RELATIVIZATION, CATEGORICITY, AND DIMENSION

A Dissertation

Submitted to the Graduate School

of the University of Notre Dame

in Partial Fulfillment of the Requirements

for the Degree of

Doctor of Philosophy

by

Charles Frederick Dymphna McCoy, B.S., M.S.

Peter Cholak, Director

Julia Knight, Director

Department of Mathematics

Notre Dame, Indiana

April 2000

RELATIVIZATION, CATEGORICITY, AND DIMENSION

Abstract

by

Charles Frederick Dymphna McCoy

We examine questions in computable model theory by studying relativized coun-

terparts. Specifically, our study centers on computable and higher-level categoricities

and computable dimension.

A computable structure is ∆0α-categorical if every computable copy is isomorphic

by a ∆0α isomorphism. Goncharov used a priority construction to give a syntac-

tic characterization of ∆01-categoricity, under added effectiveness hypotheses. Ash,

Knight, Manasse, and Slaman, and independently Chisholm, showed that the syntac-

tic property actually is equivalent to the stronger notion of relativized ∆01-categoricity.

Their proofs use forcing constructions.

A computable structure’s computable dimension is the number of equivalence

classes of computable copies under the relation of being computably isomorphic.

Goncharov produced, for each finite n, a structure with computable dimension n;

however, no “natural” characterization of dimension n had been discovered in any

class of structures.

Could studying relative dimension yield a characterization, perhaps a syntactic

one like that for ∆01-categoricity? We show that, in fact, finite computable dimension

does not relativize: any structure with relative finite computable dimension is rela-

tively ∆01-categorical. Our argument uses a forcing construction related to the one

by Ash et al.

Charles Frederick Dymphna McCoy

Goncharov, Dzgoev, Remmel, and LaRoche characterized the ∆01-categorical lin-

ear orderings and Boolean algebras. We examine ∆02-categoricity in these structures.

Unlike those who studied ∆01-categoricity, we begin with the relativized notion; us-

ing the result by Ash et al, we give noneffective syntactic arguments to provide a

complete description. We then classify, under added effectiveness hypotheses, the

∆02-categorical Boolean algebras and linear orderings. Beginning with the relativized

results proves useful in two ways. First, since relativized and unrelativized cate-

goricities often coincide, the relativized results provide a probable statement of the

unrelativized results, modulo extra hypotheses. Second, the proofs of the unrela-

tivized results give a basic organization for the priority arguments and an idea of

what extra hypotheses will make these arguments valid.

Next, we study ∆03-categoricity. We provide insight into why a classification for

the ∆03-categorical linear orderings might be impossible, and we classify the relatively

∆03-categorical Boolean algebras.

To my mom and my dad,

who challenge me and help me

to be more than I was “cut out” to be.

ii

CONTENTS

ACKNOWLEDGMENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v

CHAPTER 1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1 The basics of categoricity and dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 A brief description of results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

CHAPTER 2 COMPUTABLE DIMENSION n DOES NOT RELATIVIZE . . . 6

2.1 The Transition Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.2 Forcing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.3 Relative dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

CHAPTER 3 RELATIVIZED ∆02-CATEGORICITY IN BOOLEAN ALGEBRAS

AND LINEAR ORDERINGS. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3.2 The relativized results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

CHAPTER 4 UNRELATIVIZED ∆02-CATEGORICITY . . . . . . . . . . . . . . . . . . . . . . 35

4.1 ∆02-categoricity in Boolean algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

4.2 ∆02-categoricity in linear orderings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

4.3 Some open questions about ∆02-categoricity . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

CHAPTER 5 PARTIAL RESULTS ∆03-CATEGORICITY . . . . . . . . . . . . . . . . . . . . 78

5.1 Linear orderings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

iii

5.2 Boolean algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

BIBLIOGRAPHY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

iv

ACKNOWLEDGMENTS

Many people have provided me with support and encouragement without which

this task would have been literally impossible. I certainly cannot provide proper

acknowledgment to all of these people. To those I do not explicitly mention below,

I sincerely apologize; I just didn’t think it seemed right to have this section longer

than the math.

First, I give my thanks to my advisors, Peter Cholak and Julia Knight. Their

dedicated mentoring gave me knowledge, and their constant encouragement gave me

confidence. I hope that they never interpreted my reluctance to call them by their

first names as a lack of affection, for they are both very dear to me. I would be remiss

if I did not also thank Zabe, David, and Nathan Cholak and Bill Knight for sharing

Peter and Julia’s time with me throughout these five years and for showing me such

kindness when I see them.

I cannot give enough thanks to my mom, my dad, Liz, and Mad. They touch

everything I do. I especially thank my dad, who supported my decision to go to grad

school more than I myself did.

Eric and Mary Jo Stromberg have become my second family. Roxi Smarandache,

Charlie Miller, Andy Arana, Rebekah Arana, and Michael Bergan also made my years

at Notre Dame much richer and happier.

Sergey Goncharov, Valentina Harizanov, and Michael Moses provided important

insights into problems I was studying and generously shared their problems with me.

My defense committee, Steven Buechler, Sergey Goncharov, and Valentina Harizanov,

suggested great improvements to my thesis.

v

The staff in our department helped me with many projects related to my teaching

and my research. They always handled my last-minute requests with great cheer and

skill.

The Arthur J. Schmitt Foundation provided tremendous financial support through-

out my graduate years.

Finally, I thank Joshua, Mary, Joe, Therese, D., Al, and Ed for their eternal aid.

vi

CHAPTER 1

INTRODUCTION

1.1 The basics of categoricity and dimension

We define the notions of categoricity and dimension relevant to our study, and we

provide earlier results which motivate our questions or are used in our proofs.

Definition 1.1 A computable structure A is computably categorical if for any com-

putable copy B ∼= A there is a ∆01 isomorphism ϕe : B ∼= A.

The relativized notion is defined as follows.

Definition 1.2 A computable structure A is relatively computably categorical if for

any copy B ∼= A there is a ∆01(D(B)) isomorphism ϕD(B)

e : B ∼= A.

In Chapter 5, we will generalize this definition for structures A which are not com-

putable and may not even have a computable copy.

We give some examples of computably categorical structures.

1. A linear order is computably categorical iff it has finitely many pairs of direct

successors ([15], [24]).

2. A Boolean algebra is computably categorical iff it has finitely many atoms ([15],

[25], [23]).

1

3. An Abelian p-group is computably categorical iff it can be written in one of the

following two forms: (Z(p∞))α ⊕G1, where α ∈ N or α =∞, and G1 is finite;

or (Z(p∞))l⊕G1⊕ (Zpk)∞, where k, l ∈ N and G1 is finite ([13]). (The notation

is that of [16].)

Note that relative computable categoricity implies computable categoricity; we

will discuss below why the converse is not true. In the above examples, however, the

computably categorical structures are in fact all relatively computably categorical.

Definition 1.3 For a computable ordinal α, a computable structure A is ∆0α-categorical

if for any computable copy B ∼= A there is a ∆0α isomorphism ϕ

∆0α

e : B ∼= A.

The relativized notion is defined as follows.

Definition 1.4 A computable structure A is relatively ∆0α-categorical if for any copy

B ∼= A there is a ∆0α(D(B)) isomorphism ϕ

∆0α(D(B))

e : B ∼= A.

Goncharov generalized the notion of computable categoricity by defining com-

putable dimension.

Definition 1.5 A computable structure A has computable dimension m < ω if

1. there exist m computable copies B1, . . . ,Bm ∼= A with no two computably iso-

morphic; and

2. for any m+1 computable copies B1, . . . ,Bm+1∼= A, there are at least two which

are computably isomorphic.

(Therefore, a computable structure has dimension 1 iff it is computably categorical.)

Definition 1.6 A computable structure A has computable dimension ω if there ex-

ists a sequence of computable copies (Bi)i<ω so that each Bi ∼= A, but no two are

computably isomorphic.

2

In [14] Goncharov proved that for any finite m ≥ 2 there is a computable structure of

computable dimension m. In [15], Goncharov and Dzgoev defined a condition called

branching which guarantees that a structure has computable dimension ω.

Goncharov and Ventsov defined the relativized notion of finite computable dimen-

sion.

Definition 1.7 A computable structure A has relative computable dimension m < ω

if

1. there exist m copies (with computable universes) B1, . . . ,Bm ∼= A with no two

∆01((D(B1)⊕ . . .⊕D(Bm)) isomorphic; and

2. for any m+ 1 copies (with computable universes) B1, . . . ,Bm+1∼= A, there are

at least two which are ∆01((D(B1)⊕ . . .⊕D(Bm+1)) isomorphic.

Goncharov proved a syntactic characterization of computable categoricity that

provides the starting point for many of the results in this thesis. Before we can state

this result precisely, we need to review some material on infinitary formulas.

Definition 1.8 Fix a computable language L. We define infinitary and computable

infinitary formulas inductively for all computable ordinals α.

1. A Σ0 or Π0 formula is a finitary, quantifier-free (q.f.) formula. We can assign

these formulas Godel numbers in the standard way.

2. An infinitary q.f. formula is constructed from Σ0 formulas using only disjunc-

tion and conjunction.

3. A Σα infinitary formula is a countable disjunction of the form ψ(~x) =∨∨i∈I ∃~uiθi(~x, ~ui), where each θi is a Πβ formula for some β < α. A Πα formula

is defined analogously, with universal quantifiers replacing existential quanti-

fiers, and conjunctions replacing disjunctions.

3

4. A computable Σα infinitary formula, or Σcα formula, is a Σα formula involving

disjunctions over only computably enumerable sets of formulas. That is, a Σcα

formula ψ(~x) is of the form ψ(~x) =∨∨

i∈I ∃~uiθi(~x, ~ui), where each θi is a Πcβ

formula for some β < α, and the set of Godel numbers for the θi’s is c.e. (We

assume, by induction, that we can assign Godel numbers to the computable

formulas of lower complexity.)

A more complete treatment of computable infinitary formulas is found in [1].

Definition 1.9 Let A be an L-structure for some computable language L. A for-

mally Σ0α Scott family for A is an ordered pair 〈~c,Θ〉 where ~c is a tuple of distinct

parameters; and Θ is a c.e. set of Σcα formulas of L with parameters among ~c so that

1. for each tuple ~a ∈ A, there is θ(~x) ∈ Θ with A |= θ(~a); and

2. for any θ in Θ and ~a,~a ′ ∈ A, if A |= θ(~a) and A |= θ(~a ′), then 〈A,~a,~c〉 ∼=

〈A,~a ′,~c〉.

For structures satisfying a stronger decidability condition, Goncharov provided a

syntactic characterization of computable categoricity in [11].

Definition 1.10 A structure is 1-decidable if its ∃-diagram is computable.

Definition 1.11 A structure is 2-decidable if its ∀∃-diagram is computable.

Theorem 1.12 (Goncharov) Let A be 2-decidable. Then A is computably categorical

iff it has a formally Σ01 Scott family.

Goncharov and others wondered, however, what relationship existed between the

existence of a formally Σ01 Scott family and computable categoricity under relaxed

hypotheses. Was the above connection merely a by-product of the strong decidability

4

conditions on A? Ash, Knight, Manasse, and Slaman in [4], and independently,

Chisholm in [6] proved that there is indeed a strong connection between computable

categoricity and the existence of a formally Σ01 Scott family.

Theorem 1.13 (Ash et al, Chisholm) A computable structure A is relatively ∆0α-

categorical iff it has a formally Σ0α Scott family.

In [12] Goncharov himself produced a structure that is computably categorical

but has no formally Σ01 Scott family, thus demonstrating that the relatively com-

putably categorical structures constitute a proper subset of the computably categor-

ical structures. In fact, Kudinov, using a characterization of computable categoricity

in 1-decidable structures which he proved in [20], produced a 1-decidable structure

that is computably categorical but has no no formally Σ01 Scott family. Consequently,

the extra decidability condition in Theorem 1.12 is as weak as possible.

1.2 A brief description of results

In Chapter 2 we prove that finite computable dimension does not relativize. In

Chapter 3 we use Theorem 1.13 to characterize relativized ∆02-categoricity in Boolean

algebras and linear orderings. In Chapter 4 we prove the analogous unrelativized

results. Finally, in Chapter 5 we discuss ∆03-categoricity in these two classes of

structures.

5

CHAPTER 2

COMPUTABLE DIMENSION n DOES NOT RELATIVIZE

Goncharov and Ventsov questioned whether computable dimension actually rela-

tivized to anything meaningful; if so, perhaps a syntactic characterization could be

established to produce more “natural” examples of structures of dimension n > 1.

During a talk in the summer of 1997, Knight stated the result that no computable

structure has a finite relative computable dimension greater than 1. The main pur-

pose of this chapter is to provide a proof of this result. Theorem 1.13 is critical to

our proof: we demonstrate that any structure of finite relative computable dimension

has a formally Σ01 Scott family and thus has relative computable dimension 1.

The outline of this chapter closely follows that of [4]. In the first section we

define products of structures and prove the definability of ∆01((D(B1)⊕ . . .⊕D(Bn))

functions. In the next section we develop forcing machinery to define a finite number

of generic copies of a structure. Finally, in the last section we establish our main

result by using a forcing construction to produce a formally Σ01 Scott family for a

structure assumed to have finite relative computable dimension.

2.1 The Transition Lemma

To simplify certain definitions and arguments, we include the logical constants T

(truth) and F (falsity) in addition to the usual logical symbols. For the rest of this

chapter we fix a computable language L with only relation symbols and a computable

6

L-structureAwith universe A = N. IfX ⊂ N is computable, we write L(X) to denote

the augmented language obtained by adding a constant symbol for each element of

X.

Definition 2.1 Let B1, . . . ,Bn be L-structures with respective pairwise disjoint uni-

verses B1, . . . , Bn. Let L′ be the new language obtained from L by adding new unary

predicate symbols S1, . . . , Sn. Then the cardinal sum B1⊕· · ·⊕Bn is an L′-structure

defined as follows (recall that L has only relation symbols):

1. the universe of B1 ⊕ · · · ⊕ Bn is⋃ni=1Bi;

2. for any relation symbol R in L, RB1⊕···⊕Bn =⋃ni=1R

Bi;

3. for i = 1, . . . , n, SB1⊕···⊕Bni = Bi.

The cardinal sum is one of a general kind of product structure considered by Feferman

and Vaught in [10].

Note: Throughout the rest of the chapter we consider only cardinal sums where the

universes B1, . . . , Bn are computable, infinite and disjoint; and⋃ni=1Bi = N.

For L-structures B1, . . . ,Bn, we notice the following about the satisfaction of

atomic sentences γ of L(N):

If all the parameters of γ come from a single Bi, then B1 ⊕ · · · ⊕ Bn |= γ iff Bi |= γ.

If γ has parameters from Bi and Bj , where i 6= j, then B1 ⊕ · · · ⊕ Bn 6|= γ.

We now give definitions and results analogous to some of those in the first two

sections of [4].

Notation 1) For any tuples ~u = (u1, . . . , um) and ~v = (v1, . . . , vm) and any function

f , we write f(~u) = ~v if f(ui) = vi for each i ∈ {1, . . . ,m}. (We think of the ui’s as

distinct.)

7

2) When we write D(B1 ⊕ · · · ⊕ Bn), we mean the atomic diagram of B1 ⊕ · · · ⊕ Bn

as an L-structure, not an L′-structure.

3) We denote the atomic sentence of L(N) with Godel number x by θx.

Lemma 2.2 (Transition Lemma): Let B1, . . . , Bn have the properties stated after

Definition 2.1, and let and m > 0. From an m-tuple ~u of distinct elements, an m-

tuple ~v, and e ∈ ω we may effectively obtain an index for a quantifier-free Σc1 sentence

ψe,~u,~v of L(N) such that for all structures B1, . . . ,Bn with respective universes Bi we

have ϕD(B1⊕···⊕Bn)e (~u) = ~v ↔ B1 ⊕ · · · ⊕ Bn |= ψe,~u,~v. Specifically, the sentence ψe,~u,~v

has the form∨∨σ∈S γσ,1 ∧ . . . ∧ γσ,n

where each γσ,i is a finitary conjunction of atomic sentences and negations of atomic

sentences from L(Bi).

Proof: We must recall two very basic but important facts about computations. First,

the program for the eth Turing machine with oracle does not depend on the oracle

we use in a particular computation. Second, any computation uses only a finite

initial segment of the oracle. Therefore, ϕD(B1⊕···⊕Bn)e (~u) = ~v iff there is a string

σ ∈ 2<ω with σ ⊂ D(B1 ⊕ · · · ⊕ Bn) and ϕσe (~u) = ~v. By the note above about the

satisfaction of atomic sentences in a product structure, we note that the property

“σ ⊂ D(B1⊕· · ·⊕Bn)” can be expressed by a conjunction of atomic sentences, where

each atomic sentence has parameters from a single Bi.

More precisely, we let σ be an element of 2<ω. For any B1, . . . ,Bn, σ is an initial

segment of D(B1 ⊕ · · · ⊕ Bn) iff for each x ∈ domσ we have the following:

1. if θx has parameters from Bi, Bj , i 6= j, then σ(x) = 0;

2. if θx has parameters from only one Bi, then σ(x) = 0 iff Bi |= ¬θx.

8

Define the c.e. set S = {σ : ϕσe (~u) = ~v and ∀x ∈ domσ[(σ(x) = 1) ⇒ (all the

parameters in θx come from a single Bi)]}.

For each i = 1, . . . , n and each σ ∈ S, let

Rσ,i0 = {x: x ∈ domσ[σ(x) = 0, and θx involves only parameters from Bi]}; and

Rσ,i1 = {x: x ∈ domσ[σ(x) = 1, and θx involves only parameters from Bi]}.

If either Rσ,i0 or Rσ,i

1 is nonempty, then we let γσ,i :=∧∧

x∈Rσ,i0¬θx ∧

∧∧x∈Rσ,i1

θx.

Otherwise, we let γσ,i := T.

Then for all B1, . . . ,Bn, ϕD(B1⊕···⊕Bn)e (~u) = ~v iff B1⊕· · ·⊕Bn |=

∨∨σ∈S γσ,1∧ . . .∧γσ,n.

QED

2.2 Forcing

Throughout the rest of this chapter, we fix n > 1 and sets B1, . . . , Bn with the

properties stated after Definition 2.1. Furthermore, we assume that any structure

called Bi has universe Bi.

Notation We add function symbols f1, . . . , fn to L and denote the augmented lan-

guage by L?.

Definition 2.3 For a q.f. sentence ψ of L(N) we define ψ′ of L?(N) inductively:

1. if ψ is atomic and contains parameters from more than one Bi, then ψ′ = F;

2. if ψ is atomic and contains parameters from only one Bi, then ψ′ results from

replacing the occurrence of each bi in Bi with fi(bi);

3. (¬ψ)′ = ¬(ψ′), (∨∨

ψi)′ =∨∨

(ψ′i), etc.

Then the following is immediate.

Lemma 2.4 Let B1, . . . ,Bn be L-structures with gi : Bi ∼= A. Extend each gi : Bi →

A to a total function hi : A → A. (We need hi to be total to interpret the function

9

symbol fi. The precise way of extending gi is irrelevant; for instance, we could just

map everything in the complement of Bi to 0.) Let hi interpret the function symbol

fi to obtain the L?-structure 〈A, h1, . . . , hn〉. Then for a q.f. sentence ψ of L(N),

B1 ⊕ · · · ⊕ Bn |= ψ iff 〈A, h1, . . . , hn〉 |= ψ′.

The forcing construction we describe below will give us bijections gi : Bi → A

(Proposition 2.11). This bijection gi immediately induces an L-structure Bi ∼=gi A.

Lemma 2.4 guarantees that L? is a language in which we can talk about these induced

structures.

Note: When we actually use the forcing concepts we now develop, the sentences of

L?(N) we consider will be q.f. (finitary or infinitary) and have terms of the form a

for a ∈ A and fi(ai) for ai ∈ Bi. Therefore, these are the only sentences we consider

here. That is, when we refer to a collection of L?(N) sentences, we automatically

assume they are of this form.

Definition 2.5 Let ψ be a sentence of L?(N) (with the properties described above).

Let p = 〈p1, . . . , pn〉 be an n-tuple of finite partial 1-1 functions pi : Bi → A. (These

tuples are called forcing conditions.) We write q = 〈q1, . . . , qn〉 ⊇ p iff qi ⊇ pi for all

i. We define the relation p ‖− ψ (read “p forces ψ”) inductively:

1. If ψ is a finitary sentence with the above properties, then p ‖− ψ iff

(a) for each fi(ai) appearing in ψ, pi(ai) is defined; and

(b) A |= ψ where ψ is obtained by substituting pi(ai) for fi(ai).

2. If ψ is not finitary, and ψ =∧∧

n ψn, p ‖− ψ iff ∀n∀q ⊇ p∃r ⊇ q(r ‖− ψn).

3. If ψ is not finitary, and ψ =∨∨

n ψn, p ‖− ψ iff ∃n(p ‖− ψn).

4. If ψ is not finitary, and ψ = ¬θ, p ‖− ψ iff ∀q ⊇ p(q 6 ‖− θ).

10

We establish some very simple, very important properties of forcing.

Proposition 2.6 Let ψ be a sentence of L?(N) (with the properties described above)

and p be a forcing condition. Then

1. p cannot force both ψ and ¬ψ;

2. if p ‖− ψ and q ⊃ q, then q ‖− ψ;

3. there is q ⊃ p with q ‖− ψ or q ‖− ¬ψ.

Proof: 1) Assume p ‖− ψ. If ψ is finitary, then the definition of satisfaction precludes

p from forcing ¬ψ. If ψ is infinitary, then since p ⊆ p, we cannot have p ‖− ¬ψ.

2) Fix q ⊃ p. We argue by induction on the complexity of ψ that p ‖− ψ implies

q ‖− ψ.

Let p ‖− ψ, a finitary sentence. Then the definition of forcing implies the result,

because qi(ai) = pi(ai) for each fi(ai) appearing in ψ.

Let p ‖− ψ =∧∧

n ψn, an infinitary conjunction. Suppose n ∈ N and r ⊇ q. Then

r ⊇ p, so there is s ⊇ r with s ‖− ψn. Hence, q ‖− ψ.

Let p ‖− ψ =∨∨

n ψn, an infinitary disjunction. Since p ‖− ψ, there is n with

p ‖− ψn. By induction hypothesis, q ‖− ψn, so q ‖− ψ.

Let p ‖− ψ = ¬θ, an infinitary negation. Suppose r ⊇ q. Then r ⊇ p, so r 6 ‖− θ.

Thus, q ‖− ψ.

3) Try to define q ⊃ p to force ψ. If we cannot, and ψ is finitary, then we define

q = 〈q1, . . . , qn〉 so that qi(ai) exists if fi(ai) appears in ψ. Then q ‖− ¬ψ. If we

cannot, and ψ is infinitary, then p ‖− ¬ψ by definition. QED

The following definition of a fragment of sentences is not standard.

Definition 2.7 A subset F of (L?(N))ω1ω is called a fragment of q.f. sentences iff

it has the following properties:

11

1. if θ ∈ F , θ = ¬ψ, then ψ ∈ F ;

2. if θ ∈ F , θ =∧∧

n∈R ψn, then {ψn : n ∈ R} ⊂ F , and∨∨

n∈R ¬ψn ∈ F ;

3. if θ ∈ F , θ =∨∨

n∈R ψn, then {ψn : n ∈ R} ⊂ F ;

4. for each i = 1, . . . , n, b ∈ Bi, and a ∈ N,∨∨

c∈N fi(b) = c,∨∨

d∈Bi fi(d) = a ∈ F .

Note: Any countable subset S of (L?(N))ω1ω can be extended to a countable fragment

F by a standard closing procedure.

Definition 2.8 Let F be fragment of (L?(N))ω1ω. A sequence (pm)m∈ω of forcing

conditions is F-complete if for all m, pm ⊆ pm+1, and for each ψ in F , there is

m ∈ ω so that pm ‖− ψ or pm ‖− ¬ψ.

Proposition 2.9 For any countable fragment F of (L?(N))ω1ω, there exists an F-

complete sequence of forcing conditions.

Proof: Let the sentences of F be listed as (ψm)m∈ω. We let p0 = 〈∅, . . . , ∅〉. At each

stage m > 0, we define pm ⊇ pm−1 so that pm ‖− ψm−1 or pm ‖− ¬ψm−1. We can do

this by part 3 of Proposition 2.6. QED

Proposition 2.10 For any countable fragment F and any forcing condition p, we

can include p in an F-complete forcing sequence.

Proof: Let p0 = p and proceed as in the previous proof. QED

Proposition 2.11 Let F be a fragment and (pm)m∈ω be an F-complete sequence of

forcing conditions. Then for each i = 1, . . . , n, gi =⋃m∈ω p

im : Bi → A is a bijection.

Proof: This follows directly from the definitions of forcing, fragments, and complete

sequences. The formulas of the first type in property 4 of Definition 2.7 guarantee

12

totality; those of the second type guarantee surjectivity. Since each pim is injective,

so is⋃m∈ω p

im. QED

For each i = 1, . . . , n, let Bi be the structure on Bi induced by the bijection gi.

The product B1⊕· · ·⊕Bn obtained from a complete forcing sequence is called generic.

We now establish the most important fact about forcing.

Lemma 2.12 (Forcing Lemma) Let F be a fragment of sentences of the type de-

scribed at the beginning of this section, and (pm)m∈ω be an F-complete sequence. For

each i = 1, . . . , n let hi extend gi as described in Lemma 2.4. Then for any ψ in F

〈A, h1, . . . , hn〉 |= ψ iff ∃m(pm ‖− ψ).

Sketch of proof: The argument is standard, so we give only a brief outline. It proceeds

by induction on the complexity of the sentences in our fragment. If ψ is finitary, then

the very definition of forcing yields the desired conclusion. If ψ is a disjunction or

negation, then the induction is relatively straightforward. If ψ is a conjunction, then

the argument is a bit more complicated; the inclusion of the second set in property

2 of Definition 2.7 is crucial. QED

As we shall see in Section 4, we often use Lemma 2.4, which connects satisfaction

in B1⊕· · ·⊕Bn with satisfaction in 〈A, h1, . . . , hn〉, in conjunction with Lemma 2.12,

which connects satisfaction in 〈A, h1, . . . , hn〉 with forcing.

2.3 Relative dimension

2.3.1 A different definition of formally Σ01 Scott families suffices

We give a weaker definition of a formally Σ01 Scott family than that appearing in the

introduction.

13

Definition 2.13 Let A be an L-structure for some computable language L. A for-

mally Σ01 Scott family for A is an ordered pair 〈~c,Θ〉 where ~c is a tuple of distinct

parameters; and Θ is a c.e. set of Σc1 formulas of L with parameters among ~c so that

1. for each tuple ~a ∈ A of distinct elements with those from ~c listed first, there is

θ(~x) ∈ Θ with A |= θ(~a); and

2. for any θ in Θ and ~a,~a ′ ∈ A, if A |= θ(~a) and A |= θ(~a ′), then 〈A,~a〉 ∼= 〈A,~a ′〉.

The definition of a formally Σ01 Scott family stated in Chapter 1 contains a stronger

first and second condition:

1. for each tuple there must be a formula, not just those tuples of distinct elements

with those from ~c listed first;

2. if ~a and ~a ′ satisfy the same formula, there must be an isomorphism between

〈A,~a,~c〉 and 〈A,~a ′,~c〉.

With some rather long but straightforward technical arguments, we can show that

the two definitions are equivalent in the following sense: A has a formally Σ01 Scott

family according to one definition iff it has a formally Σ01 Scott family according to

the other definition. However, in this chapter, our sole purpose in constructing the

formally Σ01 Scott family is to invoke Theorem 1.13. Therefore, rather than argue

for equivalence directly, we reprove Theorem 1.13 for our definition of a formally Σ01

Scott family. Since we have altered the definition by weakening conditions, we need

only check one direction of the result.

Theorem 1.13 A computable structure A is relatively computably categorical iff it

has a formally Σ01 Scott family (as defined in Definition 2.13).

Proof: (⇐) Assume that A has a formally Σ01 Scott family 〈~c,Θ〉, and let B ∼= A.

We must produce a ∆01(D(B))-isomorphism f between them.

14

First there is some ~d with 〈A,~c〉 ∼= 〈B, ~d〉. Let f(~c) = ~d. Let a1 ∈ A−{~c}. There is

θ1(~x, y) ∈ Θ so that A |= θ1(~c, a1). So there is b1 ∈ B−{~d} with B |= θ1(~d, b1). Define

f(a1) = b1. We argue, in fact, that 〈A,~c, a1〉 ∼= 〈B, ~d, b1〉. Of course, for some a′1, we

have 〈A,~c, a′1〉 ∼= 〈B, ~d, b1〉. Then A |= θ1(~c, a′1), so 〈A,~c, a1〉 ∼= 〈A,~c, a′1〉 ∼= 〈B, ~d, b1〉.

Let b2 ∈ B − {~d, b1}. There is θ2(~x, y1, y2) ∈ Θ so that B |= θ2(~d, b1, b2). Thus

there is a2 ∈ A − {~c, a1} so that A |= θ2(~c, a1, a2). Define f(a2) = b2. We can again

argue that 〈A,~c, a1, a2〉 ∼= 〈B, ~d, b1, b2〉. For some a′2, 〈A,~c, a1, a′2〉 ∼= 〈B, ~d, b1, b2〉.

Then A |= θ2(~c, a1, a′2), so 〈A,~c, a1, a2〉 ∼= 〈A,~c, a1, a

′2〉 ∼= 〈B, ~d, b1, b2〉.

We continue this back-and-forth argument to obtain the desired isomorphism.

QED

2.3.2 An important lemma

Lemma 2.14 For each e ∈ ω and i, j ∈ {1, . . . , n}, there is a sentence ρe,i,j of L?(N)

so that for B1∼=g1 A, . . . ,Bn ∼=gn A, ϕ

D(B1⊕···⊕Bn)e : Bi ∼= Bj iff 〈A, h1, . . . , hn〉 |=

ρe,i,j, where each hi is defined from gi as usual. (Note that in viewing ϕD(B1⊕···⊕Bn)e

as an isomorphism from Bi onto Bj, we are concerned only with its behavior on the

computable subdomain Bi.)

Proof: By the Transition Lemma, we know that for all e, ~x, ~y, there is a sentence

ψe,~x,~y such that, regardless of the structure of B1, . . . ,Bn, ϕD(B1⊕···⊕Bn)e (~x) = ~y iff

B1⊕· · ·⊕Bn |= ψe,~x,~y. Therefore, by Lemma 2.4 we conclude that for B1, . . . ,Bn ∼= A,

ϕD(B1⊕···⊕Bn)e (~x) = ~y iff 〈A, h1, . . . , hn〉 |= ψ′e,~x,~y. Consequently, the sentence∧∧x∈Bi

∨∨y∈Bj ψ

′e,x,y expresses the totality of ϕ

D(B1⊕···⊕Bn)e ;

∧∧y∈Bj

∨∨x∈Bi ψ

′e,x,y, sur-

jectivity;∧∧x1,x2∈Bi,x1 6=x2

∧∧y∈Bj ¬ψ

′e,x1,y

∨ ¬ψ′e,x2,y, injectivity; and∧∧

R∈L,~x∈Bi,~y∈Bj of R′s arity(¬ψ′e,~x,~y ∨ [R(fi(~x)) ↔ R(fj(~y))]), preservation of relations.

The conjunction of these four sentences is ρe,i,j. QED

15

2.3.3 Finite computable dimension does not relativize

Theorem 2.15 If A has finite relative computable dimension, then it has a formally

Σ01 Scott family, and therefore has dimension 1 by Theorem 1.13.

Proof: Assume that A has relative computable dimension < n. We will build generic

copies B1, . . . ,Bn ∼= A by forcing conditions, as described in the previous section.

Our fragment F should contain all q.f. Σc1 sentences; and ¬ρe,i,j for each e ∈ ω and

each i, j ∈ {1, . . . , n} with i < j. If we construct an F -complete sequence (pm)m∈ω,

then the fact that A has relative computable dimension < n and Lemma 2.14 imply

that for some e and i < j, 〈A, h1, . . . , hn〉 |= ρe,i,j. Therefore, one direction of the

Forcing Lemma implies that for some m, pm ‖− ρe,i,j. The other direction implies

that no matter how we would have extended pm to an F -complete sequence, we

still would have 〈A, h1, . . . , hn〉 |= ρe,i,j. Therefore, any generic copies B1, . . . ,Bn

constructed by a complete forcing sequence extending pm would have the property

that ϕD(B1⊕···⊕Bn)e : Bi ∼= Bj . We are now in a position to construct the formally Σ0

1

Scott family.

For the rest of this discussion, we let p denote pm. (Therefore, any subsequent

use of the character m has nothing to do with this index.) Let p = 〈p1, . . . , pn〉; ~bk =

dom(pk); and ~ak = pk(~bk). Fix

1. ~t = ~t1~t2 an m-tuple of distinct elements of A with ~t1 an l-tuple from ~aj , and ~t2

an (m− l)-tuple disjoint from ~aj ;

2. ~u = ~u1~u2 an m-tuple of distinct elements of Bi with ~u1 an o-tuple from ~bi and

~u2 an (m− o)-tuple disjoint from ~bi;

3. ~v = ~v1~v2 an m-tuple of distinct elements of Bj with pj(~v1) = ~t1 and ~v2 an

(m− l)-tuple disjoint from ~bj .

16

Fix ~x, an m-tuple of distinct variables. We will construct a formula θ~t,~u,~v(~ai, ~x) of

L with the following property:

for m-tuples ~a of distinct elements of A, A |= θ~t,~u,~v(~ai,~a) iff ∃q =

〈q1, . . . , qn〉 ⊇ p [qi(~u) = ~a, qj(~v) = ~t, and q ‖− ψ′e,~u,~v].

We could include such a q in an F -complete forcing sequence. Then qi would be

a piece of an isomorphism gi : Bi ∼= A; qj would be a piece of an isomorphism

gj : Bj ∼= A; and q ‖− ρe,i,j, since p does. Thus, such a q would “force” the following

diagram to hold. (Note that when we include q in our forcing sequence, we have

defined only a finite part of each gk, and thus only a finite part of the structure of

each Bk. Nevertheless, the diagram necessarily holds no matter how we complete the

construction after q.)

&%'$

&%'$

&%'$

~u ~v

~a ~t

AAAAAAAAAAAAAAAAAAU

��������������������

-

gi ∼= gj∼=

ϕD(B1⊕···⊕Bn)e

∼=Bi Bj

AFigure 4.3

By the Transition Lemma, we know that ψe,~u,~v is of the form

17

∨∨σ∈S γσ,1(~b1,~b

′σ,1)∧. . .∧γσ,i(~bi, ~u2,~b

′σ,i)∧. . .∧γσ,j(~bj , ~v2,~b

′σ,j)∧. . .∧γσ,n(~bn,~b

′σ,n) where

1. ~b′σ,k is a tuple of distinct elements of Bk distinct from ~bk (and from ~u2 when

k = i, ~v2 when k = j); and

2. each γσ,k is a conjunction of atomic sentences and negations of atomic sentences

of L(Bk).

The Transition Lemma, Lemma 2.4, and the Forcing Lemma imply that for any

generic structures defined by an F -complete sequence, ϕD(B1⊕···⊕Bn)e (~u) = ~v iff B1 ⊕

· · ·⊕Bn |= ψe,~u,~v iff 〈A, h1, . . . , hn〉 |= ψ′e,~u,~v iff for some q in our sequence, q ‖− ψ′e,~u,~v.

Therefore, we consider all σ ∈ S where A |= {∃~zσ,1[~zσ,1 distinct and disjoint from ~a1

and γσ,1(~a1, ~zσ,1)]∧. . .∧∃~zσ,j [~zσ,j distinct and disjoint from ~aj , ~t2 and γσ,j(~aj ,~t2, ~zσ,j)]∧

. . . ∧ ∃~zσ,n[~zσ,n distinct and disjoint from ~an and γσ,n(~an, ~zσ,n)]}. Let S ′ = {σ ∈ S : σ

has the above property}. Since A is computable and S is c.e., S′ is c.e.

For an m-tuple ~x of variables, let ~x1 be its first o variables, and ~x2 be the next (m−o)

variables. Consider θ~t,~u,~v(~ai, ~x) =∨∨

σ∈S′ [(~x distinct) ∧ (~x1 = pi(~u1)) ∧ (~x2 disjoint

from ~ai) ∧ (∃~yσ(~yσ distinct and disjoint from ~ai, ~x2 and γσ,i(~ai, ~x2, ~yσ))].

No longer view ~t, ~u,~v as fixed.

Let T = {〈~t, ~u,~v〉: ~t ∈ A,~u ∈ Bi, ~v ∈ Bj are as described after the first paragraph of

this proof for some m, l, o ∈ N with m ≥ l, o, }.

Let Θ = {θ~t,~u,~v(~ai, ~x) : 〈~t, ~u,~v〉 ∈ T}.

Finally, let our proposed Scott family be 〈~ai,Θ〉.

Claim 1: For each tuple ~a ∈ A of distinct elements with those of ~ai (if any) appearing

first, there is θ~t,~u,~v ∈ Θ such that A |= θ~t,~u,~v(~ai,~a).

Sketch of proof: For each a ∈ ~a, F contains the sentence∨∨

x∈Bi fi(x) = a, and

we define the sequence of forcing conditions to be F -complete. Thus, for some q =

18

〈q1, . . . , qn〉 ⊇ p and some ~u, qi(~u) = ~a. Since p ‖− ρe,i,j, we may assume, without

loss of generality, that q ‖− ψ′e,~u,~v for some ~v. Finally, since for each v ∈ ~v, F contains

the sentence∨∨

x∈N fj(v) = x, we may also assume that qj(~v) = ~t for some ~t.

Claim 2: If for some 〈~t, ~u,~v〉 ∈ T , A |= θ~t,~u,~v(~ai,~a) and A |= θ~t,~u,~v(~ai,~a′), then

〈A,~a〉 ∼= 〈A,~a ′〉.

Sketch of proof: Fix 〈~t, ~u,~v〉 ∈ T . If A |= θ~t,~u,~v(~ai,~a), then we can extend p to a

forcing condition q with qi(~u) = ~a, q ‖− ψ′e,~u,~v, and qj(~v) = ~t. Furthermore, since

q ⊇ p and p ‖− ρe,i,j, q ‖− ρe,i,j by Proposition 2.6 (ii). By Proposition 2.9, we can

include q in an F -complete sequence. Hence, we can induce structures Bi and Bj with

〈A,~a〉 ∼= 〈Bi, ~u〉 ∼= 〈Bj , ~v〉 ∼= 〈A,~t〉. If A |= θ~t,~u,~v(~ai,~a′) as well, then 〈A,~a ′〉 ∼= 〈A,~t〉,

so 〈A,~a〉 ∼= 〈A,~a ′〉.

Therefore, we indeed have defined a formally Σ01 Scott family for A. QED

19

CHAPTER 3

RELATIVIZED ∆02-CATEGORICITY IN BOOLEAN

ALGEBRAS AND LINEAR ORDERINGS

3.1 Introduction

Recall that Goncharov and others characterized computable categoricity in many

classes of structures, among them Boolean algebras and linear orderings.

Theorem 3.1 (Goncharov, La Roche, Remmel) A Boolean algebra is computably

categorical iff it has finitely many atoms.

Theorem 3.2 (Goncharov-Dzgoev, Remmel) A linear ordering is computably cate-

gorical iff it has finitely many pairs of direct successors.

In fact, using Theorem 1.13, we can easily show that these theorems remain true

if “computably categorical” is replaced with “relatively computably categorical.” In

computable model theory there have been several other instances where a priority

argument was used to prove a result, and later a forcing argument was used to prove

the analogous relativized result. In many cases, the forcing argument often is actually

simpler than the priority construction. In this thesis, we examine the notions of ∆02-

categoricity and relativized ∆02-categoricity in Boolean algebras and linear orderings

by first studying the relativized notion. We shall see not only that here the arguments

for the relativized notion are again easier than those for the unrelativized notion, but

also that the former provide a helpful guide for the latter.

20

By Theorem 1.13, in order to examine relativized ∆02-categoricity, we must under-

stand the satisfaction of Σ2 formulas in the structures we examine. In [1], Ash and

Knight described back-and-forth relations that give us the necessary tools to char-

acterize the satisfaction of Σ2 formulas in Boolean algebras and linear orders quite

easily.

Definition 3.3 Let A be any structure, and ~a, ~b tuples of the same length from A.

Then ~a ≤1~b iff the Σ1 formulas true of ~b are true of ~a iff the Π1 formulas true of ~a

are true of ~b.

Proposition 3.4 Let A be a Boolean algebra and ~a, ~b be tuples from A so that the

Σ0 formulas true of ~a and ~b are the same. Then ~a ≤1~b iff each atom of the finite

algebra determined by ~a is infinite or at least as large as the corresponding atom of ~b.

Proposition 3.5 Let A be a linear ordering and ~a, ~b be tuples from A so that the

ordering of ~a is the same as that of ~b. Then ~a ≤1~b iff each interval in A determined

by ~a is infinite or at least as large as the corresponding interval determined by ~b.

3.2 The relativized results

In order to state our result for Boolean algebras, we must recall the definition of a

1-atom.

Definition 3.6 Let ∼ be the equivalence relation in a Boolean algebra A where a ∼ b

iff a∩ b and b∩ a are both either empty or a union of finitely many atoms of A. The

set of equivalence classes A/ ∼ is again a Boolean algebra. An element a ∈ A is a

1-atom of A iff the equivalence class [a] is an atom of A/ ∼.

Theorem 3.7 A Boolean algebra A is relatively ∆02-categorical iff it can be expressed

as a finite direct sum c1 ∨ · · · ∨ cn, where each ci is either atomless, an atom, or a

1-atom.

21

Proof: (⇐) If the summands are all either atomless or atoms, then it is ∆01-categorical

by Theorem 3.1. If A has some 1-atoms, then absorb all of the atoms into one of

the 1-atoms, so we write A = c1 ∨ · · · ∨ cn, where c1 is atomless, and c2, . . . , cn are

1-atoms. Consider the collection of parameters ~c = c1, . . . , cn; with these we will

construct a c.e. Scott family of Σc2 formulas.

Let ~a = a1, . . . , aj ∈ A, and let b1, . . . , b2j the atoms in the formal finite subalge-

bra generated by ~a. (Some of the bi’s might equal 0.) For each bi, we construct the

following formulas:

1. θbi1 (yi, c1) is “yi ∩ c1 = 0” if bi ∩ c1 = 0; “yi ∩ c1 = c1” if bi ∩ c1 = c1; or

“yi ∩ c1 6= 0 ∧ yi ∩ c1 6= c1” if bi ∩ c1 is a proper subset of c1, respectively.

2. For each k ∈ {2, . . . , n}, exactly one of bi ∩ ck, bi ∩ ck is empty or a finite

join of atoms (of A), precisely because each ck is a 1-atom. The formula γm(y) =

∃z1 · · · zm[z1 6= 0∧ · · · ∧ zm 6= 0∧ z1 ∪ · · · ∪ zm = y ∧ ∀w(¬(0 < w < z1)∧ · · · ∧ ¬(0 <

w < zm))] expresses that y is a join of m atoms. For each k ∈ {2, . . . , n}, let nbi,k be

the size of yi ∩ ck or yi ∩ ck, and let θbi1 (yi, ck) be γnbi,k(yi ∩ ck) if |bi ∩ ck| = nbi,k; or

γnbi,k(yi ∩ ck) if |bi ∩ ck| = nbi,k.

For a tuple of variables ~x = x1, . . . , xj , let ~y = y1, . . . , y2j be the terms in the

formal finite subalgebra determined by ~x. Let ψ~a(~x,~c) =∧∧

i∈{1,... ,2j} θbi1 ∧ · · · ∧ θbin .

Of course, A |= ψ~a(~a,~c). Furthermore, if for some ~a ′, A |= ψ~a(~a′,~c), then we

immediately have (A,~a,~c) ∼= (A,~a ′,~c). Consequently, {ψ~a|~a ∈ A} is a c.e. Scott

family of Σc2 formulas.

(⇒) Assume that A is not of the described form, and let ~c be parameters in a putative

formally Σ02 Scott family. Consider the finite subalgebra determined by ~c. One of the

atoms a of this algebra must be such that

1. a is not a 1-atom; and

22

2. a contains infinitely many atoms (of A).

Why must this be true? Otherwise, each atom of the subalgebra determined by ~c

is either a 1-atom or is a finite join of atoms and an atomless algebra. Therefore,

contrary to hypothesis, A can be written as a finite join of 1-atoms, atoms, and an

atomless algebra.

Let a be the disjoint union of a1, a2, where a1 is an infinite Boolean algebra and

a2 contains infinitely many atoms. We claim that for any Σ2 formula γ(x,~c) satisfied

by a1, there is a′1 composed of only finitely many atoms of A so that a′1 also satisfies

γ. Since obviously (A, a1,~c) 6∼= (A, a′1,~c), A cannot have the supposed Scott family.

We can assume, without loss of generality, that the formula γ is of the form

∃~uδ(x, ~u,~c), where δ is the conjunction of all finitary Π1 formulas satisfied by a1,~c,

and some ~u. Consequently, we must show that, given a tuple ~u, there are a′1, ~u′ so

that every Π1 formula satisfied by a1, ~u,~c is satisfied by a′1, ~u′,~c, or that (a1, ~u,~c) ≤1

(a′1, ~u′,~c). By Proposition 3.4 we must show that there are a′1, ~u

′ so that each atom

of the finite algebra determined by a1, ~u,~c is infinite or at least as big as the corre-

sponding atom determined by a′1, ~u′,~c.

Consider the finite subalgebra determined by a1, ~u,~c; let its atoms be z1, . . . , zm.

For each zi with zi ∩ a = 0, let z′i = zi. In this subalgebra, the element a1 is divided

into zj1 , . . . , zjs and a2 into zk1, . . . , zkt. One of the zk’s must be infinite; without

loss of generality, let it be zk1 . In a2 find atoms (of A) z′j1, . . . , z′js, z

′k2, . . . , z′kt; and

let z′k1= a − [

⋃p∈{1,... ,s} z

′jp ∪

⋃q∈{2,... ,t} z

′kq

]. Define a′1, ~u′, and ~c ′ in the subalgebra

generated by the z′ elements in the same way that a1, ~u, and ~c are generated by the z

elements. For instance, a′1 =⋃p∈{1,... ,s} z

′jp, so a′1 is a finite join of atoms of A. Note

that ~c ′ = ~c, since⋃p∈{1,... ,s} z

′jp ∪

⋃q∈{1,... ,t} z

′kq =

⋃p∈{1,... ,s} zjp ∪

⋃q∈{1,... ,t} zkq = a,

and for each zi with zi ∩ a = 0, we let z′i = zi. Notice that for each i = 1, . . . ,m, zi

23

is a Boolean algebra of size at least as large as z′i, and that a′1 is finite. Thus we have

proven our claim. QED

Throughout the rest of the chapter, we will assume that a linear ordering A has a

greatest and a least element; this assumption makes the statements of certain results

simpler. However, it should be clear that a linear ordering without a greatest or least

element is (relatively) ∆02-categorical iff the same ordering with a greatest and least

element “attached” is. Hence, our assumption is merely one of convenience.

Theorem 3.8 Let A = (A,<A) be a sum of finitely many intervals, each of type n,

ω, ω∗, Z, or n · η, so that each interval of type n · η has a supremum and infimum.

Then A is relatively ∆02-categorical.

Proof: We use Theorem 1.13. For each interval of type ω or ω∗, name the “0”; for

interval of type Z, pick a single element and name it; for each interval of type n · η,

name the supremum and infimum; finally, name all of the remaining elements, of

which there are a finite number. These will be the parameters appearing in our Scott

family.

Let ~a = a1, . . . , ak ∈ A. We explicitly construct a formula γ~a(x1, . . . , xk,~c) such

that A |= γ~a(a1, . . . , ak,~c) and if A |= γ~a(b1, . . . , bk,~c), then

(A, a1, . . . , ak,~c) ∼= (A, b1, . . . , bk,~c).

First, express the ordering of ~a,~c in a formula θ(~x,~c). We may assume without

loss of generality that a1, . . . , ak are listed in increasing order. Next, let j1 be the

least number and j2 the greatest number so that 1 ≤ j1 ≤ j2, aj1 , . . . , aj2 all fall in

the same interval of A of the types described above, and neither aj1 nor aj2 is an

element of ~c. We describe how to find a formula ψ(xj1 , . . . , xj2 ,~c) which characterizes

aj1, . . . , aj2 up to isomorphism. The formula γ~a will be a conjunction of such formulas.

If aj1, . . . aj2 lie in the same interval of type ω with c = “0”, then aj1 = “p1,”

. . . , aj2 = “pj2−j1+1” in this copy of ω. Therefore, aj1 satisfies the formula ψ1(x1, c) =

24

∃y1 · · · yp1−1[c < y1 < · · · < yp1−1 < x1 ∧ ∀z(¬(c < z < y1) ∧ . . .¬(yp1−1 < z < x1))].

We have similar formulas ψ2(x2, c), . . . , ψj2−j1+1(xj2−j1+1, c). Let ψ(x1, . . . , xj2−j1+1, c)

be the conjunction.

If aj1 , . . . , aj2 lie in an interval of type ω∗ or Z, we similarly obtain a formula

ψ(x1, . . . , xj2−j1+1, c), where c is the named element of the interval.

Assume aj1, . . . , aj2 lie in the same interval of type n · η with aj1, . . . , am1 in the

same n interval of n · η; am1+1, . . . , am2 in the same n interval that is different from

a1’s; . . . ; amt+1, . . . , aj2 in the same n interval that is different from the t previous

n intervals. For instance, consider the case where n = 5, m1 = j1 + 1, aj1 = “1,”

and am1 = “3.” Then aj1 , am1 satisfy the formula ψ1(x1, x2) = ∃y0y1y2[(y0 < x1 <

y1 < x2 < y2) ∧ ∀z(¬(y0 < z < x1) ∧ · · · ¬(x2 < z < y2))]. We have similar

formulas ψ2(xm1+1, . . . , xm2), . . . , ψt+1(xmt+1, . . . , xj2). Furthermore, am1 and am1+1

satisfy the formula ρ1(xm1 , xm1+1) = ∃y0 · · ·yn(xm1 < y0 < · · · < yn < xm1+1).

Similarly, there are formulas ρ2, . . . , ρt. Let ψ(x1, . . . , xj) be the conjunction of all

of ψ1, . . . , ψt+1, ρ1, . . . , ρt.

By design, A |= γ~a(a1, . . . , ak,~c). If A |= γ~a(b1, . . . , bk,~c), then the ordering of

~b, ~c is the same as that of ~a, ~c. Furthermore, because of the way in which we chose

parameters, the distribution among the intervals of A with named parameters must

be the same for ~a and ~b. Finally for ap, . . . , aq and bp, . . . , bq all falling in the same

interval with a named parameter, we can draw the following conclusions:

1. If they all fall in an interval of type ω, ω∗, or Z, then γ~a actually guarantees

that ap = bp, . . . , aq = bq.

2. If they all fall in an interval of type n · η, then γ~a guarantees that

(a) for all u, v with p ≤ u, v ≤ q, au and av fall in the same interval of type n

exactly if bu and bv do;

25

(b) the position within the intervals of type n of au and bu are the same;

(c) the ordering of ap, . . . , aq matches that of bp, . . . , bq.

Therefore, the ordering properties of η imply that in the second case, there is

an automorphism of this entire interval taking ap, . . . , aq onto bp, . . . , bq. It follows

that (A,~a,~c) ∼= (A,~b,~c). Hence, it is clear that the set {γ~a|~a is a tuple from A} is a

formally Σ02 Scott family. QED

We now give a proof of the converse of Theorem 3.8.

Theorem 3.9 Let A = (A,<A) be a relatively ∆02-categorical linear ordering. Then

A is a sum of finitely many intervals, each of type n, ω, ω∗, Z, or n · η, so that each

of the n · η interval has a supremum and infimum.

We prove this theorem by dividing it into several lemmas and smaller propositions.

Definition 3.10 Let A be a linear ordering. A maximal interval with property P is

an interval I of A so that

1. I has property P ; and

2. If a is any other element of A, then I ∪ {a} is not contained in an

interval with property P .

We will often use expressions such as “maximal ω-interval” to denote a maximal

interval of order type ω.

Proposition 3.11 A relatively ∆02-categorical linear orderingA must have only finitely

many maximal intervals of order type ω, ω∗,Z, and cannot have arbitrarily large max-

imal finite intervals.

26

Proof: Assume thatA either has infinitely many maximal intervals of one of the above

order types, or has arbitrarily large maximal finite intervals. Let ~c = c0, . . . , cn be the

parameters in a putative formally Σ2 Scott family, and let A = I0 +c0 +I1 + · · ·+In+

cn + In+1. For some k = 0, . . . , n+ 1, Ik contains infinitely many maximal intervals

of one of the above order types, or has arbitrarily large maximal finite intervals. (For

simplicity, we write I for this Ik. There are a1 < a2 in I so that

1. there are infinitely many elements in I to the right of a1;

2. there are infinitely many elements between a1 and a2; and

3. there are infinitely many elements in I to the left of a2.

We claim that for any Σ2 formula γ(x1, x2,~c) satisfied by a1, a2, there exist a′1, a′2 so

that a′1, a′2 also satisfy γ but have only finitely many elements of A between them.

Since obviously (A, a1, a2,~c) 6∼= (A, a′1, a′2,~c),A cannot have the supposed Scott family.

We can assume, without loss of generality, that the formula γ is of the form

∃~uδ(x1, x2, ~u,~c), where δ is the conjunction of all finitary Π1 formulas satisfied by

a1, a2, some ~u, and ~c. Consequently, we must show that, given a tuple ~u, there are

a′1, a′2, ~u

′ so that every Π1 formula satisfied by a1, a2, ~u,~c is satisfied by a′1, a′2, ~u

′,~c,

or that (a1, a2, ~u,~c) ≤1 (a′1, a′2, ~u

′,~c). Therefore, by Proposition 3.5, in order to verify

our claim, we must demonstrate the following

for any ~u containing the parameters ~c, there exist a′1, a′2, ~u

′ so that u′i = ui for

ui ∈ ~c, and the intervals determined by a′1, a′2, ~u′ are of size no greater than

the corresponding intervals of a1, a2, ~u.

Assume, without loss of generality, that ~u = u1, . . . , un is arranged in increasing

order, and that the ordering is as follows:

1. u1, . . . , uj1 lie to the left of I;

27

2. uj1+1, . . . , uj2 lie in the interval I and to the left of a1, but they are in the same

successor chain as an element of ~c;

3. uj2+1, . . . , uj3 lie in the interval I and to the left a1, and they are not in the

same successor chain as an element of ~c;

4. a1 < uj3+1, . . . , uj4 < a2;

5. uj4+1, . . . , uj5 lie in the interval I and to the right of a2, and they are not in

the same successor chain as an element of ~c;

6. uj5+1, . . . , uj6 lie in the interval I and to the right of a2, but they are in the

same successor chain as an element of ~c;

7. uj6+1, . . . , un lie to the right of the interval I.

Note that the intervals (uj2, uj2+1) and (uj5, uj5+1) are infinite.

Find a successor chain (not necessarily a maximal successor chain) in I of length

n+ 2 so that no element in this chain is in a successor chain with any element of ~c.

Define a′1, a′2, ~u

′ as follows:

1. for i ≤ j2, u′i = ui;

2. u′j2+1, . . . , u′j3, a′1, u

′j3+1, . . . , u

′j4, a′2, u

′j4+1, . . . , u

′j5

is a sequence of successors in

the chain of length n+ 2;

3. for i ≥ j5+1, u′i = ui.

The desired property of a′1, a′2~u′ follows immediately from the fact we noted above.

QED

28

Lemma 3.12 If B is a countable linear ordering containing no ω, ω∗, or Z intervals,

then B is of the form n1 + Σq∈ηBq + n2, where n1 or n2 could be 0 and each Bq is

finite with |Bq| ≥ 1.

Proof: If B has a least element a, then we know that either a doesn’t have an

immediate successor, or a+ doesn’t, or a++ doesn’t, etc., because B doesn’t contain

an ω interval. Therefore, B has an initial maximal finite discrete interval of order

type n1. (Of course, n1 = 0 if B has no least element.) Similarly B has a terminal

maximal finite discrete interval of order type n2. (Of course, n2 = 0 if B has no

greatest element.) We show that B has the form n1 +B′+ n2, where B′ has the form

Σq∈ηBq.

Let c1 ∈ B′. By repeatedly applying the successor and predecessor function we

can obtain only a finite interval {c1,0, . . . , c1,n1}. Let this finite interval be Bq1 for

some q1 ∈ η. Note that c1,0 must have no immediate predecessor and c1,n1 must have

no immediate successor. Furthermore, c1,0 is greater than any element of n1 and c1,n1

is less than any element of n2.

Let q2 ∈ η − {q1}. If q2 < q1, then find some c2 greater than any element of

n1 and less than c1,0. Repeatedly apply the successor and predecessor function to

obtain a finite interval {c2,0, . . . , c2,n2}. Let this finite interval be Bq2. Note that c2,0

must have no immediate predecessor and c2,n2 must have no immediate successor.

Furthermore, c2,0 is greater than all the elements of n1 and c2,n2 < c1,0. We make the

appropriate changes to our construction and our argument if q2 > q1.

We can continue this back-and-forth argument to construct a Σq∈ηBq∼= B′. Con-

sequently, B is of the form n1 + Σq∈ηBq + n2. QED

Convention Hereafter, when we write Σq∈ηBq, we assume that each Bq is finite with

|Bq| ≥ 1.

29

Corollary 3.13 If A is a relatively ∆02-categorical linear ordering, then there is an

n so that A is a finite sum of intervals, each of the form ω, ω∗,Z,m,Σq∈ηBq, where

each |Bq| < n.

Proposition 3.14 If A is a relatively ∆02-categorical linear ordering (with endpoints),

then any maximal interval of the form Σq∈ηBq has an infimum and a supremum.

Proof: We prove the fact for infimums; the argument for supremums is similar.

Assume that A is a relatively ∆02-categorical linear ordering with ~c the parameters

in a putative formally Σ2 Scott family. Let A have a maximal interval P of the form

Σq∈ηBq. By Corollary 3.13, there must be an interval directly to the left of one of

the following forms:

1. finite;

2. ω∗;

3. ω.

In (1) and (2), P has an infimum. We give the argument for (3). Fix a ∈ P so a

has no immediate successor, and there are no members ~c in P to the left of a. We

claim that for any Σ2 formula γ(x,~c) satisfied by a, there exists a′ in the adjacent

ω-interval so that a′ also satisfies γ(x,~c). Again we must show that

for any ~u containing ~c, there exist a′, ~u ′ so that u′i = ui for ui ∈ ~c, and the

intervals determined by a′, ~u ′ are of size no greater than the corresponding

intervals of a, ~u.

Assume, without loss of generality, that ~u = u1, . . . , un is in increasing order, and

that the ordering is as follows:

1. u1, . . . , uj1 lie to the left of the ω-interval;

30

2. uj1+1, . . . , uj2 lie in the ω-interval;

3. uj2+1, . . . , uj3 lie to the left of a in P ;

4. a < uj3+1, . . . , ujn.

Note that (a, uj3+1) is infinite. Define a′, ~u ′ as follows, where “x+i” is the ith successor

of x in the ω-interval:

1. for 1 ≤ i ≤ j2, u′i = ui;

2. for 1 ≤ i ≤ j3 − j2, u′j2+i = uj2 + i;

3. a′ = u′j3 + 1;

4. for j3 + 1 ≤ i ≤ n, u′i = ui.

The desired property of a′~u ′ follows immediately from the fact we noted above.

QED

Definition 3.15 A finite partition of η is a finite sequence q1 < . . . < qj ∈ η and

open intervals J1, . . . , Jj+1 so that η = J1 + q1 + J2 + . . .+ qj + Jj+1. A partition P1

is finer than a partition P2 exactly if every element in the finite sequence for P2 is in

the sequence for P1.

Definition 3.16 Let B be a linear ordering of the form Σq∈ηBq.

1. B satisfies Property 1 iff there is a finite partition of η so that for each open

interval Ji in the partition there is ni so that for all q ∈ Ji, Bq has order type ni.

That is, there are n1, n2, . . . , nm ≥ 1 so that B = n1 · η + n2 + n3 · η + . . .+ nm · η.

2. B satisfies Property 2 iff there exist m1 6= m2 so that for any finite partition of

η there is an open interval J in the partition so that for infinitely many q ∈ J, Bq is

of order type m1 and for infinitely many q′ ∈ J, Bq′ is of order type m2.

31

Lemma 3.17 Fix a linear ordering B of the form Σq∈ηBq with n as in Corollary 3.13.

If B does not satisfy Property 2, then it satisfies Property 1.

Proof: Let m0, . . . ,mk be the complete list of natural numbers so that for each

0 ≤ i ≤ k, there is q ∈ η with Bq of order type mi. For each mi 6= mj , there is a

finite partition of η so that no interval of the partition contains infinitely many copies

of mi and infinitely many copies of mj . Choose a finite partition of η finer than all

of these partitions. Now in each open interval of the partition, there exists only one

mi so that Bq is of order type mi for infinitely many q in that interval. Make the

partition finer to obtain finite partition that satisfies the requirements in Property 1.

QED

Proposition 3.18 If A is a relatively ∆02-categorical linear order, then no interval

of the form Σq∈ηBq satisfies Property 2.

Proof: Assume that A does have an interval of the form Σq∈ηBq with m1 and m2

as in Property 2, and let ~c be the parameters in a putative c.e. Scott family of Σc2

formulas. Let m1 be the largest number for which there is such an m2. Define a finite

partition of η so that for each m > m1 and each open interval J of the partition

1. for all q ∈ J , Bq has type m; or for all q ∈ J , Bq does not have type m; and

2. for all q ∈ J , no parameter appears in Bq.

For one of these open intervals J , there are infinitely many q ∈ J with Bq of type

m1, and there are infinitely many q′ ∈ J with Bq′ of type m2. Consider q1 ∈ J and

a ∈ A with a ∈ Bq1 of type m2. We claim that for any Σ2 formula γ(x,~c) satisfied by

a, there exists q′1 ∈ J and a′ ∈ A so that a′ ∈ Bq′1of type m1, and a′ satisfies γ(x,~c).

Again, we must show that

for any ~u containing the parameters ~c there exist a′, ~u ′ so that u′i = ui for

32

ci ∈ ~c and the intervals determined by a′, ~u ′ are of size no greater than

the corresponding intervals of a, ~u.

Assume, without loss of generality, that ~u = u1, . . . , un is in increasing order, and

that the ordering is as follows:

1. for all q ∈ J and all i with 1 ≤ i ≤ j1 and all b ∈ Bq, ui < b;

2. there exists r1 < r2 < · · · < rk < q1 in J so that uj1+1, . . . , uj2 ∈ Br1 ;

uj2+1, . . . , uj3 ∈ Br2; . . . ; ujk+1, . . . , ujk+1∈ Brk ;

3. ujk+1+1, . . . , ujk+2lie to the left of a in Bq1;

4. ujk+2+1, . . . , ujk+3lie to the right of a in Bq1;

5. there exist q1 < s1 < s2 < · · · < st in J so that ujk+3+1, . . . , ujk+4∈ Bs1 ;

ujk+4+1, . . . , ujk+5∈ Bs2; . . . ; ujk+t+2+1, . . . , ujk+t+3

∈ Bst;

6. for all q ∈ J and all i with jk+t+3 < i < n and all b ∈ Bq, b < ui.

Note that there are infinitely many elements between uj1 and uj1+1; between uj2 and

uj2+1; . . . ; between ujk+1and ujk+1+1; between ujk+3

and ujk+3+1; . . . ; and between

ujk+t+3and ujk+t+3+1. Furthermore, note that none of the intervals Bq, Br, or Bs has

more than m1 elements, because of the construction of our partition.

Now find r′1 < r′2 < · · · < r′k < q′1 < s′1 < · · · < s′t in J so that Bq′1has type m1

and each Br′ , Bs′ has type m1. Define a′, ~u ′ as follows:

1. for i ≤ j1, u′i = ui;

2. for 1 ≤ i ≤ k, u′ji+1, . . . , u′ji+1

are ordered in Br′iexactly as uji+1, . . . , uji+1 are

ordered in Bri;

3. u′jk+1+1, . . . , u′jk+2

, a′, u′jk+2+1, . . . , u′jk+3

are ordered inBq′1exactly as ujk+1+1, . . . , ujk+2

, a, ujk+2+

are ordered in Bq1;

33

4. for 1 ≤ i ≤ t u′jk+2+i+1, . . . , u′jk+3+i

are ordered inBs′iexactly as ujk+2+i+1, . . . , ujk+3+i

are ordered in Bsi;

5. for jk+t+3 < i < n, u′i = ui.

The desired property of a′, ~u ′ follows immediately from the facts we noted above.

QED

Lemma 3.17 immediately implies the following corollary.

Corollary 3.19 If A is a relatively ∆02-categorical linear order, then any interval of

the form Σq∈ηBq satisfies Property 1.

We now complete the proof of Theorem 3.9. Assume that A is relatively ∆02-

categorical. By Corollary 3.13 and Proposition 3.14, A can be written as a finite sum

of intervals, each of the form n, ω, ω∗, Z, Σq∈ηBq, so that each interval of the form

Σq∈ηBq has a supremum and infimum. By Corollary 3.19, for each interval of the form

Σq∈ηBq there are n1, n2, . . . , nm ≥ 1 so that Σq∈ηBq = n1 ·η+n2 +n3 ·η+ . . .+nm ·η.

QED

We note that our characterization of ∆02-categorical linear orderings is exactly the

same as Michael Moses’ characterization of computably categorical 1-decidable linear

orderings in [22].

Theorem 3.20 For a 1-decidable linear ordering A (with greatest and least element)

the following are equivalent:

1. Every 1-decidable linear ordering isomorphic to A is isomorphic by a

computable isomorphism;

2. A is of the form given in Theorem 3.9.

The “reason” for this correlation has not yet been established. Some further obser-

vations will appear in Chapter 5.

34

CHAPTER 4

UNRELATIVIZED ∆02-CATEGORICITY

4.1 ∆02-categoricity in Boolean algebras

In this chapter we offer priority constructions that give, under some extra assump-

tions, results on ∆02-categoricity (unrelativized) for Boolean algebras and linear or-

derings. We shall see that with these assumptions, the realtivized and unrelativized

notions define the same class of structures. Furthermore, we shall attempt to highlight

where the syntactic arguments from the previous chapter provide some insight into

the organization, strategies, and hypotheses needed for the priority constructions.

Notation Let A be a Boolean algebra.

i) the unary predicate Atom(x) denotes that x is an atom of A; i.e.,

(A, Atom) |= Atom(x) iff A |= x 6= 0 ∧ ¬(∃z∃y(z 6= x ∧ y 6= x ∧ z ∪ y = x));

ii) the unary predicate Atomless(x) denotes that x contains no atoms;

i.e., (A, Atom,Atomless) |= Atomless(x) iff (A, Atom) |= ∀z[∃y(y ∪ z = x) →

¬Atom(z)].

4.1.1 Our first priority construction

Theorem 4.1 Let A be a Boolean algebra so that (A, Atom,Atomless) is a com-

putable structure. If A is ∆02-categorical, then it is a finite direct sum of atoms,

1-atoms, and an atomless algebra, then A.

35

Proof: Assume A is not of the described form. We attempt to construct a computable

B ∼= A so that there is no ∆02 isomorphism between them. We use ϕ

∆02

e to denote the

eth ∆02 function, and ϕ

∆02

e,s to denote the computable approximation of ϕ∆0

2e at stage

s. (Recall that the limit lemma tells us that any total ∆02 function is the pointwise

limit of computable approximation functions.)

Requirements and the true path

We employ a tree construction where each node has two outcomes, I (inactive) and

A (active), with I < A. Nodes of length e work on requirement

Re : ϕ∆0

2e is not an isomorphism from A onto B.

Moreover, the construction must determine an isomorphism g : B ∼= A.

At stage s we inductively define δs, a node of length s−1 approximating the true

path f through the tree. If α ⊂ δs, then α decides its outcome o (o = I or o = A)

by determining if ϕ∆0

2e threatens to be an isomorphism. In addition, α ⊂ δs defines

gα,s, its contribution to the stage s approximation of the function g, so that |α| is in

the domain and range. The construction will ensure that for β ⊆ α ⊂ δs, gβ,s ⊆ gα,s;

therefore, δs defines the stage s approximation gs ⊇⋃α⊂δs gα,s. (Note: at stage s, a

node α ⊂ δs will often first determine g′α,s, a preliminary version of gα,s, then decide

its outcome, and finally determine gα,s itself.)

Finally, at the end of stage s, we use gs to fix a Boolean algebra Bs containing s

atoms and containing the elements 0, 1, . . . 2s − 1. Since B must be computable, all

of the relations true among the elements of Bs must be true in Bs+1.

The true path will be defined in the standard way for 0′′ constructions: α ⊂ f iff

α is the left-most node of length |α| so that α ⊂ δs for infinitely many s. We must

show that nodes along the true path succeed in defining an isomorphism g : B ∼= A

and in meeting the requirements Re.

36

Challenging pairs and candidate pairs

Consider α ⊂ δs working on requirement Re. If α ⊃ λ, the empty string, then it

receives from its predecessor node β the function gβ,s that it believes is a correct

approximation of g, because all requirements Ri with i < e seem to be met. Let ~d be

the domain of gβ,s with gβ,s(~d) = ~c. If α = λ, then there is no β, and ~c = 0A, 1A.

The node α considers the finite algebra determined by ~c and attempts to determine

the first pair 〈a1α,s, a

2α,s〉 so that a1

α,s seems infinite, a2α,s seems to have infinitely atoms,

and for some atom a in the finite subalgebra determined by ~c, a1α,s ∪ a2

α,s = a; α

includes a1α,s, a

2α,s in the range of gα,s. Consider a′, the atom corresponding to a in

the subalgebra determined by the preimage of ~c under gα,s ◦ϕ∆02

e,s . The pair 〈a1α,s, a

2α,s〉

challenges this a′ to contain elements a1α,s′, a2

α,s′ which have the same properties that

a1α,s, a

2α,s seem to have. Hence we call it a challenging pair. A pair a1

α,s′, a2

α,s′ which

seems to meet the challenge is called a candidate pair.

We should note that α may make incorrect guesses about which pairs have the

above properties, because it cannot computably determine which elements are infinite

or have infinitely many atoms. However, the properties of A do guarantee that such

a pair indeed exists, and eventually a node α along the true path will determine its

final challenging pair. Similarly, a pair may appear to be a candidate pair until the

enumeration of the diagram of 〈A, Atom,Atomless〉 reveals otherwise.

If α is along the true path and its ultimate challenge is never actually met, then

the final outcome of α is I. If a true candidate pair a1α,s′, a2

α,s′ is found at stage t,

then α attempts to change its approximation so that the image under gα,t ◦ ϕ∆02

e,t of

a1α,s′ is finite; we thereby meet requirement Re.

Relationships among nodes

Throughout the construction, a node α ⊂ δs may initialize another node γ; i.e., if

γ ⊆ δt for some t < s, then α cancels the assignment of γ’s challenging pair and

37

the approximation gt. There will be three circumstances when a node will initialize

another:

1. if α redefines its challenging pair at stage s, then α initializes each γ with α ⊂ γ;

2. if α ⊂ δs determines its outcome to be I, then α initializes all γ with α 〈I〉 <L γ;

3. if α diagonalizes against ϕ∆0

2e,s at stage s, then α initializes all γ with α 〈A〉 ⊆ γ.

In turn, as α defines gα,s, it must respect nodes σ <L α 〈o〉. What exactly does

“respect” mean? Consider σ <L α 〈o〉 and some stage t < s so that

1. σ ⊆ δt;

2. gt(~v) = ~u; and

3. gt has not been canceled by stage s.

If at stage s, α defines gα,s(~v) = ~u ′, then as we determine the Boolean algebra Bs

based on α’s approximation of g, we may give an element of the finite algebra de-

termined by ~v a certain number of subsets. However, this definition at stage s must

allow for the possibility that at some later stage t′ with σ ⊂ δt′ , gt′(~v) = ~u. How

can the construction ensure this compatibility? Assume that each atom of the finite

algebra determined by ~u ′ contains no more elements than corresponding atom of ~u.

Then at stage s we do not risk giving some element in the algebra determined by ~v

more subsets than its image under gt actually contains. Consequently, at stage t′,

δt′ can redefine gt′(~v) = ~u and remain in agreement with the algebra Bt′−1 we have

chosen. In other words, ~u ≤1 ~u′. Therefore, the relation crucial to classifying rela-

tivized ∆02-categoricity also play a critical role in our examination of the unrelativized

notion.

38

Construction

Stage 1: Fix, for the rest of the construction, g(0) = 0A and g(1) = 1A. For each σ

in the tree, σ is initialized and δ1 = λ. The algebra B1 is the trivial algebra.

Stage s + 1: Of course, λ ⊂ δs+1. Let α ⊂ δs+1 be a node of length e. Let β be

the predecessor node of α, and let dom(gβ,s) = ~d with gβ,s+1(~d) = ~c. (If α = λ, then

there is no β, and ~c = 0A, 1A.) The node α uses the computability of the relations

Atom and Atomless in A to search for the first pair 〈a1α,s+1, a

2α,s+1〉 so that

1. for an atom a of the finite algebra determined by ~c, a1α,s+1 ∪ a2

α,s+1 = a;

2. a1α,s+1 has s + 1 proper subsets;

3. a2α,s+1 contains s+ 1 atoms of A; and

4. we have not yet discovered either that a1α,s+1 is finite or that a2

α,s+1 has only

finitely many atoms of A.

This is the challenging pair of α at stage s + 1.

Case I: The node α has been initialized since its last active stage or has defined a

new challenging pair since its last active stage.

The node α initializes all γ with α ⊂ γ ⊆ δt for some t < s, and it chooses

outcome I. (Notice, in particular, that any approximation gt defined in part by α at

an earlier stage t is canceled. However, of course, α does not cancel the definition of

its challenging pair.) It must define gα,s+1 so that the challenging pair is in the range.

We may assume by induction on the construction that ~d includes all elements which

are in the domain of an uncanceled approximation gt where t < s+1 and δt <L α. Let

~b be the set of elements in Bs but not in the domain of gβ,s+1. Again, we may assume

by induction on the construction that gβ,s+1 is compatible with the algebra Bs; i.e.,

bβ,s+1 can be extended to be a relation-preserving map between Bs and A. Therefore,

39

α can extend gβ,s+1 so that it includes ~b and e in the domain and the challenging pair

and e in the range. Let b1α,s+1, b

2α,s+1 be such that gα,s+1(b1

α,s+1, b2α,s+1) = a1

α,s+1, a2α,s+1.

[[Summary: After redefining its challenging pair, the node α restarts its work on the

requirement Re.]]

Case II: α has not been initialized since its last active stage, and its challenging

pair is the same as it was at its last active stage.

There is a stage t < s+ 1 so that

1. α ⊂ δt;

2. either α has been initialized since the previous stage r in which α ⊂ δr, or the

definition of the challenging pair changed during stage t;

3. α has not been initialized since stage t; and

4. 〈a1α,s+1, a

2α,s+1〉 = 〈a1

α,t, a2α,t〉.

(Throughout the rest of this construction, we will refer to this particular t.)

We assume by induction on the construction that since α has not been initialized

since stage t, the portion of gs+1 defined by nodes above α is the same as the portion

of gt defined by nodes above α. The node α defines g′α,s+1, its tentative version of

gα,s+1, to be the same as gα,t. We now consider the following subcases.

Case IIa: One of the following is true:

1. An element of ~d or one of the elements b1α,s+1, b

2α,s+1 is not in the range of

ϕ∆0

2e,s+1. (Otherwise, let ~c ′, a1

α,s+1′, a2

α,s+1′ be such that ϕ

∆02

e,s+1(~c′) = ~d and

ϕ∆0

2e (a1

α,s+1′, a2

α,s+1′) = b1

α,s+1, b2α,s+1.)

2. Some relation realized by ~c, a1α,s+1, a

2α,s+1 is not realized by ~c ′, a1

α,s+1′, a2

α,s+1′,

or vice-versa.

40

3. If α enumerates the diagram of 〈A, Atom,Atomless〉 for s+1 steps, then it sees

that one of the atoms of the finite algebra determined by ~c, a1α,s+1, a

2α,s+1 has

a different size or a different number of atoms (of A) than the corresponding

atom in the finite algebra determined by ~c ′, a1α,s+1

′, a2α,s+1

′.

4. If t′ < s+1 is the greatest stage so that α ⊆ δt′ , then our guess at the preimage

of ~d or one of b1α,s+1, b

2α,s+1 under ϕ

∆02

e has changed since t′. (Otherwise, the pair

〈a1α,s+1

′, a2α,s+1

′〉 is a candidate pair.)

The outcome of α is I; the node α initializes all γ with α 〈I〉 <L γ; and gα,s+1 = g′α,s+1.

[[Summary: If the approximation g′α,s+1 defined so far is part of the isomorphism g

itself, then ϕ∆0

2e doesn’t appear to be an isomorphism, because it does not seem total,

as in (1) and (4), or we can readily see that g′α,s+1 ◦ ϕ

∆02

e,s+1 does not appear to be an

automorphism of A, as in (2) and (3).]]

Case IIb: Not Case IIa and all of the following are true:

1. there is a stage r with t < r < s + 1 so that α 〈A〉 ⊂ δr, and α 〈A〉 has not

been initialized since r;

2. α diagonalized against Re at stage r;

3. if y is an atom of the finite subalgebra determined by elements in the range of

higher priority gw, and during the diagonalization α guessed that y was infinite,

then α finds 2s+1 subsets of y and still guesses that y is infinite.

The outcome of α remains A, and gα,s+1 = gα,r.

[[Summary: The node α seems already to have diagonalized successfully against Re

at an earlier stage, and nothing has injured this work.]]

41

Case IIc: Neither Case IIa nor Case IIb

First, α initializes each node γ with α 〈A〉 ⊆ γ. The node α attempts to perform

the following diagonalization on the candidate pair 〈a1α,s+1

′, a2α,s+1

′〉. If it completes

the diagonalization, then the outcome of α is A.

Diagonalization: Let w be the greatest number so that t ≤ w < s + 1, α 〈I〉 ⊆ δw,

and gw is uncanceled. Let dom(gw) = ~v with gw(~v) = ~u. By our construction, the

approximation gw must extend g′α,s+1.

Let ~b be the set of elements in Bs but not in the domain of gw. By induction

on the construction, we can assume that gw is compatible with this algebra, and α

can define an extension of gw whose domain includes ~b. Let the image of ~b be ~a.

(Note that the atom a of the algebra determined by ~c is not necessarily an element

of ~a.) Throughout the diagonalization, α must leave fixed the mapping gβ,s+1(~d) = ~c.

However, α will attempt to alter the mapping of the rest of ~v,~b, b1α,s+1, b

2α,s+1 so that

it meets requirement Re while it respects gw.

This diagonalization follows closely the argument for relativized ∆02-categoricity

given in Theorem 3.7: ~c, the range of gβ,s, corresponds to the parameters ~c in the

potential formally Σ2 Scott family; ~u, the range of gw, corresponds to the extra tuple

~u in the back-and-forth relation. Here, however, we also need to consider ~a, the image

of the extra elements of Bs. Nevertheless, note that we are not concerned with the

atom size in the finite algebra determined by ~u ∪ ~a, but only in that determined by

~u.

Recall that g′α,s+1 is α’s tentative contribution to gs+1; a

1α,s+1, a

2α,s+1 is a chal-

lenging pair with a1α,s+1 ∪ a2

α,s+1 = a, and α has defined g′α,s+1(b

1α,s+1, bα,s+1) =

a1α,s+1, a

2α,s+1. Furthermore, 〈a1

α,s+1′, a2

α,s+1′〉 seemingly has met the challenge, be-

cause ϕ∆0

2e (aα,s+1

′, a2α,s+1

′) = b1α,s+1b

2α,s+1, and a1

α,s+1′ has 2s+1 proper subsets and still

appears infinite, and a2α,s+1

′ has 2s+1 atoms ofA. The goal is to define gα,s+1(b1α,s+1) =

42

a1α,s+1

′′ so that a1α,s+1

′′ is a finite Boolean algebra, thus ensuring that ϕ∆0

2e is not an

isomorphism if ϕ∆0

2e,s+1 is a correct approximation of ϕ

∆02

e and a1α,s+1

′ truly is infinite.

Consider the finite subalgebra determined by ~u. Let k = ((s+ 1) + |~u|+ |~a|). Let

y1, y2, . . . , ym be atoms of this finite algebra with y1 ∪ · · · ∪ ym = a2α,s+1. Determine

the first among y1, . . . , ym which contains 2k subsets and still appears to be infinite.

(If no such yj exists, then a2α,s+1is finite, and hence, α must choose a new challenging

pair and enter Case I.) Assume, without loss of generality, that it is ym.

Now consider the finite subalgebra determined by ~u ∪ ~a. Let

z01 , . . . , z

0p0

, z11 , . . . , z

1p1, . . . , zm1 , . . . , z

mpm be atoms of this algebra so that z0

1∪· · ·∪z0p0

=

a1α,s+1 and zj1∪· · ·∪zjpj = yj for j = 1, . . . ,m. In a2

α,s+1 find atoms (of A) z01′, . . . , z0

p0

′

and zj1′ . . . zjpj

′ for j = 1, . . . ,m− 1. Furthermore, in a2α,s+1 find other atoms (of A)

zm1′, . . . , zmpm−1

′, and let zmpm′ = a− [[

⋃j=0,... ,m−1(∪i=1,... ,pjz

ji′)]∪ (zm1

′∪· · ·∪zmpm−1′)].

Notice that⋃z-elements =

⋃z′-elements = a.

Now we are ready to redefine gα,s+1. Let v1 be an element of ~v ∪ ~b currently

mapped to u1. If u1∩a = 0, then gα,s+1(v1) = u1. Otherwise, u1 is a union of some x

disjoint from a and some union of z elements. Let gα,s+1(v1) = the union of x and the

corresponding z′ elements. In particular, assume v1 is an element of ~d. If u1 is disjoint

from a, then gα,s+1(v1) = u1. If u1 is not disjoint from a, then u1 = x ∪ a, since a is

an atom of the algebra determined by ~c. Consequently, gβ,s+1(v1) = x∪⋃z′-elements

= u1. The node α finishes the definition of gα,s+1 by including e in the domain and

range. This completes the diagonalization.

After each node α ⊂ δs+1 has determined gα,s+1, we complete the definition

of gs+1 and define the finite Boolean algebra Bs+1. The algebra Bs has s atoms,

which we designate b1, b2, . . . , bs, and therefore has 2s distinct elements. The func-

tion⋃α⊂δs+1

gα,s+1 contains all of these 2s elements in its domain; we designate the

image of each bi as ai.

43

If the function⋃α⊂δs+1

gα,s+1 has no more than 2s elements in its domain, then

determine the first ai so that ai has a nontrivial subset. Extend⋃α⊂δs+1

gα,s+1 to

contain this subset and its complement with respect to ai in the range; this extension

is gs+1.

If⋃α⊂δs+1

gα,s+1 has more than 2s elements in its domain, then it must map

some b∗ to a nontrivial subset of some ai, which we designate a∗. If necessary, extend⋃α⊂δs+1

gα,s+1 so that every element in the finite algebra with atoms a1, . . . , ai−1, a∗, ai∩

a∗, ai+1, . . . , as is included in the range; this extension is gs+1. Using gs+1, define Bs+1

so that bi is no longer an atom. This concludes the construction.

4.1.2 Supporting lemmas

Lemma 4.2 For each s, Bs and δs satisfy the following properties:

1. If β is the predecessor node of α ⊂ δs, then gβ,s ⊆ gα,s.

2. Let α 〈o〉 ⊂ δs, and let t < s be the greatest stage so that δt <L α 〈o〉. If ~v is

the set of all elements in the domain of gt, then ~v ⊆ dom(gα,s).

3. Let o be a fixed outcome, α 〈o〉 ⊂ δs, and t < s be the last stage with α 〈o〉 ⊆ δt.

If gα,s 6= gα,t, then every node γ ⊇ α 〈o〉 is initialized at some stage t′ with

t < t′ ≤ s.

4. Let α, ~v, and gt be as in (2). If y is an atom of the finite algebra determined

by gt(~v) and y′ is the corresponding atom of the finite subalgebra determined by

gα,s(~v), then |y| ≥ min{|y′|, 2s}.

5. Bs extends Bs−1 as a finite Boolean algebra.

6. Bs is compatible with each gt not canceled by stage s.

44

Proof: We use induction on the stage s. For stage 0, (1) - (6) are trivially true.

Assume (1) - (6) are true for stage s. We must show them true for stage s+ 1.

We use induction on the length of the node α ⊂ δs+1. Assume that (1) - (4) are

true for all β ⊂ α. We must show them true for α.

1) If the outcome of α is I at stage s + 1, then α falls either into Case I or Case

IIa. If Case I, then (1) is true by construction. If Case IIa, then there is a stage

t < s + 1 so that

a) α ⊂ δt, and α defined a challenging pair at t;

b) α has not been initialized at some t′ with t < t′ ≤ s+ 1; and

c) The challenging pair defined at t is the challenging pair at s+ 1.

At stage t, gα,t ⊇ gβ,t by construction. If gβ,t 6= gβ,s+1, then by the induction hy-

potheses on stages and nodes, α was initialized at some stage t′ with t < t′ ≤ s+ 1,

a contradiction. Therefore, gβ,s+1 = gβ,t ⊆ gα,t = gα,s+1.

2) If the outcome of α at stage s + 1 is I, then (2) is true by the induction

hypothesis on nodes and (1). If the outcome is A, then let r ≤ s+ 1 be the greatest

stage at which α actually performed a diagonalization. If r = s+ 1, then (2) is true

by construction. If r < s + 1, then (2) is true for r by induction hypothesis, and

gα,s+1 = gα,r. If any more elements were added to ~v between r and s+ 1, then α 〈A〉

would have been initialized between r and s+ 1, contradicting the definition of r.

3) If o = I and gα,s+1 6= gα,t, then α redefines its challenging pair at s+ 1, so (3)

is true by construction. If o = A and gα,s 6= gα,t, then by construction α 〈A〉 was

initialized at some stage t′ with t < t′ ≤ s+ 1; therefore, so was any node γ ⊇ α 〈A〉.

4) If the outcome at stage s+ 1 is I, then (4) is true by induction on nodes and

(1) for α. If the outcome of α is A, let r ≤ s+ 1 be the last stage at which α actually

performed a diagonalization. If r = s + 1, then the diagonalization guarantees that

45

only one atom y of the algebra determined by gt(~v) is possibly made larger when we

define gα,s+1(~v); however, such a y must have size at least 2s+1. If r < s + 1, then

at stage s + 1 we have seen that the y from stage r does indeed have 2s+1 proper

subsets.

5) Since the algebra Bs+1 is determined entirely by gs+1, we need only verify that

for each α ⊂ δs, gα,s+1 is a relation-preserving map between a subset of Bs and A. If

α is in Case I, then the construction guarantees that gα,s+1 respects Bs as it defines

gα,s+1. If Case IIa or IIb, then gα,s+1 is part of an uncancelled gt. By induction

hypothesis on (6), gt, and hence gα,s+1 is compatible with Bs. Finally, if α is in Case

IIc, then the diagonalization ensures that gα,s+1 maps atoms of the algebra Bs to

nonzero disjoint elements of A whose union is 1, and gα,s+1 maps combinations of

these atoms to the analogous combinations of these disjoint elements.

6) Let t < s + 1. First, if δt ⊂ δs+1, then by (3) either gt is canceled or

gs+1 ⊇ gt. If δs+1 <L δt, then gt is canceled. If δt <L δs+1, then by (2) we know

that dom(gt) ⊆ dom(gs+1). Furthermore, gt completely determines Bt, and gs+1

completely determines Bs+1, and by (5), Bt and Bs+1 are consistent. Consequently,

we know that the finite algebras determined by gt(dom(gt))and gs+1(dom(gt)) look

the same. However, at the end of stage s + 1, there may be other elements not in

dom(gt) that appear in the algebra Bs+1. Can gt be extended to a relation-preserving

map between Bs+1 and A? In short, are the atoms of gt(dom(gt)) big enough to

accommodate the images of the new elements of Bs+1?

For each atom y determined by gt(dom(gt)), let y′ be the corresponding atom

determined by gs+1(dom(gt)). For each such y, the construction guarantees that

|y| ≥ min{|y′|, 2s+1}. Since Bs+1 is determined entirely by gs+1, and Bs+1 contains

only s + 1 atoms, each atom y can be expressed as a finite algebra which contains

46

either more atoms than y′ has atoms of A, or more atoms than appear in the entire

algebra Bs+1. In short, the desired compatibility is guaranteed. QED

Lemma 4.3 There is a true path f with the following features:

1. If α ⊂ f , then α is the left-most node of length |α| so that α ⊂ δs for

infinitely many s.

2. If α ⊂ f , then α does not define challenging pairs infinitely often.

3. If α 〈A〉 ⊂ f , then α does not diagonalize infinitely often.

4. If α 〈o〉 ⊂ f and S = {s : α 〈o〉 ⊂ δs}, then lims∈S

gα,s = gα; i.e., there are

t ∈ S and a tuple ~d so that for all s ∈ S with s ≥ t, dom(gα,s) = dom(gα,t) =

dom(gα) = ~d, and gα,s(~d) = gα,t(~d) = gα(~d).

5. If g =⋃α⊂f gα, then g : B ∼= A.

Proof: We show (2) - (4) by simultaneous induction on the length of α ⊂ f . Assume

(2) - (4) are true for all β ⊂ α. Let t be the least stage and ~d, ~c be the tuples so that

a) For all stages s ≥ t, α ≤L δs.

b) For all nodes β ⊂ α and all s ≥ t, 〈a1β,s, a

2β,s〉 = 〈aβ,t−1, aβ,t−1〉.

c) If β 〈A〉 ⊂ f , then β does not diagonalize after stage t− 1.

d) If β is the predecessor node of α, then t and ~d witness that (4) is true for β,

and gβ(~d) = ~c.

e) Let 〈a1, a2〉 be the least pair so that for some atom a of the finite algebra

determined by ~c, a1 ∪ a2 = a; a1 is infinite; and a2 has infinitely many atoms

of A. Then all lesser pairs have been discovered not to have this property by

stage t in the enumeration of 〈A, Atom,Atomless〉.

47

First, we claim that α is not initialized at any s ≥ t. A node α can be initialized at

stage s for one of three reasons:

a) δs <L α;

b) a node β ⊂ α redefines its challenging pair;

c) a node β with β 〈A〉 ⊆ α actively diagonalizes at stage s.

However, none of these can occur at a stage s ≥ t by the induction hypothesis.

Therefore, by the construction, for all stages s ≥ t, 〈a1α,s, a

2α,s〉 = 〈a1

α,t, a2α,t〉 = 〈a1, a2〉.

If α 〈I〉 ⊂ f , then by the construction gα = gα,t.

If α 〈A〉 ⊂ f , let r ≥ t be the first stage so that α 〈A〉 ⊂ δr, and for all s ≥ r,

α 〈I〉 6⊆ δs. Recall that w < r is the greatest number so that t ≤ w < s + 1,

α 〈I〉 ⊆ δw, and gw is uncanceled by r; ~v = dom(gw); and gw(~v) = ~u. Let r′ ≥ r be

the first stage where α 〈A〉 ⊂ δr′, and y, the atom of the finite subalgebra determined

by ~u which we guessed at stage r′ to be infinite, is indeed infinite. By the construction

gα = gα,r′.

Part (5) now follows almost immediately from Lemma 4.2. First, by part (1) of

Lemma 4.2 and (4) of this lemma, g =⋃α⊂f gα is a well-defined function. Since each

gs is 1-1, g is 1-1. Since each gα contains |α| in its domain and range, g is total and

onto. Finally, by parts (5) and (6) of Lemma 4.2, B is a computable Boolean algebra,

and each finite piece gα (for α ⊂ f) is a relation-preserving function. Consequently,

g is an isomorphism. QED

Lemma 4.4 Each requirement Re is satisfied.

Proof: Let α ⊂ f be of length e . If e > 0, then let β be the predecessor node of α,

and let ~c consist of the range of gβ and the elements in the challenging pair eventually

defined by α.

48

If the final outcome of α is I, then one of the following is true:

1. not every element of ~c is in ran(g ◦ ϕ∆02

e );

2. the algebra determined by ~c does not satisfy the same relations as the algebra

determined by its preimage under g ◦ ϕ∆02

e ;

3. the size of one of the elements of ~c is different from its preimage under g ◦ϕ∆02

e .

If the final outcome of α is A, then Lemma 4.2 guarantees that there is some r

so that for all s ≥ r with α ⊂ δs, the outcome of α is A, and α does not actively

diagonalize at any stage after r. Our construction then dictates that all approxima-

tions of ϕ∆0

2e including and after r map a1

α,r′, an infinite element of A, to b1

α,r, a finite

element of B. Consequently, ϕ∆0

2e is not an isomorphism. QED

4.2 ∆02-categoricity in linear orderings

Notation Let A be a linear ordering.

i) The predicate S(x, y) denotes the successor relation; i.e., (A, S) |=

S(x, y) iff A |= (x < y ∧ ∀z[(¬(x < z < y)]).

ii) The predicate L−(x) denotes that x has no immediate predecessor;

i.e., (A, S, L−) |= L−(x) iff (A, S) |= ∀y(¬S(y, x)) iff A |= ∀y < x∃z(y < z < x).

iii) The predicate L+(x) denotes that x has no immediate successor.

(Throughout the remainder of this paper, the words “successor” and “predecessor”

are intended to mean “immediate successor” and “immediate predecessor,” respec-

tively.)

The main theorem of this section is

Theorem 4.5 Let A = (A,<A) be a ∆02-categorical linear ordering so that (A, S, L−, L+)

is a computable structure. Then A = (A,<A) is a sum of finitely many intervals,

49

each of type n, ω, ω∗, Z, or n · η, so that each interval of type n · η has a supremum

and infimum.

The proof of this theorem is considerably more complicated than the one for

Boolean algebras, just as the arguments concerning relative ∆02-categoricity were.

4.2.1 Some propositions about the maximal discrete pieces

Proposition 4.6 Let A be a ∆02-categorical linear ordering so that (A, S, L−, L+) is

a computable structure. Then all of the following conditions hold:

1. A has finitely many maximal ω-intervals and maximal ω∗-intervals.

2. A does not contain arbitrarily large maximal finite discrete intervals.

3. There exist intervals I0, . . . , In so that A = I0 + c1 + I1 + · · · + cn + In and

NO interval Ik is of the form J1 + Z+ J2 + Z+ J3, where each Ji contains an

element with either no successor or no predecessor.

Proof: Assume A does not satisfy one of the three conditions. We attempt to con-

struct a computable B ∼= A so that there is no ∆02 isomorphism between them.

We employ a tree construction with the same requirements and outcomes as those

in Theorem 4.1: Re says that ϕ∆0

2e is not an isomorphism between A and B; and the

outcomes are I (inactive) and A (active). At stage s, we define the approximations

δs and gs, and we fix an ordering Bs, which orders the elements 0, 1, . . . , s− 1 in B.

Challenging pairs and candidate pairs

Consider α ⊂ δs working on requirement Re. If α ⊃ λ, the empty string, then it

receives from its predecessor node β the function gβ,s that it believes to be a correct

approximation of g, because all requirements Ri with i < e seem to be satisfied. Let

50

~d be the domain of gβ,s with gβ,s(~d) = ~c = c1 < . . . < cj. For α ⊇ λ, let c0 be the

least element of A and cj+1 be the greatest element of A.

In each interval (ci, ci+1), α searches for challenging pairs 〈a1i , a

2i 〉 so that

(ci, a1i ), (a

1i , a

2i ), and (a2

i , ci+1) each contains an element with either no successor or

no predecessor. (The assumption that A does not satisfy one of the conditions in

the statement of the theorem guarantees that α will always be able to find an in-

terval with such a pair.) The node α next includes a1i , a

2i in the range of gα,s. The

pair 〈a1i , a

2i 〉 challenges the preimage under gα,s ◦ ϕ∆0

2e,s of each of the three intervals

(ci, a1i ), (a

1i , a

2i ), (a

2i , ci+1) to contain an element with either no successor or no prede-

cessor. If all challenges seem to be met at some later stage t by candidate pairs,

then α attempts to change its approximation of g so that the image under gα,t ◦ ϕ∆02

e,t

of one of the candidate pairs is a pair of elements in the same successor chain; we

thereby meet Re.

The challenging pairs in this construction differ from those in Theorem 4.1 in

two ways. First, a node α may have multiple challenging pairs associated with it at

the same time. This feature is necessary because we cannot computably determine

which intervals of A have elements with no successor or no predecessor. The set of

challenging pairs associated with a node α (whether or not α ⊂ δs) at stage s is

designated by chα,s. There is a bit of ambiguity in this notation, as the challenging

pairs associated with a node α may change during the stage s. However, an easy

convention should dispel any confusion: if we are currently working in stage s of the

construction, then chα,s denotes the set associated with α at the actual moment; if we

are working in a later stage t, or if we are describing some feature of the construction

as a whole, then chα,s represents the pairs associated with α at the end of s.

51

Second, a node α never makes an incorrect guess about what is a challenging pair,

because we assume that the relations S, L+, L− are all computable. Therefore, only

the initialization of a node α will remove a pair’s status as a challenging pair.

Relationships among nodes

Throughout the construction, a node α ⊂ δs may initialize another node γ; i.e., if

γ ⊆ δt for some t < s, then α defines chγ,s = ∅, and it cancels the approximation gt.

There will be two circumstances when a node will initialize another:

1. if α defines new challenging pairs at stage s, then α initializes each γ with

α ⊂ γ;

2. if α ⊂ δs determines its outcome to be I, then α initializes all γ with α 〈I〉 <L γ.

(We will see that here, unlike in the proof of Theorem 4.1, a node will make no

incorrect decisions about how to diagonalize after it ceases to be initialized and has

defined all of its possible challenging pairs. Therefore, there is no need for the third

occasion of initialization given in Theorem 4.1.)

In turn, as α defines gα,s, it must respect nodes σ <L α 〈o〉. What exactly does

“respect” mean? Consider σ <L α 〈o〉 and some stage t < s so that

1. σ ⊆ δt;

2. gt(v1, v2) = u1, u2; and

3. gt has not been canceled by stage s.

If at stage s, α defines gα,s(v1, v2) = u′1, u′2, then as we decide the ordering Bs based on

α’s approximation of g, we may order a certain number of elements between v1 and

v2. However, this ordering must allow for the possibility that at some later stage t′

with σ ⊂ δt′ , gt′(v1, v2) = u1, u2. How can the construction ensure this compatibility?

52

Assume that the interval (u′1, u′2) contains no more elements than the interval (u1, u2).

Then at stage s we do not risk placing more elements between v1 and v2 than actually

exist between u1 and u2. Consequently, at stage t′, δt′ can redefine gt′(v1, v2) = u1, u2

and remain in agreement with the ordering Bt′−1 we have chosen. Again, we note the

significance of the ≤1 relation in our priority construction.

Construction and supporting lemmas

Stage 0: For each σ in the tree, chσ,0 = ∅. g0 = ∅. δ0 = λ.

Stage s+ 1: For each σ in the tree, we define chσ,s+1 = chσ,s.

Of course, λ ⊂ δs+1. Let α ⊂ δs+1 be a node of length e. Let β be the predecessor

node of α, and let dom(gβ,s+1) = ~d with gβ,s+1(~d) = ~c = c1 < . . . < cj. (If α = λ, then

there is no β and gβ,s+1 = ∅.) For α ⊇ λ, let c0 be the least element of A and cj+1 be

the greatest element of A. If α has no associated challenging pairs, then it searches

until it finds a pair. Furthermore, if any interval (ci, ci+1) is without a challenging

pair associated with α, then α searches s + 1 steps to see if the interval has a pair

〈a1i , a

2i 〉 to add to chα,s+1.

Case I: α finds a pair to add to chα,s+1.

The node α initializes all γ with α ⊂ γ ⊆ δt for some t < s, and it chooses

outcome I. (Notice, in particular, that any approximation gt defined in part by α at

an earlier stage t is canceled. However, of course, α does not define chα,s+1 to be ∅.)

It must define gα,s+1 so that these challenging pairs are in the range. We may assume

by induction on the construction that ~d includes all elements which are in the domain

of an uncanceled approximation gt where t < s + 1 and δt <L α. Let ~b be the set of

elements in Bs but not in the domain of gβ,s+1. Again, we may assume by induction

on the construction that gβ,s+1 is compatible with the ordering Bs; therefore, α can

53

extend gβ,s+1 so that it includes ~b and e in the domain and all challenging pairs and

e in the range. Let b1i , b

2i be such that gα,s+1(b1

i , b2i ) = a1

i , a2i .

[[Summary: After redefining chα,s+1, the node α restarts its work on the requirement

Re.]]

Case II: α finds no such challenging pair to add to chα,s+1.

There is a stage t < s+ 1 so that

1. α ⊂ δt;

2. α defined a challenging pair at stage t;

3. α has not been initialized since stage t; and

4. chα,s+1 = chα,t.

(Throughout the rest of this construction, we will refer to this particular t.)

We assume by induction on the construction that since α has not been initialized

since t, the portion of gs+1 defined by nodes above α is the same as the portion of gt

defined by nodes above α. The node α defines g′α,s+1, its tentative version of gα,s+1,

to be the same as gα,t. We now consider the following subcases.

Case IIa: One of the following is true

1. An element of ~d is not in the range of ϕ∆0

2e,s+1, or for some i, b1

i , b2i is not in the

range of ϕ∆0

2e,s+1. (Otherwise, let ~c ′, a1

i′, a2

i′ be such that ϕ

∆02

e,s+1(~c′) = ~d and

ϕ∆0

2e,s+1(a

1i′, a2

i′) = b1

i , b2i .)

2. The ordering of ~c and the challenging pairs does not match the ordering of the

inverse images under g′α,s+1 ◦ ϕ

∆02

e,s+1.

54

3. The size of some interval determined by ~c and the challenging pairs does not

match the sizes of corresponding interval determined by their inverse images

under g′α,s+1 ◦ ϕ

∆02

e,s+1.

4. If t′ < s+1 is the greatest stage so that α ⊆ δt′ , then our guess at the preimage

of ~d or b1i , b

2i for some i under ϕ

∆02

e has changed since t′.

5. For some i where (ci, ci+1) has a challenging pair, after enumerating the open

diagram of (A, S, L−, L+) for s+1 steps, one of the intervals (c′i, a1i′), (a1

i′, a2

i′),

(a2i′, c′i+1) appears to have only elements with successors and predecessors.

(Otherwise, the pairs 〈a1i′, a2

i′〉 are candidate pairs.)

The outcome of α is I, and gα,s+1 = g′α,s+1.

[[Summary: If the approximation g′α,s+1 defined so far is part of the isomorphism g

itself, then ϕ∆0

2e doesn’t appear to be an isomorphism, because it does not seem total,

as in (1) and (4), or we can readily see that g′α,s+1 ◦ ϕ

∆02

e,s+1 does not appear to be an

automorphism of A, as in (2), (3), and (5).]]

Case IIb: Not Case IIa and both of the following are true:

1. there is a stage r with t < r < s + 1 so that α 〈A〉 ⊂ δr, and α 〈A〉 has not

been initialized since r;

2. at stage r, α diagonalized against ϕ∆0

2e by defining gα,r so that gα,r◦ϕ∆0

2e,r (a1

i′, a2

i′)

is a pair in the same successor chain.

The outcome of α remains A, and gα,s+1 = gα,r.

[[Summary: The node α has already successfully diagonalized against ϕ∆0

2e at an

earlier stage, and nothing has injured this work.]]

55

Case IIc: Neither Case IIa nor Case IIb

The node α attempts to perform the following diagonalization for each interval

(ci, ci+1) containing a challenging pair. If there are intervals containing no challenging

pairs, then α simultaneously searches for challenging pairs in them. If it finds a

challenging pair in an interval without one before completing a diagonalization, then α

adds it to chα,s+1 and enters Case I. If it completes a single one of the diagonalizations,

then the outcome of α is A.

Diagonalization: Let w be the greatest number so that t ≤ w < s + 1, α 〈I〉 ⊆ δw,

and gw is uncanceled. Let dom(gw) = ~v with gw(~v) = ~u. By our construction, the

approximation gw must extend g′α,s+1.

Let ~b be the set of elements in Bs but not in the domain of gw. By induction

on the construction, we can assume that gw is compatible with this ordering, and

α can define an extension of gw whose domain includes ~b. Let the image of ~b be

~a. Throughout the diagonalization, α must leave fixed the mapping gβ,s+1(~d) = ~c.

However, α will attempt to alter the mapping of the rest of b1i , b

2i , ~v,

~b so that it meets

requirement Re while it respects gw.

Note that, as in Theorem 4.1, the diagonalization follows closely the argument

for relativized ∆02-categoricity given in Proposition 3.11.

Recall that g′α,s+1 is α’s tentative contribution to gs+1, and for each challenging

pair 〈a1i , a

2i 〉 α has defined g

′α,s+1(b1

i , b2i ) = a1

i , a2i . Furthermore, 〈a1

i′, a2

i′〉 has met

the challenge, because ϕ∆0

2e,s+1(a

1i′, a2

i′) = b1

i , b2i , and each of the intervals (c′i, a

1i′),

(a1i′, a2

i′), (a2

i′, c′i+1) has an element with either no successor or no predecessor. The

goal is to define gα,s+1(b1i , b

2i ) = a1

i′′, a2

i′′ so that a1

i′′, a2

i′′ are in the same successor

chain, thus ensuring that ϕ∆0

2e is not an isomorphism if ϕ

∆02

e,s+1 is a correct approxima-

tion of ϕ∆0

2e .

56

Let v1, . . . , vk be the elements of ~v ∪~b currently mapped to elements u1, . . . , uk

in (ci, ci+1). Assume, without loss of generality, that the ordering of these elements

in A is of the form

ci < u1 < · · · < ul < a1i < ul+1 < · · · < um < a2

i < um+1 < · · · < uk < ci+1.

By the very definition of challenging pairs, there is an element x that has no successor

or no predecessor and thereby guarantees one of the following intervals is infinite:

1. (ci, u1);

2. (uj, uj+1) for some j ∈ {1, l − 1};

3. (ul, a1i ).

Similarly, there is an element y that has no successor or no predecessor and thereby

guarantees that one of the following intervals is infinite:

1. (a2i , um+1);

2. (uj, uj+1), for some j ∈ {m+ 1, . . . , k − 1};

3. (uk, ci+1).

Find in (ci, ci+1) a successor chain of k + 2 elements so that to the left of this

chain there is an element x′ in (ci, ci+1) with no successor or no predecessor, and

to the right of this chain there is an element y′ in (ci, ci+1) with no successor or no

predecessor. (“To the right” and “to the left” are not intended to mean “directly to

the right” or “directly to the left.”)

Let x′′ = min{x, x′} and y′′ = max{y, y′}. Either ci or some uj is the left endpoint

of an interval guaranteed to be infinite by x′′. For each un up to and including this

endpoint, leave the mapping of vn as it is. Similarly, either ci+1 or some uj is the

right endpoint of an interval guaranteed to be infinite by y′′. For each un including

57

and after this endpoint, leave the mapping of vn as it is. Thus we are left with a

sequence of r ≤ k + 2 elements in the universe of A. The node α defines gα,s+1 so

as to remap the current preimages of these elements to appropriate elements in the

successor chain and finishes the definition of gα,s+1 by including e in the domain and

range. This concludes the diagonalization.

After each α ⊂ δs+1 has determined gα,s+1, we define gs+1 = ∪α⊂δs+1gα,s+1 and

use it to define Bs+1. This concludes the construction.

Lemma 4.7 For each s, δs and Bs satisfy the following properties:

1. If β is the predecessor node of some α ⊂ δs, then gβ,s ⊆ gα,s.

2. Let α 〈o〉 ⊂ δs, and let t < s be the greatest stage so that δt <L α 〈o〉.

If ~v is the set of all elements in the domain of gt, then ~v ⊆ dom(gα,s).

3. Let o be a fixed outcome, α 〈o〉 ⊂ δs,and t < s be the last stage with α 〈o〉 ⊆ δt.

If gα,s 6= gα,t, then every node γ ⊇ α 〈o〉 is initialized at some stage t′ with

t < t′ ≤ s.

4. Let α, ~v, and gt be as in (2). Then the intervals of the partition of A determined

by gt(~v) are at least as large as those of the partition determined by gα,s(~v).

5. Bs extends Bs−1 as a finite ordering.

6. Bs is compatible with each gt not canceled by stage s.

Proof: We use induction on the stage s. For stage 0, (1) - (6) are trivially true.

Assume that (1) - (6) are true for stage s. We must show them true for stage s + 1.

We use induction on the length of the node α ⊂ δs+1. The arguments for (1) - (4)

are essentially the same as those in Lemma 4.2.

58

5) Since the algebra Bs+1 is determined entirely by gs+1, we need only verify that

for each α ⊂ δs, gα,s+1 is an order-preserving map between a subset of Bs and A. If

α is in Case I, then the construction guarantees that gα,s+1 respects Bs as it defines

gα,s+1. If Case IIa or IIb, then gα,s+1 is part of an uncancelled gt. By induction

hypothesis on (6), gt, and hence gα,s+1 is compatible with Bs. Finally, if α is in Case

IIc, then the diagonalization ensures that gα,s+1 preserves the order of the elements

of Bs.

6) Let t < s + 1. If δt ⊂ δs+1, then by (3) either gt is canceled or gs+1 ⊇ gt.

If δs+1 <L δt, then gt is canceled. If δt <L δs+1, then by (2) and (4) we know that

dom(gt) ⊆ dom(gs+1) and the intervals determined by gs+1(dom(gt)) are no larger

than those determined by gt(dom(gt)). Since the ordering of Bs+1 is determined

entirely by gs+1, the desired compatibility is guaranteed. QED

Lemma 4.8 There is a true path f with the following features

1. If α ⊂ f , then α is the left-most node of length |α| so that α ⊂ δs for

infinitely many s.

2. If α ⊂ f , then α does not define challenging pairs infinitely often.

3. If α 〈o〉 ⊂ f and S = {s : α 〈o〉 ⊂ δs}, then lims∈S

gα,s = gα; i.e., there are

t ∈ S and a tuple ~d so that for all s ∈ S with s ≥ t, dom(gα,s) =

dom(gα,t) = dom(gα) = ~d, and gα,s(~b) = gα,t(~d) = gα(~d).

4. If g =⋃α⊂f gα, then g : B ∼= A.

Proof: We show (2) and (3) by simultaneous induction on the length of α ⊂ f .

Assume (2) and (3) are true for all β ⊂ α. Let t be the least stage and ~b be the tuple

so that

59

a) for all stages s ≥ t, α ≤L δs;

b) for all nodes β ⊂ α and all s ≥ t, chβ,s = chβ,t−1;

c) if β is the predecessor node of α, then t and ~d witness that (3) is true for β;

and

d) for each interval determined by gβ(~d) containing a challenging pair, α has found

such a pair by stage t.

First, we claim that α is not initialized at any stage s ≥ t. A node can be initialized

at stage s for one of two reasons:

a) δs <L α;

b) a node β ⊂ α defines a challenging pair at s.

However, neither of these can occur at a stage s ≥ t, by the induction hypothesis.

Therefore, by the construction, for all s ≥ t, chα,s = chα,t. If α 〈I〉 ⊂ f , then by the

construction gα = gα,t. If α 〈A〉 ⊂ f , then let r ≥ t be the first stage so that

a) α 〈A〉 ⊂ δr, and α 〈A〉 is not initialized after r;

b) α performed a diagonalization on a candidate pair 〈a1i′, a2

i′〉 at stage r.

By the construction, gα = gα,r.

The argument for (4) is the same as that given for (5) of Lemma 4.3. QED

Lemma 4.9 Each requirement Re is satisfied.

Proof: Let α ⊂ f be of length e . If e > 0, then let β be the predecessor node of α,

and let ~c consist of the range of gβ and all challenging pairs eventually defined by α.

60

If the final outcome of α is I, then one of the following is true:

1. not every element of ~c is in ran(g ◦ ϕ∆02

e );

2. the ordering of ~c is not consistent with the ordering of its preimage under g◦ϕ∆02

e ;

3. the size of some interval determined by ~c is different from its preimage under

g ◦ ϕ∆02

e ;

4. g ◦ϕ∆02

e maps an interval without candidate pairs to one with challenging pairs.

If the final outcome of α is A, then there is some r so that for all s ≥ r with

α ⊂ δs, the outcome of α is A. Our construction then dictates that all approximations

of ϕ∆0

2e including and after r map a candidate pair to elements in the same successor

chain in B. Consequently, ϕ∆0

2e is not an isomorphism. QED

Definition 4.10 A maximal Z-cluster is a maximal interval I of the form Z · C for

some linear ordering C.

Corollary 4.11 Let A be a ∆02-categorical linear ordering so that (A, S, L−, L+) is

computable. Then A has finitely many maximal Z-clusters.

Proof: Let c1, . . . , cn partition A so that condition (iii) of Proposition 4.6 holds.

Each of the open intervals determined by the must contain at most three maximal

Z-clusters. Therefore, A contains less than 4n+ 4 maximal Z-clusters. QED

Corollary 4.12 Let A be a ∆02-categorical linear ordering so that (A, S, L−, L+) is

computable. Then there is an n so that A can be written as a finite sum of intervals

of the following forms:

1. m for some m < n;

2. maximal ω-intervals, maximal ω∗-intervals;

61

3. maximal Z-clusters;

4. maximal intervals of the form Σq∈ηBq, where each Bq < n.

Proof: Order the finitely many maximal ω-intervals, maximal ω∗-intervals, and max-

imal Z-clusters in A. There remain finitely many intervals, each of which contains no

ω-intervals, no ω∗-intervals, and no Z intervals. Furthermore, there is n so that all

maximal finite discrete intervals have order type < n. By Lemma 3.12, each of these

intervals is either finite or of the form m1 + Σq∈ηBq +m2, where 0 ≤ m1,m2 < n, and

each Bq < n. QED

4.2.2 Linear orderings with intervals of the form Σq∈ηBq

The propositions in this subsection are analogous to Proposition 3.14 and Proposi-

tion 3.18. The diagonalizations employed are relatively straightforward implementa-

tions of the strategies described in these two results, so only sketches are provided.

Proposition 4.13 Let A be a linear ordering so that (A, S, L−, L+) is computable.

Let n be such that one of the following is true:

1. A has the form ω + Σq∈ηBq + 1, and for all q ∈ η, |Bq| < n;

2. A has the form ω + Z-cluster +Σq∈ηBq + 1, and for all q ∈ η, Bq < n.

Then A is not ∆02-categorical.

Sketch: We will discuss the proof for (2), as it is the more difficult case. The argu-

ment will involve a priority construction very much like the one in Proposition 4.6.

Differences will appear only in the details of candidate assignment and diagonaliza-

tion.

Consider α ⊂ δs working on requirement Re. If α ⊃ λ, the empty string, then

it receives from its predecessor node β the function gβ,s that it believes is a correct

62

approximation of g. Let ~d be the domain of gβ,s with gβ,s(~d) = ~c = c1, . . . , cj. For

α ⊇ λ, let c0 = the “0” of the ω summand and cj+1 = the “0” to the right of the

Σq∈ηBq summand.

The node α first determines which interval (ci, ci+1) contains both elements from

the Z-cluster and from the Σq∈ηBq summand. This can be done computably, since,

given an element a, α can see if there is a successor chain of at least n+ 1 elements

directly to the left of or to the right of a. The node α then determines if the preimage

of this interval under gβ,s ◦ ϕ∆02

e,s has the same property. If so, then α includes in the

range of gα,s a challenging element ai which is in the Σq∈ηBq summand but not

in the same successor chain as ci+1. The node α determines if a′i, the preimage of ai

under the function gα,s ◦ ϕ∆02

e,s , is a true candidate element; i.e., if it has properties

analogous to those of ai. If so, then α changes its approximation of g so that the

image under gα,s ◦ ϕ∆02

e,s of a′i is in ω or the Z-cluster. We thereby meet requirement

Re. QED

Corollary 4.14 Let A be a ∆02-categorical linear ordering so that (A, S, L−, L+) is

computable. Then every maximal Σq∈ηBq interval has a supremum and an infimum.

Recall Definition 3.16 and Lemma 3.17, which are about linear orderings with inter-

vals of the form Σq∈ηBq.

Proposition 4.15 Let A be a ∆02-categorical linear ordering so that (A, S, L−, L+)

is computable. Any maximal Σq∈ηBq interval satisfies Property 1.

Sketch: We must show that an ordering satisfying Property 2 cannot be ∆20-categorical.

First, as in Proposition 3.18, let m1 be the largest number for which there is an m2

as in Property 2. Define a partition of η so that for each m > m1 and each open

interval J of η defined by the partition, either Bq = m for all q ∈ J , or Bq 6= m for

all q ∈ J .

63

There is an open interval J = (q1, q2) (q1 might be −∞, and q2 might be ∞) of η

so that for all m > m1 and all q ∈ J , Bq 6= m, and Σq∈JBq satisfies Property 2 with

m1 and some other m2. We will build a computable copy B and g : B ∼= A so that

1. for all a 6∈ Σq∈JBq, g(a) = a; and

2. there is no ∆20 isomorphism between B and A.

Again, the argument will involve a priority construction very much like the one

in Proposition 4.6, and differences will appear mainly in the details of candidate

assignment and diagonalization.

Consider α ⊂ δs working on requirement Re. If α ⊃ λ, the empty string, then

it receives from its predecessor node β the function gβ,s that it believes is a correct

approximation of g. Let ~d be the domain of gβ,s with gβ,s(~d) = ~c = c1, . . . , cj. For

α ⊇ λ, define c0 as follows:

1. c0 = the greatest element of Bq1 if q1 6= −∞;

2. c0 = the greatest bound of Σq∈ηBq in A if q1 = −∞.

We define the element cj+1 similarly.

In each interval (ci, ci+1), α searches for challenging pairs 〈a1i , a

2i 〉 where a1

i is in

a maximal successor chain of length m1 but not in the same successor chain as ci or

ci+1; and a2i is in a maximal successor chain of length m2 but not in the same successor

chain as ci or ci+1. (The hypotheses on Σq∈ηBq guarantee that α will always be able

to find an interval with such a pair.) The node α then includes a1i , a

2i in the range of

gα,s. The node α then determines if the pair a1i′, a2

i′, the preimage of the pair a1

i , a2i

under gα,s ◦ ϕ∆02

e,s , is a true candidate pair; i.e., if it has analogous properties. If so,

then α attempts to change its approximation of g so that the image under gα,t ◦ ϕ∆02

e,t

of a2i′ is an element in a maximal successor chain of length m1. We thereby meet Re.

64

Corollary 4.16 Let A be a ∆02-categorical linear ordering so that (A, S, L−, L+) is

computable. Then there is n ∈ ω so that A can be written as a finite sum of maximal

intervals of the following forms:

1. m for some m < n;

2. m · η for some m < n;

3. ω, ω∗;

4. Z-clusters,

where each maximal m · η interval has an infimum and a supremum.

4.2.3 Z-clusters

When we examined relativized ∆02-categoricity, we showed, in a single proposition,

that a relatively ∆02-categorical linear ordering could have only finitely many maximal

ω, ω∗,Z intervals. There we were not concerned with computability issues in our

arguments; we merely needed to demonstrate the existence of certain tuples without

describing how to find them. Throughout our study of ∆02-categoricity, however, such

computability concerns are central, because the copy B we produce via the priority

construction must be computable.

Each maximal ω and ω∗ interval has an element with either no successor or no

predecessor, so under the extra decidability assumptions on A, we can easily identify

certain infinite intervals. This property played a key role in the proof of Propo-

sition 4.6. A Z-cluster, of course, has no such element. Consequently, it may be

impossible to identify the infinite intervals computably, and thus it is harder to for-

mulate a diagonalization to prove that A has only finitely many Z-intervals.

65

Proposition 4.17 Let A be a ∆02-categorical linear ordering so that (A, S, L−, L+) is

computable. Then any Z-cluster contained in A actually contains only finitely many

Z-intervals.

The argument must be broken down into a few subpropositions, one of which will

require a slightly more complicated priority construction in its proof.

Proposition 4.18 Let A be a linear ordering so that (A, S, L−, L+) is computable.

Furthermore, let A contain a maximal Z-cluster I of the form Z·C, where C is infinite

and discrete. Then A is not ∆02-categorical.

Proof: Since C is infinite and discrete, I must be a sum of the form c1(Z · ω) +

c2[(Z ·Z) ·D] + c3(Z ·ω∗), where D is some linear ordering and each c is either 0 or 1.

Furthermore, if there is anything in A to the left of I, then by Corollary 4.16, either

I has an infimum, or I is directly to the right of a maximal ω-interval. Similarly, if

there is anything in A to the right of I, then either I has a supremum, or I is directly

to the left of a maximal ω∗-interval.

Case I: c2 = 0. Then A contains an interval K with endpoints of one of the following

forms:

a) 1 + c1(Z · ω) + c3(Z · ω∗) + 1;

b) 1 + c1(Z · ω) + c3(Z · ω∗) + ω∗;

c) ω + c1(Z · ω) + c3(Z · ω∗) + 1;

d) ω + c1(Z · ω) + c3(Z · ω∗) + ω∗.

In any case, we can construct A′, a computable copy of A, where K ′, the image of

this interval K, is such that

66

1. K ′ contains infinitely many Z-intervals; and

2. for a, b ∈ K ′ we can computably decide if the interval (a, b) is infinite or finite.

We can show thatA′, and hence A, is not ∆02-categorical. The argument is almost ex-

actly like that given in Proposition 4.6, except that we replace searching for elements

with no successor or predecessor with searching for intervals which are infinite.

Case II: c2 = 1. If the linear ordering D = 1 or η, then we can argue that there

is A′ and K ′ with the properties i) and ii) listed above. However, if the linear

ordering D has a successor pair, then A has an interval, with endpoints, of the type

ω+ (Z ·ω) + (Z ·ω∗) +ω∗. Hence, we can again argue for the existence of A′ and K ′.

Therefore, A is not ∆02-categorical.

Proposition 4.19 Let A be a ∆02-categorical linear ordering so that (A, S) is com-

putable, and let I = Z · C be a maximal Z-cluster, where C is not discrete. Then A is

not ∆02-categorical.

Proof: This argument requires a bit more than the minor adjustments to Proposi-

tion 4.6 needed by Proposition 4.13 and 4.15. Therefore, we provide a more detailed

construction and a subsequent list of lemmas.

Of course, we attempt to construct a computable B ∼= A so that there is no ∆02

isomorphism between them. We employ a tree construction where each node has

two outcomes, I (inactive) and A (active), with I < A. Nodes of length e work on

requirement Re. Finally, the construction must determine an isomorphism g : B ∼= A.

At stage s we will define the approximations δs and gs, and the ordering Bs.

First, without loss of generality, let J be a Z-interval in I so that there is a Z-

interval in I to the right of J , but there is no next Z-interval to the right of it. We

fix c−∞, some element in J , and c∞, some element in I to the right of J . For all

67

a ∈ A − (c−∞, c∞), we define g(a) = a. Throughout the construction, we tacitly

assume that any tuple ~a from A or ~b from B has elements only from (c−∞, c∞).

Challenging elements and initialization

Consider α ⊂ δs working on requirement Re. If α ⊃ λ, the empty string, then

it receives from its predecessor node β the function gβ,s that it believes is a correct

approximation of g. Let ~d be the domain of gβ,s with gβ,s(~d) = ~c = c1 < . . . < cj . For

α ⊇ λ, let c0 = c−∞ and cj+1 = c∞.

The node α guesses at which is the first interval (ci, ci+1) to be infinite. (Thus,

c0, c1, . . . , ci are all in the same Z-interval.) In this interval α locates a challenging

element aα,s so that (ci, aα,s) and (aα,s, ci+1) both seem infinite. The node α includes

it in the range of gα,s. The element aα,s challenges (c′i, c′i+1), the preimage under g◦ϕ∆0

2e

of (ci, ci+1), to contain an element a′α,s with the analogous property. If the challenge

seems to be met at stage t, then α changes its approximation so that the image under

gα,t ◦ ϕ∆02

e,t of a′α,s is in the same Z-interval as c0; we thereby meet requirement Re.

In the definition of challenging elements, this construction resembles Theorem 4.1

more than Proposition 4.6. First, at stage s, α will define at most a single chal-

lenging element in one interval, rather than challenging pairs in multiple intervals.

Second, α will make incorrect guesses as to which element has the above properties.

In Proposition 4.6, α could determine computably that an element had no successor

or no predecessor, and so it never mistakenly assigned pairs; any cancelling of a chal-

lenging pair was done by a node of higher priority. Here, however, α may incorrectly

guess the first element of ~c to be in a Z-interval different from c0. Furthermore,

even when α eventually guesses the correct interval (ci, ci+1), it may incorrectly guess

that an element is a challenging element when it actually is in the same Z-interval

as either ci or ci+1. Nevertheless, as in Theorem 4.1 both of these incorrect guesses

should introduce only finitely much injury, and a 0′′ construction still suffices.

68

A node α ⊂ δs may initialize another node γ: if γ ⊆ δt for some t < s, then

α cancels the assignment of aγ,t and the approximation gt. There will be three

circumstances when a node will initialize another:

1. if α redefines its challenging element at stage s, then α initializes each γ with

α ⊂ γ;

2. if α determines its outcome to be I, then α initializes all γ with α 〈I〉 <L γ;

3. if α diagonalizes against ϕ∆0

2e at stage s, then it initializes all γ with α 〈A〉 ⊆ γ.

Construction and supporting lemmas

Stage 0: g0 = ∅. δ0 = λ. Assign no challenging elements.

Stage s + 1: Of course, λ ⊂ δs+1. Let α ⊂ δs+1 be a node of length e. Let β be

the predecessor node, and let dom(gβ,s) = ~b, with gβ,s(~d) = ~c = c1 < . . . < cj . (If

α = λ, then there is no β and gβ,s+1 = ∅.) For α ⊇ λ, let c0 = c−∞ and cj+1 = c∞ as

elements of A. As elements of B let d0 = c−∞ and dj+1 = c∞. The node α searches

for the first interval (ci, ci+1), and the first element aα,s+1 so that

1. after s+ 1 steps of enumerating the diagram of (A, S), ci+1 appears to be in a

different Z-interval from c0;

2. aα,s+1 ∈ (ci, ci+1) appears to be in a Z-interval different from both ci and ci+1;

3. in the course of this searching for aα,s+1, α does not discover that ci+1 is in the

same interval as c0.

This element aα,s+1 is the challenging element for α at stage s+ 1.

Case I: The node α has been initialized since its last active stage or has defined a

new challenging element since its last active stage.

69

The node α initializes all γ with α ⊂ γ and chooses outcome I. It must include

aα,s+1 in the range of gα,s+1. We may assume by induction on the construction that

~d includes all elements which are in the domain of an uncanceled approximation

gt where t < s + 1 and δt <L α. Let ~b be the set of elements in Bs but not in

the domain of gβ,s+1. Again, we may assume by induction on the construction that

gβ,s+1 is compatible with the ordering Bs; therefore, α can extend gβ,s+1 so that it

includes ~b and e in the domain, and aα,s+1 and e in the range. Let bα,s+1 be such that

gα,s+1(bα,s+1) = aα,s+1.

[[Summary: After redefining aα,s+1, the node α restarts its work on the requirement

Re.]]

Case II: α does not need to redefine its challenging pair.

There is a stage t < s+ 1 so that

1. α ⊂ δt;

2. either α has been initialized since the previous stage r in which α ⊂ δr, or the

definition of aα,t changed during stage t;

3. α has not been initialized since stage t; and

4. aα,s+1 = aα,t;

(Throughout the rest of this construction, we will refer to this particular t.)

We assume by induction on the construction that since α has not been initialized

since t, the portion of gs+1 defined by nodes above α is the same as the portion of gt

defined by nodes above α. The node α defines g′α,s+1, its tentative version of gα,s+1,

to be the same as gα,t. We now consider the following subcases.

70

Case IIa: One of the following is true:

1. An element of ~d or one of d0, dj+1, bα,s+1 is not in the range of ϕ∆0

2e,s+1. (Otherwise,

let ~c ′, c′0, c′j+1, aα,s+1

′ be such that ϕ∆0

2e,s+1(~c

′) = ~d, ϕ∆0

2e (aα,s+1

′) = bα,s+1, and

ϕ∆0

2e (c′0, c

′j+1) = d0, dj+1.)

2. The ordering of ~c, c0, cj+1, aα,s+1 does not match that of ~c ′, aα,s+1′, c′0, c

′j+1.

3. The size of some interval determined by ~c, c0, cj+1, and aα,s+1 does not match

the size of the corresponding interval determined by ~c ′, c′0, c′j+1, aα,s+1

′.

4. If t′ < s+1 is the greatest stage so that α ⊆ δt′ , then our guess at the preimage

of ~d or one of bα,s+1, d0, dj+1 under ϕ∆0

2e has changed since t′.

(Otherwise, the element aα,s+1′ is a candidate element.)

The outcome of α is I, and gα,s+1 = g′α,s+1.

[[Summary: If the approximation g′α,s+1 defined so far is part of the isomorphism g

itself, then ϕ∆0

2e doesn’t appear to be an isomorphism, because it does not seem total,

as in (1) and (4), or g′α,s+1 ◦ ϕ

∆02

e,s+1 does not appear to be an automorphism of A, as

in (2) and (3).]]

Case IIb: Not Case IIa and all of the following are true:

1. there is a stage r with t < r < s + 1 so that α 〈A〉 ⊂ δr, and α 〈A〉 has not

been initialized since r;

2. α performed a diagonalization on a candidate element (aα,s+1)′ at stage r;

3. if K is an interval determined by aα,s+1 and elements in the range of higher

priority gw, and during the diagonalization α guessed that K was infinite, then

α can find s+ 1 elements in K and still guesses that K is infinite.

71

The outcome of α remains A, and gα,s+1 = gα,r.

[[Summary: The node α seems already to have diagonalized successfully against Re

at an earlier stage, and nothing has injured this work.]]

Case IIc: Neither Case IIa nor Case IIb

First, α initializes each node γ with α 〈A〉 ⊆ γ. The node α attempts to perform

the following diagonalization on the candidate element aα,s+1′. If it completes the

diagonalization, then the outcome of α is A.

Diagonalization: Let w be the greatest number so that t ≤ w < s + 1, α 〈I〉 ⊆ δw,

and gw is uncanceled. Let dom(gw = ~v with gw(~v) = ~u. By our construction, the

approximation gw must extend g′α,s+1.

Let ~b be the set of elements in Bs but not in the domain of gw. By induction

on the construction, we can assume that gw is compatible with this ordering, and

α can define an extension of gw whose domain includes ~b. Let the image of ~b be

~a. Throughout the diagonalization, α must leave fixed the mapping gβ,s+1(~d) = ~c.

However, α will attempt to alter the mapping of the rest of ~v,~b, bα,s+1 so that it meets

requirement Re while it respects gw.

Recall that g′α,s+1 is α’s tentative contribution to gs+1, and for the challenging

element aα,s+1, α has defined g′α,s+1(bα,s+1) = aα,s+1. Furthermore, aα,s+1

′ seem-

ingly has met the challenge, because ϕ∆0

2e (aα,s+1

′) = bα,s+1, and each of the intervals

(c′i, aα,s+1′), (aα,s+1

′, c′i+1) appears to be infinite. The goal is to define gα,s+1(bα,s+1) =

aα,s+1′′ so that ci, aα,s+1

′′ are in the same Z-interval, thus ensuring that ϕ∆0

2e is not

an isomorphism if ϕ∆0

2e,s+1 is a correct approximation of ϕ

∆02

e .

Let v1, . . . , vk be the elements of ~v currently mapped to elements u1, . . . , uk in

(ci, ci+1). Assume, without loss of generality, that the ordering of these elements in

A is of the form

ci < u1 < . . . < ul < aα,s+1 < ul+1 < . . . < uk < ci+1.

72

In each of the following intervals, the node α attempts to find s + 1 elements or to

determine that the interval is finite:

1. (aα,s+1, ul+1);

2. (uj, uj+1) for some j ∈ {l + 1, . . . , k − 1};

3. (uk, ci+1).

Let u∗j be the right endpoint of the leftmost interval in which α can locate s +

1 elements without discovering that it is finite. (There must be such an interval.

Otherwise α discovers that (aα,s+1, ci+1) is finite, redefines its challenging element

and enters Case I.) Let ~u ′′ be the list of elements of ~u ∪ ~a which are in (ci, u∗j), and

~u ′′′ the list of elements of ~u ∪ ~a which are in [u∗j , ci+1).

For each element of ~u ′′′, leave the mapping of the corresponding element of ~v∪~b as

it is. The node α then defines gα,s+1 so that for every element of ~u ′′, the corresponding

element of ~v∪~b is now mapped to an element in the same Z-interval as ci (and hence,

as c0). Finally, α finishes the definition of gα,s+1 by including e in the domain and

range. This concludes the diagonalization.

After each α ⊂ δs+1 has determined gα,s+1, we define gs+1 = ∪α⊂δs+1gα,s+1 and

use it to define Bs+1. This concludes the construction.

Lemma 4.20 For each s, δs and Bs satisfy the following properties:

1. If β is the predecessor node of α ⊂ δs, then gβ,s ⊆ gα,s.

2. Let α 〈o〉 ⊂ δs, and let t < s be the greatest stage so that δt <L α 〈o〉. If ~v is

the set of all elements in the domain of gt, then ~v ⊆ dom(gα,s).

3. Let o be a fixed outcome, α 〈o〉 ⊂ δs, and t < s be the last stage with α 〈o〉 ⊆ δt.

If gα,s 6= gα,t, then every node γ ⊇ α 〈o〉 is initialized at some stage t′ with

t < t′ ≤ s.

73

4. Let α, ~v, and gt be as in (2). If K is an interval determined by gt(~d) and K ′ is

the corresponding interval determined by gα,s(~d), then |K| ≥ min{|K ′|, s}.

5. Bs extends Bs−1.

6. Bs is compatible with each gt not canceled by stage s.

Proof: We again use induction on the stage s. For stage 0, (1) - (6) are trivially true.

Assume (1) - (6) are true for stage s. We must show them true for stage s+ 1.

We use induction on the length of the node α ⊂ δs+1. Assume that (1) - (4) are

true for all β ⊂ α. We must show them true for α.

The arguments for (1) - (4) are essentially the same as those in Lemma 4.2.

The argument for (5) is the same as in Lemma 4.7.

6) Let t < s + 1. First, if δt ⊂ δs+1, then by (3) either gt is canceled or

gs+1 ⊇ gt. If δs+1 <L δt, then gt is canceled. If δt <L δs+1, then by (2) we know

that dom(gt) ⊆ dom(gs+1). Furthermore, gt completely determines Bt, and gs+1 com-

pletely determines Bs+1, and by (5), Bt and Bs+1 are consistent. Consequently, we

know that the ordering of gt(dom(gt))and gs+1(dom(gt)) are the same. However, at

the end of stage s + 1, there may be other elements not in dom(gt) that are ordered

in Bs+1. Can gt be extended to an order-preserving map between Bs+1 and A? In

short, are the intervals determined by gt(dom(gt)) big enough to accommodate the

images of the new elements of Bs+1?

For each interval K determined by gt(dom(gt)), let K ′ be the corresponding in-

terval determined by gs+1(dom(gt)). For each such |K|, the construction guarantees

that |K| ≥ min{|K ′|, s + 1}. Since Bs+1 is determined entirely by gs+1, and only

0, . . . , s + 1 are ordered in Bs+1, each interval K is big enough to accommodate the

images of all elements of Bs+1. In short, the desired compatibility is guaranteed.

QED

74

Lemma 4.21 There is a true path f with the following features:

1. If α ⊂ f , then α is the left-most node of length |α| so that α ⊂ δs for

infinitely many s.

2. If α ⊂ f , then α does not define challenging pairs infinitely often.

3. If α 〈A〉 ⊂ f , then α does not actively diagonalize infinitely often.

4. If α 〈o〉 ⊂ f and S = {s : α 〈o〉 ⊂ δs}, then lims∈S

gα,s = gα; i.e., there are

t ∈ S and a tuple ~d so that for all s ∈ S with s ≥ t, dom(gα,s) =

dom(gα,t) = dom(gα) = ~d, and gα,s(~d) = gα,t(~d) = gα(~d).

5. If g =⋃α⊂f gα, then g : B ∼= A.

Proof: We show (2) - (4) by simultaneous induction on the length of α ⊂ f . Assume

(2) - (4) are true for all β ⊂ α. Let t be the least stage and ~d, ~c be the tuples so that

a) for all stages s ≥ t, α ≤L δs;

b) for all nodes β ⊂ α and all s ≥ t, aβ,s = aβ,t−1;

c) if β 〈A〉 ⊂ f , then β does not diagonalize after stage t− 1;

d) if β is the predecessor node of α, then t and ~d witness that (3) is true for β,

and gβ(~d) = ~c; and

e) aα,t+1 is in (ci, ci+1), ci+1 is the least element of ~c not in the same Z-interval as

c0, and aα,t+1 is not in the same Z-interval as either ci nor ci+1.

First, we claim that α is not initialized at any stage s ≥ t. A node can be initialized

at stage s for one of three reasons:

75

a) α <L δs;

b) a node β ⊆ α redefines aβ,s;

c) a node β 〈A〉 ⊂ α actively diagonalizes at stage s.

However, none of these can occur at s ≥ t, by the induction hypothesis. Therefore,

by the construction, for all s ≥ t, aα,s = aα,t. If α 〈I〉 ⊂ f , then by the construction

gα = gα,t.

If α 〈A〉 ⊂ f , let r ≥ t be the first stage so that α 〈A〉 ⊂ δr, and for all s ≥ r,

α 〈I〉 6⊆ δs. Recall that w < r is the greatest number so that t ≤ w < s + 1,

α 〈I〉 ⊆ δw, and gw is uncanceled by r; ~v is the domain of gw; and gw(~v) = ~u. Let

r′ ≥ r be the first stage where α 〈A〉 ⊂ δr′ , and u∗j chosen by α is the right endpoint

of the leftmost infinite interval determined by aα,r and the elements of ~u between

aα,r′ and ci+1. By the construction, gα = gα,r′.

The argument for (5) is the same as that given in Lemma 4.3. QED

Lemma 4.22 Each requirement Re is satisfied.

Proof: Let α ⊂ f be of length e . If e > 0, then let β be the predecessor node of α,

and let ~c consist of the range of gβ and all challenging elements eventually defined by

α.

If the final outcome of α is I, then one of the following is true:

1. not every element of ~c is in ran(g ◦ ϕ∆02

e );

2. the ordering of ~c is not consistent with the ordering of its preimage under g◦ϕ∆02

e ;

3. the size of one of the intervals determined by ~c is different from its preimage

under g ◦ ϕ∆02

e .

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If the final outcome of α is A, then there is some r so that for all s ≥ r with

α ⊂ δs, the outcome of α is A, and α does not actively diagonalize at any stage after

r. Our construction then dictates that all approximations of ϕ∆0

2e including and after

r map c′−∞ and a′α,r, two elements not in the same Z-interval in A, to d∞ and bα,r,

two elements in the same Z-interval in B. Consequently, ϕ∆0

2e is not an isomorphism.

QED

4.3 Some open questions about ∆02-categoricity

We pose a few questions about ∆02-categoricity, suggested by our results and the

works of others cited here:

1. Can the hypotheses about the effectiveness of one copy of our structure be

weakened? Can we produce examples to show what the weakest hypotheses

might be?

2. The extra hypotheses we formulated are implied if we have 2-decidability. Gon-

charov showed that the existence of a single 2-decidable copy of a structure is

enough to establish the equivalence of relativized and unrelativized computable

categoricity. What is true for the ∆02 level?

3. What exactly is the relationship between 1-decidable computable categoricity

and ∆02-categoricity? (See Theorem 3.20.) Is the correspondence here an in-

stance of a more general result?

4. Is there a general notion for ∆02-categoricity similar to Goncharov and Dzgoev’s

branching?

77

CHAPTER 5

PARTIAL RESULTS ∆03-CATEGORICITY

We discuss results which suggest that ∆03-categoricity may be very difficult to charac-

terize for linear orderings and possible to characterize for Boolean algebras. In fact,

we characterize relativized ∆03-categoricity for Boolean algebras.

5.1 Linear orderings

In every definition of categoricity and dimension we have given thus far, we have

assumed that our original structure A is computable. The definition of relative

computably categoricity and the syntactic characterization given in [4] actually apply

to structures that are not computable, and even to those with no computable copy.

Definition 5.1 A structure A is relatively ∆0α-categorical if for every B ∼= A, there

is a ∆0α(D(B),D(A)) isomorphism ϕ

∆0α(D(B)⊕D(A))

e : B ∼= A.

Definition 5.2 A relatively computable Σα formula for A, or Σcα(D(A)) formula, is

a Σα formula involving disjunctions and conjunctions over sets computably enumer-

able relative to D(A).

Definition 5.3 A relatively formally Σ0α Scott family for A is a c.e. family of

Σcα(D(A)) formulas with the same two properties as in Definition 1.9.

Theorem 5.4 A structure A is relatively ∆0α-categorical iff it has a relatively for-

mally Σ0α Scott family.

78

When we proved that a linear ordering or Boolean algebra A with certain proper-

ties was not ∆02-categorical, we used neither the assumption that A was computable,

nor that the supposed Scott family had computable Σ2 formulas. Rather, what we

really proved was that any A with these properties had no Scott family consisting

only of Σ2 formulas. Similarly, when we proved that a linear ordering or boolean al-

gebra A of a certain isomorphism type had a formally Σ02 Scott family, we never used

the assumption that it was a computable copy of this structure. Consequently, we

completely characterized the relatively ∆02-categorical linear orderings and Boolean

algebras under this new definition as well. It is simply the case that each of the

orderings in our class has a computable copy.

Using this more general definition of relativized computable categoricity, Knight

established the following result.

Theorem 5.5 1) For each subset S of ω, there exists a relatively ∆03-categorical

linear ordering AS so that AS 6∼= AT if S 6= T . Therefore, there are 2ℵ0 different

relatively ∆03-categorical linear orderings.

2) For each properly Σ03 subset S of ω, there exists a computable linear ordering AS

so that

a) AS is a shuffle sum of finite discrete linear orderings of size 2n+ 1 for each

n ∈ ω, and 2n+ 2 for each n ∈ S;

b) AS is relatively ∆03-categorical; and

c) AS has no 2-decidable copy.

Proof: Part 1): Let S be an arbitrary subset of ω. Construct for this set S the shuffle

sum AS as described in Part 2) for a Σ03 set. First, if S 6= T , then AS 6∼= AT , because

if n ∈ S − T , then AS has a maximal successor chain of length 2n+ 2, and AT does

79

not. Furthermore, AS has a formally Σ03 Scott family, because we can say all of the

following with finitary Σ3 formulas:

1. two elements x1 and x2 have an element between them with no successor;

2. the elements x1, . . . , xk are the mth1 , . . . ,m

thk elements in a maximal successor

chain of length exactly n.

Part 2): We construct the linear ordering stage-wise: at the end of each stage t, we

define the finite linear ordering At, and A =⋃t∈ωAt.

Let ρ be a ∆03 enumeration of S. By the Limit Lemma there are there are a ∆0

2

function f(k, s) so that lims→∞ f(k, s) = ρ(k), and a ∆01 function g(k, s, t) so that

limt→∞ g(k, s, t) = f(k, s).

Each element that we include in the ordering at stage s will be part of a string of

elements designated as a chain. The idea is that we guess that elements in the same

chain at stage s actually comprise a maximal successor chain in the structure A we

build.

Construction

Stage t: Let the chains at the end of stage t− 1 be listed c1, . . . , cl. (The chains are

considered to be listed by the order in which they are designated, not by their order

in A. Therefore, the chain ci at this stage is also the ith chain at any subsequent

stage; however, as we shall see, the exact elements in the chain ci may change in later

stages.)

For each k, s ≤ t, compute g(k, s, t). For each k ≤ t, find the least number s ≤ t

so that

1. g(k, s, t) = n; and

2. the set Ss,n,k,t = {r : s ≤ r ≤ t and for all q with s ≤ q ≤ r, g(k, q, t) = n} is

larger than Ss,n,k,u for all u < t.

80

Denote this s by sk,t. (Note that sk,t always exists, because if n = g(k, t, t), then

St,n,k,t = {t}, and St,n,k,u = ∅ for all u < t by definition.)

Case I: There is already a chain ci, i ≤ l, marked with 〈k, sk,t, n〉. For each such

chain, do the following:

1. if the chain is of length 2n+ 2, then leave it alone;

2. if the chain is of length 2n + 3, then split the last element from the chain, to

make ci a chain of length 2n+ 2 and to make a new chain cj of length 1.

Case II: There is not a chain marked with 〈k, sk,t, n〉. Add a new chain cj of length

2n+ 2 to the ordering and mark it with 〈k, sk,t, n〉.

For a chain ci, i ≤ l, with marking 〈k, r,m〉 do the following:

1. if r > sk,t, then make ci a chain of length 2m + 3 (if it is not already so) by

adding a new element to the end of ci, and remove any markings;

2. if r = sk,t and m 6= n, then make ci a chain of length 2m+3 (if it is not already

so) by adding a new element to the end of ci, and remove any markings;

3. if r < sk,t, then make ci a chain of length 2m + 3 (if it is not already so) by

adding a new element to the end of ci, but do NOT remove any markings.

This concludes the work we do for each k ≤ t.

For each m ≤ t, let a new chain of length of 2m + 1 be added to the ordering.

“Shuffle” the chains currently in the ordering; if a chain is marked with 〈k, sk,t, n〉,

then so is every copy of it added to the ordering. This concludes the construction of

At.

81

We note the following obvious but important facts about our construction:

1. If an element a is in chain ci at some stage t and in cj, j 6= i, at another stage

u > t, then:

(a) at stage t, the element a is the last element in ci, which has odd length at

this stage;

(b) and at some stage v ≤ u the element a is removed from the marked chain

ci to form the unmarked chain cj of length 1.

2. Chains never change or regain markings; i.e., if a chain ci is marked with 〈k, s, n〉

at stage t, and this marking is removed at a stage u ≥ t, then it is unmarked

at all stages v ≥ u.

3. If a chain ci is unmarked at any stage u, then at the end of stage u it is a

chain of odd length 2m+ 1; it remains unmarked forever; and, in A, it forms a

maximal discrete chain of length 2m+ 1.

4. If an element a is the last element in a chain ci of length 2m+ 1 at some stage

u, then either a is the last element of a maximal discrete chain of length 2m+1

in A, or a has neither predecessor nor successor in A.

Lemma 5.6 Fix k. By the limit lemma, there is a least number s so that f(k, r) =

ρ(k) for all r ≥ s. This number s is also the least number so that sk,t = s for infinitely

many t.

Proof: Let s be as stated, and let u ≥ s, k be such that for all t ≥ u and all q ≤ s,

g(k, q, t) = f(k, q).

Let r < s. Therefore, r ≤ s − 1 < s and f(k, s − 1) 6= f(k, s). For all t ≥ u,

g(k, r, t) = f(k, r), and Sr,f(k,r),k,t ⊆ {r, . . . , s−1}. Therefore, there must be a t′ ≥ u

so that |Sr,f(k,r),k,t′| ≥ |Sr,f(k,r),k,t| for all t ≥ t′. For all t > t′, sk,t 6= r.

82

Assume that there is some t′ ≥ u so that for all t ≥ t′, |Ss,f(k,s),k,t′| ≥ |Ss,f(k,s),k,t|.

But |Ss,f(k,s),k,t′| ≤ t′ + 1 by definition. Let u′ ≥ u, (s + t′ + 1) be such that for

all t ≥ u′ and all q with s ≤ q ≤ (s + t′ + 1), g(k, q, t) = f(k, q) = f(k, s). Then

g(k, s, t) = f(k, s) and Ss,f(k,s),k,u′ ⊇ {s, . . . , s + t′ + 1}, so |Ss,f(k,s),k,u′| ≥ (t′ + 2),

a contradiction. Therefore, there are infinitely many stages t at which g(k, s, t) =

f(k, s) and Ss,f(k,s),k,t is bigger than Ss,f(k,s),k,v for all v ≤ t. And so, by the definition

of sk,t and the above paragraph, s is the least number for which there are infinitely

many stages t with sk,t = s. QED

Lemma 5.7 An element a ∈ A is in a maximal discrete chain of length 2n + 2 iff

there are k, s, n, u, t and a chain ci so that

1. s is the least number for which f(k, r) = ρ(k) = n for all r ≥ s;

2. t is a stage for which

(a) sk,t = s;

(b) for all r < s and v > t, sk,v 6= r;

(c) for all v ≥ t, g(k, s, v) = f(k, s); and

3. at the end of some stage u ≥ t, the chain ci has length at least 2n+2; the element

a is one of the first 2n+ 2 elements in ci; and ci is marked with 〈k, s, n〉.

Proof: (⇒) Assume that a is in a maximal chain of length 2n+ 2. The facts at the

end of the construction guarantee that there exist a chain ci, a stage u, and a mark

〈k, s, n〉 so that the element a is one of the first 2n+2 elements in ci, and ci is marked

with 〈k, s, n〉 at all stages v ≥ u.

If there is r < s so that at some stage v > u, r = sk,v, then all marks 〈k, s, n〉

are removed at stage v. Thus, ci becomes a chain of length 2n + 3 and its mark is

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removed at stage v; therefore, in A, a is in a maximal discrete chain of length 2n+ 3,

a contradiction.

If there is a stage u′ ≥ u so that sk,v 6= s for all v ≥ u′, then ci is chain of

length 2n + 3 at stage u′, and the last element is never “removed” from the chain.

Consequently, in A, a is in a maximal discrete chain of length 2n+3, a contradiction.

Thus, s is the least number so that sk,v = s for infinitely many v. By the previous

lemma, s is the least number so that f(k, r) = ρ(k) for all r ≥ s. Therefore, there is

a stage t which satisfies condition 2, and we can, if necessary, redefine u to be greater

than or equal to t.

Finally, if n 6= ρ(k), then let v ≥ u so that sk,v = s, and g(k, s, v) = f(k, s) =

ρ(k) 6= n. Then ci, which is marked with 〈k, s, n〉, becomes a chain of length 2n+3 (if

it is not so already), and its mark is removed. Consequently, in A, a is in a maximal

discrete chain of length 2n+ 3, a contradiction. Therefore, n = ρ(k).

(⇐) Assume there are k, s, n, u, t, and ci satisfying the above conditions for a ∈ A.

By the construction, ci cannot lose its marking after stage u. By the previous lemma,

there is t′ ≥ u with sk,t′ = s; at this stage, the chain ci = a1 < · · · < a2n+1, and

one of these elements is a. At later stages, we possibly add elements only to the end

of this chain, not to the beginning or between elements, and we “shuffle” chains at

the end of every stage. Therefore, a1 definitely has no immediate predecessor, and

ai+1 is the direct successor of ai for i = 1, . . . , 2n. And so, we need only verify that

a2n+2 has no immediate successor. If b > a2n+2 is in another chain some stage, then

certainly it is not an immediate successor, because we never combine chains in our

construction, and we shuffle chains at the end of every stage. If b > a2n+2 is in the

same chain at some stage u′ ≥ t′, then at the first stage t′′ ≥ u′ in which sk,t′′ = s,

this element b is removed from the chain with a2n+1 and placed in a chain of length

1. QED

84

We are now ready to finish our proof. Assume that n ∈ S. Then there are k, s, n, t

so that

1. s is the least number for which f(k, r) = ρ(k) = n for all r ≥ s; and

2. t is a stage for which

(a) sk,t = s;

(b) for all r < s and v > t, sk,v 6= r;

(c) for all v ≥ t, g(k, s, v) = f(k, s).

At the end of this stage t, we define a chain ci of exactly 2n + 2 and mark it with

〈k, s, n〉. By the previous lemma and its proof, we know that this chain is, in A, a

maximal discrete chain of length 2n+ 2.

By condition 1 of the previous lemma and the properties of the functions f(k, r)

and ρ(k), if there is a maximal discrete chain of length 2n+ 2 in A, then n ∈ S.

The actions performed at the end of each stage guarantee that there are maximal

discrete chains of every odd length and that A is indeed a shuffle sum of the desired

discrete chains.

Finally, we note that A has no 2-decidable copy. If it did, then we could com-

putably list the elements of S, since we could determine the exact length of any

maximal finite chain in A. Of course, this is a contradiction if S is a properly Σ03 set.

QED

The first part of this theorem shows that the relativized and unrelativized ∆03-

categorical linear orderings comprise two very different classes. The second part

suggests that perhaps the task of characterizing the ∆03-categorical linear orderings

will prove extremely difficult, perhaps impossible.

85

In addition, the second part reveals that ∆03-categoricity and computable cat-

egoricity of 2-decidable linear orderings do not define the same class of orderings.

Therefore, there is not, in general, a correspondence between the ∆0n+1-categorical

linear orderings and the n-decidable computably categorical linear orderings, as The-

orem 3.20 and Theorem 4.5 might have suggested.

5.2 Boolean algebras

Recall the definition of A/ ∼ given in Definition 3.6.

Definition 5.8 A Boolean algebra A is rank 1 iff A/ ∼ is a nontrivial atomless

Boolean algebra.

Definition 5.9 A Boolean algebra A is atomic iff for each a ∈ A, there is an atom

b ≤ a.

Proposition 5.10 Any countable rank 1 atomic Boolean algebra is isomorphic to

I(2 · η).

Proof: Let A be a rank 1 atomic Boolean algebra, and let B = I(2 · η). Let p0(0A) =

0B, and p0(1A) = 1B. Assume pn is given and isomorphically maps a finite subalgebra

An onto a finite subalgebra Bn so that pn preserves the size of elements. Let one of the

atoms a of An be split to obtain a new finite subalgebra whose atoms include a1, a2,

where a1 ∪ a2 = a. Because a and pn(a) have the same size, and any infinite element

of A or B has infinitely many atoms and can be split into two infinite elements, we

can easily find b1, b2 so that b1 ∪ b2 = pn(a), b1 has the same size as a1, and b2 has

the same size as a2. The “back” argument works in precisely the same way. QED

Proposition 5.11 Let A be a countable Boolean algebra so that

1. A has infinitely many atoms, but cannot be split into 2 algebras each having

infinitely many atoms; and

86

2. any element a ∈ A bounding infinitely many atoms must bound an atomless

element, as well.

Then A is isomorphic to I(ω + η).

Proof: Let A have the above properties, and let B = I(ω+ η). Let p0(0A) = 0B, and

p0(1A) = 1B. Assume pn is given and isomorphically maps a finite subalgebra An onto

a finite subalgebra Bn so that pn preserves the size of elements and the number of

atoms bound by elements. Let one of the atoms a of An be split to obtain a new finite

subalgebra whose atoms include a1, a2, where a1 ∪ a2 = a. If a is finite or contains

only finitely many atoms, then we can easily find images b1, b2 so that b1∪b2 = pn(a),

and each bi matches ai both in size and number of atoms. If a contains infinitely

many atoms, then exactly one of a1, a2 does; assume that it is a1. If a2 is finite or

infinite, we can find images b1, b2 so that so that b1 ∪ b2 = pn(a) and each bi matches

ai both in size and number of atoms. The “back” argument works the same way.

QED

Theorem 5.12 Let A be a Boolean algebra which can be expressed as a direct sum

of finitely many algebras of the following form: 1) an atom; 2) an atomless element;

3) a 1-atom; 4) a rank 1 atomic element; 5) an element isomorphic to the interval

algebra I(ω+η). Then A has a formally Σ03 Scott family, and hence is ∆0

3-categorical.

Proof: If the summands are all either atomless or atoms, then it is ∆01-categorical by

Goncharov. If A has summands of type 3), 4), or 5), then absorb all of the atoms

into one of these, so we write A = c1 ∨ · · · ∨ cn, where each ci is of type 2) - 5). Let

~c consist of the summands of the type listed above. Let ~a = a1, . . . , aj ∈ A, and let

b1, . . . , b2j the atoms in the formal finite subalgebra generated by ~a. (Some of the

bi’s might equal 0.) For each bk and each ci, we construct the following formulas:

1) If ci is atomless or a 1-atom, then θbki (yk, ci) is constructed as in Chapter 3.

87

2) If ci is a rank 1 atomic element, then θbki (yk, ci) is constructed to say that yk ∩ ci is

finite of size n, or cofinite in ci with complement of size n, or infinite and coinfinite

in ci, depending on what is true of bk. As we have seen, there is a finitary Σ2 formula

θn(x) saying that x is of size n. Therefore, the formula∧∧

n∈N ¬θn(x) is a Πc2 formula

which expresses that x is infinite.

3) If ci is an element isomorphic to the interval algebra I(ω + η), then θbki (yk, ci)

is constructed to say that yk ∩ ci is finite of size n; contains all but n atoms of ci,

but is not cofinite in ci; is sum of n atoms and an atomless element; or is cofinite

in ci, depending on what is true of bk. First, we note that the formula which says

“z is an atom” is finitary Π1. Therefore, the formula γn(x) = ∃z1 · · · ∃zn(∧∧

m=1...n

“zm is an atom” ∧ zm < ci ∧ zm 6< x) is a finitary Σ2 formula expressing that there

are n atoms of ci not contained in x. Consequently, γn(x) ∧ ¬γn+1(x) is equivalent

to a finitary Σ3 formula which expresses that x contains all but n of the atoms of

ci. By our argument in 2) above, there is a Πc2 which expresses that x is coinfinite

in ci. Finally, the formula ρ(z) = ∀w(w 6< z ∨ ∃v(v < w)) is a finitary Π2 formula

expressing that z is atomless. Consequently, ∃z1 · · · ∃zn+1(∧∧

m6=n+1 “zm is an atom”

∧ “zn+1 is atomless” ∧z1 ∪ . . . ∪ zn+1 = x is a finitary Σ3 formula which expresses

that x is a join of n atoms and an atomless piece.

For a tuple of variables ~x = x1, . . . , xj , let ~y = y1, . . . , y2j be the terms in the

formal finite subalgebra determined by ~x. Let ψ~a(~x,~c) =∧∧

i∈{1,... ,2j} θbi1 ∧ · · · ∧ θbin .

Of course, A |= ψ~a(~a,~c). Furthermore, if for some ~a ′, A |= ψ~a(~a′,~c), then we

immediately have (A,~a,~c) ∼= (A,~a ′,~c). Consequently, {ψ~a|~a ∈ A} is a c.e. Scott

family of Σc3 formulas. QED

Lemma 5.13 Let A be a Boolean algebra. Suppose that for each ~c, we can find an

element a, bound by of one of the atoms c of the subalgebra determined by ~c, so that

the following is true:

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Let ~u be given. Consider the subalgebra determined by a, ~u,~c; let α1, . . . , αn

be the atoms whose union is a, and let β1, . . . , βm be the atoms whose union

is c− a. Then we can find α′1, . . . , α′n, β

′1, . . . , β

′m so that

1. if αi bounds finitely many atoms, then α′i bounds the same number of atoms;

same for βj;

2. αi is finite iff α′i is;

3. α′1 ∪ · · · ∪ α′n ∪ β′1 ∪ β′m = c; and

4. if a′ = α′1 ∪ . . . ∪ α′n, then (A, a,~c) 6∼= (A, a′,~c).

Then A has no formally Σ03 Scott family.

Proof: In order to show that A has no formally Σ03 Scott family, we fix ~c. Then we

must show that there is an ~a so that

∀~u∃~a ′ ~u ′∀~v ′∃~v(each atom of the finite subalgebra determined by ~a ′, ~u ′, ~v ′,~c is at

least as large as the corresponding atom of ~a, ~u,~v,~c, but (A, a,~c) 6∼= (A, a′,~c)).

Fix ~c, and let a be as in the hypothesis of the lemma. Then if ~u is given, let

α′1, . . . , α′n, β

′1, . . . , β

′m be as in the hypothesis, and define a′, ~u ′ accordingly. Let ~v ′

be given, and consider the finite subalgebra determined by α′1, . . . , α′n, β

′1, . . . , β

′m, ~v

′.

Assume α′i = δ′i1 ∪ · · · ∪ δ′ini, where the δ′ elements are atoms of the subalgebra. If

we consider only the δ′ik elements which are finite, then we conclude that αi must

bound at least as many atoms of A as compose the δik elements, and we define

δi1, . . . , δini accordingly. We make the same argument and construction for any β′j .

We can then define ~v based on the designation of these atoms of the finite subalgebra.

Consequently, each atom of the subalgebra determined by a′, ~u′, ~v′,~c is at least as large

as the corresponding atom in the subalgebra determined by a, ~u,~v,~c. However, the

89

hypothesis guarantees that (A, a,~c) 6∼= (A, a′,~c). So A has no formally Σ30 Scott

family. QED

Proposition 5.14 Let A be a Boolean algebra. If A has infinitely many disjoint

1-atoms, then it has no formally Σ03 Scott family.

Proof: Let ~c be given. One of the atoms c of the subalgebra determined by ~c must

bound infinitely many disjoint 1-atoms. Let a be one of these 1-atoms. For any ~u as

in the statement of Lemma 5.13, exactly one of the αi bounds infinitely many atoms.

Let α′i be a union of two 1-atoms contained in c. We can define α′ and β′ elements

to satisfy the hypotheses of Lemma 5.13. QED

Proposition 5.15 Let A be a Boolean algebra. Suppose that for any ~c, there is an

atom c of the finite subalgebra determined by ~c so that c bounds an element a where

1. a bounds infinitely many atoms and an atomless element; and

2. c− a bounds infinitely many atoms and an atomless element.

Then A has no formally Σ03 Scott family.

Proof: Fix ~c, and consider a as the hypothesis. Let ~u be given, and consider

α1, . . . , αn, β1, . . . , βm as in Lemma 5.13. Define each α′i to bound finitely many

atoms, and an atomless element if αi is infinite. Find βj which bounds infinitely

many atoms. For k 6= j, define βk to bound finitely many atoms, and an atomless

element if βk is infinite. Let β′j = c − (⋃

1≤i≤n α′i ∪⋃j 6=k β

′k). So for each ~c, there is

an a as in Lemma 5.13. QED

Proposition 5.16 Let A be a countable Boolean algebra. If A has a rank 1 atomic

element but no maximal rank 1 atomic element, then A has no formally Σ03 Scott

family.

90

Proof: Let ~c be the tuple of the Scott family. Assume, without loss of generality,

that ~c includes all of the disjoint 1-atoms of A in the following sense: if d is a 1-

atom of A, then it is equal to one of the elements of ~c up to a finite difference.

(By Proposition 5.14 there can be only finitely many disjoint 1-atoms in A.) Thus,

every atom of the finite subalgebra determined by ~c must satisfy one of the following

conditions:

1. it is bound by one of the 1-atoms of ~c;

2. it is finite;

3. it is rank 1.

Otherwise, an infinite atom c, disjoint from of all of the 1-atoms of A, bounds a b so

that [b] is an atom of A/ ∼. By definition, b is a 1-atom, a contradiction.

If one of the atoms bounds a rank 1 atomic element and an atomless element, then

by Proposition 5.15, ~c is not the tuple in a formally Σ03 Scott family. Consequently,

any atom of the subalgebra bounding a rank 1 atomic element is itself rank 1 atomic.

The union of all such rank 1 atomic elements ofA is a maximal rank 1 atomic element.

QED

Theorem 5.17 Let A be a countable Boolean algebra with a formally Σ03 Scott family.

Then A can be expressed as a direct sum of finitely many algebras of the following

form: 1)finite; 2) atomless; 3) 1-atoms; 4) rank 1 atomic; 5) I(ω + η).

Proof: Let ~c be the tuple of the Scott family. Assume, without loss of generality,

that ~c includes all of the 1-atoms of A and the maximal rank 1 atomic element, if it

exists. Every atom of the finite subalgebra determined by ~c must satisfy one of the

following conditions:

1. it is bound by one of the 1-atoms or the maximal rank 1 atomic element;

91

2. it is finite;

3. it is rank 1 but bounds no rank 1 atomic element.

Any atom c satisfying 3) must bound either finitely many atoms or infinitely many.

If finitely many, then we are done. If c bounds infinitely many, then any d ≤ c which

bounds infinitely many atoms must also bound an atomless element; otherwise, d

is rank 1 atomic, a contradiction. If c can be split into two disjoint elements each

bounding infinitely many atoms, then by the previous statement, each one of these

elements bounds infinitely many atoms and an atomless element. By Proposition 5.15,

~c cannot be the tuple in a formally Σ03 Scott family. Consequently, c cannot be split

into two such elements, each having infinitely many atoms. By Proposition 5.11, each

such c is an algebra isomorphic to I(ω + η). QED

92

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