Δ20-categoricity in Boolean algebras and linear orderings

Annals of Pure and Applied Logic 119 (2003) 85 – 120

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02-categoricity in Boolean algebras and linear orderings Charles F.D. McCoy1 University of Wisconsin, Madison, WI 53706, USA Received 26 September 2000; received in revised form 25 September 2001; accepted 11 December 2001 Communicated by R.I. Soare

Abstract We characterize
02 -categoricity in Boolean algebras and linear orderings under some extra e3ectiveness conditions. We begin with a study of the relativized notion in these structures. c 2002 Elsevier Science B.V. All rights reserved.
MSC: 03D45; 03C57 Keywords: Boolean algebra; Linear ordering; Computable categoricity; Scott family

1. The basics of categoricity In computable model theory there have been several 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 paper, we examine the notions of
20 categoricity and relativized
20 -categoricity in Boolean algebras and linear orderings by 9rst 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. We now de9ne the notions of categoricity relevant to our study, and we provide earlier results which motivate our questions or are used in our proofs. Denition 1.1. A computable structure A is computably categorical if for any computable copy B ∼ = A there is a
10 isomorphism ’e : B ∼ = A. 1

Partially supported by NSF grant DMS95-04594. E-mail address: [email protected] (C.F.D. McCoy).

c 2002 Elsevier Science B.V. All rights reserved. 0168-0072/03/$ - see front matter
PII: S 0 1 6 8 - 0 0 7 2 ( 0 2 ) 0 0 0 3 5 - 0

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The relativized notion is de9ned as follows. Denition 1.2. A computable structure A is relatively computably categorical if for D(B) any copy B ∼ :B∼ = A there is a
10 (D(B)) isomorphism ’e = A, where D(B) is the atomic diagram of the structure B. We give some examples of computably categorical structures. 1. A linear order is computably categorical if and only if it has 9nitely many direct successivities [10,17]. 2. A Boolean algebra is computably categorical if and only if it has 9nitely many atoms [10,18,16]. 3. An Abelian p-group is computably categorical if and only if it can be written in one of the following two forms: (Z(p∞ )) ⊕ G1 , where  ∈ N or  = ∞, and G1 is 9nite; or (Z(p∞ ))l ⊕ G1 ⊕ (Zp k )∞ , where k; l ∈ N and G1 is 9nite [8]. (The notation is that of [11].) 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. Denition 1.3. For a computable ordinal , a computable structure A is
0 -categorical
0 if for any computable copy B ∼ = A there is a
0 isomorphism ’e  : B ∼ = A. 

The relativized notion is de9ned as follows. Denition 1.4. A computable structure A is relatively
0 -categorical if for any copy
0 (D(B)) B∼ :B∼ = A there is a
0 (D(B)) isomorphism ’e  = A. In [6] Goncharov proved, under an added e3ectiveness hypothesis, a syntactic characterization of computable categoricity that provides the starting point for the main results of this paper. Before we can state this result precisely, we need to review some material on in9nitary formulas. Denition 1.5. Fix a computable language L. We de9ne in9nitary and computable in9nitary formulas inductively for all computable ordinals . 1. A 0 or 0 formula is a 9nitary, quanti9er-free (q.f.) formula. We can assign these formulas GDodel numbers in the standard way.

2. A  in4nitary formula is a countable disjunction of the form (˜x) = i∈I ∃˜ui i (˜x; ˜ui ), where each i is a  formula for some ¡. A  formula is de9ned analogously, with universal quanti9ers replacing existential quanti9ers, and conjunctions replacing disjunctions. 3. A computable  in4nitary formula, or c formula, is a  formula involving disjunctions over only computably enumerable sets of formulas. That is, a c formula

(˜x) is of the form (˜x) = i∈I ∃˜ui i (˜x;˜ui ); where each i is a c formula for some

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¡, and the set of GDodel numbers for the i ’s is c.e. (We assume, by induction, that we can assign GDodel numbers to the computable formulas of lower complexity.) A more complete treatment of computable in9nitary formulas is found in [1]. Denition 1.6. Let A be an L-structure for some computable language L. A formally 0 Scott family for A is an ordered pair ˜c;  where ˜c is a tuple of distinct parameters; and  is a c.e. set of c formulas of L with parameters among ˜c so that 1. for each tuple ˜a ∈ A, there is (˜x) ∈  with A |= (˜a); and 2. for any  in  and ˜a;˜a ∈ A, if A |= (˜a) and A |= (˜a ), then A;˜a;˜c  ∼ = A;˜a ;˜c. Denition 1.7. A structure is 1-decidable if its ∃-diagram is computable. Denition 1.8. A structure is 2-decidable if its ∀∃-diagram is computable. Theorem 1.9 (Goncharov). Let A be 2-decidable. Then A is computably categorical if and only if it has a formally 10 Scott family. Goncharov and others wondered, however, what relationship existed between the existence of a formally 10 Scott family and computable categoricity under relaxed hypotheses. Was the above connection merely a by-product of the strong decidability conditions on A? Ash et al. in [2], and independently, Chisholm in [4] proved that there is indeed a strong connection between computable categoricity and the existence of a formally 10 Scott family. Theorem 1.10 (Ash et al., Chisholm). A computable structure A is relatively
0 categorical if and only if it has a formally 0 Scott family. In [7] Goncharov himself produced a structure that is computably categorical but has no formally 10 Scott family, thus demonstrating that the relatively computably categorical structures constitute a proper subset of the computably categorical structures. In fact, Kudinov, using a characterization of computable categoricity in 1-decidable structures which he proved in [14], produced a 1-decidable structure that is computably categorical but has no formally 10 Scott family. Consequently, the extra decidability condition in Theorem 1.9 is as weak as possible. 2. Relativized
20 -categoricity in Boolean algebras and linear orderings By Theorem 1.10, in order to examine relativized
20 -categoricity, we must understand the satisfaction of 2 formulas in the structures we examine. In [1], Ash and Knight described back-and-forth relations that give us the necessary tools to characterize the satisfaction of 2 formulas in Boolean algebras and linear orders quite easily. Denition 2.1. Let A be any structure, and ˜a; ˜b tuples of the same length from A. Then ˜a 61 ˜b if and only if the 1 formulas true of ˜b are true of ˜a if and only if the 1 formulas true of ˜a are true of ˜b.

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Proposition 2.2. Let A be a Boolean algebra and ˜a; ˜b be tuples from A so that the

0 formulas true of ˜a and ˜b are the same. Then ˜a 61 ˜b if and only if each atom of the 4nite algebra determined by ˜a is in4nite or at least as large as the corresponding atom of ˜b. Proposition 2.3. 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 61 ˜b if and only if each interval in A determined by ˜a is in4nite or at least as large as the corresponding interval determined by ˜b. In order to state our result for Boolean algebras, we must recall the de9nition of a 1-atom. Denition 2.4. Let ∼ be the equivalence relation in a Boolean algebra A where a ∼ b if and only if a ∩ bJ and b ∩ aJ are both either empty or a union of 9nitely many atoms of A. The set of equivalence classes A=∼ again forms Boolean algebra. An element a ∈ A is a 1-atom of A if and only if the equivalence class [a] is an atom of A=∼. Theorem 2.5. A Boolean algebra A is relatively
20 -categorical if and only if it can be expressed as a 4nite direct sum c1 ∨ · · · ∨ cn , where each ci is either atomless, an atom, or a 1-atom. Proof. (⇐) If the summands are all either atomless or atoms, then it is
10 -categorical by the second example after De9nition 1.2. 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 2c formulas. Let ˜a = a1 ; : : : ; aj ∈ A, and let b1 ; : : : ; b2j the atoms in the formal 9nite subalgebra generated by ˜a. (Some of the bi ’s might equal 0.) For each bi , we construct the following formulas: 1. 1bi (yi ; c1 ) is “yi ∩ c1 = 0” if bi ∩ c1 = 0; “yi ∩ c1 = c1 ” if b1 ∩ c1 = c1 ; or “yi ∩ c1 = 0 ∧ yi ∩ c1 = c1 ” if bi ∩ c1 is a proper subset of c1 . 2. For each k ∈ {2; : : : ; n}, exactly one of bi ∩ ck ; bJ i ∩ ck is empty or a 9nite join of atoms (of A), precisely because each ck is a 1-atom. The formula m (y) = ∃z1 · · · zm [z1 = 0 ∧ · · · ∧ zm = 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 bi ∩ ck or bJ i ∩ ck , and let kbi (yi ; ck ) be nbi ; k (yi ∩ ck ) if |bi ∩ ck | = nbi ; k ; or nbi ; k (yJi ∩ ck ) if | bJ i ∩ ck | = nbi ; k . For a tuple of variables ˜x = x1 ; : : : ; xj , let ˜y = y1 ; : : : ; y2 j , be the terms in the formal  9nite subalgebra determined by ˜x. Let ˜a (˜x;˜c) = i∈{1;:::;2 j } 1bi ∧ · · · ∧ nbi . Of course, A |= ˜a (˜a;˜c). Furthermore, if for some ˜a ; A |= ˜a (˜a ;˜c), then we immediately have (A;˜a;˜c) ∼ = (A;˜a ;˜c). Consequently, { ˜a |˜a ∈ A} is a c.e. Scott family of 2c formulas.

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(⇒) Assume that A is not of the described form, and let ˜c be parameters in a putative formally 20 Scott family. Consider the 9nite subalgebra determined by ˜c. One of the atoms a of this algebra must be such that 1. a is not a 1-atom; and 2. a contains in9nitely 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 9nite join of atoms and an atomless algebra. Therefore, contrary to hypothesis, A can be written as a 9nite join of 1-atoms, atoms, and an atomless algebra. Let a be the disjoint union of a1 ; a2 , where a1 is an in9nite Boolean algebra and a2 contains in9nitely many atoms. We claim that for any 2 formula (x;˜c) satis9ed by a1 , there is a1 composed of only 9nitely many atoms of A so that a1 also satis9es . Since obviously (A; a1 ;˜c)  (A; a1 ;˜c), A cannot have the supposed Scott family. We can assume, without loss of generality, that the formula  is of the form ∃˜u#(x;˜u;˜c), where # is the conjunction of all 9nitary 1 formulas satis9ed by a1 ;˜c, and some ˜u. Consequently, we must show that, given a tuple ˜u, there are a1 ;˜u so that every 1 formula satis9ed by a1 ;˜u;˜c is satis9ed by a1 ;˜u ;˜c, or that (a1 ;˜u;˜c) 61 (a1 ;˜u ;˜c). By Proposition 2.2 we must show that there are a1 ;˜u so that each atom of the 9nite algebra determined by a1 ;˜u;˜c is in9nite or at least as big as the corresponding atom determined by a1 ;˜u ;˜c. Consider the 9nite subalgebra determined by a1 ;˜u;˜c; let its atoms be z1 ; : : : ; zm . For each zi with zi ∩ a = 0, let zi = zi . In this subalgebra, the element a1 is divided into zj1 ; : : : ; zjs and a2 into zk1 ; : : : ; zkt . One of the zk ’s must be in9nite; without loss of      generality, let it be z k1 . In a2 9nd atoms (of A) zj1 ; : : : ; zjs ; zk2 ; : : : ; zkt ; and let zk1 =       a − [ p∈{1;:::; s} zjp ∪ q∈{2;:::; t} zkq ]. De9ne a1 ; ˜u and ˜c in the subalgebra generated by the z  elementsin the same way that a1 ;˜u, and ˜c are generated by the z elements. For instance, a1 = p∈{1;:::; s} zjp , so a1 is a 9nite join of atoms of A. Note that ˜c =˜c,     since p∈{1;:::; s} zjp ∪ q∈{1;:::; t} zk q = p∈{1;:::; s} zjp ∪ q∈{1;:::; t} zkq = a, and for each zi with zi ∩ a = 0, we let zi = zi . Notice that for each i = 1; : : : ; m; zi is a Boolean algebra of size at least as large as zi , and that a1 is 9nite. Thus we have proven our claim. Throughout the rest of the paper, 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)
20 -categorical if and only if the same ordering with a greatest and least element “attached” is. Hence, our assumption is merely one of convenience. Theorem 2.6. Let A = (A; ¡A ) be a sum of 4nitely many intervals, each of type m; !; !∗ ; Z, or n · (, so that each interval of type n · ( has a supremum and in4mum. Then A is relatively
20 -categorical. Proof. We use Theorem 1.10. For each interval of type ! or !∗ , name the “0”; for interval of type Z, pick a single element and name it; for each interval of type n · (, name the supremum and in9mum; 9nally, name all of the remaining elements, of

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which there are a 9nite number. These will be the parameters appearing in our Scott family. Let ˜a = a1 ; : : : ; ak ∈ A. We explicitly construct a formula ˜a (x1 ; : : : ; xk ;˜c) such that A |= ˜a (a1 ; : : : ; ak ;˜c) and if A |= ˜a (b1 ; : : : ; bk ;˜c), then (A; a1 ; : : : ; ak ;˜c) ∼ = (A; b1 ; : : : ; bk ;˜c). First, express the ordering of ˜a;˜c in a formula (˜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 16j1 6j2 ; aj1 ; : : : ; aj2 all fall in the same interval of A of one of the types described above, and neither aj1 nor aj2 is an element of ˜c. We describe how to 9nd a formula (xj1 ; : : : ; xj2 ;˜c) which characterizes aj1 ; : : : ; aj2 up to isomorphism. The formula ˜a will be a conjunction of such formulas. If aj1 ; : : : ; aj2 lie in the same interval of type ! with c = “0”, then aj1 = “p1 ”; : : : ; aj2 = “pj2 −j1 +1 ” in this copy of !. Therefore, aj1 satis9es the formula 1 (x1 ; c) = ∃y1 · · · yp1 −1 [c¡y1 ¡ · · · ¡yp1 −1 ¡x1 ∧ ∀z(¬(c¡z¡y1 ) ∧ · · · ∧ ¬(yp1 −1 ¡z¡x1 ))]. We have similar formulas (x1 ; : : : ; xj2 −j1 +1 ; c) be the 2 (x2 ; c); : : : ; j2 −j1 +1 (xj2 −j1 +1 ; c). Let conjunction. If aj1 ; : : : ; aj2 lie in an interval of type !∗ or Z, we similarly obtain a formula (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 · ( with aj1 ; : : : ; am1 in the same n interval of n · (; am1 +1 ; : : : ; am2 in the same n interval that is di3erent from a1 ’s; : : : ; amt +1 ; : : : ; aj2 in the same n interval that is di3erent 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 1 (x1 ; x2 ) = ∃y0 y1 y2 [(y0 ¡x1 ¡y1 ¡x2 ¡y2 ) ∧ ∀z(¬(y0 ¡z¡x1 ) ∧ · · · ¬(x2 ¡z¡y2 ))]. We have similar formulas 2 (xm1 +1 ; : : : ; xm2 ); : : : ; t+1 (xmt +1 ; : : : ; xj2 ). Furthermore, am1 and am1 +1 satisfy the formula )1 (xm1 ; xm1 +1 ) = ∃y0 · · · yn (xm1 ¡y0 ¡ · · · ¡yn ¡xm1 +1 ). Similarly, there are formulas )2 ; : : : ; )t . Let (x1 ; : : : ; xj ) be the conjunction of all of 1 ; : : : ; t+1 ; )1 ; : : : ; )t . By design, A |= ˜a (a1 ; : : : ; ak ;˜c). If A |= ˜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 !; !∗ , or Z, then ˜a actually guarantees that ap = bp ; : : : ; aq = bq . 2. If they all fall in an interval of type n · (, then ˜a guarantees that (a) for all u; v with p6u; v6q; au and av fall in the same interval of type n exactly if bu and bv do; (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 ( 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 {˜a |˜a is a tuple from A} is a formally 20 Scott family. We now give a proof of the converse of Theorem 2.6.

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Theorem 2.7. Let A = (A; ¡A ) be a relatively
20 -categorical linear ordering. Then A is a sum of 4nitely many intervals, each of type m; !; !∗ ; Z, or n · (, so that each interval of type n · ( has a supremum and in4mum. We prove this theorem by dividing it into several lemmas and smaller propositions. Denition 2.8. 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 !-interval” to denote a maximal interval of order type !. Proposition 2.9. A relatively
20 -categorical linear ordering A must have only 4nitely many maximal intervals of order type !; !∗ ; Z and cannot have arbitrarily large maximal 4nite intervals. Proof. Assume that A either has in9nitely many maximal intervals of one of the above order types, or has arbitrarily large maximal 9nite intervals. Let ˜c = c0 ; : : : ; cn be the parameters in a putative formally 2 Scott family, and let A = I0 + c0 + I1 + · · · + In + cn + In+1 . For some k = 0; : : : ; n + 1; Ik contains in9nitely many maximal intervals of one of the above order types, or has arbitrarily large maximal 9nite intervals. (For simplicity, we write I for this Ik .) There are a1 ¡a2 in I so that 1. there are in9nitely many elements in I to the left of a1 ; 2. there are in9nitely many elements between a1 and a2 ; and 3. there are in9nitely many elements in I to the right of a2 . We claim that for any 2 formula (x1 ; x2 ; ˜c ) satis9ed by a1 ; a2 , there exist a1 ; a2 so that a1 ; a2 also satisfy  but have only 9nitely many elements of A between them. Since obviously (A; a1 ; a2 ; ˜c )  (A; a1 ; a2 ; ˜c ); A cannot have the supposed Scott family. We can assume, without loss of generality, that the formula  is of the form ∃˜u#(x1 ; x2 ; ˜u; ˜c ), where # is the conjunction of all 9nitary 1 formulas satis9ed by a1 ; a2 , some ˜u, and ˜c. Consequently, we must show that, given a tuple ˜u, there are a1 ; a2 ; ˜u  , so that every 1 formula satis9ed by a1 ; a2 ; ˜u; ˜c is satis9ed by a1 ; a2 ; ˜u  ; ˜c or that (a1 ; a2 ; ˜u; ˜c)61 (a1 ; a2 ; ˜u  ; ˜c). Therefore, by Proposition 2.3, in order to verify our claim, we must demonstrate the following: for any ˜u containing the parameters ˜c, there exist a1 ; a2 ; ˜u  so that ui = ui for ui ∈ ˜c, and the intervals determined by a1 ; a2 ; ˜u  , are of size no greater than those determined by a1 ; a2 ; ˜u. Assume, without loss of generality, that ˜u = u1 ; : : : ; un is arranged in increasing order, and that the ordering is as follows:

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1. u1 ; : : : ; uj1 lie to the left of I ; 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 in9nite. 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. De9ne a1 ; a2 ; ˜u  as follows: 1. for i6j2 ; ui = ui ; 2. uj2 +1 ; : : : ; uj3 ; a1 ; uj3 +1 ; : : : ; uj4 ; a2 ; uj4 +1 ; : : : ; uj5 is a sequence of successors in the chain of length n + 2; 3. for i¿j5+1 ; ui = ui . The desired property of a1 ; a2 ; ˜u  follows immediately from the fact we noted above. ∗ Lemma 2.10. If B is a countable  linear ordering containing no !; ! , or Z intervals, then B is of the form n1 + q∈( Bq + n2 , where n1 or n2 could be 0 and each Bq is 4nite 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 ! interval. Therefore, B has an initial maximal 9nite discrete interval of order type n1 . (Of course, n1 = 0 if B has no least element.) Similarly B has a terminal maximal 9nite discrete interval of order type n2 . (Of course, n2 = 0 if B has no great  est  element.) We show that B has the form n1 + B + n2 , where B has the form q∈( Bq . Let c1 ∈ B . By repeatedly applying the successor and predecessor function we can obtain only a 9nite interval {c1; 0 ; : : : ; c1; n1 }. Let this 9nite interval be Bq1 for some q1 ∈ (. 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 ∈ ( − {q1 }. If q2 ¡q1 , then 9nd some c2 greater than any element of n1 and less than c1; 0 . Repeatedly apply the successor and predecessor function to obtain a 9nite interval {c2; 0 ; : : : ; c2; n2 }. Let this 9nite 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 .

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We can continue this back-and-forth argument to construct a  quently, B is of the form (1 + q∈( Bq + (2 . Convention. Hereafter, when we write with |Bq |¿1.

 q∈(

 q∈(

93

Bq ∼ = B . Conse-

Bq , we assume that each Bq is 9nite

Corollary 2.11. If A is a relatively
02 -categorical linear ordering, then  there is an n so that A is a 4nite sum of intervals, each of the form !; !∗ ; Z; m; q∈( Bq , where each |Bq |¡n. Proposition 2.12. If A is a relatively
02 -categorical linear ordering (with endpoints), then any maximal interval of the form q∈( Bq has an in4mum and a supremum. Proof. We prove the fact for in9mums; the argument for supremums is similar. Assume that A is a relatively
02 -categorical linear ordering with ˜c the parameters in a putative formally 2 Scott family. Let A have a maximal interval P of the form q∈( Bq . By Corollary 2.11, there must be an interval directly to the left of one of the following forms: 1. 9nite; 2. !∗ ; 3. !. In (1) and (2), P has an in9mum. We give the argument for (3). Fix a ∈ P so a has no immediate successor, and there are no members of ˜c in P to the left of a. We claim that for any 2 formula (x; ˜c ) satis9ed by a, there exists a in the adjacent !-interval so that a also satis9es (x; ˜c). Again we must show that for any ˜u containing ˜c, there exist a ; ˜u  so that ui = 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 !-interval; 2. uj1 +1 ; : : : ; uj2 lie in the !-interval; 3. uj2 +1 ; : : : ; uj3 lie to the left of a in P; 4. a¡uj3 +1 ; : : : ; ujn . Note that (a; uj3 +1 ) is in9nite. De9ne a ; ˜u  as follows, where “x+i” is the ith successor of x in the !-interval: 1. for 16i6j2 ; ui = ui ; 2. for 16i6j3 − j2 ; uj2 +i = uj2 + i; 3. a = uj3 + 1; 4. for j3 + 16i6n; ui = ui . The desired property of a ; ˜u  follows immediately from the fact we noted above.

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Denition 2.13. A 4nite partition of ( is a 9nite sequence q1 ¡ · · · ¡qj ∈ ( and open intervals J1 ; : : : ; Jj+1 so that ( = J1 +q1 +J2 +· · ·+qj +Jj+1 . A partition P1 is 4ner than a partition P2 exactly if every element in the 9nite sequence for P2 is in the sequence for P1 .  Denition 2.14. Let B be a linear ordering of the form q∈( Bq . 1. B satis9es Property 1 if and only if there is a 9nite partition of ( 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 · ( + n2 + n3 · ( + n4 + n5 · ( + · · · + nm−1 + nm · (. 2. B satis9es Property 2 if and only if there exist m1 = m2 so that for any 9nite partition of ( there is an open interval J in the partition so that for in9nitely many q ∈ J; Bq is of order type m1 and for in9nitely many q ∈ J; Bq
is of order type m2 .  Lemma 2.15. Fix a linear ordering B of the form q∈( Bq with n as in Corollary 2:11. If B doesn’t satisfy Property 2, then it satis4es Property 1. Proof. Let m0 ; : : : ; mk be the complete list of natural numbers so that for each 06i6k, there is q ∈ ( with Bq of order type mi . For each mi = mj , there is a 9nite partition of ( so that no interval of the partition contains in9nitely many copies of mi and in9nitely many copies of mj . Choose a 4nite partition of ( 9ner 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 in9nitely many q in that interval. Make the partition 9ner to obtain a 9nite partition that satis9es the requirements in Property 1. 0 Proposition 2.16. If A is a relatively
2 -categorical linear order, then no interval of the form q∈( Bq satis4es Property 2.

 Proof. Assume that A does have an interval of the form q∈( Bq with m1 and m2 as in Property 2, and let ˜c be the parameters in a putative c.e. Scott family of 2c formulas. Let m1 be the largest number for which there is such an m2 . De9ne a 9nite partition of ( 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 doesn’t have type m; and 2. for all q ∈ J , no parameter appears in Bq . For one of these open intervals J , there are in9nitely many q ∈ J with Bq of type m1 , and there are in9nitely 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 2 formula (x; ˜c ) satis9ed by a, there exists q1 ∈ J and a ∈ A so that a ∈ Bq1
of type m1 , and a satis9es (x; ˜c). Again, we must show that for any ˜u containing the parameters ˜c there exist a ; ˜u  so that ui = ui for ui ∈ ˜c and the intervals determined by a ; ˜u  are of size no greater than the corresponding intervals of a; ˜u.

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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 16i6j1 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+2 lie to the left of a in Bq1 ; 4. ujk+2 +1 ; : : : ; ujk+3 lie 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 in9nitely many elements between uj1 and uj1 +1 ; between uj2 and uj2 +1 ; : : :; between ujk+1 and ujk+1 +1 ; between ujk+3 and ujk+3 +1 ; : : :; and between ujk+t+3 and 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 9nd r1 ¡r2 ¡ · · · ¡rk ¡q1 ¡s1 ¡ · · · ¡st in J so that Bq1
has type m1 and each Br
; Bs
has type m1 . De9ne a ; ˜u  as follows: 1. for i6j1 ; ui = ui ; 2. for 16i6k; uji +1 ; : : : ; uji+1 are ordered in Bri
exactly as uji +1 ; : : : ; uji+1 are ordered in Bri ; 3. ujk+1 +1 ; : : : ; ujk+2 ; a ; ujk+2 +1 ; : : : ; ujk+3 are ordered in Bq1
exactly as ujk+1 +1 ; : : : ; ujk+2 , a; ujk+2 +1 ; : : : ; ujk+3 are ordered in Bq1 ; 4. for 16i6t ujk+2+i +1 ; : : : ; ujk+3+i are ordered in Bsi
exactly as ujk+2+i +1 ; : : : ; ujk+3+i are ordered in Bsi ; 5. for jk+t+3 ¡i¡n; ui = ui . The desired property of a ; ˜u  follows immediately from the facts we noted above. Lemma 2.15 immediately implies the following corollary. Corollary 2.17. If A is a relatively
02 -categorical linear order, then any interval of the form q∈( Bq satis4es Property 1. We now complete the proof of Theorem 2.7. Assume that A is relatively
02 categorical. By Corollary 2.11 and Proposition  2.12, A can be written as a 9nite sum of intervals, each of the form m; !; !∗ ; Z; q∈( Bq , so that each interval of the form  Bq has a supremum and in9mum. By Corollary 2.17, for each interval of the q∈(  form q∈( Bq there are n1 ; n2 ; : : : ; nm ¿1 so that q∈( Bq = n1 ·(+n2 +n3 ·(+· · ·+nm ·(. We note that our characterization of relatively
02 -categorical linear orderings is exactly the same as Michael Moses’ characterization of computably categorical 1-decidable linear orderings in [15]. Theorem 2.18. For a 1-decidable linear ordering A (with greatest and least element) the following are equivalent:

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1. Every 1-decidable linear ordering isomorphic to A is isomorphic by a computable isomorphism; 2. A is of the form given in Theorem 2.7. The “reason” for this correlation has not yet been established. 3. Unrelativized
02 -categoricity 3.1.
02 -categoricity in Boolean algebras In this section we o3er priority constructions that give, under some extra assumptions, results on
02 -categoricity (unrelativized) for Boolean algebras and linear orderings. We shall see that with these assumptions, the relativized and unrelativized notions de9ne the same class of structures. Furthermore, we shall attempt to highlight where the syntactic arguments from the previous section provide some insight into the organization, strategies, and hypotheses needed for the priority constructions. Notation. Let A be a Boolean algebra. 1. The predicate Atom(x) denotes that x is an atom of A; i.e., (A; Atom) |= Atom(x) if and only if A |= x = 0 ∧ ¬ (∃z ∃y(z = x ∧ y = x ∧ z ∪ y = x)); 2. The unary predicate Atomless(x) denotes that x contains no atoms; i.e., (A; Atom; Atomless) |= Atomless(x) if and only if (A; Atom) |= ∀z[∃y(y ∪ z = x) → ¬ Atom(z)]. 3.1.1. Our 4rst priority construction Theorem 3.1. Let A be a Boolean algebra so that (A; Atom; Atomless) is a computable structure. If A is
02 -categorical, then it is a 4nite direct sum of atoms, 1-atoms, and an atomless algebra. Proof. Assume A is not of the described form. We attempt to construct a computable
0 B∼ = A so that there is no
02 isomorphism between them. We use ’e 2 to denote the
0


0

eth
02 function, and ’e;2s to denote the computable approximation of ’e 2 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
0

Re : ’e 2 is not an isomorphism from A onto B: Moreover, the construction must determine an isomorphism g : B ∼ = A.

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At stage s we inductively de9ne #s , a node of length s−1 approximating the true path f through the tree. If  ⊂ #s and || = e, then  decides its outcome o (o = I or o = A)
0

by determining if ’e 2 threatens to be an isomorphism. In addition,  ⊂ #s de9nes g; s , its contribution to the stage s approximation of the function g, so that || is in the domain and range. The construction will ensure  that for  ⊆  ⊂ #s ; g; s ⊆ g; s ; therefore, #s de9nes the stage s approximation gs ⊇ ⊂#s g; s . (Note: at stage s, a node  ⊂ #s will  often 9rst determine g; s , a preliminary version of g; s , then decide its outcome, and 9nally determine g; s itself.) Finally, at the end of stage s, we use gs to 4x 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 de9ned in the standard way for 0 constructions:  ⊂ f if and only if  is the left-most node of length || so that  ⊂ #s for in9nitely many s. We must show that nodes along the true path succeed in de9ning an isomorphism g:B ∼ = A and in meeting the requirements Re . Challenging pairs and candidate pairs. Consider  ⊂ #s working on requirement Re . If  ⊃ 4, the empty string, then it receives from its predecessor node  the function g; s that it believes is a correct approximation of g, because all requirements Ri with ˜ = ˜c. If  = 4, then there ˜ be the domain of g; s with g; s (d) i¡e seem to be met. Let d is no , and ˜c = 0A ; 1A . The node  considers the 9nite algebra determined by ˜c and attempts to determine the 9rst pair a1; s ; a2; s  so that a1; s seems in9nite, a2; s seems to have in9nitely many atoms, and for some atom a in the 9nite subalgebra determined by ˜c; a1; s ∪ a2; s = a;  includes a1; s ; a2; s in the range of g; s . Consider a , the atom corresponding to a in
0 the subalgebra determined by the preimage of ˜c under g; s ◦ ’e;2s . The pair a1; s ; a2; s  challenges this a to contain elements ;1 s  ; a2; s  which have the same properties that a1; s ; a2; s seem to have. Hence we call it a challenging pair. A pair a1; s  ; a2; s  which seems to meet the challenge is called a candidate pair. We should note that  may make incorrect guesses about which pairs have the above properties, because it cannot computably determine which elements are in9nite or have in9nitely many atoms. However, the properties of A do guarantee, as in Theorem 2.5, that such a pair indeed exists. Moreover, the extra assumption that Atom is computable relation implies that the unary relation “x is in9nite” is 10 ; similarly, the computability of Atomless guarantees that the unary relation “x has in9nitely many atoms” is 10 . Therefore, eventually a node  along the true path will determine its 9nal 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  is along the true path and its ultimate challenge is never actually met, then the 9nal outcome of  is I . If a true candidate pair a1; s  ; a2; s  is found at stage t, then
0

 attempts to change its approximation so that the image under g; t ◦ ’e; 2t of a1; s  is 9nite; we thereby meet requirement Re . Relationships among nodes. Throughout the construction, a node  ⊂ #s may initialize another node ; i.e., if  ⊆#t for some t¡s, then  cancels the assignment of ’s

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challenging pair and the approximation gt . There will be three circumstances when a node will initialize another: 1. if  rede9nes its challenging pair at stage s, then  initializes each  with  ⊂ ; 2. if  ⊂ #s determines its outcome to be I , then  initializes all  with  ˆI ¡L ;
0

3. if  diagonalizes against ’e; 2s at stage s, then  initializes all  with  ˆA ⊆. In turn, as  de9nes g; s , it must respect nodes 7¡L  ˆo. What exactly does “respect” mean? Consider 7¡L  ˆo and some stage t¡s so that 1. 7 ⊆#t ; 2. gt (˜v) =˜u; and 3. gt has not been canceled by stage s. If at stage s;  de9nes g; s (˜v) =˜u , then as we determine the Boolean algebra Bs based on ’s approximation of g, we may give an element of the 9nite algebra determined by ˜v a certain number of subsets. However, this de9nition at stage s must allow for the possibility that at some later stage t  with  ⊂ #t
; gt
(˜v) =˜u. How can the construction ensure this compatibility? Assume that each atom of the 9nite 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  ; #t
can rede9ne gt
(˜v) =˜u and remain in agreement with the algebra Bt
−1 we have chosen. In other words, ˜u61˜u . Therefore, the relation crucial to classifying relativized
02 -categoricity also plays a critical role in our examination of the unrelativized notion. Moreover, notice that the computability of Atom makes 61 a 10 relation between tuples. Consequently, while we cannot actually know that ˜u61˜u , we would discover that ˜u1˜u before destroying compatibility. Construction. Stage 1: Fix, for the rest of the construction, g(0) = 0A and g(1) = 1A . For each 7 in the tree, 7 is initialized and #1 = 4. The algebra B1 is the trivial algebra. Stage s + 1: Of course, 4 ⊂ #s+1 . Let  ⊂ #s+1 be a node of length e. Let  be the ˜ with g; s+1 (d) ˜ = ˜c. (If  = 4, then there predecessor node of , and let dom(g; s+1 ) = d is no , and ˜c = 0A ; 1A :) The node  uses the computability of the relations Atom and Atomless in A to search for the 9rst pair a1; s+1 ; a;2 s+1  so that 1. for an atom a of the 4nite algebra determined by ˜c; a1; s+1 ∪ a;2 s+1 = a; 2. a1; s+1 has s + 1 proper subsets; 3. a;2 s+1 contains s + 1 atoms of A; and 4. we have not yet discovered either that a1; s+1 is 9nite or that a;2 s+1 has only 9nitely many atoms of A. This is the challenging pair of  at stage s + 1. Case I: The node  has been initialized since its last active stage or has de9ned a new challenging pair since its last active stage. The node  initializes all  with  ⊂ ⊆#t for some t6s, and it chooses outcome I . (Notice, in particular, that any approximation gt de9ned in part by  at an earlier stage t is canceled. However, of course,  doesn’t cancel the de9nition of its challenging pair.) It must de9ne g; s+1 so that the challenging pair is in the range. We may assume ˜ includes all elements which are in the domain by induction on the construction that d of an uncanceled approximation gt where t¡s + 1 and #t ¡L  (see Lemma 3.2).

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Let ˜b be the set of elements in Bs but not in the domain of g; s+1 . Again, we may assume by induction on the construction that g; s+1 is compatible with the algebra Bs ; i.e., g; s+1 can be extended to be a relation-preserving map between Bs and A (see Lemma 3.2). Therefore,  can extend g; s+1 so that it includes ˜b and e in the domain and the challenging pair and e in the range. Let b;1 s+1 ; b;2 s+1 be such that g; s+1 (b;1 s+1 ; b;2 s+1 ) = a1; s+1 ; a;2 s+1 . (Summary: After rede9ning its challenging pair, the node  restarts its work on the requirement Re .) Case II:  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.  ⊂ #t ; 2. either  has been initialized since the previous stage r in which  ⊂ #r , or the de9nition of the challenging pair changed during stage t; 3.  has not been initialized since stage t; and 4. a1; s+1 ; a;2 s+1 =a1; t ; a2; t . (Throughout the rest of this construction, we will refer to this particular t.) We assume by induction on the construction that since  has not been initialized since stage t, the portion of gs+1 de9ned by nodes above  is the same as the portion of  gt de9ned by nodes above  (see Lemma 3.2). The node  de9nes g; s+1 , its tentative version of g; s+1 , to be the same as g; t . We now consider the following subcases. Case IIa: One of the following is true:
02 ˜ or one of the elements b1 ; b2 1. An element of d ; s+1 0; s+1 is not in the 0range of ’e; s+1 .
˜ and ’e
2 (a1  ; a 2  ) (Otherwise, let ˜c ; a1; s+1  ; a2; s+1  be such that ’e; 2s+1 (˜c ) = d ; s+1 ; s+1 = b;1 s+1 ; b;2 s+1 :) 2. Some relation realized by ˜c; a1; s+1 ; a;2 s+1 is not realized by ˜c ; a1; s+1  ; a2; s+1  , or vice versa. 3. If  enumerates the diagram of A; Atom; Atomless for s + 1 steps, then it sees that one of the atoms of the 9nite algebra determined by ˜c; a1; s+1 ; a;2 s+1 has a di3erent size or a di3erent number of atoms (of A) than the corresponding atom in the 9nite algebra determined by ˜c ; a1; s+1  ; a2; s+1  . 4. If t  ¡s + 1 is the greatest stage so that  ⊆#t
, then our guess at the preimage
02  ˜ or one of b1 ; b2 of d ; s+1 ; s+1 under ’e has changed since t . (Otherwise, the pair 1  2  a; s+1 ; a; s+1  is a candidate pair.)  The outcome of  is I ; the node  initializes all  with  ˆI ¡L ; and g; s+1 = g; s+1 .  (Summary: If the approximation g; s+1 de9ned so far is part of the isomorphism g
0 itself, then ’e 2 doesn’t appear to be an isomorphism, because it doesn’t seem total,
0 as in (1) and (4), or we can readily see that g; s+1  ◦ ’e; 2s+1 doesn’t 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  ˆA ⊂ #r , and  ˆA has not been initialized since r; 2.  diagonalized against Re at stage r;

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3. if y is an atom of the 9nite subalgebra determined by elements in the range of higher priority gw , and during the diagonalization  guessed that y was in9nite, then  9nds 2s+1 subsets of y and still guesses that y is in9nite (see Case IIc). The outcome of  remains A, and g; s+1 = g; r . (Summary: The node  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,  initializes each node  with  ˆA ⊆. The node  attempts to perform the following diagonalization on the candidate pair a1; s+1  ; a2; s+1  . If it completes the diagonalization, then the outcome of  is A. Diagonalization: Let w be the greatest number so that t6w¡s+1;  ˆI  ⊆#w , and gw is uncanceled. Let dom(gw ) =˜v with gw (˜v) =˜u. By our construction, the approximation  gw must extend g; 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  can de9ne an extension of gw whose domain includes ˜b (see Lemma 3.2). Let the image of ˜b be ˜a. (Note that the atom a of the algebra determined by ˜c is unrelated to the ˜ = ˜c. tuple ˜a.) Throughout the diagonalization,  must leave 9xed the mapping g; s+1 (d) 1 2 ˜ However,  will attempt to alter the mapping of the rest of ˜v; b; b; s+1 ; b; s+1 so that it meets requirements Re while it respects gw . This diagonalization follows closely the argument for relativized
02 -categoricity given in Theorem 2.5: ˜c, the range of g; s , corresponds to the parameters ˜c in the potential formally 2 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 9nite algebra determined by ˜u ∪ ˜a, but only in that determined by ˜u.  1 2 Recall that g; s+1 is ’s tentative contribution to gs+1 ; a; s+1 ; a; s+1 is a challenging 1 2  1 pair with a; s+1 ∪ a; s+1 = a, and  has de9ned g; s+1 (b; s+1 ; b; s+1 ) = a1; s+1 ; a;2 s+1 . Fur
0

thermore, a1; s+1  ; a2; s+1   seemingly has met the challenge, because ’e 2 (a; s+1 ; a2; s+1  ) = b;1 s+1 b;2 s+1 , and a1; s+1  has 2s+1 proper subsets and still appears in9nite, and a2; s+1  has 2s+1 atoms of A. The goal is to de9ne g; s+1 (b;1 s+1 ) = a1; s+1  so that a1; s+1  is
0


0

a 9nite Boolean algebra, thus ensuring that ’e 2 is not an isomorphism if ’e; 2s+1 is a
0

correct approximation of ’e 2 and a1; s+1  truly is in9nite. Consider the 9nite subalgebra determined by ˜u. Let k = ((s + 1) + |˜u| + |˜a |). Let y1 ; y2 ; : : : ; ym be atoms of this 9nite algebra with y1 ∪ · · · ∪ ym = a;2 s+1 . Determine the 9rst among y1 ; : : : ; ym which contains 2k subsets and still appears to be in9nite. (If no such yj exists, then a;2 s+1 is 9nite, and hence,  must choose a new challenging pair and enter Case I.) Assume, without loss of generality, that it is ym . Now consider the 9nite subalgebra determined by ˜u ∪ ˜a. Let z10 ; : : : ; zp00 ; z11 ; : : : ; zp11 ; : : : ; z1m ; : : : ; zpmm be atoms of this algebra so that z10 ∪ · · · ∪zp00 = a1; s+1 and z1j ∪ · · · ∪ zpkj = yj for j = 1; : : : ; m. In a;2 s+1 9nd atoms (of A) z10  ; : : : ; zp00  and z1j : : : zpjj for j = 1; : : : ; m − 1. Furthermore, in a;2 s+1 9nd other atoms (of A) z1m ; : : : ;

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  zpmm−1 , and let zpmm = a − [[ j=0;:::; m−1 ( i=1;:::;pj zij )] ∪ (z1m ∪ · · · ∪ zpmm−1 )]. Notice that    z-elements = z -elements = a. Now we are ready to rede9ne g; s+1 . Let v1 be an element of ˜v ∪ ˜b currently mapped to u1 . If u1 ∩ a = 0, then g; s+1 (v1 ) = u1 . Otherwise, u1 is a union of some x disjoint from a and some union of z elements. Let g; s+1 (v1 ) = the union of x and the corre˜ If u1 is disjoint from sponding z  elements. In particular, assume v1 is an element of d. a, then g; 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, g; s+1 (v1 ) = x ∪ z  -elements = u1 . The node  9nishes the de9nition of g; s+1 by including e in the domain and range. This completes the diagonalization. After each node  ⊂ #s+1 has determined g; s+1 , we complete the de9nition of gs+1 and de9ne the 9nite Boolean algebra Bs+1 . The algebra Bs has s atoms, which we des ignate b1 ; b2 ; : : : ; bs , and therefore has 2s distinct elements. The function ⊂#s+1 g; s+1 contains all of these2s elements in its domain; we designate the image of each bi as ai . If the function  ⊂ #s+1 g; s+1 has no more than 2s elements  in its domain, then determine the 9rst ai so that ai has a nontrivial subset. Extend  ⊂ #s+1 g; s+1 to contain this subset and its complement with respect to ai in the range; this extension is gs+1 .  If  ⊂ #s+1 g; 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 ∗  ⊂ #s+1 g; s+1 so that every element in the 9nite 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 , de9ne Bs+1 so that bi is no longer an atom. This concludes the construction. 3.1.2. Supporting lemmas Lemma 3.2. For each s; Bs and #s satisfy the following properties: 1. If  is the predecessor node of  ⊂ #s , then g; s ⊆g; s . 2. Let  ˆo ⊂ #s , and let t¡s be the greatest stage so that #t ¡L  ˆo. If ˜v is the set of all elements in the domain of gt , then ˜v ⊆dom(g; s ). 3. Let o be a 4xed outcome,  ˆo ⊂ #s , and t¡s be the last stage with  ˆo ⊆#t . If g; s = g; t , then every node  ⊇  ˆo is initialized at some stage t  with t¡t  6s. 4. Let ; ˜v, and gt be as in (2). If y is an atom of the 4nite algebra determined by gt (˜v) and y is the corresponding atom of the 4nite subalgebra determined by g; s (˜v), then |y|¿ min{|y |; 2s }. 5. Bs extends Bs−1 as a 4nite Boolean algebra. 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 (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  ⊂ #s+1 . Assume that (1) – (4) are true for all  ⊂ . We must show them true for . (1) If the outcome of  is I at stage s + 1, then  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)  ⊂ #t , and  de9ned a challenging pair at t;

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(b)  has not been initialized at some t  with t¡t  6s + 1; and (c) The challenging pair de9ned at t is the challenging pair s + 1. At stage t; g; t ⊇ g; t by construction. If g; t =g; s+1 , then by the induction hypotheses on stages and nodes,  was initialized at some stage t  with t¡t  6s+1, a contradiction. Therefore, g; s+1 = g; t ⊆g; t = g; s+1 . (2) If the outcome of  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 r6s + 1 be the greatest stage at which  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 g; s+1 = g; r . If any more elements were added to ˜v between r and s + 1, then  ˆA would have been initialized between r and s + 1, contradicting the de9nition of r. (3) If o = I and g; s+1 =g; t , then  rede9nes its challenging pair at s + 1, so (3) is true by construction. If o = A and g; s =g; t , then by construction  ˆA was initialized at some stage t  with t¡t  6s + 1; therefore, so was any node  ⊇  ˆA. (4) If the outcome at stages s + 1 is I , then (4) is true by induction on nodes and (1) for . If the outcome of  is A, let r6s + 1 be the last stage at which  actually performed a diagonalization. If r = s + 1, then the diagonalization guarantees that only one atom y of the algebra determined by gt (˜v) is possibly made larger when we de9ne g; 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  ⊂ #s ; g; s+1 is a relation-preserving map between a subset of Bs and A. If  is in Case I, then the construction guarantees that g; s+1 respects Bs as it de9nes g; s+1 . If Case IIa or IIb, then g; s+1 is part of an uncanceled gt . By induction hypothesis on (6), gt , and hence g; s+1 is compatible with Bs . Finally, if  is in Case IIc, then the diagonalization ensures that g; s+1 maps atoms of the algebra Bs to nonzero disjoint elements of A whose union is 1, and g; s+1 maps combinations of these atoms to the analogous combinations of these disjoint elements. (6) Let t¡s + 1. First, if #t ⊂ #s+1 , then by (3) either gt is canceled or gs+1 ⊇ gt . If #s+1 ¡L #t , then gt is canceled. If #t ¡L #s+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 9nite 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 9nite algebra which contains 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.

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Lemma 3.3. There is a true path f with the following features: 1. if  ⊂ f, then  is the left-most node of length || so that  ⊂ #s for in4nitely many s. 2. If  ⊂ f, then  doesn’t de4ne challenging pairs in4nitely often. 3. If  ˆA ⊂f, then  doesn’t diagonalize in4nitely often. 4. If  ˆo ⊂f and S = {s:  ˆo ⊂#s }, then lims∈S g; s = g ; i.e., there are t ∈ S and ˜ and ˜ so that for all s ∈ S with s¿t; dom(g; s ) = dom(g; t ) = dom(g ) = d, a tuple d ˜ ˜ ˜ g; s (d) = g; t (d) = g (d). 5. If g =  ⊂ f g , then g : B ∼ = A. Proof. We show (2) – (4) by simultaneous induction on the length of  ⊂ f. Assume ˜ ˜c be the tuples so (2) – (4) are true for all  ⊂ . Let t be the least stage and d; that (a) For all stages s¿t; 6L #s . (b) For all nodes  ⊂  and all s¿t; a1; s ; a2; s  = a; t−1 ; a; t−1 . (c) If  ˆA ⊂f, then  doesn’t diagonalize after stage t − 1. (d) If  is the predecessor node of , then t and ˜d witness that (4) is true for , and g (˜d) = ˜c. (e) Let a1 ; a2  be the least pair so that for some atom a of the 9nite algebra determined by ˜c, a1 ∪a2 = a; a1 is in9nite; and a2 has in9nitely 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. First, we claim that  is not initialized at any s¿t. A node  can be initialized at stage s for one of three reasons: (a) #s ¡L ; (b) a node  ⊂ rede9nes its challenging pair; (c) a node  with ˆA ⊆ 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, a1; s ; a2; s  = a1; t ; a2; t  = a1 ; a2 . If ˆI ⊂f, then by the construction g = g; t . If ˆA ⊂f, let r¿t be the 9rst stage so that ˆA ⊂#r , and for all s¿r, ˆI *#s . Recall that w¡r is the greatest number so that t6w¡s + 1, ˆI  ⊆#w , and gw is uncanceled by r; ˜v = dom(gw ); and gw (˜v) =˜u. Let r  ¿r be the 9rst stage where ˆA⊂#r
, and y, the atom of the 9nite subalgebra determined by ˜u which we guessed at stage r  to be in9nite, is indeed in9nite. By the construction g = g; r
. Part (5) now follows almost immediately from Lemma 3.2. First, by part (1) of  Lemma 3.2 and (4) of this lemma, g = ⊂f g is a well-de9ned function. Since each gs is 1-1, g is 1-1. Since each g contains || in its domain and range, g is total and onto. Finally, by parts (5) and (6) of Lemma 3.2, B is a computable Boolean algebra, and each 9nite piece g (for a ⊂f) is a relation-preserving function. Consequently, g is an isomorphism. Lemma 3.4. Each requirement Re is satis4ed.

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Proof. Let  ⊂f be of length e. If e¿0, then let  be the predecessor node of , and let ˜c consist of the range of g and the elements in the challenging pair eventually de9ned by . If the 9nal outcome of  is I , then one of the following is true: 0 1. not every element of ˜c is in ran(g ◦’e
2 ); 2. the algebra determined by ˜c doesn’t satisfy the same relations as the algebra deter0 mined by its preimage under g ◦’e
2 ; 0 3. the size of one of the elements of ˜c is di3erent from its preimage under g ◦ ’e
2 . If the 9nal outcome of  is A, then Lemma 3.2 guarantees that there is some r so that for all s¿r with  ⊂#s , the outcome of  is A, and  does not actively diagonalize 0 at any stage after r. Our construction then dictates that all approximations of ’e
2 including and after r map a1; r  , an in9nite element of A, to b;1 r , a 9nite element of 0 B. Consequently, ’e
2 is not an isomorphism. 3.2.
02 -categoricity in linear orderings Notation. Let A be a linear ordering. 1. The predicate S(x; y) denotes the successor relation; i.e., (A; S) |= S(x; y) if and only if A |= (x¡y ∧ ∀z[(¬(x¡z¡y)]). 2. The predicate L− (x) denotes that x has no immediate predecessor; i.e., (A; S; L− ) |= L− (x) if and only if (A; S) |= ∀y(¬S(y; x)) if and only if A |= ∀y¡x∃z(y¡z¡x). 3. 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”, respectively.) Recall that throughout this paper, we assume, without loss of generality, that a linear ordering has a greatest and a least element. The main theorem of this subsection is Theorem 3.5. 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 4nitely many intervals, each of type m; !; !∗ ; Z, or n · (, so that each interval of type n · ( has a supremum and in4mum. The proof of this theorem is considerably more complicated than the one for Boolean algebras, just as the arguments concerning relative
02 -categoricity were. Notice that the computability of S implies that the binary relation “(a; b) is in9nite” and the relation 61 among tuples are both 10 . Furthermore, we shall see that in Propositions 3.6, 3.13, and 3.15, the additional assumptions of the computability of L+ and L− give us considerably more power: we can assign challenging pairs and diagonalize without making any of the mistakes that were possible in proof of Theorem 3.1. However, the proof we give of Proposition 3.19 to 9nish the proof of Theorem 3.5 resembles that given in Theorem 3.1 more closely: we make mistakes in assigning challenging pairs and diagonalizing, but the extra decidability assumptions allow us to see these mistakes “in time”.

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3.2.1. Some propositions about the maximal discrete pieces Proposition 3.6. 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 4nitely many maximal !-intervals and maximal !∗ -intervals. 2. A doesn’t contain arbitrarily large maximal 4nite 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 doesn’t satisfy one of the three conditions. We attempt to construct 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 0 in Theorem 3.1: Re says that ’e
2 is not an isomorphism between A and B; and the outcomes are I (inactive) and A (active). At stage s, we de9ne the approximations #s and gs , and we 4x an ordering Bs , which orders the elements 0; 1; : : : ; s − 1 in B. Challenging pairs and candidate pairs. Consider  ⊂#s working on requirement Re . If  ⊃ 4, the empty string, then it receives from its predecessor node  the function g; s that it believes to be a correct approximation of g, because all requirements Ri with i¡e seem to be satis9ed. Let ˜d be the domain of g; s with g; s (˜d) = ˜c = c1 ¡ · · · ¡cj . For  ⊇ 4, let c0 be the least element of A and cj+1 be the greatest element of A. (Recall that we always assume, without loss of generality, that A has a least and a greatest element.) In each interval (ci ; ci+1 ),  searches for challenging pairs a1i ; a2i  so that (ci ; a1i ), 1 2 (ai ; ai ), and (a2i ; ci+1 ) each contains an element with either no successor or no predecessor. (The assumption that A doesn’t satisfy one of the conditions in the statement of the theorem guarantees that  will always be able to 9nd an interval with such a pair.) The node  next includes a1i ; a2i in the range of g; s . The pair a1i ; a2i  challenges
0

the preimage under g; s ◦ ’e; 2s of each of the three intervals (ci ; a1i ); (a1i ; a2i ); (a2i ; ci+1 ) to contain an element with either no successor or no predecessor. If all challenges seem to be met at some later stage t by candidate pairs, then  attempts to change its
0

approximation of g so that the image under g; t ◦ ’e; 2t 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 di3er from those in Theorem 3.1 in two ways. First, a node  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  (whether or not  ⊂#s ) at stage s is designated by ch; s . There is a bit of ambiguity in this notation, as the challenging pairs associated with a node  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 ch; s denotes the set associated with  at the actual moment; if we are working in a later

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stage t, or if we are describing some feature of the construction as a whole, then ch; s represents the pairs associated with  at the end of s. Second, a node  never makes an incorrect guess about what is a challenging pair, because we assume that the relations L+ , L− are all computable. Therefore, only the initialization of a node  will remove a pair’s status as a challenging pair. Relationships among nodes. Throughout the construction, a node  ⊂#s may initialize another node ; i.e., if  ⊆#t for some t¡s, then  de9nes ch ; s = ∅, and it cancels the approximation gt . There will be two circumstances when a node will initialize another: 1. if  de9nes new challenging pairs at stage s, then  initializes each  with  ⊂; 2. if  ⊂#s determines its outcome to be I , then  initializes all  with  ˆI ¡L . (We will see that here, unlike in the proof of Theorem 3.1, a node will make no incorrect decisions about how to diagonalize after it ceases to be initialized and has de9ned all of its possible challenging pairs. Therefore, there is no need for the third occasion of initialization given in Theorem 3.1.) In turn, as  de9nes g; s , it must respect nodes  ¡L ˆo. What exactly does “respect” mean? Consider 7 ¡L ˆo and some stage t¡s so that 1. 7 ⊆#t ; 2. gt (v1 ; v2 ) = u1 ; u2 ; and 3. gt has not been canceled by stage s. If at stage s,  de9nes g; s (v1 ; v2 ) = u1 ; u2 , then as we decide the ordering Bs based on ’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 7 ⊂#t
, gt
(v1 ; v2 ) = u1 ; u2 . How can the construction ensure this compatibility? Assume that the interval (u1 ; u2 ) 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  , #t
can rede9ne gt
(v1 ; v2 ) = u1 ; u2 and remain in agreement with the ordering Bt
−1 we have chosen. Again we note the signi9cance of the 61 relation in our priority construction. Construction and supporting lemmas. Stage 0: For: each 7 in the tree, ch7; 0 = ∅. g0 = ∅. #0 = 4. Stage s + 1: For each 7 in the tree, we de9ne ch7; s+1 = ch7; s . Of course, 4 ⊂#s+1 . Let  ⊂#s+1 be a node of length e. Let  be the predecessor node of , and let dom(g; s+1 ) = ˜d with g; s+1 (˜d) = ˜c = c1 ¡ · · · ¡cj . (If  = 4, then there is no  and g; s+1 = ∅.) For  ⊇ 4, let c0 be the least element of A and cj+1 be the greatest element of A. If  has no associated challenging pairs, then it searches until it 9nds a pair. Furthermore, if any interval (ci ; ci+1 ) is without a challenging pair associated with , then  searches s + 1 steps to see if the interval has a pair a1i ; a2i  to add to ch; s+1 . Case I:  9nds a pair to add to ch; s+1 . The node  initializes all  with  ⊂⊆#t for some t¡s, and it chooses outcome I . (Notice, in particular, that any approximation gt de9ned in part by  at an earlier stage t is canceled. However, of course,  doesn’t de9ne ch; s+1 to be ∅.) It must de9ne g; s+1 so that these challenging pairs are in the range. We may assume by induction on

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the construction that ˜d includes all elements which are in the domain of an uncanceled approximation gt where t¡s + 1 and #t ¡L . Let ˜b be the set of elements in Bs but not in the domain of g; s+1 . Again, we may assume by induction on the construction that g; s+1 is compatible with the ordering Bs ; therefore,  can extend g; s+1 so that it includes ˜b and e in the domain and all challenging pairs and e in the range. Let b1i ; b2i be such that g; s+1 (b1i ; b2i ) = a1i ; a2i . (Summary: After rede9ning ch; s+1 , the node  restarts its work on the requirement Re .) Case II:  9nds no such challenging pair to add to ch; s+1 . There is a stage t¡s + 1 so that 1.  ⊂#t ; 2.  de9ned a challenging pair at stage t; 3.  has not been initialized since stage t; and 4. ch; s+1 = ch; t . (Throughout the rest of this construction, we will refer to this particular t.) We assume by induction on the construction that since  has not been initialized since t, the portion of gs+1 de9ned by nodes above  is the same as the portion of gt de9ned by nodes  above . The node  de9nes g; s+1 , its tentative version of g; s+1 , to be the same as g; t . We now consider the following subcases. Case IIa: One of the following is true:
0 1. An element of d˜ is not in the range of ’e; 2s+1 , or for some i, b1i ; b2i is not in
0
0 the range of ’ 2 . (Otherwise, let ˜c  , a1i  ; a2i  be such that ’ 2 (˜c  ) = ˜d and e; s+1
02 1 2 ’e; s+1 (ai ; ai ) = b1i ; b2i .)

e; s+1

2. The ordering of ˜c and the challenging pairs doesn’t match the ordering of the inverse
0

 2 images under g; s+1 ◦ ’e; s+1 . 3. The size of some interval determined by ˜c and the challenging pairs doesn’t match  the sizes of corresponding interval determined by their inverse images under g; s+1 ◦
0

’e; 2s+1 . 4. If t  ¡s + 1 is the greatest stage so that  ⊆#t
, then our guess at the preimage of 0 ˜d or b1i ; b2i for some i under ’e
2 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 (ci ; a1i  ); (a1i  ; a2i  ); (a2i  ;  ci+1 ) appears to have only elements with successors and predecessors. (Otherwise, the pairs a1i  ; a2i   are candidate pairs.)  The outcome of  is I , g; s+1 = g; s+1 .  (Summary: If the approximation g; s+1 de9ned so far is part of the isomorphism g
0

itself, then ’e 2 doesn’t appear to be an isomorphism, because it doesn’t seem total,
0

 2 as in (1) and (4), or we can readily see that g; s+1 ◦ ’e; s+1 doesn’t 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 ˆA ⊂#r , and ˆA has not been initialized since r;

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0


0

2. at stage r,  diagonalized against ’e 2 by de9ning g; r so that g; r ◦ ’e; 2r (a1i  ; a2i  ) is a pair in the same successor chain. The outcome of  remains A, and g; s+1 = g; r .
0

(Summary: The node  has already successfully diagonalized against ’e 2 an earlier stage, and nothing has injured this work.) Case IIc: Neither Case IIa nor Case IIb. The node  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  simultaneously searches for challenging pairs in them. If it 9nds a challenging pair in an interval without one before completing a diagonalization, then  adds it to ch; s+1 and enters Case I. If it completes a single one of the diagonalizations, then the outcome of  is A. Diagonalization: Let w be the greatest number so that t6w¡s+1, ˆI  ⊆#w , and gw is uncanceled. Let dom(gw ) =˜v with gw (˜v) =˜u. By our construction, the approximation  gw must extend g; 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  can de9ne an extension of gw whose domain includes ˜b. Let the image of ˜b be ˜a. Throughout the diagonalization,  must leave 9xed the mapping g; s+1 (˜d) = ˜c. However,  will attempt to alter the mapping of the rest of b1i ; b2i ;˜v; ˜b so that it meets requirement Re while it respects gw . Note that, as in Theorem 3.1, the diagonalization follows closely the argument for relativized
02 -categoricity given in Proposition 2.9.  Recall that g; s+1 is ’s tentative contribution to gs+1 , and for each challenging  1 2 1 2 1 2 pair a1i ; a2i   has de9ned g; s+1 (bi ; bi ) = ai ; ai . Furthermore, ai ; ai  has met the
0

challenge, because ’e; 2s+1 (a1i  ; a2i  ) = b1i ; b2i , and each of the intervals (ci ; a1i  ); (a1i  ; a2i  );  (a2i  ; ci+1 ) has an element with either no successor or no predecessor. The goal is to de9ne g; s+1 (b1i ; b2i ) = a1i  ; a2i  so that a1i  ; a2i  are in the same successor chain, thus
0


0


0

ensuring that ’e 2 is not an isomorphism if ’e;2s+1 is a correct approximation of ’e 2 . 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 ¡a2i ¡um+1 ¡ · · · ¡uk ¡ci+1 . By the very de9nition of challenging pairs, there is an element x that has no successor or no predecessor and thereby guarantees one of the following intervals is in9nite: 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 in9nite: 1. (a2i ; um+1 ); 2. (uj ; uj+1 ), for some j ∈ {m + 1; : : : ; k − 1}; 3. (uk ; ci+1 ).

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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 in9nite 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 in9nite by y . For each un including and after this endpoint, leave the mapping of vn as it is. Thus we are left with a sequence of r6k + 2 elements in the universe of A. The node  de9nes g; s+1 so as to remap the current preimages of these elements to appropriate elements in the successor chain and 9nishes the de9nition of g; s+1 by including e in the domain and range. This concludes the diagonalization.  After each  ⊂ #s+1 has determined g; s+1 , we de9ne gs+1 = ⊂#s+1 g; s+1 and use it to de9ne Bs+1 . This concludes the construction. Lemma 3.7. For each s; #s and Bs satisfy the following properties: 1. if  is the predecessor node of some  ⊂ #s , then g; s ⊆ g; s . 2. Let ˆo ⊂ #s , and let t¡s be the greatest stage so that #t ¡L ˆo. If ˜v is the set of all elements in the domain of gt , then ˜v ⊆ dom(g; s ). 3. Let o be a 4xed outcome, ˆo ⊂ #s , and t¡s be the last stage with ˆo ⊆ #t . If g; s = g; t , then every node  ⊇ ˆo is initialized at some stage t  with t¡t  6s. 4. Let ; ˜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 g; s (˜v ). 5. Bs extends Bs−1 as a 4nite 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  ⊂ #s+1 . The arguments for (1) – (4) are essentially the same as those in Lemma 3.2. (5) Since the ordering Bs+1 is determined entirely by gs+1 , we need only verify that for each  ⊂ #s , g;s+1 is an order-preserving map between a subset of Bs and A. If  is in Case I, then the construction guarantees that g; s+1 respects Bs as it de9nes g; s+1 . If Case IIa or IIb, then g; s+1 is part of an uncanceled gt . By induction hypothesis on (6), gt , and hence g; s+1 is compatible with Bs . Finally, if  is in Case IIc, then the diagonalization ensures that g; s+1 preserves the order of the elements of Bs . (6) Let t¡s + 1. If #t ⊂ #s+1 , then by (3) either gt is canceled or gs+1 ⊇ gt . If #s+1 ¡L #t , then gt is canceled. If #t ¡L #s+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.

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Lemma 3.8. There is a true path f with the following features: 1. if  ⊂ f, then  is the left-most node of length || so that  ⊂ #s for in4nitely many s. 2. If  ⊂ f, then  doesn’t de4ne challenging pairs in4nitely often. 3. If ˆo ⊂ f and S = {s: ˆo ⊂ #s }, then lims∈S g; s = g ; i.e., there are t ∈ S and ˜ so that for all s ∈ S with s¿t, dom(g; s ) = dom(g; t ) = dom(g ) = d, ˜ and a tuple d ˜ ˜ ˜ g; s (b) =  g;t (d) = g (d). 4. If g = ⊂f g , then g : B ∼ = A. Proof. We show (2) and (3) by simultaneous induction on the length of  ⊂ f. Assume (2) and (3) are true for all  ⊂ . Let t be the least stage and ˜b be the tuple so that: (a) for all stages s¿t, 6L #s ; (b) for all nodes  ⊂  and all s¿t, ch; s = ch; t−1 ; ˜ witness that (3) is true for ; and (c) if  is the predecessor node of , then t and d ˜ containing a challenging pair,  has found (d) for each interval determined by g (d) such a pair by stage t. First, we claim that  is not initialized at any stage s¿t. A node can be initialized at stage s for one of two reasons: (a) #s ¡L ; (b) a node  ⊂  de9nes 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, ch; s = ch; t . If ˆI  ⊂ f, then by the construction g = g; t . If ˆA ⊂ f, then let r¿t be the 9rst stage so that (a) ˆA ⊂ #r , and ˆA is not initialized after r; (b)  performed a diagonalization on a candidate pair a1i  ; a2i   at stage r. By the construction, g = g; r . The argument for (4) is the same as that given for (5) of Lemma 3.3. Lemma 3.9. Each requirement Re is satis4ed. Proof. Let  ⊂ f be of length e. If e¿0, then let  be the predecessor node of , and let ˜c consist of the range of g and all challenging pairs eventually de9ned by . If the 9nal outcome of  is I , then one of the following is true:
0

1. not every element of ˜c is in ran(g ◦ ’e 2 ); 2. the ordering of ˜c is not consistent with the ordering of its preimage
0

under g ◦ ’e 2 ; 3. the size of some interval determined by ˜c is di3erent from its preimage under
0

g ◦ ’e 2 ;
0 4. g ◦ ’e 2 maps an interval without candidate pairs to one with challenging pairs. If the 9nal outcome of  is A, then there is some r so that for all s¿r with  ⊂ #s ,
0 the outcome of  is A. Our construction then dictates that all approximations of ’e 2 including and after r map a candidate pair to elements in the same successor chain in
0

B. Consequently, ’e 2 is not an isomorphism.

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Denition 3.10. A maximal Z-cluster is a maximal interval I of the form Z · C for some linear ordering C. Corollary 3.11. Let A be a
02 -categorical linear ordering so that (A; S; L− ; L+ ) is computable. Then A has 4nitely many maximal Z-clusters. Proof. Let c1 ; : : : ; cn partition A so that condition (3) of Proposition 3.6 holds. Each of the open intervals determined by the partition must contain at most three maximal Z-clusters. Therefore, A contains less than 4n + 4 maximal Z-clusters. Corollary 3.12. 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 4nite sum of intervals of the following forms: 1. m for some m¡n; 2. maximal !-intervals, maximal !∗ -intervals; 3. maximal Z-clusters;  4. maximal intervals of the form q∈( Bq , where each Bq ¡n. Proof. Order the 9nitely many maximal !-intervals, maximal !∗ -intervals, and maximal Z-clusters in A. There remain 9nitely many intervals, each of which contains no !-intervals, no !∗ -intervals, and no Z intervals. Furthermore, there is n so that all maximal 9nite discrete intervals have order type  ¡n. By Lemma 2.10, each of these intervals is either 9nite or of the form m1 + q∈( Bq + m2 , where 06m1 ; m2 ¡n, and each Bq ¡n.  3.2.2. Linear orderings with intervals of the form q∈( Bq The propositions below are analogous to Propositions 2.12 and 2.16. The diagonalizations employed are relatively straightforward implementations of the strategies described in these two results, so only sketches are provided. Proposition 3.13. 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 ! + q∈( Bq + 1, and for all q ∈ (, |Bq |¡n;  2. A has the form ! + Z-cluster + q∈( Bq + 1, and for all q ∈ (, Bq ¡n;  3. A has the form 1 + q∈( Bq + !∗ , and for all q ∈ (, |Bq |¡n;  4. A has the form 1 + q∈( Bq + Z-cluster + !∗ , and for all q ∈ (, |Bq |¡n. Then A is not
02 -categorical. Sketch. We will discuss the proof for (2), as it is the more diRcult case. The argument will involve a priority construction very much like the one in Proposition 3.6. Di3erences will appear only in the details of candidate assignment and diagonalization. Consider  ⊂ #s working on requirement Re . If  ⊃ 4, the empty string, then it receives from its predecessor node  the function g; s that it believes is a correct

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˜ be the domain of g; s with g; s (d) ˜ = ˜c = c1 ; : : : ; cj . For  ⊇ 4, approximation of g. Let d  let c0 = the “0” of the ! summand and cj+1 = the “0” to the right of the q∈( Bq summand. The node  9rst determines  which interval (ci ; ci+1 ) contains both elements from the Z-cluster and from the q∈( Bq summand. This can be done computably, since, given an element a,  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  then determines if the preimage
0

of this interval under g; s ◦ ’e; 2s has the same property.  If so, then  includes in the range of g; s a challenging element ai which is in the q∈( Bq summand but not in the same successor chain as ci+1 . The node  determines if ai , the preimage of ai under
0

the function g; s ◦ ’e; 2s , is a true, candidate element; i.e., if it has properties analogous to those of ai . If so, then  changes its approximation of g so that the image under
0

g; s ◦ ’e; 2s of ai is in ! or the Z-cluster. We thereby meet requirement Re . Corollary 3.14. Let A be a
02 -categorical linear ordering so that (A; S; L− ; L+ ) is  computable. Then every maximal q∈( Bq interval has a supremum and an in4mum. Recall De9nition  2.14 and Lemma 2.15, which are about linear orderings with intervals of the form q∈( Bq . Proposition 3.15. Let A be a
02 -categorical linear ordering so that (A; S; L− ; L+ ) is computable. Any maximal q∈( Bq interval satis4es Property 1. Sketch. We must show that an ordering satisfying Property 2 cannot be
02 categorical. First, as in Proposition 2.16, let m1 be the largest number for which there is an m2 as in Property 2. De9ne a partition of ( so that for each m¿m1 and each open interval J of ( de9ned by the partition, either Bq = m for all q ∈ J , or Bq = m for all q ∈ J . There is an open interval J = (q1 ; q2 ) (q1 might be −∞, and q2 might be ∞) of ( so that for all m¿m1 and all q ∈ J , Bq = m, and q∈J Bq satis9es 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 ∈= B , g(a) = a; and q∈J q 2. there is no
02 isomorphism between B and A. Again, the argument will involve a priority construction very much like the one in Proposition 3.6, and di3erences will appear mainly in the details of candidate assignment and diagonalization. Consider  ⊂ #s working on requirement Re . If  ⊃ 4, the empty string, then it receives from its predecessor node  the function g; s that it believes is a correct ap˜ = ˜c = c1 ; : : : ; cj . For  ⊇ 4, ˜ be the domain of g; s with g; s (d) proximation of g. Let d de9ne c0 as follows: 1. c0 = the greatest element ofBq1 if q1 = −∞; 2. c0 = the greatest bound of q∈( Bq in A if q1 = −∞. We de9ne the element cj+1 similarly.

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In each interval (ci ; ci+1 ),  searches for challenging pairs a1i ; a2i  where a1i 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 q∈( Bq guarantee that  will always be able to 9nd an interval with such a pair.) The node  then includes a1i ; a2i in the range of g; s . The node  then determines if the pair a1i  ; a2i  , the preimage of the pair a1i ; a2i
0

under g; s ◦ ’e; 2s , is a true candidate pair; i.e., if it has analogous properties. If so, then
0

 attempts to change its approximation of g so that the image under g; t ◦ ’e; 2t of a2i  is an element in a maximal successor chain of length m1 . We thereby meet Re . Corollary 3.16. Let A be a
02 -categorical linear ordering so that (A; S; L− ; L+ ) is computable. Then there is n ∈ ! so that A can be written as a 4nite sum of maximal intervals of the following forms: 1. m for some m¡n; 2. m · ( for some m¡n; 3. !; !∗ ; 4. Z-clusters, where each maximal m · ( interval has an in4mum and a supremum. 3.2.3. Z-clusters When we examined relativized
02 -categoricity, we showed, in a single proposition, that a relatively
02 -categorical linear ordering could have only 9nitely many maximal !; !∗ ; 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 9nd 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 ! and !∗ interval has an element with either no successor or no predecessor, so under the extra decidability assumptions on A, we can easily identify certain in9nite intervals. This property played a key role in the proof of Proposition 3.6. A Z-cluster, of course, has no such element. Consequently, it may be impossible to identify the in9nite intervals computably, and thus it is harder to formulate a diagonalization to prove that A has only 9nitely many Z-intervals. Proposition 3.17. 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 4nitely 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 3.18. 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 in4nite and discrete. Then A is not
02 -categorical.

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Proof. Since C is in9nite and discrete, I must be a sum of the form c1 (Z · !) + c2 [(Z · Z) · D] + c3 (Z · !∗ ), 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 3.16, either I has an in9mum, or I is directly to the right of a maximal !-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 !∗ -interval. Case I: c2 = 0. Then A contains an interval K with endpoints of one of the following forms: (a) 1 + c1 (Z · !) + c3 (Z · !∗ ) + 1; (b) 1 + c1 (Z · !) + c3 (Z · !∗ ) + !∗ ; (c) ! + c1 (Z · !) + c3 (Z · !∗ ) + 1; (d) ! + c1 (Z · !) + c3 (Z · !∗ ) + !∗ . In any case, we can construct A , a computable copy of A, where K  , the image of this interval K, is such that 1. K  contains in9nitely many Z-intervals; and 2. for a; b ∈ K  we can computably decide if the interval (a; b) is in9nite or 9nite. We can show that A , and hence A, is not
02 -categorical. The argument is almost exactly like that given in Proposition 3.6, except that we replace searching for elements with no successor or predecessor with searching for intervals which are in9nite. Case II: c2 = 1. If the linear ordering D = 1 or (, then we can argue that there is A and K  with 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 ! + (Z · !) + (Z · !∗ ) + !∗ . Hence, we can again argue for the existence of A and K  . Therefore, A is not
02 -categorical. Proposition 3.19. Let A be a
02 -categorical linear ordering so that (A; S) is computable, 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 Proposition 3.6 needed by Propositions 3.13 and 3.15. 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 de9ne the approximations #s and gs , and the ordering Bs . First, without loss of generality, let J be a Z-interval in I so that there is a Zinterval in I to the right of J , but there is no next Z-interval to the right of it. We 9x c−∞ , some element in J , and c∞ , some element in I to the right of J . For all a ∈ A − (c−∞ ; c∞ ), we de9ne 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  ⊂ #s working on requirement Re . If  ⊃ 4, the empty string, then it receives from its predecessor node  the function

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˜ be the domain of g; s with g; s that it believes is a correct approximation of g. Let d ˜ g; s (d) = ˜c = c1 ¡ · · · ¡cj . For  ⊇ 4, let c0 = c−∞ and cj+1 = c∞ . The node  guesses at which is the 9rst interval (ci ; ci+1 ) to be in9nite. (Thus, c0 ; c1 ; : : : ; ci are all in the same Z-interval.) In this interval  locates a challenging element a; s so that (ci ; a; s ) and (a; s ; ci+1 ) both seem in9nite. The node  includes it
0

 in the range of g; s . The element a; s challenges (ci ; ci+1 ), the preimage under g ◦ ’e 2  of (ci ; ci+1 ), to contain an element a; s with the analogous property. If the challenge seems to be met at stage t, then  changes its approximation so that the image under
0

g; t ◦ ’e; 2t of a; s is in the same Z-interval as c0 ; we thereby meet requirement Re . In the de9nition of challenging elements, this construction resembles Theorem 3.1 more than Proposition 3.6. First, at stage s;  will de9ne at most a single challenging element in one interval, rather than challenging pairs in multiple intervals. Second,  will make incorrect guesses as to which element has the above properties. In Proposition 3.6,  could determine computably that an element had no successor or no predecessor, and so it never mistakenly assigned pairs; any canceling of a challenging pair was done by a node of higher priority. Here, however,  may incorrectly guess the 9rst element of ˜c to be in a Z-interval di3erent from c0 . Furthermore, even when  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 3.1 both of these incorrect guesses should introduce only 9nitely much injury, and a 0 construction still suRces. A node  ⊂ #s may initialize another node : if  ⊆ #t for some t¡s, then  cancels the assignment of a; t and the approximation gt . There will be three circumstances when a node will initialize another: 1. if  rede9nes its challenging element at stage s, then  initializes each  with  ⊂ ; 2. if  determines its outcome to be I , then  initializes all  with  ˆI ¡L ;
0

3. if  diagonalizes against ’e 2 at stage s, then it initializes all  with  ˆA ⊆ . Construction and supporting lemmas. Stage 0: g0 = ∅: #0 = 1. Assign no challenging elements. Stage s + 1: Of course, 4 ⊂ #s+1 . Let  ⊂ #s+1 be a node of length e. Let  be ˜ with g; s (d) ˜ = ˜c = c1 ¡ · · · ¡cj . (If  = 4, the predecessor node, and let dom(g; s ) = d, then there is no  and g; s+1 = ∅.) For  ⊇ 4, let c0 = c−∞ and cj+1 = c∞ as elements of A. As elements of B let d0 = c−∞ and dj+1 = c∞ . The node  searches for the 9rst interval (ci ; ci+1 ), and the 9rst element a; s+1 so that 1. after s + 1 steps of enumerating the diagram of (A; S); ci+1 appears to be in a di3erent Z-interval from c0 ; 2. a; s+1 ∈ (ci ; ci+1 ), and a; s+1 ; ci and ci+1 appear to be in three distinct Z-intervals. This element a; s+1 is the challenging element for  at stage s + 1. Case I: The node  has been initialized since its last active stage or has de9ned a new challenging element since its last active stage. The node  initializes all  with  ⊂  and chooses outcome I . It must include a; s+1 ˜ includes in the range of g; s+1 . We may assume by induction on the construction that d all elements which are in the domain of an uncanceled approximation gt where t¡s+1 and #t ¡L . Let ˜b be the set of elements in Bs but not in the domain of g; s+1 . Again,

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we may assume by induction on the construction that g; s+1 is compatible with the ordering Bs ; therefore,  can extend g; s+1 so that it includes ˜b and e in the domain, and a; s+1 and e in the range. Let b; s+1 be such that g; s+1 (b; s+1 ) = a; s+1 . (Summary: After rede9ning ; s+1 , the node  restarts its work on the requirement Re .) Case II:  doesn’t need to rede9ne its challenging pair. There is a stage t¡s + 1 so that 1.  ⊆ #t ; 2. either  has been initialized since the previous stage r in which  ⊂ #r , or the de9nition of a; t changed during stage t; 3.  has not been initialized since stage t; and 4. a; s+1 = a; t ; (Throughout the rest of this construction, we will refer to this particular t.) We assume by induction on the construction that since  has not been initialized since t, the portion of gs+1 de9ned by nodes above  is the same as the portion of gt de9ned by nodes  above . The node  de9nes g; s+1 its tentative version of g; s+1 , to be the same as g; t . We now consider the following subcases. Case IIa: One of the following is true: 0 ˜ or one of d0 ; dj+1 ; b; s+1 is not in the range of ’
2 . (Other1. An element of d e; s+1 0
0 ˜ ’
e 2 (a; s+1  ) = b; s+1 , and wise, let ˜c  ; c ; c ; a; s+1  be such that ’ 2 (˜c  ) = d; 0


0

j+1

e; s+1

 ’e 2 (c0 ; cj+1 ) = d0 ; dj+1 .)  2. The ordering of ˜c; c0 ; cj+1 ; a; s+1 doesn’t match that of ˜c  ; a; s+1  ; c0 ; cj+1 . 3. The size of some interval determined by ˜c; c0 ; cj+1 and a; s+1 doesn’t match the size  of the corresponding interval determined by ˜c  ; ˜c0  ; cj+1 ; a; s+1 .  4. If t ¡s + 1 is the greatest stage so that  ⊆ #t
, then our guess at the preimage of 0 ˜ or one of b; s+1 ; d0 ; dj+1 under ’
e 2 has changed since t  . d (Otherwise, the element a; s+1  is a candidate element.)  The outcome of  is I , and g; s+1 = g; s+1 .  (Summary: If the approximation g; s+1 de9ned so far is part of the isomorphism g
0

itself, then ’e 2 doesn’t appear to be an isomorphism, because it doesn’t seem total, as
0

 2 in (1) and (4), or g; s+1 ◦ ’e; s+1 doesn’t 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 that  ˆA ⊂ #r , and  ˆA has not been initialized since r; 2.  performed a diagonalization on a candidate element (a; s+1 ) at stage r; 3. if K is an interval determined by a; s+1 and elements in the range of higher priority gw , and during the diagonalization  guessed that K was in9nite, then  can 9nd s + 1 elements in K and still guesses that K is in9nite. The outcome of  remains A, and g; s+1 = g; r . (Summary: The node  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.

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First,  initializes each node  with  ˆA ⊆ . The node  attempts to perform the following diagonalization on the candidate element a; s+1  . If it completes the diagonalization, then the outcome of  is A. Diagonalization: Let w be the greatest number so that t6w¡s + 1;  ˆI  ⊆ #w and g w is uncanceled. Let dom(g w ) = ˜v with g w (˜v) = ˜u. By our construction, the approxi mation g w must extend g; s+1 . ˜ Let b be the set of elements in Bs but not in the domain of g w . By induction on the construction, we can assume that g w is compatible with this ordering, and  can de9ne an extension of g w whose domain includes ˜b. Let the image of ˜b be ˜a. Throughout the ˜ = ˜c. However,  will attempt diagonalization,  must leave 9xed the mapping g; s+1 (d) to alter the mapping of the rest of ˜v; ˜b; b; s+1 so that it meets requirement Re while it respects g w .  Recall that g; s+1 is ’s tentative contribution to gs+1 , and for the challenging element   a; s+1 ,  has de9ned g; s+1 (b; s+1 ) = a; s+1 . Furthermore, a; s+1 seemingly has met the
0

challenge, because ’e 2 (a; s+1  ) = b; s+1 , and each of the intervals (ci ; a; s+1  ); (a; s+1  ;  ci+1 ) appears to be in9nite. The goal is to de9ne g; s+1 (b; s+1 ) = a; s+1  so that
0

ci ; a; s+1  are in the same Z-interval, thus ensuring that ’e 2 is not an isomorphism
0


0

if ’e;2s+1 is a correct approximation of ’e 2 . 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 ¡ · · · u1 ¡ a;s+1 ¡ ul+1 ¡ · · · ¡ uk ¡ ci+1 : In each of the following intervals, the node  attempts to 9nd s + 1 elements or to determine that the interval is 9nite: 1. (a; s+1 ; ul+1 ); 2. (uj ; uj+1 ) for some j ∈ {l + 1; : : : ; k − 1}; 3. (uk ; ci+1 ). Let uj∗ be the right endpoint of the leftmost interval in which  can locate s + 1 elements without discovering that it is 9nite. (There must be such an interval. Otherwise  discovers that (a; s+1 ; ci+1 ) is 9nite, rede9nes its challenging element and enters Case I.) Let ˜u  be the list of elements of ˜u ∪ ˜a which are in (ci ; uj∗ ), and ˜u  the list of elements of ˜u ∪ ˜a which are in [uj∗ ; ci+1 ). For each element of ˜u  , leave the mapping of the corresponding element of ˜v ∪ ˜b as it is. The node  then de9nes g; 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,  9nishes the de9nition of g; s+1 by including e in the domain and range. This concludes the diagonalization.  After each  ⊂ #s+1 has determined g; s+1 , we de9ne gs+1 = ⊂#s+1 g; s+1 and use it to de9ne Bs+1 . This concludes the construction. Lemma 3.20. For each s; #s and Bs satisfy the following properties: 1: If  is the predecessor node of  ⊂ #s , then g; s ⊆ g; s .

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2: Let  ˆo ⊂ #s , and let t¡s be the greatest stage so that #t ¡L  ˆo. If ˜v is the set of all elements in the domain of gt , then ˜v ⊆ dom(g; s ). 3: Let o be a 4xed outcome,  ˆo ⊂ #s , and t¡s be the last stage with  ˆo ⊆ #t . If g; s = g; t , then every node  ⊇  ˆo is initialized at some stage t  with t¡t  6s. ˜ and K  is 4: Let ; ˜v, and gt be as in (2). If K is an interval determined by gt (d)  ˜ the corresponding interval determined by g; 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  ⊂ #s+1 . Assume that (1) – (4) are true for all  ⊂ . We must show them true for . The arguments for (1) – (4) are essentially the same as those in Lemma 3.2. The argument for (5) is the same as in Lemma 3.7. (6) Let t¡s + 1. First, if #t ⊂ #s+1 , then by (3) either gt is canceled or gs+1 ⊇ gt . If #s+1 ¡L #t , then gt is canceled. If #t ¡L #s+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 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 interval 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. Lemma 3.21. There is a true path f with the following features: 1: If  ⊂ f, then  is the left-most node of length || so that  ⊂ #s for in4nitely many s. 2: If  ⊂ f, then  doesn’t de4ne challenging pairs in4nitely often. 3: If  ˆA ⊂ f, then  doesn’t actively diagonalize in4nitely often. 4. If  ˆo ⊂ f and S = {s :  ˆo ⊂ #s }, then lims∈S g; s = g ; i.e., there are ˜ so that for all s ∈ S with s¿t; dom(g; s ) = dom(g; t ) = dom(g ) t ∈ S and a tuple d ˜ ˜ ˜ ˜ g (d). = d, and  g; s (d) = g; t (d) =∼ 5: If g = ⊂f g , then g : B = A. Proof. We show (2) – (4) by simultaneous induction on the length of  ⊂ f. Assume ˜ ˜c be the tuples so that (2) – (4) are true for all  ⊂ . Let t be the least stage and d; (a) for all stages s¿t; 6L #s ; (b) for all nodes  ⊂  and all s¿t; a; s = a; t−1 ;

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(c) if  ˆA ⊂ f, then  doesn’t diagonalize after stage t − 1; ˜ witness that (3) is true for , and (d) if  is the predecessor node of , then t and d ˜ g (d) = ˜c and (e) a; t+1 is in (ci ; ci+1 ); ci+1 is the least element of ˜c not in the same Z-interval as c0 , and a; t+1 is in the same Z-interval as neither ci nor ci+1 . First, we claim that  is not initialized at any stage s¿t. A node can be initialized at stage s for one of three reasons: (a) ¡L #s ; (b) a node  ⊆  rede9nes a; s ; (c) a node  ˆA ⊂  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; a; s = a; t . If  ˆI  ⊂ f, then by the construction g = g; t . If  ˆA ⊂ f, let r¿t be the 9rst stage so that  ˆA ⊂ #r , and for all s¿r;  ˆI  * #s . Recall that w¡r is the greatest number so that t6w¡s + 1;  ˆI  ⊆ #w , and g w is uncanceled by r; ˜v is the domain of g w ; and g w (˜v) = ˜u. Let r  ¿r be the 9rst stage where  ˆA ⊂ #r
, and uj∗ chosen by  is the right endpoint of the leftmost in9nite interval determined by a; r and the elements of ˜u between a; r
and ci+1 . By the construction g = g; r
. The argument for (5) is the same as that given in Lemma 3.3. Lemma 3.22. Each requirement Re is satis4ed. Proof. Let  ⊂ f be of length e. If e¿0, then let  be the predecessor node of , and let ˜c consist of the range of g and all challenging elements eventually de9ned by . If the 9nal outcome of  is I , then one of the following is true:
0

1. not every element of ˜c is in ran(g ◦ ’e 2 );


0

2. the ordering of ˜c is not consistent with the ordering of its preimage under g ◦ ’e 2 ; 3. the size of one of the intervals determined by ˜c is di3erent from its preimage under
0

g ◦ ’e 2 . If the 9nal outcome of  is A, then there is some r so that for all s¿r with  ⊂ #s , the outcome of  is A, and  does not actively diagonalize at any stage after r. Our
0

construction then dictates that all approximations of ’e 2 including and after r map  c−∞ and a; r , two elements not in the same Z-interval in A, to d∞ and b; r , two
0

elements in the same Z-interval in B. Consequently, ’e 2 is not an isomorphism. 3.3. 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 e3ectiveness 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. Goncharov showed that the existence of a single 2-decidable copy of a structure is

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enough to establish the equivalence of relativized and unrelativized computable categoricity. What is true in general for the
02 level? 3. What exactly is the relationship between 1-decidable computable categoricity and
02 -categoricity? (See Theorem 2.18.) Is the correspondence here an instance of a more general result? 4. In [10] Goncharov and Dzgoev proved that results about computable categoricity in structures such as Boolean algebras, linear orderings, and Abelian p-groups were particular instances of a general result about a notion called branching. Is there a similar notion for
02 -categoricity and a result which would supercede the results of this paper? References [1] C.J. Ash, J.F. Knight, Computable Structures and the Hyperarithmetical Hierarchy, North-Holland Publishing Company, Amsterdam, 2000. [2] C.J. Ash, J.F. Knight, M. Manasse, T. Slaman, Generic copies of countable structures, Ann. Pure Appl. Logic 42 (1989) 195–205. [3] C.C. Chang, H.J. Keisler, Model Theory, North-Holland, Amsterdam, 1973. [4] J. Chisholm, E3ective model theory vs. recursive model theory, J. Symbolic Logic 55 (1990) 1168–1191. [5] R. Downey, Computability theory and linear ordering, in: Y.L. Ershov, S.S. Goncharov, A. Nerode, J.B. Remmel, V.W. Marek (Eds.), Recursive Algebra, Analysis and Combinatorics, Handbook of Recursive Mathematics, Vol. 2, Elsevier, Amsterdam, 1998, pp. 823–976. [6] S.S. Goncharov, Autostability and computable families of constructivizations. Algebra Logic 14 (1975) 647– 680 (Russian), 392– 409 (English). [7] S.S. Goncharov, The quantity of non-autoequivalent constructivizations, Algebra Logic 16 (1977) 257– 282 (Russian), 169 –185 (English). [8] S.S. Goncharov, Autostability of models and abelian groups, Algebra Logic 19 (1980) 23– 44 (Russian), 13–27 (English). [9] S.S. Goncharov, The problem of the number of non-self-equivalent constructivizations, Algebra Logic 19 (1980) 621– 639 (Russian), 401– 414 (English). [10] S.S. Goncharov, V.D. Dzgoev, Autostability of models, Algebra Logic 19 (1980) 45 –58 (Russian), 28–36 (English). [11] T. Hungerford, Algebra, Springer, Berlin, 1974. [12] J.F. Knight, M. Stob, Computable Boolean Algebras, preprint. [13] S. Koppelberg, Handbook of Boolean Algebras, North-Holland, Amsterdam, 1989. [14] O.V. Kudinov, A criterion for the autostability of 1-decidable models, Algebra Logic 31 (1992) 479 – 492 (Russian), 284 –292 (English). [15] M. Moses, Recursive linear orderings with recursive successivities, Ann. Pure Appl. Logic 27 (1984) 253–264. [16] J.B. Remmel, Recursive isomorphism types of recursive Boolean algebras, J. Symbolic Logic 46 (1981) 596–616. [17] J.B. Remmel, Recursively categorical linear orderings, Proc. AMS 83 (1981) 387–391. [18] P. La Roche, Recursively presented Boolean algebras, Notices AMS 24 (1977) A552–A553. [19] R.I. Soare, Recursively Enumerable Sets and Degrees, Springer, Berlin, 1987.