Timeline for True by accident (and therefore not amenable to proof)
Current License: CC BY-SA 4.0
11 events
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| Nov 27, 2022 at 4:00 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 4.0 |
typo
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| Aug 12, 2022 at 7:57 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
http -> https (the question was bumped anyway)
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| Mar 11, 2022 at 13:15 | comment | added | M. Winter | Is it possible that the reconstruction conjecture is false in every non-standard model, but true in the standard model? If so, we could argue that the standard model has some "accidental" properties. | |
| Sep 1, 2011 at 15:55 | comment | added | Pace Nielsen | ... we cannot tell whether or not P is true. To give a concrete example, take a model sufficiently complex enough so we can express the statement P="The Riemann Hypothesis" and ask whether it is true in the given model. What prevents the statement from being accidentally true in that model (so it is true, but we just cannot see that)? Further, if that question is even reasonable, what would then prevent statements from being accidentally independent (i.e., two models exist where P is true in one and false in the other, but we can't see that). | |
| Sep 1, 2011 at 15:50 | comment | added | Pace Nielsen | Joel, I freely admit to not being an expert in model theory, so please excuse this question. Why can you not move the "accident" to the construction of the models? Or, to put it another way, let P be a statement we want to prove. It might be the case that we can simply prove P. It might be the case that we can simply prove not(P). It might be the case that we can easily construct a model where P is true, and another model where P is false, thus showing that P is independent. However, aren't there other cases? For example, what prevents the existence of a model where (continued...) | |
| Aug 25, 2011 at 21:48 | comment | added | Joel David Hamkins | Timothy, yes, I am arguing that for statements in a first-order language, the (profound) fact that true-in-all-models is the same as provable means that there is no accidental truth. A statement in the language of graph theory is true in all graphs if and only if it is provable from the axioms of a graph. So we don't need another distinction besides provable, negated-provable, and independent. In the case of statements of arithmetic, such a view is intimately connected with a considerations of nonstandard models of arithmetic, but I maintain that it still answers the question. | |
| Aug 25, 2011 at 20:13 | comment | added | François G. Dorais | Congrats on your 500th! | |
| Aug 25, 2011 at 20:08 | comment | added | Timothy Chow | @Joel: I confess that I don't understand how this answers the OP's question. Are you saying that there aren't any statements that are true by accident? | |
| Aug 25, 2011 at 19:31 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
Added incidental remark about 500th answer.; deleted 1 characters in body
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| Aug 25, 2011 at 13:00 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
deleted 38 characters in body
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| Aug 25, 2011 at 12:15 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |