Timeline for proof that "small" sets in an extension by iterated forcing already appear in an earlier stage
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 14, 2014 at 22:31 | vote | accept | user40921 | ||
| Oct 14, 2014 at 22:13 | comment | added | user40921 | Very interesting, thank you! It seems the second part of Joel's answer solves the problem, although one would expect a much simpler explanation (and indeed there is, somehow evading my notice until now). by the way, I didn't quite understand (in the first part of Joel's answer) if (and how) the new predicate $\check{M}$ can be used to define M in M[G] (and not just refer to M in the forcing language) | |
| Oct 14, 2014 at 22:13 | answer | added | Joel David Hamkins | timeline score: 3 | |
| Oct 14, 2014 at 21:59 | review | Close votes | |||
| Oct 14, 2014 at 22:17 | |||||
| Oct 14, 2014 at 21:50 | answer | added | user40921 | timeline score: 3 | |
| Oct 14, 2014 at 20:59 | comment | added | Yair Hayut | see mathoverflow.net/questions/78751/name-for-the-ground-model | |
| Oct 14, 2014 at 20:54 | comment | added | user40921 | (it is, if e.g. M satisfies V=L, but I don't think it's generally true) | |
| Oct 14, 2014 at 20:48 | comment | added | user40921 | and why is M definable in M[G]? | |
| Oct 14, 2014 at 20:42 | comment | added | Yair Hayut | Since $M$ is definable in $M[G]$, everything that is definable in $M$ will be also definable in $M[G]$. | |
| Oct 14, 2014 at 20:28 | history | asked | user40921 | CC BY-SA 3.0 |