Timeline for Can a model of $V\neq L$ contain a class giving the $L$-ordering on all its sets?
Current License: CC BY-SA 3.0
12 events
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| May 1, 2021 at 17:39 | vote | accept | jonasreitz | ||
| Apr 27, 2021 at 9:17 | answer | added | Farmer S | timeline score: 5 | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jan 29, 2013 at 22:41 | comment | added | Danielle Ulrich | This is exactly my motivation for the question! Basically I'm trying to figure out, sitting in a model of $V \not= L$, how much we can reason about a hypothetical extension of the universe in which $V = L$. In my mind its analogous to reasoning about hypothetical generic filters | |
| Jan 29, 2013 at 16:48 | comment | added | jonasreitz | @Joel, yes, I had intended $M$ to be transitive. However, I know if we allow non-transitive models we get some very interesting and strange results, as in your "Multiverse Perspective on the Axiom of Constructibility" -- so I guess I'm also interested in the nontransitive case... | |
| Jan 29, 2013 at 16:45 | comment | added | Joel David Hamkins | Nice question, Jonas! I assume throughout that you want $M$ to be transitive? | |
| Jan 29, 2013 at 16:40 | comment | added | jonasreitz | Douglas, just wanted to say I love your question. In contemplating these odd models in which every element is constructible, but $V \neq L$, I am always struck by the possibility that every set is, in fact, constructible, if we are willing to continue the $L$-construction far enough beyond $ORD$. This idea is addressed more rigorously by Joel Hamkins in his A Multiverse Perspective on the Axiom of Constructibility | |
| Jan 29, 2013 at 14:53 | comment | added | Emil Jeřábek | Never mind, I misread it thinking it referred to $<_L$ inside the model rather than the external one. | |
| Jan 29, 2013 at 14:05 | comment | added | Andreas Blass | @Emil: Yes, Jonas wants the ordering of all of $M$ in order of constructibility. That ordering exists because he assumed $M\in L$. | |
| Jan 29, 2013 at 14:00 | comment | added | Danielle Ulrich | This is not an answer- this would be a comment if I had enough reputation points for comments. I just wanted to mention that I am no longer convinced by the answer to my question (to which you linked). | |
| Jan 29, 2013 at 3:43 | history | edited | jonasreitz | CC BY-SA 3.0 |
added 82 characters in body
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| Jan 29, 2013 at 3:03 | history | asked | jonasreitz | CC BY-SA 3.0 |