Timeline for Can ZFC prove "false theorems", and still be consistent? (was "Junk Theorems" follow up)
Current License: CC BY-SA 3.0
12 events
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| Mar 14, 2012 at 19:07 | comment | added | Will Sawin | A better example than Goldbach is Twin Primes. There could be finitely many twin primes and a proof that there are infinitely many without contradiction, or infinitely many twin primes and a proof that there are finitely many without contradiction. Of course, the proofs couldn't be constructive, but a non-constructive proof is entirely possible. | |
| Mar 14, 2012 at 18:53 | comment | added | Yaakov Baruch | I think the comment stream to the answer below is relevant to this discussion: mathoverflow.net/questions/28806/… especially the final comments by Joel David Hamkins and Carl Mummert | |
| Mar 14, 2012 at 12:10 | comment | added | Not Mike | Goldbach's conjecture? Seriously? If there is a counter-example, then it exists for reasons beyond coding. And will in no way be independent. Not even close. | |
| Mar 14, 2012 at 11:37 | comment | added | user13113 | In ZFC+~GC, there is a finite counterexample. | |
| Mar 14, 2012 at 4:05 | comment | added | Steven Gubkin | If GC is independent of ZFC then there is no finite counterexample to ZFC (if you can produce a counterexample, that counterexample can be checked in ZFC). If there is no finite counterexample, GC is true. | |
| Mar 14, 2012 at 3:35 | comment | added | user13113 | In fact, isn't it true that ZFC+~GC can prove ~GC provable by PA? | |
| Mar 14, 2012 at 3:28 | comment | added | user13113 | @Steven: To give an example of a differing opinion, I would consider that argument a type-error. If GC is unprovable in ZFC, that means GC is true in certain models of PA (as constructed in certain models of ZFC) -- to say GC is true in an absolute sense goes too far. | |
| Mar 13, 2012 at 21:44 | comment | added | Will Sawin | Depends on who you ask. If you ask someone who agrees with all the axioms of ZF + determinacy + Con(ZFC), then he will say "Yes, of course. ZFC proves that there exist non-measurable sets. But, much like even numbers not the sum of two primes, it provides no procedure to produce such a set, and indeed, no such sets exist! Therefore, ZFC is false. But it's still consistent, as there is no proof in ZFC of this obvious falsehood." | |
| Mar 13, 2012 at 14:27 | vote | accept | Steven Gubkin | ||
| Mar 13, 2012 at 14:27 | comment | added | Steven Gubkin | Is it still possible that our means of encoding the natural numbers into ZFC allows proofs of some false theorems? I mean say goldbach conjecture turns out to be unprovable in ZFC. Then GC is true, and ZFC + (~GC) is consistent. So ZFC + (~GC) proves a false theorem about the natural numbers. Is it possible that there are some false theorems lurking in ZFC alone, even given its consistency? | |
| Mar 13, 2012 at 14:23 | comment | added | Steven Gubkin | Great! So these kinds of type differences will never butt head with each other so much that you end up producing false theorems. | |
| Mar 13, 2012 at 5:57 | history | answered | Andrej Bauer | CC BY-SA 3.0 |