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Nov 2, 2011 at 15:59 comment added Emil Jeřábek I see. Well, I find inconsistent large cardinals rather unpleasant, but whatever. Large cardinals are not the point, the point is that you can take as strong a theory as you want, as long as it does not prove things that are too obviously false.
Nov 2, 2011 at 15:43 comment added Joel David Hamkins Emil, I took Andreas to mean that your comment suggests that all the large cardinals that please us are actually also consistent...
Nov 2, 2011 at 14:13 comment added Emil Jeřábek (In case I was unclear, I’m referring to the arithmetical hierarchy, not Levy hierarchy.) Under normal circumstances, the $\Sigma^0_1$-soundness of a particular large cardinal is implied by the existence of a larger large cardinal (or even of two specimens of the same cardinal). If the large cardinals in whose consistency you believe can be put in an increasing chain, you are covered.
Nov 2, 2011 at 12:59 comment added Andreas Blass @Emil: The combination of "$\Sigma_1$-sound" near the beginning of your message and "whatever large cardinals you please" near the end suggests considerable confidence in my taste in large cardinal axioms. Thank you.
Nov 2, 2011 at 11:38 comment added Emil Jeřábek The actual McAloon theorem is much stronger. It says that for every $\Sigma_1$-sound recursively axiomatized theory $T\supseteq I\Sigma_1$, you can find (arbitrarily short) nonstandard initial segments that are models of $T$. For example, one can take for $T$ the set of all arithmetical consequences of ZFC (plus whatever large cardinals you please). Moreover, the theorem also holds for nonstandard models of the weaker theory $IE_1$ in place of $I\Delta_0$, as shown by Wilmers.
Nov 2, 2011 at 9:34 history edited Joel David Hamkins CC BY-SA 3.0
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Nov 2, 2011 at 9:23 history answered Joel David Hamkins CC BY-SA 3.0