A note on counterexamples to the Vaught conjecture

Greg Hjorth

August 29, 2005

1 Introduction

In [2] it was shown that:

Theorem 1.1 There is a countable structure M with S_{∞} dividing Aut(M) and whose Scott sentence has models of size ℵ_{1} but no higher.

This note makes the following corollary of that theorem explicit.

Corollary 1.2 If there is some σ ∈ L_{ω_{1}ω} providing a counterexample to Vaught’s conjecture for L_{ω_{1}ω}, then there is some σ′ ∈ L_{ω_{1}ω} which again provides a counterexample and has no model of cardinality bigger than ℵ_{1}.

Leo Harrington has shown that any counterexample to Vaught’s conjecture will have, necessarily uncountable, models with arbitrarily high Scott ranks below ω_{2}.

2 Proof

For convenience, we work under ¬CH. A similar argument works for the version of the Vaught conjecture which asks for uncountably many non-isomorphic countable models without a perfect set of non-isomorphic countable models.

Definition S_{∞} is said to divide a Polish group G if there is a closed subgroup H < G and a continuous, onto, homomorphism

π : H ↠ S_{∞}.

Notation For L a countable language we let X_{L} be the space of all L-structures with underlying set ℕ and with the topology having basic open sets

{N : N ⊧ ψ(a⃗)}

for ψ quantifier free and a⃗ a finite sequence from ℕ.

It is well known and easily checked that X_{L} is a Polish space in the indicated topology. Compare §2.7 [1].

Let M be as in the statement of the theorem and let σ ∈ L_{ω_{1}ω} have exactly ℵ_{1} many non-isomorphic models. For convenience I will assume that M has ℕ as its underlying set. Fix H a closed subgroup of Aut(M) which maps continuously and homomorphically onto S_{∞}.

In particular we have a continuous action of S_{∞} on a Polish space X with exactly ℵ_{1} many orbits. This obviously can be lifted up to a continuous action of the Polish group H with the same orbit structure, and then following [1] 2.3.5 we may lift the action of H on X up to an action of Aut(M) on a Polish space Y ⊃ X such that, among other things, Y has the same number of orbits under Aut(M) as X does under H.

That is to say, Y/Aut(M) has size ℵ_{1}.

Then by 2.7.4 [1] we can find a richer countable language L′ ⊃ L such that the expansions of M to L′, Y_{L′}^{M}, form a universal Polish Aut(M) space. In particular there will be a injective and Borel Aut(M)-map

ρ : Y → Y_{L′}^{M}.

Since the injective images of Borel sets are Borel, there will in particular be an invariant Borel set

B ⊂ Y_{L′}^{M}

which contains exactly ℵ_{1} many orbits under the action of Aut(M).

Now let C be the set of all N ∈ X_{L′} such that there exists N′ ≅ N with N′ ∈ B. Note that C is then Δ ∼_{1}^{1}, and hence Borel, since, by invariance of B, we can equivalently describe C as the set of N ∈ X_{L′} which satisfy φ_{M}, the Scott sentence of M, and for which any N′ ∈ Y_{L′}^{M} isomorphic to N is in B.

Thus by [3], there is some σ′ ∈ L′_{ω_{1}ω} whose models with underlying set ℕ are exactly the elements of C. Since σ′ ⇒ φ_{M}, σ′ has no model of size ℵ_{2}. By construction, σ′ has exactly ℵ_{1} many non-isomorphic models.

References

[1] H. Becker and A. S. Kechris, The descriptive set theory of Polish group actions, London Mathematical Society Lecture Note Series, 232, Cambridge University Press, Cambridge, 1996.

[2] G. Hjorth, Knight’s model, its automorphism group, and characterizing the uncountable cardinals, Journal of Mathematical Logic, vol. 2 (2002), 113-144.

[3] E.K.G. Lopez-Escobar, An interpolation theorem for denumerably long formulas, Fundamenta Mathematicae, vol. 57 (1957), 253-272.

greg@math.ucla.edu

MSB 6363 UCLA CA 90095-1555