Timeline for The type of nondefinable elements
Current License: CC BY-SA 3.0
15 events
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| Nov 16, 2013 at 21:51 | comment | added | user38200 | Ok, that is sufficient for my purposes, thank you very much. | |
| Nov 16, 2013 at 21:48 | comment | added | Joel David Hamkins | No, because those hypotheses don't rule out the case that $X$ is finite. But if $X$ is also infinite, then yes, the non-definability type will not be principal, because by compactness there will be an elementary extension of the pointwise definable model in which $X$ now has non-definable elements, and the arguments from my answer kick in. | |
| Nov 16, 2013 at 21:43 | comment | added | user38200 | Assume $X$ is definable by some formula $\Psi(x)$. Can we deduce from the mere existence of pointwise definable models of set theory (which satisfy furthermore the property that $X \neq \emptyset$) that the non-definability type is not definable? | |
| Nov 16, 2013 at 21:07 | comment | added | Joel David Hamkins | There is no one answer for "general" $X$---although I'm not sure what that would mean---because the answer depends on $X$ and $\mathcal{M}$. Even if $X$ is infinite, if $X$ has only finitely many non-definable elements, then the non-definability type for elements of $X$ will be principal. Even if $X$ is not definably well-ordered, still if the non-definable elements of $X$ all lie within a set with a definable well-order, then the non-definability type will not be definable. | |
| Nov 16, 2013 at 21:01 | comment | added | user38200 | What about the case where $X$ is general, i.e. not finite and not necessarily definably well-orderable in $\mathcal{M}$? | |
| Nov 14, 2013 at 16:14 | comment | added | Joel David Hamkins | I suppose another interpretation of the question would be: in a given model, is the set of non-definable elements a definable class? My examples show that sometimes they are (such as in a pointwise definable model), but sometimes they aren't (for example, if there are non-definable ordinals). | |
| Nov 14, 2013 at 15:42 | comment | added | Joel David Hamkins | So far, I haven't really considered $X$, but a fuller answer would be that it depends on $X$. If $X$ is finite, then clearly the non-definability type will be principal. If $X$ is definably well-orderable in $\mathcal{M}$ (such as if $X$ is a set of ordinals or if $V=HOD$ holds in $\mathcal{M}$), then my argument kicks in to show the non-definability type is not principal over $\text{Th}(\mathcal{M})$. | |
| Nov 14, 2013 at 15:30 | comment | added | Joel David Hamkins | See my update . | |
| Nov 14, 2013 at 15:29 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Nov 14, 2013 at 15:23 | comment | added | Joel David Hamkins | I was mainly answering the first question. For the second, I take the question to be considering the non-definability type over ZFC, but then asking whether it is principal (and I argue it is not). But I suppose one could also ask about the non-definability type over the theory of a fixed model $\mathcal{M}$, and in this case, another argument would be needed. | |
| Nov 14, 2013 at 15:18 | comment | added | Emil Jeřábek | The answer makes much more sense than the question, nevertheless I’m struggling to figure out the exact connection between the two. In particular, the question seems to have the form “given $\mathcal M$, is xxx principal?” whereas you seem to argue that “there exists $\mathcal M$ such that xxx is not principal”. | |
| Nov 14, 2013 at 15:08 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Nov 14, 2013 at 15:01 | vote | accept | user38200 | ||
| Nov 14, 2013 at 15:00 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Nov 14, 2013 at 14:47 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |