[FOM] inverse forcing

Monroe Eskew meskew at math.uci.edu
Thu Aug 26 14:35:47 EDT 2010

In fact, x+y = z, where z codes the automorphism mapping x to y, so
the solutions are identical.

On Wed, Aug 25, 2010 at 4:07 PM, Ashutosh <ashu1559 at gmail.com> wrote:
> 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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