Timeline for "Almost all" quantifier
Current License: CC BY-SA 3.0
6 events
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| Sep 7, 2011 at 18:30 | comment | added | Carl Mummert | Re Kaveh's question. This answer is perfectly valid, but I want to point out that usually, in the context of arithmetic, when we say "finitely many" we mean "for a bounded set in the model", and if I saw $\forall^\infty$ or $\exists^\infty$ in this setting I would assume the quantifiers are the usual definitional extension of the object language. It's not surprising that it should be impossible to define metafiniteness within an arbitrary model of arithmetic; it it was possible, it would mean that either every element of the model is metafinite, or else some induction axiom should fail. | |
| Sep 7, 2011 at 15:15 | comment | added | Emil Jeřábek | ... as it only makes sense for models of arithmetic. | |
| Sep 7, 2011 at 15:13 | comment | added | Emil Jeřábek |
The OP is somewhat sloppy, but gives a definition of the semantics: $M\models G_x\phi(x)$ iff $M\models\phi(a)$ for all but finitely many $a\in M$. Obviously, if you change the definition of any notion involved in the question, it may change the answer. In some contexts it might make sense to simply define the almost-all quantifier as $\exists u\,\forall x\,(u\le x\to\phi(x))$, but (1) this is not what the question says, (2) it would make the question pointless as the definition would already give a trivial answer, and (3) it would not give an extension of first-order logic in general, ...
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| Sep 7, 2011 at 14:41 | comment | added | Kaveh | Emil, doesn't this depend on how you define the semantics for G in arbitrary models? | |
| Sep 7, 2011 at 14:00 | history | edited | Emil Jeřábek | CC BY-SA 3.0 |
clarification
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| Sep 7, 2011 at 12:39 | history | answered | Emil Jeřábek | CC BY-SA 3.0 |