Timeline for Why are model theorists free to use GCH and other semi-axioms?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Oct 10, 2016 at 7:31 | comment | added | arsmath | But it's still a waste of time to look for a proof in ZFC. If they think that CH is false, they have to actually suppose it. | |
| Oct 10, 2016 at 0:20 | comment | added | Andrés E. Caicedo | @arsmath The thing I don't like about your last comment is that, in a Platonist setting, we may very well think that, say, $\mathsf{MA}+\lnot{CH}$ is true, and so we may very well be able to disprove model-theoretic consequences of $\mathsf{GCH}$. The point is that, for a Platonist, it is artificial to limit the assumptions of their theorems to $\mathsf{ZFC}$. | |
| Oct 8, 2016 at 22:40 | comment | added | arsmath | There's also the fact that if GCH implies a model theoretic result, say, then it's a waste of time to look for a proof that it's false. | |
| Oct 8, 2016 at 22:36 | comment | added | arsmath | They're still true, they're just relative to some model different from the one that you intended. This follows from the completeness theorem of first-order logic -- if you have a consistent theory, then that theory has a model. Whether or not that model is interesting is a matter of taste, not philosophy. | |
| Oct 8, 2016 at 22:27 | comment | added | user99445 | Thanks! Assume that one is a platonist and thinks that every set-theoretic assertion is either true or false. Do I understand you correctly that you think that in this case, one can still "for fun" look at the consequences of ZFC + GCH? When I think about that idea, it makes sense, thanks! But isn't it then a little bit different from working with axioms all of which one is sure that they are true? | |
| Oct 8, 2016 at 22:20 | history | answered | arsmath | CC BY-SA 3.0 |