10
Characterisation of ℵ0 -categorical theories
Recall that the complete theory T is ℵ0 -categorical if there is, up to isomorphism, only one countably infinite model of T . The following theorem characterises these theories.
Theorem 10.1. (Ryll-Nardzewski) Suppose that T is a complete theory, in a
countable language L, without a finite model. Then the following are equivalent.
(i) T is ℵ0 -categorical.
(ii) For every integer n, the set SnT (∅) of n-types is finite.
(iii) For every integer n, every type in SnT (∅) is principal.
(iv) For every tuple x of variables, there are only finitely many formulas with
free variables x up to T -equivalence (where we say that φ(x) is T -equivalent to
ψ(x) if φ(M) = ψ(M) for every, equivalently any, model M of T ).
(v) For every countable model M of T and every integer n, there are only finitely
many orbits of Aut(M) on n-tuples of elements of M .
Proof. (i)⇒(iii) If there were some non-principal type then, by [5.11], there
would be a countable model which realises that type and, by the Omitting Types
Theorem [5.15], a countable model which omits that type. These two models
cannot be isomorphic, by [9.14] (or [3.13]).
(iii)⇒(ii) For each p ∈ SnT (∅) choose (assuming (iii)) some formula φp which
generates that type. Then consider the set of formulas Φ(x) = {¬φp (x) : p ∈
SnT (∅)}. If SnT (∅) is infinite then it follows easily that Φ(x) is finitely satisfiable
in M, where M is any chosen model of T . Therefore, by [5.12], Φ(x) extends
to some member, say q, of SnT (∅). But then q would contain both φq and, since
it contains Φ, ¬φq - contradiction, as required.
(ii)⇒(iv) Fix n and let p1 , . . . , pk be the types in the finite set SnT (∅). Let
ψ(x) be a formula in n free variables x. Define Sψ = {i : ψ ∈ pi }. It is easy to
show that for formulas ψ and ψ 0 , ψ(x) and ψ 0 (x) are T -equivalent iff Sψ = Sψ0 .
Since there are only finitely many possibilities for Sψ there are, therefore, only
finitely many formulas in n free variables up to T -equivalence.
(iv)⇒(iii) Let ψ1 , . . . , ψk be representatives ofVthe finitely many T -equivalence
classes of formulas. Given p ∈ SnT (∅) set φp = {ψi : ψi ∈ p}. So φp ∈ p and
any formula in p is T -equivalent to one of the ψi and hence is a consequence
(modulo T ) of φp . Thus φp is a generator of p and p is principal, as required.
(iii)⇒(i) This is the longest part of the proof; it is a back-and-forth argument where we use the fact that types are principal. Here’s the start of
the argument. Take countable models, M and N of T ; we have to show that
they are isomorphic. So list the elements of M as a1 , a2 , . . . and similarly list
the elements of N . Consider a1 ; we must find an element b of N such that
tpN (b) = tpM (a1 ). By assumption, there is a formula φ(x) which generates
tpM (a1 ). Certainly M |= ∃x φ(x) so, since M ≡ N , N |= ∃x φ(x), say b ∈ N
is such that N |= φ(b). By [9.17], tpN (b) = tpM (a1 ), as required. We set
α1 (a1 ) = b.
For the induction step, in the “forth” direction, suppose that we have defined
the partial map αn , sending a1 , . . . , ak in M to b01 , . . . , b0k in N and such that
tpM (a1 , . . . , ak ) = tpN (b01 , . . . , b0k ). Where do we send ak+1 ? We need to find
an element b0k+1 such that tpM (a1 , . . . , ak , ak+1 ) = tpN (b01 , . . . , b0k , b0k+1 ). Let
ψ(x1 , . . . , xn , y) be a generator for tpM (a1 , . . . , ak , ak+1 ). Then ∃y ψ(x, y) ∈
tpM (a1 , . . . , ak ) which, by the inductive assumption, equals tpN (b01 , . . . , b0k ).
1
So N |= ∃y ψ(b01 , . . . , b0k , y), so N |= ∃y ψ(b01 , . . . , b0k , b) for some b; set b0k+1 = b.
Then, by construction and [9.17], tpN (b01 , . . . , b0k , b0k+1 ) = tpM (a1 , . . . , ak , ak+1 ),
as required.
As usual, half the time we use the “forth” argument and the other half, use
the “back” argument (in other words, the “forth” argument using αn−1 ) and
build up an isomorphism between M and N .
(v)⇒(ii) If there are infinitely many types then choose countably many of
them - p1 , p2 , . . . , pi , . . . . By [5.10] used repeatedly, there is some model N of T
which realises each of these, say tpN (ai ) = pi . By the downwards LöwenheimSkolem Theorem, there is a countable model of T which contains all (the entries
of) the ai and hence which realises all of the infinitely many types pi . But, by
[9.14], these must lie in different orbits of Aut(M).
(ii),(iii)⇒(v) We have to show that if a, b are n-tuples in M with the same
type then there is α ∈ Aut(M) with αa = b. To get such an α we just apply
the proof of part (iii)⇒(i) but, instead of starting at the empty function, we
begin with the partial automorphism which takes a to b and extend that by a
back-and-forth argument. 2