Timeline for If there is a non-constructible real, is there an $L$-generic real?
Current License: CC BY-SA 3.0
13 events
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| Aug 3, 2018 at 11:23 | comment | added | Asaf Karagila♦ | @jonas: I don't know the exact reference. But if that helps, I know that in Grigorieff's treasure trove of a paper "Intermediate Submodels and Generic Extensions" he mentions that result on page 471. | |
| Aug 2, 2018 at 16:17 | comment | added | jonasreitz | Wonderful - yes, that’s exactly what I’m looking for. I’ll think through the argument and do some looking into Solovays result, too. Many thanks - this is a big help! | |
| Aug 2, 2018 at 6:43 | comment | added | Asaf Karagila♦ | @jonasreitz: Well, if I understand correctly, then yes. If $\kappa$ is a cardinal in $V$, then $\operatorname{Add}(\kappa,1)^L$ is an $L$-$\kappa$-closed forcing, which only has $\kappa$ many subsets, so you should be able to construct generics. If my memory serves me even better, Solovay proved that from $0^\#$ there is a class generic for a class forcing over $L$ which violates GCH on a proper class of cardinals. I imagine that was the sort of arguments used... | |
| Aug 2, 2018 at 0:59 | comment | added | jonasreitz | Thanks, Asaf - that’s great! Do you know if it holds for cardinals not countable in $V$? I don’t see immediately how to generalize the argument, but it would be great if $L[0^\#]$ had L-generics for all posets $Add(\kappa,1)^L$. (Ps. Thanks also for the markup advice - I’m posting from my phone so we’ll see how it comes out). | |
| Aug 1, 2018 at 23:06 | comment | added | Asaf Karagila♦ |
(Also, as a side note, \# produces $\#$, just for future reference. So you wanted to write $0^\#$, @jonasreitz.)
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| Aug 1, 2018 at 23:05 | comment | added | Asaf Karagila♦ | @jonasreitz: That's easy. If $0^\#$ exists, then $\omega_1^L$ is countable, therefore $\operatorname{Add}(\omega,1)^L$ has only countably many dense open sets (in $L$), so there is a generic filter meeting them. The same argument holds for any $V$-countable $L$-cardinal. | |
| Aug 1, 2018 at 23:03 | comment | added | jonasreitz | Hi Asaf - I came across this old question while researching other things, and I'm very intrigued by your side comment in the question above, that the existence of 0# implies the existence of L-Cohen generics. Could you provide a reference? Many thanks! | |
| Aug 16, 2013 at 2:13 | history | edited | Asaf Karagila♦ | CC BY-SA 3.0 |
added 233 characters in body
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| Aug 16, 2013 at 0:04 | vote | accept | Asaf Karagila♦ | ||
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| Aug 15, 2013 at 21:32 | answer | added | Joel David Hamkins | timeline score: 8 | |
| Aug 15, 2013 at 20:58 | comment | added | Steven Stadnicki | My knowledge of generics is pretty weak, but from the bit of recursion theory I know, and what I know of the analogies, it seems like some version of the usual means for constructing reals of intermediate degree ought to provide such an $x$? | |
| Aug 15, 2013 at 20:51 | history | edited | Asaf Karagila♦ | CC BY-SA 3.0 |
edited title
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| Aug 15, 2013 at 20:45 | history | asked | Asaf Karagila♦ | CC BY-SA 3.0 |