Reverse mathematics and theories with finitely
many models
David Belanger
9 May 2013
at the ASL North American Annual Meeting
EMAIL: [email protected]
WEB: http://www.math.cornell.edu/∼dbelange
Department of Mathematics
Cornell University
Reverse mathematics
We consider three of the ‘Big Five’ subsystems of second-order
arithmetic:
ACA0
⇓
WKL0
⇓
RCA0
Arithmetic Comprehension Axiom
Weak König0 s Lemma
Recursive Comprehension Axiom
and some of their negations.
‘Basic model theory’
By basic model theory we mean things like:
• The completeness theorem
• The compactness theorem
• Type-omitting
• Saturated models, atomic models, homogeneous models,
ℵ0 -categorical models, . . .
• etc.
Model theory in reverse math
We work in the language of second-order arithmetic, typically
within a model (M, S) of RCA0 , where M is the first-order part,
and S is the second-order part.
Model-theoretic definitions in this language:
• A theory is a set T ∈ S of first-order sentences over some
language L ∈ S.
• A model A ∈ S of T is an elementary diagram containing T .
• An n-type p ∈ S of T is a maximal set of n-ary formulas
consistent with T .
• Two models A, B ∈ S are isomorphic if there is an f ∈ S
which is an isomorphism between them.
Model theory in reverse math
Simpson in Subsystems of second-order arithmetic develops some
of first-order logic in this setting, including:
Theorem (Completeness; RCA0 )
Every complete consistent theory has a model.
Theorem (Completeness; WKL0 )
Every consistent theory has a model.
Theorem (Compactness; WKL0 )
Every finitely satisfiable theory has a model.
Proofs can be similar to the classical versions, or to those in
effective mathematics:
Theorem (Effective Completeness Theorem)
Every complete consistent decidable theory has a decidable model.
Some well-known theorems
Theorem (Ehrenfeucht)
For each n = 1 and n ≥ 3, there is a complete theory having
exactly n countable models.
Theorem (Vaught)
There is no complete theory having exactly two countable models.
Theorem (Millar; Kudaibergenov)
For each n ≥ 1, there is a complete decidable theory having exactly
n decidable models, up to recursive isomorphism.
The proofs of Ehrenfeucht’s and Vaught’s can be carried out in
ACA0 ; the proof of Millar’s can be carried out in RCA0 + ¬WKL0 .
Some well-known theorems, translated
Theorem (ACA0 )
For each n = 1 and n ≥ 3, there is a complete theory having
exactly n models.
Theorem (ACA0 )
There is no complete theory having exactly two models.
Theorem (RCA0 + ¬WKL0 )
For each n ≥ 1, there is a complete theory having exactly n models.
So what about the intermediate case: WKL0 + ¬ACA0 ?
Yes, what about the case of WKL0 + ¬ACA0 ?
The obvious approach: Alter the proofs of Ehrenfeucht’s,
Vaught’s, or Millar’s to work in WKL0 + ¬ACA0 .
Unfortunately:
• These proofs each make crucial use of a nonprincipal n-type.
• In WKL0 + ¬ACA0 , any theory with a nonprincipal type has
infinitely many models up to isomorphism.
Theorem
Over WKL0 , the following are equivalent:
(i) ACA0
(ii) ‘There is a complete consistent theory T with a nonprincipal
type and with only finitely many models.’
Extended version and proof
Theorem
Over WKL0 , the following are equivalent:
(i) ACA0
(ii) ‘There is a complete consistent theory T with a nonprincipal
type and with only finitely many models.’
(iii) ‘There is a complete consistent theory T with infinitely many
n-types for some n and with only finitely many models.’
(iv) ‘There is a complete consistent theory T with infinitely many
n-types for some n and with an elementary-universal model.’
• The direction (i ⇒ iv) holds because ACA0 is strong enough
to carry out the classical proof of (iv).
• Prove the direction (iv ⇒ i) by means of its contrapositive
(¬i ⇒ ¬iv ).
Proof idea
Suppose that:
• (M, S) is a model of WKL0 + ¬ACA0 ;
• T ∈ S is a complete consistent theory with infinitely many
n-types; and
• A ∈ S is a model of T .
Wish to construct:
• second model B ∈ S of T such that no f ∈ S is an
elementary embedding of B into A.
Achieve this by:
• fixing a set Z ∈ S whose jump Z 0 is not a member of S; and
• constructing B so that, if f is an elementary embedding of B
into A, then f computes Z 0 .
Proof idea
• Suppose e enters Z 0 at stage s. Look at the first s elements
of A:
a0 , a1 , . . . , as−1 .
• There are only finitely many ways to choose an n-tuple from
among these. Since T has infinitely many n-types, there is an
n-type not realized here.
• So there is an n-ary formula φ consistent with T and not
satisfied here.
• Make φ satisfied in B by an n-tuple in the short initial part:
b0 , . . . , b2en−1 .
• Then any embedding from B to A maps some bi with i < 2en
to an aj with j > s.
• Such an embedding allows us to compute Z 0 by way of its
modulus function.
Constructing B
Build an infinite binary tree of Henkin-style constructions whose
paths each encode a suitable B. Then Weak König’s Lemma
finishes the proof. The steps:
• In a Henkin construction, we sometimes can choose between
adding φ or ¬φ to the diagram. Our choices give rise to a
binary tree H.
• We specify a subtree H∗ of H by removing, at each stage s,
those constructions that don’t satisfy the formulas we wanted
on the tuples we wanted.
• Verify that H∗ is infinite, and apply WKL.
Strengthened version
By dovetailing the above construction, we get:
Theorem
Over WKL0 , the following are equivalent:
(i) ¬ACA0
(ii) ‘If T has infinitely many n-types, then T has an infinite
sequence hA0 , A1 , . . .i ∈ S of pairwise-nonisomorphic
models.’
The Henkin trick gives a variety of other results over WKL0 ,
including some about (strongly) homogeneous models, saturated
models, elementary chains, and so on.
Another well-known theorem
Theorem (Ryll-Nardzewski; Engeler; Svenonius)
Let T be a complete consistent countable theory. The following
are equivalent:
# 1 T has only one countable model up to isomorphism.
# 2 T has only finitely many n-types for each n.
# 3 All types of T are principal.
Theorem
WKL0 ` (#1 ⇒ #2)
Proof.
Nonuniform.
• If ACA0 holds, the usual proof of #1 ⇒ #2 works.
• If WKL0 + ¬ACA0 holds, instead apply the theorem: ‘If T
has infinitely many n-types, then T has infinitely many
models’.
The Ryll-Nardzewski Theorem and the requisite table
# 1 T has only one model.
# 2 T has only finitely many n-types for each n.
# 3 All types of T are principal.
# 4 There is a function f : M → M such that, for all n ∈ M, T
has exactly f (n) distinct n-types.
# 5 There is a function f : M → M such that, for all n ∈ M, T
has no more than f (n) distinct n-types.
Strength of (#row ⇒ #column) over RCA0 :
(1)
(2)
(3)
(4)
(5)
(1)
WKL0 RCA0
??
??
(2) ACA0
RCA0 ACA0 ACA0
(3) ACA0 ACA0
ACA0 ACA0
(4) RCA0 RCA0 RCA0
RCA0
(5) ACA0 RCA0 RCA0 ACA0
Some directions
independently due
to U. Andrews
and A. Kach
THE END