FOM: Boolean valued models

Carsten BUTZ butz at scylla.math.mcgill.ca
Wed Jan 19 17:26:06 EST 2000


Dear Joe,

you raised the question whether there are Boolean valued models of ZFC
such that CH has truth valued different from 0 and 1 (see below). The
answer is yes. It follows from the following (fairly well-known) fact:

  For every consistent first order theory T there exists a
  concervative Boolean valued model M_T.

Here conservative refers to the fact that a (closed) formula phi is
deducible from T if and only if it holds in that particular model M_T.

An immediate corollary is that every sentence phi _independent_ from T 
(both T+phi and T+ neg phi are consistent) has
in M_T a truth value different from 0 and 1. Apply this to ZFC and CH.

The reason for the above theorem is roughly that there are enough ordinary
models to test derivability (= Goedel's completeness theorem), and a set
of jointly conservative models can be "glued together" to one Boolean
valued model. For a simple proof of a much stronger result (you can find a
conservative Boolean valued model M_T such that every Boolean valued
subset of M_T which is invariant under all automorphisms of M_T is
definable by a first order formula) see the recent paper

  C. Butz and I. Moerdijk. 
  An elementary definability theorem for first-order logic. 
  J. Symbolic Logic, 64(3):1028-1036, 1999. 

  Best regards,

  Carsten Butz 
 
Joe Shipman wrote: 
> 
> Technical question here for logicians: in the Boolean-valued model
> approach to independence proofs, is there ever a way to "project" the
> algebra of "truth values" into the probability space [0,1] so that
> (some) statements with intermediate truth values can be assigned a
> probability strictly between 0 and 1in a consistent way?  Is there a
> Boolean-valued model in which the axioms of ZFC have value 1 but CH has
> a value strictly between 0 and 1?  (I already know there are BVM's in
> which it has value 0, showing it is independent, and BVM's in which it
> has value 1, showing it is consistent).
> 
> -- Joe Shipman
> 
> 





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