A note on bounded hyperimaginaries

Martin Ziegler

June 13, 2000

The purpose of this note is to prove the following:

Theorem 1 Every bounded hyperimaginary is a class of a bounded type–definable equivalence relation.

Let ℂ be a big saturated model. An equivalence relation E between tuples of elements of ℂ is type–definable if it is defined by a set of formulas without parameters. The tuples may be infinite, but their length should be smaller than the size of ℂ. E is bounded if it has few classes compared with the size of ℂ. Bounded type–definable equivalence relations on a type–definable subclass of ℂ are defined similarly.

A class of a type–definable equivalence relation is a hyperimaginary. A hyperimaginary is bounded if it has few conjugates under automorphisms of ℂ. Of course all classes of a bounded type–definable equivalence relation are bounded hyperimaginaries. By the proposed theorem all bounded hyperimaginaries occur in this way.

Let KP denote the finest bounded type–definable equivalence relation^{1}. We will prove Theorem 1 in the following equivalent form:

Theorem 2 Let 𝔻 a type–definable subclass of ℂ. Then KP ∩ (𝔻 × 𝔻) is the finest bounded type–definable equivalence relation on 𝔻.

To see that this implies Theorem 1 consider a bounded hyperimaginary Ea. The union 𝔻 of all conjugates of Ea is type–definable and E ∩ (𝔻 × 𝔻) is a bounded type–definable equivalence relation on 𝔻. Since KP refines E on 𝔻 and since 𝔻 is a union of KP–classes

F = KP ∪ (E ∩ (𝔻 × 𝔻))

defines an equivalence relation, which is bounded, type–definable and satisfies Ea = F a.

1KP depends of course on the length of the tuples under consideration. To ease notation we will assume all tuples to be of length 1.

Definition A reflexive and symmetric relation R on a set C is thick if there is no infinite R–anti-chain, i.e. an infinite set A ⊂ C such that ¬R(a, b) for all different a, b ∈ A.

The following lemma is well–known and easy to prove.

Lemma 3 Let E(x, y) be the intersection of a set ℜ of definable reflexive and symmetric relations on ℂ and 𝔻 be the intersection of a set 𝔓 of definable subclasses of ℂ. We assume ℜ and 𝔓 to be closed under finite intersections.

• E(x, y) is an equivalence relation iff for each R ∈ ℜ there is an S ∈ ℜ such that S^{2} ⊂ R.

• Assume E(x, y) is an equivalence relation. Then E(x, y) is bounded on 𝔻 iff each R ∈ ℜ is thick on some element P of 𝔓.

To prove Theorem 2 we fix a type–definable class 𝔻 and a bounded type– definable equivalence relation on 𝔻. Extending by equality on ℂ \ 𝔻 we see that this relation has the form E ∩ (𝔻 × 𝔻) for a type–definable equivalence relation E.

We write E as the intersection of a set ℜ of definable reflexive and symmetric relations and 𝔻 as the intersection of definable classes from 𝔓. We can assume that both ℜ an 𝔓 are closed under finite intersections. Fix a relation R ∈ ℜ. We will find a P_{0} ∈ 𝔓 and a sequence S_{0}, S_{1}, … of definable symmetric and reflexive relations with the following properties:

(1) S_{n} ∩ (P_{0} × ℂ) ⊂ R

(2) S_{n+1}^{2} ⊂ S_{n}

(3) S_{n}^{2} is thick.

The intersection of the S_{n} is then a bounded type–definable equivalence relation which refines R on 𝔻. This shows KP ∩ (𝔻 × 𝔻) ⊂ R and proves Theorem 2.

Lemma There is a P ∈ 𝔓 such that for all x ∈ P there is a y ∈ 𝔻 such that R(x, y).

Proof: Assume not. Then every P ∈ 𝔓 contains an x which is not R–related to any element of 𝔻. By compactness there must be a subset P′ of P which belongs to 𝔓 and does not contain elements R–related to x. We start with a set P_{0} ∈ 𝔓 on which R is thick. Then we choose x_{0} ∈ P_{0} and a subset P_{1} without elements R–related to x_{0}. In P_{1} we choose x_{1} and P_{2} etc. The x_{i} form an infinite R–antichain, which is impossible.

By the last two lemmas we can find two sequences R_{0} ⊃ R_{1} ⊃ ⋯ and P_{0} ⊃ P_{1} ⊃ ⋯ of elements of ℜ and 𝔓 with the following properties:

(a) R_{0}^{3} ⊂ R (b) R_{i+1}^{4} ⊂ R_{i}

(c) R_{i} is thick on P_{i}

(d) ℂ ⊧ ∀x ∈ P_{i} ∃y ∈ P_{i+1} R_{i+1}(x, y)

Let x be an element of ℂ. We define recursively

X_{0}(x) = {z ∈ P_{0} | R_{0}(x, z)}
X_{i+1}(x) = {z ∈ P_{i+1} | R_{i+1}(y, z) for some y ∈ X_{i}(x)}

and the relations S_{n} as

S_{n}(x, x′) ⟺ X_{n}(x) ∼_{R_{n}} X_{n}(x′),

where ∼_{R_{n}} is defined as follows:

Definition Let S be a reflexive and symmetric relation on a set C. For subsets of X, X′ of C define

X ∼_{S} X′ ⟺ ∀x ∈ X ∃x′ ∈ X′ S(x, x′) ∧ ∀x′ ∈ X′ ∃x ∈ X S(x, x′).

∼_{S} is a reflexive and symmetric relation on the power set of C.

Note that property (d) implies

X_{i}(x) ∼_{R_{i+1}} X_{i+1}(x).

Lemma 4 Let S and S′ be reflexive and symmetric relations on C.

• ∼_{SS′} is the composition of ∼_{S} and ∼_{S′}.

• If S is thick on C the relation ∼_{S^{2}} is thick on the power set of C.

Proof: The first claim is easy to see. Actually we will use only the trivial inclusion ∼_{S}∼_{S′} ⊂ ∼_{SS′}. For the second claim assume that S is thick. Choose a maximal S–anti-chain A in C. Then every element of C is S–related to an element of A. It follows that for every subset X of C

X ∼_{S} X_{A},

where X_{A} = {a ∈ A | S(x, a) for some x ∈ X}. This implies that an ∼_{S^{2}}–antichain cannot have more elements than there are subsets of A.

Finally we show that the S_{n} have the properties (1)–(3).

(1) Assume S_{n}(x, x′). Let T denote the composition R_{1}R_{2} ⋯ R_{n}. Then

X_{0}(x) ∼_{T} X_{n}(x) ∼_{R_{n}} X_{n}(x′) ∼_{T} X_{0}(x′),

which implies X_{0}(x) ∼_{T R_{n}T} X_{0}(x′). By property (b) we have T R_{n} ⊂ R_{0} (we use only R_{i+1}^{2} ⊂ R_{i}), whence X_{0}(x) ∼_{R_{0}^{2}} X_{0}(x′). Now assume that in addition x ∈ P_{0}. Then x ∈ X_{0}(x) and we find an y ∈ X_{0}(x′) such that R_{0}^{2}(x, y). Since R_{0}(x′, y) we can conclude that R_{0}^{3}(x, x′). Whence R(x, x′) by (a).

(2) If S_{n+1}^{2}(x, x′′) there is an element x′ such that S_{n+1}(x, x′) and S_{n+1}(x′, x′′). We have

X_{n}(x) ∼_{R_{n+1}} X_{n+1}(x) ∼_{R_{n+1}} X_{n+1}(x′) ∼_{R_{n+1}} X_{n+1}(x′′) ∼_{R_{n+1}} X_{n}(x′′).

This implies X_{n}(x) ∼_{R_{n}} X_{n}(x′′) by property (b).

(3) By property (c) and the last lemma the relation ∼_{R_{n}^{2}} is thick on the power set of P_{n}. This implies immediately the thickness of S_{n}.

Note added in June 2000

The theorem appeared already in Theorem 4.18 in Hyperimaginairies and automorphism groups by D.Lascar and A.Pillay. The proof given there is, if one circumvents the surrounding theory, the following: Let A be the set of all bounded hyperimaginaries a/E. Define the equivalence relation LP by

LP ⟺ tp(x/A) = tp(y/A).

LP is bounded and type–definable. Whence it is refined by KP. Now let E be a type–definable equivalence relation on 𝔻. The classes of E are bounded hyperimaginairies. Therefore elements of 𝔻 which are LP–equivalent are also E equivalent. It follows that LP and therefore KP refines E on 𝔻. In fact LP = KP.

That LP is type–definable is a basic observation: Let a/E be a hyperimaginary. Then x and y have the same type over a/E iff

∃a′ E(a, a′) ∧ tp(x, a) = tp(y, a′).

Mathematisches Institut Freiburg, Germany