Timeline for Is $\mathsf{ZFC+V=L}$ consistently $\omega$-complete?
Current License: CC BY-SA 4.0
17 events
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| Mar 10, 2021 at 21:09 | comment | added | Pace Nielsen | @NoahSchweber You wrote: "It's "level-dependent" - as with anything else there's an internal version and an external version." There are some external concepts that are not always internalizable (without expanding language, or logic, etc...). I see now that you define, externally, "standardness" to merely mean an isomorphism from the metatheoretical thing to the internal thing. When I've previously asked, on MO, how to define the "standard model of arithmetic", some answers suggested there was more too it (not necessarily first order expressible). So you can forgive my confusion here. | |
| Mar 10, 2021 at 3:48 | comment | added | Ali Enayat | @NoahSchweber Using Barwise compactness one can prove an appropriate version of Gödel's second incompleteness for this context to shows that for any r.e. extension $T$ of ZF, if $T$ has an $\omega$-model, then $T$ has an $\omega$-model that satisfies "$T$ has no $\omega$-model", which provides a high level explanation of what's happening in Farmer F.'s recursion-theoretic solution. I will try to the flesh out this as an alternative answer to your question "before long". | |
| Mar 8, 2021 at 21:44 | history | edited | Noah Schweber | CC BY-SA 4.0 |
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| Mar 8, 2021 at 21:41 | comment | added | Noah Schweber | @PaceNielsen It's "level-dependent" - as with anything else there's an internal version and an external version. The internal version being expressible in the object language, questions like "Is $S$ consistent?" make perfect sense - although the relevant objects arising in answers to such questions may interpret the relevant notions nonstandardly. | |
| Mar 8, 2021 at 21:39 | comment | added | Pace Nielsen | I think my misunderstanding came from interpreting "standard $\omega$" differently. I took it to mean the metamathematical statement that you are looking at a model whose $\omega$ is isomorphic to your metamathematical/preexisting $\mathbb{N}$. I gather that, instead, it was meant in an internal sense of standard. | |
| Mar 8, 2021 at 21:38 | vote | accept | Noah Schweber | ||
| Mar 8, 2021 at 21:08 | vote | accept | Noah Schweber | ||
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| Mar 8, 2021 at 21:07 | comment | added | Noah Schweber | @PaceNielsen I think if my previous comments don't address the issue satisfyingly you should ask a separate MO question (which I'll happily answer btw!) - this comment thread is getting a bit long (which, granted, is more my fault than yours :P). | |
| Mar 8, 2021 at 21:04 | comment | added | Noah Schweber | @PaceNielsen If $M$ is a model of $\mathsf{ZFC}$, it has a version of $\omega$ (denoted "$\omega^M$") internal to itself - this is in no way meta. Namely, consider what $M$ thinks is the smallest limit ordinal. We can turn this into a thing which $M$ thinks is (and in fact genuinely is, but that's beside the point) a model of $\mathsf{PA}$ (and indeed much more) by equipping it with what $M$ thinks are ordinal addition and multiplication. | |
| Mar 8, 2021 at 21:02 | history | edited | Noah Schweber | CC BY-SA 4.0 |
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| Mar 8, 2021 at 21:01 | comment | added | Noah Schweber | ($S$ being the theory above.) Similarly, each model of (the first-order theory annoyingly called) second-order arithmetic $\mathsf{Z}_2$ has an "internal" notion of "$\omega$-model of second-order arithmetic" and so on, and the same is true for other special classes of models, such as $\beta$-models. See e.g. Part B of Simpson's reverse math book for the development in arithmetic (there is no difference whatsoever between this and set theory, in terms of the general methods for expressing such notions). | |
| Mar 8, 2021 at 20:56 | answer | added | Farmer S | timeline score: 13 | |
| Mar 8, 2021 at 20:55 | comment | added | Noah Schweber | @PaceNielsen "$T$ is $\omega$-complete" is already a statement in the object language: every model $M$ of $\mathsf{ZFC}$ has a notion of "$\omega$-model of $\mathsf{ZFC}$," namely "model of $\mathsf{ZFC}$ whose $\omega$ is isomorphic to my $\omega$." There's no linguistic or otherwise "meta" issue here. (In particular, note that if $M\models\mathsf{ZFC}$ is not an $\omega$-model itself then the $\omega$-models of $\mathsf{ZFC}$ in the sense of $M$ - assuming $M$ thinks there are any at all - will not be genuine $\omega$-models.) So the $S$ of the OP is a genuine first-order theory. | |
| Mar 8, 2021 at 20:50 | comment | added | Pace Nielsen | So, if I'm understanding correctly, what you'd like is (under the assumption that $T$ has an $\omega$-model) the construction of a structure (in the language of set theory) that satisfies $T$, it has a unique proper substructure satisfying $T$, and $\omega$ is standard in the structure. Is that correct? I ask because I'm trying to figure out what "$T$ is $\omega$-complete" would mean as a statement in some language. I can see how to express it using a language that allows infinite conjunctions, but that goes beyond the language of set theory. | |
| Mar 8, 2021 at 20:39 | comment | added | Noah Schweber | @PaceNielsen Yup, an $\omega$-model is one where the naturals are standard. In general, in any context where we have a theory $T$ together with a canonical interpretation of some theory of arithmetic into $T$, we say that an $\omega$-model of $T$ is one in which this interpretation yields a structure isomorphic to $\mathbb{N}$ itself. The usual examples are: second-order (or higher-order) arithmetic together with the "first-order part" interpretation, and set theories together with the "$\omega$" interpretation. ("Categoricity" results can justify the privileging of a specific interpretation.) | |
| Mar 8, 2021 at 20:39 | comment | added | Pace Nielsen | Is an $\omega$-model one where the natural numbers are the standard ones? I tried to find a definition, and could only find: math.stackexchange.com/questions/1113639/what-is-an-omega-model | |
| Mar 8, 2021 at 19:47 | history | asked | Noah Schweber | CC BY-SA 4.0 |