Timeline for If there is a non-constructible real, is there an $L$-generic real?
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Aug 16, 2013 at 3:13 | comment | added | Asaf Karagila♦ | I'll get back to you after I get some sleep (if that is ever going to happen, although it's unlikely to happen soon). I think that I'm too sleep deprived to decide anything right now. But at the moment I feel that the example is good, but misses the point I was trying to understand. | |
| Aug 16, 2013 at 2:53 | comment | added | Joel David Hamkins | I don't really understand the objection, since Sy's model seems to answer the opening question of your post negatively, and it certainly isn't trivial. | |
| Aug 16, 2013 at 1:46 | comment | added | Asaf Karagila♦ | Again, I agree. But this example feels like "the trivial counterexample which I wanted to neglect anyway". If one is so inclined, then I can modify my question to say whether or not $V\neq L[A]$ for any set $A$ implies that there is an $L$-generic real; but this will also miss a big point (that there are plenty of interesting models of the form $L[A]$). Instead, I suppose, I can just require that $V$ is not a class forcing generic extension of any $V$ for which $\Bbb R^V=\Bbb R^L$. | |
| Aug 16, 2013 at 1:39 | comment | added | Joel David Hamkins | Well, in this case point wasn't to code the universe, but to add a real that wasn't set-generic over $L$, while also adding no other set-generic sets. This is, after all, what you asked for... | |
| Aug 16, 2013 at 1:22 | comment | added | Asaf Karagila♦ | Ah. I see! But now the restriction over class forcing makes slightly more sense, I think (and I mean, who encodes the universe when it's $L$? :-)). | |
| Aug 16, 2013 at 1:09 | comment | added | Joel David Hamkins | The point is that Sy's forcing only gives minimality over V, not over L. So if we start with V=L, then we get the desired counterexample for your question. | |
| Aug 16, 2013 at 1:08 | comment | added | Asaf Karagila♦ | Wait, if we have that $V=L[c]$ for a Cohen real $c$, then we encode this using Sy's class forcing. We have $L[R]$, but $c\in L[R]$ and $c$ is still $L$-generic. (As a side note, due to Andreas' comment I am not being pinged automatically, if it's not too much trouble, please ping me in the next comments!) | |
| Aug 16, 2013 at 0:53 | comment | added | Joel David Hamkins | Oh, I had just meant my remark about your classification follow-up question. I think my answer is correct for your question, since as I say there, we should start in L. | |
| Aug 16, 2013 at 0:44 | comment | added | Asaf Karagila♦ | I'll unaccept for now, then. But I'm hoping to see a revision! :-) | |
| Aug 16, 2013 at 0:41 | comment | added | Joel David Hamkins | Oops, my previous comment is wrong, since $V$ could easily have $L$-generics....I had neglected that part of your question... | |
| Aug 16, 2013 at 0:36 | comment | added | Asaf Karagila♦ | Hmm. I see your point, and it feels that (especially after a recent question) a "rapidly changing" question is not in place here. But I can't help and wonder if one can ask about inner models. That is, rephrase the question. Given $V$ such that $\Bbb R^V\neq\Bbb R^L$, is there an inner model $W$ and $x\in\Bbb R^W$ such that $W=L[x]$ is a generic extension of $L$? (But I don't really expect this to have a reasonable answer. Just throwing it out there!) | |
| Aug 16, 2013 at 0:31 | comment | added | Joel David Hamkins | I seriously doubt there will be a good characterization of such models, since the coding-the-universe argument shows that any model $V$ is a ground model of such an $L[x]$. So our characterization will have to include in some a way an account of all possible $V$'s. | |
| Aug 16, 2013 at 0:26 | comment | added | Asaf Karagila♦ | Joel, here's a slightly more sensitive follow up. Can we characterize (except trivially) models of the form $L[x]$ where $x$ is a real number, in which there are no $L$-generic reals? | |
| Aug 16, 2013 at 0:04 | vote | accept | Asaf Karagila♦ | ||
| Aug 16, 2013 at 0:44 | |||||
| Aug 15, 2013 at 21:58 | comment | added | Andreas Blass | Sy has also studied a concept of hyperclass forcing, but I don't know whether it provides a counterexample for the follow-up question. If it does, then there's an obvious further follow-up. | |
| Aug 15, 2013 at 21:42 | comment | added | Joel David Hamkins | Well, that seems far more difficult. First, it cannot really be formalized in ZFC. Second, if you forbid class forcing, then our hands are too-much tied for building a counterexample. | |
| Aug 15, 2013 at 21:38 | comment | added | Asaf Karagila♦ | That sure answers the question, thanks! Now comes the obvious follow up, what if we allow class-generics? | |
| Aug 15, 2013 at 21:32 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |