Timeline for How badly does compactness fail in $\mathcal{L}_{\omega_1\omega}$?
Current License: CC BY-SA 3.0
10 events
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| Dec 10, 2013 at 23:50 | comment | added | Emil Jeřábek | Sorry for a not quite on-topic comment, but in case someone reading this is wondering: that no $L_{\kappa,\omega}$ can define well-foundedness is known for fact, it was proved by Lopez-Escobar eudml.org/doc/213903 . For fascinating connections of $L_{\infty,\omega}$ extended with a means for expressing well-foundedness, see Barwise dx.doi.org/10.1016/0003-4843(72)90002-2 . | |
| May 17, 2012 at 9:57 | vote | accept | Toby Meadows | ||
| May 16, 2012 at 15:38 | comment | added | Joel David Hamkins | I'm inclined to think that $L_{\omega_1,\omega}$ has the $\kappa$-compactness property if and only if there is a strongly compact cardinal $\leq\kappa$. The reverse direction is clear. For the forward direction, my argument above shows that for all regular $\theta\geq\kappa$, there is a countably complete uniform ultrafilter on $\theta$. But I'm not clear on whether this gives a strongly compact cardinal or not. But this property certainly fails in $L[\mu]$, and so a measurable cardinal is not sufficient. I think it will similarly fail in many other canonical inner models. | |
| May 16, 2012 at 15:34 | comment | added | Joel David Hamkins | Yes, I agree with you now. It isn't important to express well-foundedness; this is a mere convenience. And it works with strong compactness also, as in my answer, using the Ketonen theorem, since in $L_{\kappa,\omega}$ one can express the $\kappa$-completeness of the ultrafilter. | |
| May 16, 2012 at 15:12 | comment | added | Asaf Karagila♦ | I checked the notes from a course about large cardinals I attended last year - my claim is true for weakly-compact it is enough to require $\mathcal L_{\kappa,\omega}$ is weakly-compact. We can use the compactness of the language to show that $\kappa$ is strongly inaccessible and has the tree property. | |
| May 16, 2012 at 14:15 | comment | added | Joel David Hamkins | If you mean $L_{\kappa,\omega_1}$, then I agree. $\kappa$ is weakly compact iff $L_{\kappa,\kappa}$ has $\kappa$-compactness iff $L_{\kappa,\omega_1}$ has $\kappa$-compactness, for theories in a language of size at most $\kappa$, and the same for strong compactness if one uses larger theories. But with $L_{\kappa,\omega}$, you cannot seem to say "well-founded", and so the usual arguments break donw. | |
| May 16, 2012 at 14:09 | comment | added | Asaf Karagila♦ | Isn't the $\kappa$-compactness of $\mathcal L_{\kappa,\kappa}$ and $\mathcal L_{\kappa,\omega}$ equivalent? I recall it is so at east in the weakly compact case. | |
| May 16, 2012 at 13:46 | comment | added | Joel David Hamkins | Correction, in the last paragraph, it should be Ketonen's theorem, not Menas'. | |
| May 16, 2012 at 13:37 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
Added theorem at end; edited body; added 123 characters in body
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| May 16, 2012 at 12:49 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |