[FOM] Questions on axiomatizations of complete theories (fwd)

John Baldwin jbaldwin at uic.edu
Tue Mar 7 09:54:02 EST 2006


Joe Shipman defined `loose axiomatizations'.

Say that A = {A1, A2, ...} is "loose" if any infinite subset of A
> implies all the axioms in A.


Discussion with brought out that `trivial' examples of loose 
axiomatizations are give by `axioms of infinitity'.

Consider the stucture - an infinite direct sum of copies of Z_2.

It's theory is axiomatized by  group such that every element has order 2

and there exist infinitely many elements.

So the axiomization is loose/


This then  works for every totally categorical or even every
aleph_0 categorical omega stable theory since they are aximatized by
a single axiom plus schema of infinitity (slightly more technical in the
omega stable  case).

Joe Shipman provided two examples of tight versus trivial avoiding mere
axioms of infinity.

An example of a "tight" axiomatization is my new axiomatization for
Algebraically closed fields of characteristic 0 (needs conjunction of
field  axioms
plus 1+1+...+1 (p times) not = 0 for each prime p plus all pth-degree
polynomials have roots for each prime p).

A nontrivial example of a "loose" axiomatization that is not simply
infinity
in disguise is to axiomatize real closed fields by having an order
relation,
all positive elements have square roots, and all odd degree  polynomials
have
roots -- the order axioms ensure characteristic 0 and that -1  is not a
square, so there is an irreducible polynomial x^2+1 of degree 2, and
any infinite
subset of "all odd degree polynomials have roots" is enough.
(Technically,
you have to append the finitely many other axioms to each of the  odd
degree
axioms to get a loose axiomatization.)  But I can't prove that  EVERY
axiomatization of RCF is essentially equivalent to a loose one (that is,
has a finite
subset of axioms A1,...,An and an infinite subset of axioms B1, B2,
..., such
that {A1&...&An&B1, A1&...&An&B2, ... } is  loose).





On Sat, 4 Mar 2006 joeshipman at aol.com wrote:

> In the following, T always refers to a complete theory which is
> recursively axiomatizable (hence decidable) but not finitely
> axiomatizable.
> 
> Say that an axiomatization A = {A1, A2, ...} of T is "tight" if every
> axiom in it is logically independent of the rest.
> 
> Say that A = {A1, A2, ...} is "loose" if any infinite subset of A
> implies all the axioms in A.
> 
> These are two extremes. Every axiomatizable theory has a "loose"
> axiomatization: convert {A1, A2, A3, ... } to {A1, A1&A2, A1&A2&A3,
> ...}.
> 
> Can anyone find nontrivial examples of T (preferably theories that are
> already considered interesting) such that
> 
> 1) No recursive axiomatization of T is tight?
> 
> 2) Every recursive axiomatization of T contains a recursive "loose"
> subset?
> 
> -- JS
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> FOM at cs.nyu.edu
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>

John T. Baldwin
Director, Office of Mathematics Education
Department of Mathematics, Statistics, and Computer Science  M/C 249
jbaldwin at uic.edu
312-413-2149
Room 327 Science and Engineering Offices (SEO)
851 S. Morgan
Chicago, IL 60607

Assistant to the director
Jan Nekola: 312-413-3750


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