FOM: axioms of infinity

Stephen G Simpson simpson at
Thu Jul 20 17:16:58 EDT 2000

A refinement of Shavrukov's result:

   Given (an r.e. index of) an r.e. set S of sentences none of which
   has a finite model, we can effectively find an axiom of infinity
   which does not interpret any of the sentences in S.

Proof.  A well known refinement of Trakhtenbrot's Theorem says: A and
B are an effectively inseparable pair of r.e. sets, where A is the set
of sentences which have no models, and B is the set of sentences which
have finite models.  It follows that the complement of B is productive
in the sense of Post.  Shavrukov's argument then gives the above

The result above suggests that we ought to be able to explicitly write
down a great many different axioms of infinity, but off-hand I don't
see how.

Joe Shipman's axiom of infinity (non-commutative division rings) is

-- Steve

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