Timeline for Does finite mathematics need the axiom of infinity?
Current License: CC BY-SA 2.5
8 events
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| Oct 15, 2009 at 4:09 | comment | added | Richard Dore | Equiconsistency isn't really what you want, you want bi-interpretability. Also, when you say something is provable in Second Order Arithmetic, you should specify which Second Order Theory you're talking about. Goodstein's Theorem isn't provable in RCA_0. | |
| Oct 15, 2009 at 4:05 | comment | added | Andrew Critch | Correction: it assumes infinity in disguise --- the comprehension axiom. Thanks, guys! | |
| Oct 15, 2009 at 4:03 | history | edited | Anton Geraschenko | CC BY-SA 2.5 |
added 121 characters in body; deleted 35 characters in body
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| Oct 15, 2009 at 4:01 | comment | added | Eric Wofsey | Correction--it uses induction of length \epsilon_0. <a href=en.wikipedia.org/wiki/…> says PA is equiconsistent with finite set theory, which suggests that their strength might be the same. | |
| Oct 15, 2009 at 3:59 | comment | added | Andrew Critch | en.wikipedia.org/wiki/Goodstein's_theorem says it is provable using Second Order Arithmetic, which doesn't assume infinity. | |
| Oct 15, 2009 at 3:56 | comment | added | Eric Wofsey | This example is usually given as a statement that cannot be proven in Peano arithmetic, because it uses induction of length \omega^\omega. ZFC without Infinity is easily seen to be at least as strong as Peano arithmetic and is probably significantly stronger, but I have no idea exactly how strong it is. | |
| Oct 15, 2009 at 3:51 | comment | added | Alon Amit | "maybe you could do it without" - I don't think you can. Goodstein's Theorem cannot be proved in PA, and I'm quite sure it cannot be proved in ZFC minus the axiom of infinity either. | |
| Oct 15, 2009 at 3:44 | history | answered | Anton Geraschenko | CC BY-SA 2.5 |