Timeline for Deducing Skolem's nonstandard integers from downward Lowenheim-Skolem?
Current License: CC BY-SA 3.0
22 events
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| Sep 2, 2013 at 16:00 | history | edited | Mikhail Katz | CC BY-SA 3.0 |
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| Aug 29, 2013 at 17:23 | comment | added | Mikhail Katz | @JoelDavidHamkins: Now's the tide :-) | |
| Aug 29, 2013 at 16:31 | history | reopened |
Mikhail Katz Daniel Moskovich Andrey Rekalo Joel David Hamkins François G. Dorais |
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| Aug 29, 2013 at 8:13 | history | edited | Mikhail Katz |
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| Aug 28, 2013 at 8:01 | history | edited | Mikhail Katz | CC BY-SA 3.0 |
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| Aug 27, 2013 at 13:45 | comment | added | Joel David Hamkins | Sure, if it re-opens, then I'll post my comments as an answer | |
| Aug 27, 2013 at 12:24 | history | edited | Mikhail Katz | CC BY-SA 3.0 |
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| Aug 26, 2013 at 13:58 | review | Reopen votes | |||
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| Aug 26, 2013 at 13:49 | history | edited | Mikhail Katz | CC BY-SA 3.0 |
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| Aug 26, 2013 at 13:19 | history | closed |
Andrés E. Caicedo Ramiro de la Vega Daniel Moskovich David White Simon Thomas |
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| Aug 26, 2013 at 12:31 | history | edited | Mikhail Katz | CC BY-SA 3.0 |
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| Aug 25, 2013 at 14:04 | comment | added | Mikhail Katz | @JoelDavidHamkins: Thanks. You made some very nice points so could I prevail upon you to format these as an "answer"? They are close to "closing" this for some reason. | |
| Aug 25, 2013 at 11:36 | comment | added | Joel David Hamkins | My point with that parenthetical remark was that the transfer relation for a given structure is itself a kind of structure (between two structures), and one can make a larger structure in which that transfer structure is explicitly part of the structure, and then get a nonstandard version of that. But this isn't necessary for the rest of your question. Two different structures, say $R$ and $R^\ast$, can be thought of as a single two-sorted structure, which admits non-standard extensions. | |
| Aug 25, 2013 at 9:07 | comment | added | Mikhail Katz | ...involve higher-order quantification? (response to @Joel). | |
| Aug 25, 2013 at 8:51 | comment | added | Mikhail Katz | Very interesting. How do you set up a "situation where you already have full transfer"? I am used to thinking of having a structure specifically over R, and relating it to *R by means of a transfer principle (e.g. by means of Los's theorem in the case of an ultrapower). Having "transfer" as part of the structure itself is a new idea to me. Is the idea to incorporate both the substructure R as well as "baby transfer" from R to *R into the theory over *R, applying "super"transfer to this new theory, and then passing to definable entities everywhere? Wouldn't formalizing the "baby" transfer... | |
| Aug 23, 2013 at 11:40 | comment | added | Joel David Hamkins | Yes, you can do this also. Just take any situation where you already have full transfer, and take a countable elementary substructure. (You could even add the transfer relation in the formal language of the structure here.) Every definable function will still have its transfer analogue in this countable substructure, by elementarity. | |
| Aug 23, 2013 at 7:33 | comment | added | Mikhail Katz | ... This would mean for example that every hyperinteger is part of a Skolem-type countable model satisfying "definable" transfer. | |
| Aug 23, 2013 at 7:32 | comment | added | Mikhail Katz | @Joel: it is apparently not literally true that one "does not need a theory" since you are speaking of an elementary substructure, so at least that much "theory" has to carry over. I am wondering in fact if one can do something stronger, namely having extensions of all definable functions, with the result of applying LS being not merely an elementary countable extension but rather, an extension satisfying a transfer principle for definable functions, as seems to be the case for Skolem's nonstandard model... | |
| Aug 22, 2013 at 15:45 | review | Close votes | |||
| Aug 23, 2013 at 7:42 | |||||
| Aug 22, 2013 at 13:34 | comment | added | Joel David Hamkins | One doesn't need a theory to apply the downward LS theorem, but only a structure. So if you have a nonstandard model $\mathcal{N}$ and a nonstandard element $H$ in it, then by LS there is a countable elementary substructure $\mathcal{M}\prec\mathcal{N}$ with $H\in\mathcal{M}$. In contrast, if you take the theory of true arithmetic (true in the standard model $\mathbb{N}$), and add those assertions as you described, then the theory is consistent and hence has a (countable) model by the completeness theorem (rather than by LS). | |
| Aug 22, 2013 at 13:24 | comment | added | Emil Jeřábek | Yes. | |
| Aug 22, 2013 at 12:55 | history | asked | Mikhail Katz | CC BY-SA 3.0 |