with GBC is conservative over ZFC, and there can be no obstacle coming just from the theory of $M$. | |
| Jan 10, 2013 at 16:35 | comment | added | Joel David Hamkins | Incidentally, since not every model $M$ admits a satisfaction class with GB, there is still room here for a better answer, which either proves it negatively for all models $M$, or which finds some $M$ to which one cannot add classes except generically. | |
| Jan 10, 2013 at 16:33 | comment | added | Joel David Hamkins | I guess not, because you can make a compactness argument that there must be a ZFC model which admits a satisfaction class with GBC, since for any standard $n$ we can have a $\Sigma_n$-satisfaction class. That is, we can write down the theory of what it means to be a satisfaction class with GBC still holding, and apply compactness. | |
| Jan 10, 2013 at 16:23 | vote | accept | jonasreitz | ||
| Jan 10, 2013 at 16:23 | comment | added | jonasreitz | What a great example, Joel! The fact that the satisfaction class is compatible with GBC but is not definable is a very nice combination. Does existence of a satisfaction class increase the consistency strength of GBC? | |
| Jan 10, 2013 at 12:33 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Jan 10, 2013 at 12:28 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Jan 10, 2013 at 12:04 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |