2
$\begingroup$

I have known that for a transitive model of ZFC, M, x is a M-generic filter of some forcing notion in M, then M[x] is least transitive model of ZFC which contains M and $\{x\}$. M[x] is just the set of all values of names in M. My question is, if I replace "a M-generic filter of some forcing notion in M" by an arbitrary subset of some element of M, how to define such a least model? i.e. M is a transitive model of ZFC, x is a set and x is a subset of some element of M. Can we define a least transitive model M[x] of ZFC such that $M\subset{M[x]}$ and $x\in{M[x]}$?

$\endgroup$
7
  • 5
    $\begingroup$ mathoverflow.net/questions/83298/… $\endgroup$
    – Ed Dean
    Apr 4, 2012 at 4:19
  • 1
    $\begingroup$ As pointed out in the answer to the question linked to above, in some cases there exists such a model M[x]. Now the question is, how could you obtain information about M[x] if you only happen to "know" what's in M? That's the magic of forcing notions and the relation of forcing, that you can actually get information about M[x] from the elements of M only. I don't know if you can do something analogous when adjoining just any set x to your model M. $\endgroup$ Apr 4, 2012 at 6:06
  • $\begingroup$ I vaguely recall that there is some result of the sort that if such an $M[x]$ exists, then it is in fact equivalent to a forcing extension. $\endgroup$ Apr 4, 2012 at 10:30
  • $\begingroup$ This is the case if x is an element of a forcing extension. $\endgroup$ Apr 4, 2012 at 10:51
  • 1
    $\begingroup$ The link above might not exactly be the answer as it talks about set models of ZFC, and here Song Li probably allows classes as well. In this case there is always such a model - it should be defined similar to definition (13.24) in Jech's "Set Theory", relative constructibility, by closing under Goedels operations. @Emil - you probably refer to Jensen's coding theorem (see "Coding the Universe" by Beller, Jensen and Welch). $\endgroup$
    – Eran
    Apr 5, 2012 at 20:29

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.