Local Computability for Ordinals

Transcription

1 Local Computability for Ordinals Johanna N.Y. Franklin 1, Asher M. Kach 2, Russell Miller 3, and Reed Solomon Auditorium Road, Univ. of Connecticut, U-3009, Storrs, CT U.S.A. johanna.franklin@uconn.edu, franklin/ david.solomon@uconn.edu, solomon/ 2 University of Chicago, 5734 S. University Avenue, Chicago, IL U.S.A. asher.kach@gmail.com, kach/ 3 Queens College of CUNY, Kissena Blvd., Queens, NY USA and CUNY Graduate Center, 365 Fifth Avenue, New York, NY USA Russell.Miller@qc.cuny.edu, rmiller Abstract. We examine the extent to which well orders satisfy the properties of local computability, which measure how effectively the finite suborders of the ordinal can be presented. Known results prove that all computable ordinals are perfectly locally computable, whereas ω CK 1 and larger countable ordinals are not. We show that perfect local computability also fails for uncountable ordinals, and that ordinals α ω1 CK are θ-extensionally locally computable for all θ < ω1 CK, but not when θ > ω CK 1, nor when θ = ω CK 1 α < ω CK 1 ω. Key words: Computability theory, computable model theory, local computability, ordinal, recursion theory. 1 Introduction Local computability represents an effort to give effective presentations of structures, such as the fields of real and complex numbers, which admit computation on their elements by simple algebraic algorithms and therefore, despite their uncountability, feel as though they ought to have computable presentations. Full definitions and much more analysis are given in [3 5], and we offer some basic definitions below. Local computability applies to linear orders as well as to fields and other structures, and the intention of this work is to investigate local computability for ordinals, the most ubiquitous linear orders in mathematical logic. We started with a particular eye on uncountable ordinals, but soon found countable ordinals to be of similar interest, particularly those countable ordinals The second author was partially supported by a grant from the Packard Foundation through a Post-Doctoral Fellowship. The third author was partially supported by grant # DMS from the National Science Foundation, by the Centre de Recerca Matemática, the Isaac Newton Institute, and the European Science Foundation, and by several PSC-CUNY grants from the Research Foundation of The City University of New York.

2 2 J.N.Y. Franklin, A.M. Kach, R. Miller, & R. Solomon large enough not to be computably presentable. Most of the questions revolve around the concept of a θ-extensional computable cover, which we now define, along with other notions necessary to this topic. Fix a theory T in a finite language L that is -axiomatizable. All structures will refer to an L-structure that is a model of T. Definition 1. A simple cover of a structure S is a countable collection A of models {A i } i I of T, each generated by a finite tuple a i, such that every finitely generated substructure of S is isomorphic to some A i and every A i embeds into S. A simple cover is computable if every A i A is a computable structure with domain an initial segment of ω. A simple cover is uniformly computable if the sequence {(A i, a i )} i I can be given uniformly computably, including a strong index for each a i. Definition 2. A cover of S consists of a simple cover A of S along with sets I A ij (for all A i, A j A) of injective homomorphisms f : A i A j satisfying: For all finitely generated substructures B C S, there exists i, j ω, f Iij A, and β : A i = B and γ : A j = C with β = γ f. For all k and m and g Ik,m A, there exist finitely generated substrutures D E S and isomorphisms δ : A k = D and ɛ : Am = E with δ = ɛ g. The Amalgamation Property: for every i, j, k and all maps f Iij A and g Iik A, there exists m and maps h IA jm and p IA km with p g = h f. A cover is computable if A is a uniformly computable simple cover of S and there exists a c.e. set W such that, for all i, j ω, I A ij = {ϕ e A i : i, j, e W }. The structure S is locally computable if it has a uniformly computable cover. Since the Amalgamation Property is not always assumed to be included in the definition of cover, we will sometimes specify it when stating our theorems, even though it is taken here as part of the definition. Definition 3. Let A be a cover of a structure S. An A i A matches a substructure B S extensionally if there is an isomorphism β : A i = B satisfying: For every finitely generated C with B C S, there exists j ω, f I A ij, and γ : A j = C with β = γ f. For every m ω and g I A i,m, there exists an E S and ɛ : A m = E with B E and β = ɛ g. The map β is termed an extensional match between A i and B. Definition 4. Let A be a cover of a structure S. Every isomorphism β : A i = B, where B S is a finitely generated substructure, is 0-extensional. For an ordinal θ > 0, an isomorphism β : A i = B is θ-extensional if:

3 Local Computability for Ordinals 3 For every finitely generated C with B C S and every ordinal ζ < θ, there exists j ω, f Iij A, and a ζ-extensional γ : A j = C with β = γ f. For every m ω, g Ii,m A, and ordinal ζ < θ, there exists E S and ζ-extensional ɛ : A m = E with B E and β = ɛ g. A uniformly computable cover A of S is θ-extensional if for every A i A there is a θ-extensional isomorphism β : A i = B to some finitely generated substructure B S and for every finitely generated substructure E S there is a θ-extensional isomorphism ɛ : A j = E from some Aj A. If such a uniformly computable cover exists, we say that S is θ-extensionally locally computable or, more simply, θ-extensional. Definition 5. Let A be a uniformly computable cover for a structure S. A set M is a correspondence system for A and S if it satisfies: Each element of M is an embedding of an A i into S. For every A i A, there exists a β M with domain A i. For every finitely generated substructure B S, there exists a β M with range B. For every A i A, every β M with domain A i, and every finitely generated substructure C S with β(a i ) C, there exists A j A, γ M with domain A j and image C, and f Iij A with β = γ f. For every A i A, every β M with domain A i, and every A m A and every g Ii,m A, there exists ɛ M with domain A m with β = ɛ g. If S has a uniformly computable cover A with a correspondence system M, then we say S is -extensionally locally computable. Local computability was originally conceived as a method of describing uncountable structures. However, the definition applies perfectly well to countable structures as well, and in that context, it was natural to ask to what extent local computability corresponded to computable presentability for countable structures. A notion called perfect local computability arose as an answer to this question, when Miller showed the following theorem (see [5, 4]). Theorem 1. For each countable structure S, the following are equivalent. S has an -extensional computable cover (with the AP); S is perfectly locally computable (according to the definition in [5], without requiring the AP); S is isomorphic to a computable structure. This helped to establish -extensionality as the ultimate goal, when one desires to prove that a particular structure of arbitrary cardinality is nicely presentable. It also justifies our decision in this article to consider only covers with the AP. (It is still unknown whether there exists a structure which is not computably presentable but has an -extensional computable cover without the AP.) We reiterate that, in the rest of this paper, our definition of cover (Definition 2 above) requires the Amalgamation Property to hold.

4 4 J.N.Y. Franklin, A.M. Kach, R. Miller, & R. Solomon 2 Failures of Extensionality for Ordinals We now show that having sufficiently high ordinal levels of extensionality is sufficient for -extensional local computability. Since -extensional local computability in turn suffices for computable presentability, this result gives negative consequences: for sufficiently large θ, an arbitrary ordinal greater than or equal to ω1 CK, not being computably presentable, cannot have a θ-extensional computable cover with the Amalgamation Property. The proof, culminating in Proposition 1, is surprisingly straightforward. Lemma 1. Let A be a cover of a structure S. Suppose A i A and ψ : A i C is a θ-extensional map onto a substructure C of S, and let h be an automorphism of S. Then h ψ is also θ-extensional. Proof. We induct on θ. For θ = 0, if ψ is 0-extensional, it is an injective homomorphism, and therefore so is h ψ. Thus h ψ is 0-extensional. For θ > 0, if f Iij A lifts to an inclusion C D via ψ and a ζ-extensional map ϕ (for any ζ < θ), then f also lifts to the inclusion h(c) h(d) via h ψ and the map h ϕ. By induction on θ, the map h ϕ is also ζ-extensional. It follows that h ψ is θ-extensional. Lemma 2. Suppose that A is a computable cover of a structure S, and that a is an n-tuple from an object A i A. If ϕ and ψ are both θ-extensional maps from A i into S, then the tuples ϕ(a) and ψ(a) satisfy exactly the same Σ θ - formulas in S. We will sometimes refer to the set of these Σ θ formulas as the Σ θ -theory of a i in A, and will speak of a i satisfying various formulas in A. The lemma can be seen as saying that this notion is well-defined: in the theory of the cover A, a i satisfies exactly those Σ θ formulas that its image, under an arbitrary θ- extensional map, satisfies in S. (Alternatively, one can define the Σ θ -theory of a i in A by using -quantifiers to refer to the existence of embeddings f I A from A i into other objects A j of A, such that (A j, f(a i )) satisfies the formula inside the -quantifier. This is natural, and is equivalent to the above definition.) Proof. For θ = 0, this follows from ϕ and ψ both being 0-extensional, i.e., being embeddings of A i into S. For θ > 0, suppose S = k ω( y) [P k (ψ(a), y)] where each P k is a Π ζk -formula with ζ k < θ. Fix k ω and y S such that S = P k (ψ(a), y). The inclusion range(ψ) range(ψ) {y} in S must be the lift of some f in some Iij A, via ψ and some ζ k-extensional ψ : A j range(ψ) {y}. But this f must also lift to an inclusion range(ϕ) D in S via ϕ and some ζ k - extensional ϕ. By induction, the Π ζk formula P k (x, y), being known to hold of (ψ (f(a), y) = (ψ(a), y), must also hold of (ϕ (f(a)), z) = (ϕ(a), z), where z := ϕ (ψ 1 (y)). Thus ϕ(a) also satisfies k ( y) [P k(x, y)]. Finally, by a symmetric argument, if ϕ(a) satisfies this Σ θ formula, then so does ψ(a).

5 Local Computability for Ordinals 5 Several different definitions of Scott rank exist in the literature. The following lemma assumes that the definition used has the following property: whenever S is a countable structure of Scott rank ζ and x and y are n-tuples of elements from S (for any n) which satisfy exactly the same Π ζ formulas in n variables, there must exist an automorphism of S mapping each x i to the corresponding y i. Lemma 3. Fix ordinals θ and ζ with θ > ζ. Let A be a θ-extensional cover of a countable structure S with Scott rank ζ. Suppose A i A and ψ : A i S is a ζ-extensional map. Then ψ is also θ-extensional. Proof. Since A is a θ-extensional cover, we know that A i is the domain of some θ-extensional map ϕ : A i S. Let a be a finite tuple generating A i. Then by Lemma 2, the tuples ϕ(a) and ψ(a) satisfy exactly the same Π ζ -formulas in S. Since S has Scott rank ζ, there must be an automorphism h of S mapping ϕ(a) onto ψ(a). But then h ϕ = ψ since a generates A i, and so by Lemma 1, the map ψ is also θ-extensional. Proposition 1. For a countable structure S, the following are equivalent. 1. The structure S is computably presentable. 2. The structure S is perfectly locally computable (as defined in [5]). 3. The structure S has an -extensional computable cover with the Amalgamation Property. 4. There is an ordinal θ strictly greater than the Scott rank of S, such that S has a θ-extensional computable cover with the Amalgamation Property. Proof. The equivalence of (1), (2), and (3) is shown in [4, Thm 6.3]; some of it was originally proven by Miller and Mulcahey in [5]. Since (3) = (4) is trivial, we need only show that (4) = (3). Fix a θ-extensional computable cover A of S. We claim that the set M of all θ-extensional maps ψ of objects A i into S must be a correspondence system. Clearly every A i is the domain of such a map and every finitely generated D S is the image of such a map. Moreover, for any ψ M, say with domain A i, and every f Iij A, we can lift f to an inclusion via ψ and a ζ-extensional ϕ (since ζ < θ), and by Lemma 3, ϕ is also in M. Likewise, every inclusion of range(ψ) is the lift of some f Iij A, for some j, via some ζ-extensional ϕ, and again, by Lemma 3, this ϕ actually lies in M. Corollary 1. For every θ > ω CK 1, the ordinal ω CK 1 is not θ-extensionally locally computable. Proof. The Scott rank of the ordinal ω1 CK (as a linear order) is exactly ω1 CK. Hence, as a countable structure with no computable presentation, ω1 CK cannot be θ-extensionally locally computable for any θ > ω1 CK, by Proposition 1. In Proposition 2 and Theorem 3, we strengthen this corollary to cover some of the case θ = ω1 CK. The situation for θ < ω1 CK will be handled in Theorem 4, and the rest of the case θ > ω1 CK in Theorem 2.

6 6 J.N.Y. Franklin, A.M. Kach, R. Miller, & R. Solomon Lemma 4 (Folklore; see e.g. [1]). For each finite sequence of ordinal α 0 < < α k < ω1 CK, there is a computable infinitary formula λ(x 0,..., x k ) (in the language of linear orders) such that for every ordinal γ, γ = λ(β 0,..., β k ) if and only if β i = α i for each i k. Proposition 2. There is no ω1 CK -extensional computable cover (with AP) of the ordinal ω1 CK itself. Proof. Suppose A were such a cover. By Theorem 1, if there were a correspondence system M for A and ω1 CK, then there would be a computable copy of ω1 CK, yielding a contradiction. Our goal is to define such a correspondence system M. For any A i A with A i = a, let ψ and ψ be ω1 CK -extensional maps from A i into ω1 CK. The tuples ψ(a) and ψ (a) satisfy the same Π ω CK 1 formulas in ω1 CK. By Lemma 4, this forces ψ and ψ to agree on a, and hence on A i. Thus, every A i A is the domain of exactly one ω1 CK -extensional map ψ i into ω1 CK. In fact, for every A i, there is a δ i < ω1 CK such that ψ i is the unique δ i -extensional map of A i into ω1 CK. (We assume δ i is least with this property.) Let M be the collection of all ω1 CK -extensional maps ψ i. We claim M is a correspondence system. The first three properties of a correspondence system follow immediately from the fact that A is an ω1 CK -extensional computable cover. To verify the fourth property, fix A i A and β M with domain A i. Since β is ω1 CK -extensional, β = ψ i. Fix a finite C such that ψ i (A i ) C ω1 CK. By Lemma 4, let θ < ω1 CK be such that C is defined by a Σ θ formula in ω1 CK. Using the fact that A is an ω1 CK -extensional cover, fix A j, f Iij A and an θ-extensional map γ such that β = ψ i = γ f. Since C is defined by a Σ θ formula, it follows that δ j θ and hence γ = ψ j. Therefore, γ is ω1 CK -extensional and hence γ M as required. The fifth property follows by an similar argument which we leave to the reader. Theorem 2. If α > ω1 CK, then α has no (ω1 CK + 1)-extensional computable cover with the Amalgamation Property. Indeed, no computable cover of such an α can have any object with an (ω1 CK + 1)-extensional map into α whose image contains ω CK 1. Proof. We show that from such a cover A, we could construct a computable presentation S of ω1 CK. Let A i0 be the object with an (ω1 CK + 1)-extensional map ϕ 0 such that ϕ 0 (x 0 ) = ω1 CK for some x 0 A i0 (hereafter fixed). The argument in a nutshell is that we can watch for embeddings g I A mapping A i0 into other objects A j of A. When we find such a g, it must lift to an inclusion in α via ϕ 0 and some ω1 CK -extensional ϕ 1, and so every element y < g(x 0 ) in A j is forced to map to some ordinal < ϕ 1 (g(x 0 )) = ω1 CK in α. Since the map is ω1 CK - extensional, Lemma 2 shows that in the theory of A, y satisfies some identifying formula from Lemma 4. We then use the AP to amalgamate A j together with the portion of S already built, and either we see that y maps to some element already in S, or else we add a new element to S to correspond to this y. Since

7 Local Computability for Ordinals 7 every ordinal < ω1 CK corresponds to some such y in some such A j, the S built this way is actually a copy of ω1 CK. We start building S by setting B 0 = {y A i0 : y < x}, letting S 0 be the (possibly empty) linear order {0, 1,..., B 0 1} under <, and defining p 0 : S 0 B 0 to be an order-isomorphism. At stage s + 1, we begin with a finite order S s and with some sequence of objects and embeddings from A, given effectively: A i0 A i1 A is, where each embedding f t : A it A it+1 lies in I A i ti t+1. By induction, we know an isomorphism p s from S s onto a suborder B s of A is, with every element of B s below the element x s = f s 1 (x s 1 ) in A is, which is the image of x 0 under (f s 1 f 0 ). We now search through I A for the first map g 0,s such that: g 0,s I A i 0,j 0,s for some j 0,s ; and A j0,s contains exactly one y < g 0,s (x) which is not in range(g 0,s ); and g 0,s has not been considered at any previous stage. Such a g 0,s must exist, since there are infinitely many elements of α lying below ω1 CK satisfying distinct computable infinitary formulas in α. Once we find the least one, we fix it and search for amalgamations: first j 1,s ω and g 1,s, h 0,s I A, then j 2,s ω and g 2,s, h 1,s I A, etc., as shown here: A i0 A i1 Ai2 Ais f 0 f 1 f 2 f s 1 g 0,s g 1,s A j0,s h 0,s h 1,s Aj2,s h 2,s h s 1,s Aj1,s g 2,s g s,s A js,s We define i s+1 = j s,s and f s = g s,s, thus adding A js,s to the sequence A i0, A i1,... previously built. If the image of B s under f s already contains the element y s+1 = (h s 1,s h s 2,s h 0,s )(g 0,s ((y)), then we set S s+1 = S s and p s+1 = f s p s. If not, then we extend S s to a larger order S s+1 by adding one new element z s+1 to S s, with p s+1 (z s+1 ) = y s+1 and p s+1 = f s p s on the rest of S s+1. The order on S s+1 is defined so that p s+1 remains an order isomorphism from S s+1 into the suborder B s+1 = B s {y s+1 } of A is+1 ; clearly this is compatible with the order on S s, and it justifies the inductive hypothesis at the next stage. This is the entire construction, building the computable linear order S = s S s. We now present the (non-effective) inductive argument that S = ω1 CK, which proceeds through the same stages just described. At stage 0, of course we have an (ω1 CK + 1)-extensional map ϕ 0 : A i0 α, and we define ψ 0 = ϕ 0 p 0, embedding S 0 isomorphically into ω1 CK within α (since ϕ 0 (x 0 ) = ω1 CK, and p 0 maps all elements of S 0 to elements below x 0 in A i0 ). Now at stage s + 1 we chose an embedding g 0,s : A i0 A j0,s from I A. Since ϕ 0 is (ω1 CK + 1)-extensional, this g 0,s lifts to an inclusion range(ϕ 0 ) C, for some finite C α, via ϕ 0 and some ω1 CK -extensional map ϕ 1,s sending A j0,s

8 8 J.N.Y. Franklin, A.M. Kach, R. Miller, & R. Solomon onto C. By Lemma 4, then, the unique y < g 0,s (x) in A j0,s range(g 0,s ) must satisfy (in the theory of the cover A) some computable infinitary formula which uniquely identifies one computable ordinal. Fix some θ s with θ s 1 < θ s < ω1 CK large enough that this formula is Σ θs. Now we proceed along the diagram above. Each object A it is the domain of some (θ s )-extensional map into α, as is each object A jt,s, such that these maps are all compatible with ϕ 0. To see this, take (θ s + s)-extensional maps with domains A i1 and A j0,s, using (ω1 CK + 1)-extensionality of ϕ 0 ; then (θ s + s 1)- extensional maps with domains A i2 and A j1,s, etc. Notice that for an A jt,s with t > 0, we may have several different such maps, depending on the path one takes through the diagram. However, every element y from any B t within A it satisfies a Σ θs formula from Lemma 4, and therefore has a unique possible image in α under these θ s -extensional maps: there is only one element in α satisfying that formula. Moreover, for such a y B t, the same holds of the element g t,s (y) of A jt,s under all θ s -extensional maps from A jt,s into α. So all of these maps agree on all elements of B s and on their images in A js,s. Indeed, this remains true even when we allow s to vary: θ s will be larger for larger s, and A jt,s may be distinct from A jt,s+1, but each element of any B t within each A it in the diagram at stage s + 1 is mapped to the same element of α by all these maps at this and all subsequent stages. So, to define ψ s (z) for z S s, we just map z into B s using p s, and then send p s (z) to its image in α under any one of these θ s -extensional maps. This defines ψ s unambiguously on S s, and each ψ s is compatible with ψ s+1, because p s+1 restricts to p s and because we noted above that the image of an element of B s below the image of x has only one possible image in α under these (sufficiently extensional) maps. So it is clear that this ψ = s ψ s is an embedding of S into ω1 CK within α. Finally, for each element γ / range(ϕ 0 ) of the linear order ω1 CK, there is some j 0 and some map g 0 : A i0 A j0 which lifts to the inclusion range(ϕ 0 ) range(ϕ 0 ) {γ}, and at some stage s this j 0 and this g 0 will be chosen as j 0,s and g 0,s. At that stage, γ will become the ψ s -image of some element of S s, and so the embedding ψ actually maps S onto ω1 CK. Thus S is a computable presentation of ω1 CK, which is impossible. It remains to decide whether an α ω CK 1 could have an ω CK 1 -extensional computable cover. In the initial cases, we can answer this. Theorem 3. If ω CK 1 α < ω CK 1 ω, then α has no ω CK 1 -extensional computable cover with AP. Proof. We sketch the proof, which mixes the techniques used for Proposition 2 and Theorem 2. Now one fixes some i 0 for which A i0 is the domain of an ω1 CK - extensional map ϕ 0 onto the finite set α {ω1 CK (n + 1) : n ω}. Consider any j and any g Ii A 0j. Now for every θ < ωck 1, every g(x) maps to ϕ 0 (x) by some θ-extensional map, and so each g(x) satisfies a Σ θ -formula in A stating that in the Cantor normal form of g(x), every ω ζ with ζ < θ has coefficient 0. Since this holds for all θ < ω1 CK, each ω1 CK -extensional map ψ with domain A j must send each of these g(x) to a nonzero multiple of ω1 CK in α. If follows that each element y A j with y < min(range(g)) has ψ(y) < ω1 CK in α. By Lemma 4, each such y

9 Local Computability for Ordinals 9 satisfies a Σ θ formula in A which, in all ordinals, can only be satisfied by ψ(y). This allows us to run the same construction that we did in Theorem 2, going systematically through maps g I A from A i0 into any A j in such a way that min(range(g)) min(a j ) and amalgamating those maps into the construction to get a computable presentation of ω1 CK, which is impossible. 3 Extensionality for Ordinals Beyond ω CK 1 Theorem 4. For each computable ordinal θ, every ordinal α has a θ-extensional computable cover. The full proof is too long to present in this context, but we can provide a number of details. We state the key lemmas (in terms of the fixed computable ordinal θ), present the proof of Theorem 4 assuming these lemmas, and end with a sketch of the proofs of the lemmas. Lemma 5. If linear orders S 0 and S 1 each have θ-extensional computable covers, then so does their sum S 0 + S 1. Lemma 6. Each ordinal multiple of ω θ of the form ω θ β (with β ω) has a θ-extensional computable cover. To prove Theorem 4, notice that every computable ordinal has a θ-extensional (even -extensional) computable cover. Therefore, fix a noncomputable ordinal α and write α = ω θ β + ρ with ρ < ω θ. Since ρ < ω θ, ρ is computable and hence has a θ-extensional cover. Since β > ω (because α is not computable), ω θ β has a θ-extensional cover by Lemma 6. Therefore, by Lemma 5, α has a θ-extensional computable cover. So Lemmas 5 and 6 imply Theorem 4. To prove Lemma 5, fix θ-extensional computable covers A 0 and A 1 of S 0 and S 1 respectively. The objects in the θ-extensional computable cover of S 0 + S 1 have the form A 0 i + A1 j where A0 i A 0 and A 1 j A 1, with the caveat that one of A 0 i or A1 j is allowed to be empty. The injective maps from A0 i + A1 j to A0 k + A1 l are defined in the obvious way, and one checks that this cover is θ-extensional. The proof of Lemma 6 is notationally cumbersome, but the fundamental idea is that θ-extensionality cannot distinguish between gaps in a linear order of length ω θ γ for varying nonzero values of γ. Each A i in our θ-extensional cover A of ω θ β, is a finite linear order of the form 1 < 2 < < n, for some n, together with an n-tuple ξ 0, ξ 1,..., ξ n 1, called its label, in which each ξ i ω θ 2. If ξ i < ω θ, then ξ indicates that the gap between i and i + 1 (or the gap to the left of 1 if i = 0) in A i should have length ξ i. If ξ i = ω θ + ρ, then it indicates that this gap has length ω θ γ + ρ for some γ 1. (Note that the labels are not formally part of A i. They are merely a denotation to help us keep track of which injective maps to include in I A ij.) For each n and each label ξ 0,..., ξ n 1, we include an object A i of length n with this label in A. Let A i have domain {1,..., n} with label ξ 0,..., ξ n 1 and A j have domain {1,..., m} with label η 0,..., η m 1. We include an order

10 10 J.N.Y. Franklin, A.M. Kach, R. Miller, & R. Solomon preserving map f : {1,..., n} {1,..., m} in Iij A if and only if the labels match in the sense that ξ 0 = η η f(1) 1 and for all 0 < k < n, ξ k = η f(k) + η f(k) η f(k+1) 1. It is straightforward to check that this process defines a computable cover of ω θ β. To check that this cover is θ-extensional takes longer and will not be presented here. The key fact is the following lemma, which can be established by induction on ζ. Lemma 7. Fix ζ θ and let ψ : A i ω θ β be an increasing map. Assume A i has domain {1,..., n} and label ξ 0,..., ξ n 1. Write each ξ k = ω ζ µ k + ρ k with ρ k < ω ζ. If there are ordinals µ k, with µ k = 0 if and only if µ k = 0, such that ψ(k) = (ω ζ µ 0 + ρ 0 ) + (ω ζ µ 1 + ρ 1 ) + + (ω ζ µ k 1 + ρ k 1 ), then ψ is a ζ-extensional map from A i into ω θ β. This completes the proof sketch for Theorem 4. The full proof is quite technical, and is relegated to Appendix A. We believe that an alternative proof of this theorem can be produced by exploiting a general connection between being θ-extensionally locally computable and the existence of effective enumerations of θ back-and-forth types. This fact, combined with the analysis of the back-andforth types of ordinals found in the literature, should yield the desired conclusion. We refer the reader to [1] for this analysis, as well as the relevant definitions, and invite her to work out the proof that way, which will likely be more enlightening than processing all the details in the proof in Appendix A. With Theorem 4, the general question of θ-extensionality of ordinals α is now settled in almost all cases. When α < ω1 CK or θ < ω1 CK, the answer is positive, by Proposition 1 and Theorem 4. When ω1 CK α < ω1 CK ω, the answer is negative for every θ ω1 CK by Proposition 2 and Theorem 3. When α ω1 CK ω, the answer is negative for every θ > ω1 CK by Theorem 2. The only case remaining open is that in which θ = ω1 CK and α ω1 CK ω. References 1. C.J. Ash & J.F. Knight; Computable Structures and the Hyperarithmetical Hierarchy (Amsterdam: Elsevier, 2000). 2. V.S. Harizanov; Pure computable model theory, Handbook of Recursive Mathematics, vol. 1 (Amsterdam: Elsevier, 1998), R.G. Miller; Locally computable structures, in Computation and Logic in the Real World - Third Conference on Computability in Europe, CiE 2007, eds. B. Cooper, B. Löwe, & A. Sorbi, LNCS 4497 (Springer-Verlag: Berlin, 2007), R.G. Miller; Local computability and uncountable structures, to appear in the ASL Lecture Notes in Logic volume Effective Mathematics of the Uncountable, eds. N. Greenberg, J.D. Hamkins, D.R. Hirschfeldt, & R.G. Miller. 5. R.G. Miller & D. Mulcahey; Perfect local computability and computable simulations, in Logic and Theory of Algorithms, Fourth Conference on Computability in Europe, CiE 2008, eds. A. Beckmann, C. Dimitracopoulos, & B. Löwe, LNCS 5028 (Berlin: Springer-Verlag, 2008), R.I. Soare; Recursively Enumerable Sets and Degrees (New York: Springer, 1987).

11 Local Computability for Ordinals 11 Appendix A We now present the proof of Theorem 4. The direct proof of this result is highly technical, and requires a sequence of lemmas. Theorem 4. For every computable ordinal θ, every ordinal α has a θ-extensional computable cover. Lemma 5. For any ordinal θ, if linear orders L and M are θ-extensional, then so is the linear order L + M. Proof. Suppose B = {B i } and C = {C j } are θ-extensional covers of L and M, respectively. We define the objects of a cover A of L + M to be the linear orders A ij := B i + C j, for all objects B i B and C j C (allowing either, but not both, of B i or C j to be empty.) For each i, j, k, m, the maps in Iij,km A are precisely the maps of the form f g, where f Iik B (or B i is empty) and g Ijm C (or C j is empty). This is clearly a computable cover of L + M, and one checks quickly that it is θ-extensional. We now give a full proof of θ-extensionality of every ordinal α for every computable θ, starting with sufficiently nice limit ordinals. Lemma 6. Let θ be any computable ordinal, and assume that α is a nonzero ordinal multiple of ω θ. Then α is θ-extensionally locally computable. Proof. Our computable cover A of α will have finite linear orders as its objects, of course. Each A i A consists of the elements 1 < 2 < < n in this order, for some n, along with an n-tuple of nonzero ordinals (ξ 0,..., ξ n 1 ), with each ξ m ω θ 2. The element ξ k is called the label of the gap between k and (k + 1) in A i, with ξ 0 labeling the gap to the left of 1. (We could also define ξ n = ω θ, to label the gap to the right of n.) Every object A i is such a finite linear order with such a labeling, and every possible labeling of every finite linear order should appear as exactly one A i. Since θ < ω1 CK, ω θ 2 is also a computable ordinal, and so it is not difficult to list out this cover computably. Before continuing, we offer some intuition into the above definition. Each A i will have a θ-extensional map ψ onto a finite subset of α, and the gaps in A i are intended to measure the interval from ψ(i) to ψ(i + 1) in α. A gap ξ i = 1 indicates that ψ(i) and ψ(i + 1) are adjacent to each other, for example. Of course, α may be arbitrarily large, and so the gap between two elements of α may also be very large, whereas all ξ i are strictly less than ω θ 2. The reason for this is that, from the point of view of a θ-extensional cover, all gaps of the form ω θ µ are equivalent; there is no θ-extensional way to distinguish one such gap from another. So the cover simply regards each such gap as having length ω θ. Do note, however, that a θ-extensional cover (for θ > 1) can tell the difference between a gap of size ω θ µ and a gap of size ω θ µ + 1. Therefore, we do care about the tail of the gap, and so our ξ i are all of the form either ω θ +ρ, or else just ρ, with ρ < ω θ. In the former case, these two consecutive elements of A i

12 12 J.N.Y. Franklin, A.M. Kach, R. Miller, & R. Solomon can map to any x < y α with x + ω θ µ + ρ = y, for any µ > 0; whereas in the latter case, the two elements of A i can map to any x < y with x + ρ = y. Next we describe the embeddings of A i into A j. Let A i have domain {1,..., n} with some labeling (ξ 0,..., ξ n ), and let A j have domain {1,..., m} with some labeling (η 0,..., η m ). Suppose that f is an order-preserving map from A i into A j. We define f to lie in Iij A if and only if all of the following hold: ξ 0 = η 0 + η η f(1) 1 ξ 1 = η f(1) + η f(1) η f(2) 1.. ξ n 1 = η f(n 1) + η f(n 1) η f(n) 1. Thus, Iij A is a decidable (not just c.e.) set of maps. Moreover, the Amalgamation Property is quickly seen to hold for these objects and maps. So we have built our computable cover A, which we must now show to be a θ-extensional cover of α, and for this we need a lemma. Lemma 7. Fix any ζ θ, and let ψ : A i α be an increasing map from any object of A into α. Let (ξ 0,..., ξ n ) be the labels of A i (where A i has domain {1,..., n}), and write each ξ k = ω ζ µ k +ρ k with ρ k < ω ζ. Suppose that for each k A i there is an ordinal µ k, with µ k = 0 iff µ k = 0, such that ψ(k) = (ω ζ µ 0 + ρ 0 ) + (ω ζ µ 1 + ρ 1 ) + + (ω ζ µ k 1 + ρ k 1 ). Then this map ψ is a ζ-extensional match between its domain A i and its image in α. Proof. The proof is by transfinite induction on ζ θ. First, when ζ = 0, every k has ψ(k + 1) = ψ(k) + ω 0 µ k (since ρ k < ω 0 = 1). But then µ k 0 (lest µ k = 0, which would contradict ξ k 0), so indeed ψ(k + 1) > ψ(k), making ψ 0-extensional. For all other ordinals ζ, we assume that the lemma holds for all ordinals ɛ < ζ, and consider ζ. Fix ψ : A i α, and suppose that there exist ordinals µ k, with µ k = 0 iff µ k = 0, such that ψ(k) = (ω ζ µ 0 + ρ 0 ) + + (ω ζ µ k 1 + ρ k 1 ). Then ψ is an order-isomorphism (just as in the ζ = 0 case). Suppose that f Iij A, as defined above, and let (η 0,..., η m ) be the labeling of A j. For an arbitrary ɛ < ζ, we define an ɛ-extensional ϕ : A j α with ϕ f = ψ as follows. First, to define ϕ(1), ϕ(2),..., ϕ(f(1)), we use the equation ξ 0 = η 0 +η 1 + +η f(1) 1. For each p < f(1), write η p in the form η p = ω ζ ν p +σ p, with σ p < ω ζ. It follows that ν ν f(1) 1 = µ 0 (because, for all nonzero β, γ ω ζ, we have β + γ = ω ζ iff γ = ω ζ ). Let q < f(1) be maximal (if it exists) such that ν q 0, and define ϕ(p) + σ p, if ν p = 0; ϕ(p + 1) = ϕ(p) + ω ζ + σ p, if ν p 0 & p < q; ϕ(p) + ω ζ ( ν q ) + σ q, if p = q

13 Local Computability for Ordinals 13 where, in the p = q case, ν q is chosen so that {p < q : ν p 0} + ν q = µ 0. If no such q exists, then we always use the ν p = 0 case. These instructions also define ϕ(1), under the convenient fiction that ϕ(0) = 1. It follows that ϕ(q + 1) = ω ζ µ 0 + σ q, and then that ϕ(f(1)) = ϕ(q + 1) + σ q σ f(1) 1 = ϕ(q + 1) + η q η f(1) 1 = ψ(1). Now we repeat the process up to f(2), then up to f(3), and so on. Given ϕ(f(k)) with 1 k < n, write η p = ω ζ ν p + σ p for each p with f(k) p < f(k + 1), with all σ p < ω ζ. Again, take the greatest q with f(k) q < f(k + 1) such that ν q 0 (or q = f(k) 1 if all those ν p = 0), and define, for p = f(k),..., f(k + 1) 1 in turn, ϕ(p) + σ p, if ν p = 0; ϕ(p + 1) = ϕ(p) + ω ζ + σ p, if ν p 0 & p < q; ϕ(p) + ω ζ ( ν q ) + σ q, if p = q where again ν q is chosen so that {p < q : ν p 0} + ν q = µ k. Hence ϕ(q + 1) = ϕ(k)+ω ζ µ k +σ q, and ϕ(f(k +1)) = ϕ(q +1)+η q+1 + +η f(k+1) 1 = ψ(k +1) (except in the case k = n, where this defines ϕ(n+1),..., ϕ(m) for the rightmost elements of A j ). Now we prove that the condition in the lemma holds for the map ϕ (with ɛ in place of ζ). By inductive hypothesis, this will show that the map ϕ is ɛ- extensional. Assume that the condition holds for ϕ(1),..., ϕ(p 1) (where 1 p m, and we use the convention that ϕ(0) = 1). If ν p = 0, then ϕ(p + 1) = ϕ(p) + σ p = ϕ(p) + η p, which clearly satisfies the condition. If ν p 0 but p < q (for the q used in this segment of the construction), then ϕ(p + 1) = ϕ(p) + ω ζ 1 + σ p = ϕ(p) + ω ɛ (ω δ + ν p) + σ p, where these new ordinals satisfy ɛ + δ = ζ and σ p = ω ɛ ν p + σ p with σ < ω ɛ. Since ν p 0 and ω δ + ν p 0, the condition is satisfied. Finally, if p = q, the analysis is similar: ϕ(p + 1) = ϕ(p) + ω ζ ν q + σ p = ϕ(p) + ω ɛ (ω δ ν p + ν p) + σ p, with δ, ν p, and σ p just as above. Again ν p 0 and ω δ ν p + ν p 0, so the condition is satisfied. Thus ϕ is ɛ-extensional, as desired. Now we prove the dual condition. Again fix any ɛ < ζ, and take any D α consisting of range(ψ) and finitely many additional elements. Set m = D, let A be the linear order on the domain {1, 2,..., m}, under <, and define ϕ : A D to be the unique isomorphism between these two finite linear orders. Set f(k) =

14 14 J.N.Y. Franklin, A.M. Kach, R. Miller, & R. Solomon ϕ 1 (ψ(k)) for each k A i. This A will become our A j, with j defined as we now determine the gap labelings. Then we will show that f Iij A and that ϕ is ɛ-extensional. To get started, we set η 0 = 1 + ϕ(1). Then, for each p < ϕ(m) in order, choose the (unique) η p such that ϕ(p)+η p = ϕ(p+1). Write each η p in the form η p = ω θ λ p + ω ζ κ p + ω ɛ ν p + σ p, with σ p < ω ɛ and (ω ɛ ν p + σ p ) < ω ζ, and (ω ζ κ p + ω ɛ ν p + σ p ) < ω ζ. (We know θ ζ > η. In case ζ = θ, let all κ p = 0, leaving λ p as the entire quotient of η p by ω θ.) The bars in σ p, etc., denote the actual values from the elements of α, and we define corresponding ordinals σ p, etc., to compute the gaps η p. Recall that ξ k = ω θ µ k + ρ k. We now decompose this further, writing ξ k = ω θ µ k + ω ζ γ k + ω ɛ δ k + τ k, where τ k < ω ɛ, δ k < ω ζ, and γ k < ω θ. We set ξ 0 = 1 + ψ(1), and choose ξ k uniquely so that ψ(k) + ξ k = ψ(k + 1) for all k < n. These are decomposed in exactly the same way: ξ k = ω θ µ k + ω ζ γ k + ω ɛ δ k + τ k, noting that each µ k is exactly the value mentioned in the lemma. Also, ω ζ γ k + ω ɛ δ k + τ k = ρ k = ω ζ γ k + ω ɛ δ k + τ k (with the first equality assumed in the lemma and the second because this is how ρ k was decomposed), so in fact γ k = γ k, δ k = δ k, and τ k = τ k. Now by assumption, η f(k) + η f(k) η f(k+1) 1 = ξ k = ω θ µ k + ω ζ γ k + ω ɛ δ k + τ k, since ψ(k) + ξ k = ψ(k + 1). We now use these elements to define the gaps η f(k),..., η f(k+1) 1 which will be needed. If ξ k < ω θ, then set all λ p = 0, define q θ = f(k), and go to the next instruction. If ξ k ω θ, then µ k > 0, so µ k > 0, and so some η p with f(k) p < f(k + 1) has λ p > 0. (Conversely, if some such λ p > 0, then also ξ k > 0.) In this case there is some (unique) q θ f(k) such that ω ζ κ qθ + ω ɛ ν qθ + σ qθ + η qθ η f(k+1) 1 = ρ k. (In particular, q θ is the greatest p < f(k + 1) for which λ p > 0.) We define λ qθ to be 1, while for all p q θ with f(k) p < f(k + 1), we set λ p = 0. Thus, if ξ k ω θ, we will have η f(k) + η f(k) η qθ = ω θ + ω ζ κ qθ + ω ɛ ν qθ + σ qθ, whereas, if ξ k < ω θ, then also η f(k) + + η f(k+1) 1 < ω θ. Next we consider those p with f(k) p < q θ. For each such p, we set σ p = σ p and κ p = λ p = 0, and take either ν p = 1 (if either κ p > 0 or ν p > 0) or ν p = 0 (if not). (If ξ k < ω θ, then q θ = f(k), and this instruction defines nothing.)

15 Local Computability for Ordinals 15 Next we find the greatest q ζ with q θ q ζ < f(k + 1) and κ qζ > 0. (If no such q ζ exists, we set q ζ = q θ and go on to the next instruction.) For each p with q θ p < q ζ, we set ν p = ν p and σ p = σ p, and take either κ p = 1 if κ p > 0, or κ p = 0 if κ p = 0. Then, for q ζ itself, we choose the unique κ qζ so that κ qθ + κ qθ κ qζ = γ k. For every p with q ζ < p < f(k + 1), we set κ p = 0. Finally, we find the greatest q ɛ with q ζ q ɛ < f(k + 1) for which ν qɛ > 0. If there is no such number, then we set q ɛ = q ζ. For each p with q ζ p < q ɛ, we set σ p = σ p, and take ν p = 1 if ν p > 0, or ν p = 0 if ν p = 0. For q ɛ itself, we set σ qɛ = σ qɛ, and choose ν qɛ uniquely so that ν qζ +ν qζ ν qɛ = δ k. Every p with q ɛ < p < f(k + 1) has σ p = σ p and ν p = 0. This completes our description of the coefficients defining each η p, which is itself given by η p = ω θ λ p + ω ζ κ p + ω ɛ ν p + σ p. Recall that λ p 1 and that κ p < ω θ, ν p < ω ζ, and σ p < ω ɛ. So for every p, we have η p < ω θ 2, and therefore there must exist some j for which A j has domain {1,..., m} with gaps η 0, η 1,..., η m 1 (and η m = ω θ by definition). Now we claim that the f defined above, which is clearly an increasing map from A i into A j, does in fact lie in Iij A. The condition which must be satisfied is that ξ k = η f(k) + +η f(k+1) 1, for each k < n. We write out this sum here, carefully grouped, with each of η qθ, η qζ, and η qɛ expressed using its coefficients. η f(k) + + η f(k+1) 1 = η f(k) + + η qθ 1 + ω θ λ qθ + ω ζ κ qθ + ω ɛ ν qθ + σ qθ + η qθ η qζ 1 + ω ζ κ qζ + ω ɛ ν qζ + σ qζ + η qζ η qɛ 1 + ω ɛ ν qɛ + σ qɛ + η qɛ η f(k+1) 1 In this expression, the first line is exactly ω θ µ k, by our choice of q θ and λ qθ in the instructions above. (If λ qθ = 1, then the ω θ λ qθ term swallows up all preceding terms, since all other λ p = 0, and the entire first line is just ω θ. Otherwise λ qθ = 0, so q θ = f(k) and the first line is 0. In both cases, the first line equals ω θ µ k.) Likewise, the second line equals ω ζ γ k, the third line equals ω ɛ δ k, and the final line is τ k. Thus the entire sum is exactly ξ k. Since this holds for every k < n, we see that f Iij A. To complete the proof of the lemma, we need to show that ϕ is ɛ-extensional. By inductive hypothesis, the lemma holds with ɛ in place of ζ, and so it suffices to prove that ϕ satisfies the condition given there (with ϕ, ɛ, A j, and the labels (η 0,..., η m ) in place of ψ, ζ, A i, and (ξ 0,..., ξ n )). We already have written each η p = ω ɛ (ω (θ ɛ) λ p + ω (ζ ɛ) κ p + ν p ) + σ p as required (with the obvious meanings for θ ɛ, etc.). Now we write π p for the coefficient of ω ɛ in the expression above. Thus π p = 0 iff λ p = κ p = ν p = 0.

16 16 J.N.Y. Franklin, A.M. Kach, R. Miller, & R. Solomon (When ζ = θ, we specified that all κ p = 0, so all κ p = 0, and our analysis remains correct.) Since ϕ(p) + η p = ϕ(p + 1) for every p, we need to show that each η p is of the form ω ɛ π p + σ p, with π p = 0 iff π p = 0. We claim that this holds once we set π p = ω (θ ɛ) λ p + ω (ζ ɛ) κ p + ν p for each p. Below we fix the k such that f(k) p < f(k + 1) (or k = 0 if p < f(1)) and the q θ, q ζ, and q ɛ corresponding to that k. Now we have seen that λ p = 1 iff η p > ω θ, which holds iff λ p = 0. (It is important here that each ξ k < ω θ 2, since this meant that among all of p = f(k), f(k)+1,..., f(k+1) 1, only one could have λ p > 0.) If f(k) p < q θ, thenκ p = 0, while ν p > 0 iff either κ p > 0 or ν p > 0, so in this case π p = 0 iff π p = 0. Also, σ p = σ p for every p q θ. For all p q θ, the instructions make it clear that κ p > 0 iff κ p > 0, that ν p > 0 iff ν p > 0, and that σ p = σ p. Thus we see that the instructions yielded π p = 0 iff π p = 0, and so ϕ is indeed ɛ-extensional, by inductive hypothesis. Therefore, the map ψ : A i α is ζ-extensional, since for every ep < ζ, every f Iij A (for every j) lifts to an inclusion in α via an ɛ-extensional map, and every finite D α containing the image of ψ is likewise the lift of some map f Iij A for some j via an ɛ-extensional map. So Lemma 7 is proven. We now continue with the proof of Lemma 6. First, let A i be any object in A, with domain {1,..., n} and labels (ξ 0,..., ξ n ). Lemma 7 makes it clear that the map sending each i n to the ordinal (ξ 0 + ξ ξ i 1 is θ-extensional, just by letting µ k = µ k for every k. So every object of A is the domain of a θ- extensional map. Dually, suppose D α is finite, say with n elements. Set ξ 0 to be the least element of D, and define ξ k+1, for each k < n, so that (ξ ξ k 1 ) is the k-th least element of D. Then write each ξ k = ω θ µ k + ρ k, and define ξ k = ω θ µ k + ρ k, where µ k = 1 if µ k > 0 and µ k = 0 if µ k = 0. Then the unique order-isomorphism ϕ from the object A i with elements {1,..., n} and labels (ξ 0,..., ξ n 1, ω θ ) onto D is θ-extensional. Thus this A is a θ-extensional cover of α. Theorem 4. For every computable ordinal θ, every ordinal α has a θ-extensional computable cover. Proof. We decompose α as α = ω θ β + γ with γ < ω θ. Since θ < ω1 CK, we also have γ < ω1 CK, and so γ has a perfect cover by Proposition 1. If β = 0, then this is also a perfect cover of α. If β > 0, then we apply Lemma 6 to see that ω θ β has a θ-extensional cover, and Lemma 5 then completes the proof.