Timeline for When can we detect forcing?
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
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| Oct 13, 2011 at 0:18 | comment | added | Norman Lewis Perlmutter | In the definition of consistent, do you mean to say "forcing extension" rather than "elementary extension"? | |
| Nov 15, 2010 at 0:09 | answer | added | Andrés E. Caicedo | timeline score: 6 | |
| Nov 8, 2010 at 19:08 | vote | accept | Noah Schweber | ||
| Nov 8, 2010 at 17:30 | answer | added | jonasreitz | timeline score: 11 | |
| Nov 8, 2010 at 16:05 | comment | added | Noah Schweber | @Amit Kumar Gupta: Yeah, I forgot to finish typing the second paragraph. That's fixed now, and I've noted that you answered that question. I hope you expand your comment into an answer! | |
| Nov 8, 2010 at 16:03 | history | edited | Noah Schweber | CC BY-SA 2.5 |
added 76 characters in body
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| Nov 8, 2010 at 14:19 | comment | added | François G. Dorais | Amit, could you repost your comment as an answer... | |
| Nov 8, 2010 at 7:20 | answer | added | Stefan Geschke | timeline score: 6 | |
| Nov 8, 2010 at 6:36 | comment | added | Amit Kumar Gupta | Reitz and Hamkins have explored a possible axiom called the Ground Axiom, which states that $V$ is not a forcing extension of any inner model $W$ by a nontrivial forcing $\mathbb{P} \in W$. In your second paragraph you seem to define a notion of detectability but it doesn't appear that you ask an actual question. A result of Laver's is that if $V=M[G]$ is a forcing extension of $M$ by a set forcing $\mathbb{P} \in M$ then $M$ is definable in $V$ from parameters in $M$. So in a sense this says that if $V$ is a forcing extension, then it can detect the ground model. | |
| Nov 8, 2010 at 5:48 | history | edited | Andrés E. Caicedo |
edited tags
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| Nov 8, 2010 at 3:33 | history | asked | Noah Schweber | CC BY-SA 2.5 |