[FOM] inverse forcing
Ashutosh
ashu1559 at gmail.com
Wed Aug 25 19:07:09 EDT 2010
Ali Enayat's solution also works: Let x, y be Cohen over M[y] and M[x]
respectively. Then x+y is Cohen over both M[x] and M[y].
Regards
Ashutosh
On Tue, Aug 24, 2010 at 7:52 PM, Monroe Eskew <meskew at math.uci.edu> wrote:
>
> On July 22, I posed the question of whether after forcing with P to
> get M[G], is the ground model M uniquely determined by the parameters
> P, G. Ali Enayat suggested on July 23 that a counterexample could be
> found by 2-step iteration of Cohen forcing. I replied that I wasn't
> sure it works with Cohen forcing. However, there is a way to do it:
>
> There exist x,y,z mutually generic Cohen reals such that M[x][y] =
> M[x][z] = M[y][z].
>
> The trick is to start with x generic over M, then z generic over M[x].
> We can use z to code an automorphism \phi of Cohen forcing (the
> binary tree). Namely the nth digit in all nodes is switched iff z(n)
> = 1. Let y be the image of x under \phi. One can show that y is
> generic over M[x]. The equalities above follow from the fact that any
> real in {x,y,z} is definable from the other two.
>
> Monroe
>
>
>
>
> On Fri, Jul 23, 2010 at 4:52 PM, Ali Enayat <ali.enayat at gmail.com> wrote:
> >
> > In a more recent posting (July 23), Eskew noted that the answer to (1)
> > in negative by choosing P as a collapsing poset.
> >
> > Alternatively, one can choose mutually generic Cohen reals r and s
> > over some M, and observe that if N = M(r,s), then N is a generic
> > extension of each of M(r) and M(s), via the same notion of forcing.
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