[FOM] inverse forcing
meskew at math.uci.edu
Sat Jul 24 13:36:24 EDT 2010
On Sat, Jul 24, 2010 at 7:52 AM, Ali Enayat <ali.enayat at gmail.com> wrote:
> 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.
This is the first thing I tried, but I don't think it works. I want
distinct M, M* such that with the same G, M[G]=M*[G]. If we take r, s
mutually generic Cohen reals then of course M[r][s]=M[s][r]=M[t],
where t is the image of the isomorphism of P with PxP. We're not
using the same G to get to the final model from the intermediate ones,
but perhaps we can do it with t. However, t is actually not generic
over either M[r] or M[s]. Even though it seems to contain "more
genercity" than either, you can use r to construct a dense set in M[r]
which does not meet t.
I should note that the idea for the example I gave came from Amit
Gupta of UC Berkeley.
> However, the answer to (2) - assuming that one is allowed to use
> parameters in the defining formula - is positive and nontrivial. This
> result is independently due to Richard Laver and Hugh Woodin; for a
> proof see Theorem 8 of the following paper:
> Reitz, Jonas , "The ground axiom", Journal of Symbolic Logic 72 (4):
> 1299–1317 (2007).
> Let me close by noting that it is easy to see that the above use of
> parameters is necessary, e.g., in the example above, M(r) is not first
> order definable by a parameter free formula in M(r,s).
Thanks for the reference. That's an interesting result: There is not
necessarily a unique such M, but every possible M is definable from
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