the computable dimension of trees of infinite height

The Computable Dimension of Trees of Infinite HeightAuthor(s): Russell MillerSource: The Journal of Symbolic Logic, Vol. 70, No. 1 (Mar., 2005), pp. 111-141Published by: Association for Symbolic LogicStable URL: http://www.jstor.org/stable/27588351 .

Accessed: 13/06/2014 05:55

Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .http://www.jstor.org/page/info/about/policies/terms.jsp

.JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range ofcontent in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new formsof scholarship. For more information about JSTOR, please contact [email protected].

.

Association for Symbolic Logic is collaborating with JSTOR to digitize, preserve and extend access to TheJournal of Symbolic Logic.

http://www.jstor.org

This content downloaded from 91.229.229.203 on Fri, 13 Jun 2014 05:55:29 AMAll use subject to JSTOR Terms and Conditions

http://www.jstor.org/action/showPublisher?publisherCode=asl

http://www.jstor.org/stable/27588351?origin=JSTOR-pdf

http://www.jstor.org/page/info/about/policies/terms.jsp


The Journal of Symbolic Logic

Volume 70. Number 1. March 2005

THE COMPUTABLE DIMENSION OF TREES OF INFINITE HEIGHT

RUSSELL MILLER

Abstract. We prove that no computable tree of infinite height is computably categorical, and indeed

that all such trees have computable dimension co. Moreover, this dimension is effectively co, in the sense

that given any effective listing of computable presentations of the same tree, we can effectively find another

computable presentation of it which is not computably isomorphic to any of the presentations on the list.

?1. Introduction. In a finite language, a countable structure s? whose universe A

is a subset of co is computable if A is a computable set and for all functions / and

relations R in the language, f^ is a computable function and R^ is a computable relation.

Any computable structure will be isomorphic to infinitely many other computable structures. It may happen, however, that two computable structures are isomorphic,

yet that the only isomorphisms between them are noncomputable (as maps from one domain to the other). If so, then these structures lie in distinct computable

isomorphism classes of the isomorphism type of the structure. On the other hand, if

there exists a computable function taking one structure isomorphically to the other, then the two structures lie in the same computable isomorphism class.

The computable dimension of a computable structure is the number of computable

isomorphism classes of that structure. The most common computable dimensions are 1 and co, but for each n e co, there do exist structures with computable dimension

?, by a result of Goncharov [7]. If the computable dimension of s? is 1, we

say that s? is computably categorical. This notion is somewhat analogous to the

concept of categoricity in ordinary model theory: a theory is categorical in a given power k if all models of the theory of power k are isomorphic. Computable

categoricity is a property of structures, not of theories: a computable structure s?

is computably categorical if every other computable structure which is isomorphic to s? is computably isomorphic to s?.

A standard example of a categorical theory is the theory of dense linear orders

without end points, which is categorical in power co. One proves this by taking two

arbitrary countable dense linear orders and building an isomorphism between them

Received April 4, 2003; revised July 24, 2004.

The first four sections of this article appeared as a chapter of a Ph. D. thesis at the University of

Chicago under the supervision of Robert I. Soare. Thanks are also due to Bakhadyr Khoussainov,

who originally posed the question of computable categoricity of trees and aided in the research, and to

Richard Shore for useful conversations. Later the author was partially supported by a VIGRE postdoc under NSF grant number 9983660 to Cornell University.

? 2005, Association for Symbolic Logic 0022-4812/05/7001-0006/S4.10

111



112 RUSSELL MILLER

by a back-and-forth construction. The same construction allows us to prove that the

structure Q is computably categorical. (More formally, let (co, -<) be a computable linear order isomorphic to (Q, <). Then (co, -<) is computably categorical.)

Characterizations of computable categoricity have been found for certain types of structures. Goncharov and Dzgoev [10] and Remmel [17] proved that a linear

order is computably categorical precisely if it contains finitely many successivities

(that is, if only finitely many elements have an immediate successor in the linear

order). Remmel also proved that a Boolean algebra is computably categorical if

and only if it contains only finitely many atoms [16]. In the present paper we consider computable categoricity of trees, and prove that

no tree of infinite height is computably categorical. The question of computable

categoricity of trees of finite height is the subject of joint work by Lempp, McCoy, Solomon, and the author [14], and appears in this issue.

To prove that a tree T is not computably categorical, we will construct a new tree

T' isomorphic to T, satisfying the following requirements 9t\\

9le : <pe total => there exists x G T' such that levels (x) ^ levelj((pe(x)).

Clearly 9le implies that <pe is not an isomorphism from T' to T. If we can establish

91 e for every e, then, we will have proven that T is not computably categorical. Our notation is standard, but our definitions demand attention. A tree consists

of a universe T with a strict partial order ^onl such that for every x <E T, the

set of predecessors of x in T is well-ordered by -<, and such that T contains a least

element under -<. (Hence the tree is computable if T is a computable set and -< a

computable relation.) In this paper, T will represent the computable tree which we

wish to prove not to be computably categorical. If two nodes x and y in T are incomparable under ~<, then we write x _1_ y. For

each node x ? T, we define the level of x in T to be the order type of the set of prede cessors of x in T. We view our trees as growing upwards, with a single element r (the root, or least element under -<) at the base. Thus the level of the root is 0, its immedi

ate successors under -< are at level 1, and so on. The height of T is defined as follows:

ht(r) =

sup(levelr(x) + 1). x?T

Thus, the height of T will be the least ordinal a such that no node of T has

level a. In this paper we only consider trees of infinite height. The level of a

node of T is generally not a computable function on T. (For computable trees of

height < co, though, it is a Si function, since there exists a computable function

f (x, s) =

\{ y < s : y ^ x }\ such that for all x ? T,

level r(x) =

\imf(x,s).) s

The reader should note that different definitions of subtree and tree homomor

phism have been used for different purposes in the literature. In this paper a homo

morphism from one tree (T, -<) to another tree (T', -<') will be a map / : T ?> T'

which respects the partial orders:

x -< y <=> fix) -<' f(y).

(An embedding is a one-to-one homomorphism.) In other papers (such as [5]), trees are sometimes defined using the infimum function A, where the infimum x Ay



THE COMPUTABLE DIMENSION OF TREES OF INFINITE HEIGHT 11 3

of x and y is the greatest z such that z < x and z -< y. For trees of height < co, these definitions are equivalent, but when the infimum function is used, all

homomorphisms are required to respect the infimum function. This is a strictly

stronger requirement: all maps respecting A respect -<:, because

x <y <^=> x A y = x,

but not conversely. Kruskal's Theorem, which we use in section 2, proves the

existence of the stronger type of embedding. If the infimum fuction is computable, then the relation ^< is computable, since it

is definable in terms of A without quantifiers. Therefore, if the computable trees

(r, -<) and iT',-<') are isomorphic but not computably isomorphic, then the cor

responding structures {T, A) and iT', A7) are also isomorphic, but not computably

isomorphic. (Notice, however, that (T, A) and iT1, A') need not be computable, since computability of -< does not guarantee that we can compute the infimum

function.) When we build T', we will ensure that not only -<' but also A; are com

putable. Thus, our theorem suffices to prove that even when tree is defined using the

infimum, no tree of infinite height is computably categorical. Section 7 is devoted

to the differences in the proof required for trees defined using the infimum. The

definitions of tree and tree homomorphism using the infimum are probably more

common in the literature. We adopt the definitions using -< because for the purposes of our proof, they will be far more useful.

Our definition of subtree arises from our definition of homomorphism. Once

again, therefore, it diverges from much of the literature: for our purposes, a tree

iT', -<f) is a subtree of iT, -<) if T' ? T and the inclusion map respects the partial orders. Thus the infimum of two elements in T may not be the same as their infimum

in T'. Also, the root of T may be distinct from the root of T', as in the case of the

subtrees T[x], which we will be considering frequently. If x is a node in T, then the

subtree T[x] is just the tree

T[x] =

{yeT:xly}.

The partial order on T[x] is the restriction to T[x] of the partial order -< on T.

Therefore T[x] is a subtree of T with root x. We define the height ofT above x by:

htxiT) = htiT[x]).

The reason for our use of -< rather than A to define homomorphism and subtree

is twofold. First, -< is the basic relation we used to define the notion of a tree; A was

derived from -<. If A were the basic function, then computability questions would

be very different. Second, during our proofs about a tree T we will be considering

many subsets of T which we will want to regard as subtrees. Under our definition,

they will be subtrees (as will any subset of T with a -<<-least element), but under the

A-definition some would not be subtrees.

A path y through T is a maximal linearly ordered subset of T. It may be finite or infinite. Any tree containing an infinite path must have infinite height. A node is

extendible if it lies on an infinite path through T, and non-extendible otherwise. The

extendible nodes of a tree T (if any exist) form a subtree of T, which we denote by rext. Notice, however, that since we allow T to be infinite-branching, the height of

T above a node may be co even if the node is nonextendible.



114 RUSSELL MILLER

?2. Kruskal's Theorem. Although our results concern infinite trees, we will need

the ability to manipulate finite subtrees. For this purpose Kruskal's Theorem is

essential.

Theorem 2.1 (Kruskal's Theorem). (See [13], [18].) Let {T? : i ? co} be an

infinite collection of finite trees. Then there exist i < j in co such that T? can be

embedded in T?.

Every version of Kruskal's Theorem which we will encounter has an analogue of

the following corollary:

Corollary 2.2. Let {T? : i ? co} be an infinite collection of finite trees. Then

there exists n ? co such that for every i > n, T? can be embedded in some Tj with

j > i, and some 7\ with k < i can be embedded in T?.

Proof. If the set

{ i ? co : (Vy > i) T? does not embed in Tj }

were infinite, it would itself contradict Kruskal's Theorem. The same is true of

{ i ? co : ( V/c < i) Tk does not embed in Ti}. H

We can extend Kruskal's Theorem to a version dealing with infinite trees.

Corollary 2.3. Let { Tt : i ? co } be an infinite collection of trees. {These trees

need not be finite, nor even finitely branching.) Then there exists an i ? co such that

for every finite subtree T ? Tt, there exists j > i for which T embeds in Tj. Proof. Suppose { Tt: : / ? co } were a collection of trees contradicting this

corollary. Then for each i, we would have some finite subtree S? ? T? which did not

embed into any T? with j > i. In particular, for each i < j, S? would not embed in

Sj, Thus the collection { S? : i ? co } would contradict Kruskal's Theorem. H

Corollary 2.4. Let {Tj : i ? co} be as in Corollary 2.3. Then there is an n ? co

such that for every i > n and every finite subtree T ? T?, there exists j > i such that

T embeds into Tj. Proof. If not. then we could find an increasing sequence ?o < h < h < such

that { Tik : k ? co} contradicted Corollary 2.3. H

In this paper we will want to embed trees in such a way that nodes with p

predecessors are mapped to nodes with more than p predecessors. That is, the level

in the tree T of the node x should be less than the level in T' of its image under the

embedding of T into T'. To map nodes to other nodes at greater levels, we need the

following stronger version of Kruskal's Theorem, in which one is allowed to "label"

nodes of each tree. For our purposes, a labelling of a tree T is simply a map from

T to co. Proofs of this result appear in [13], [18], and [5].

Theorem 2.5 (Kruskal). Let {T? : i ? co} be an infinite collection of finite trees, each with a labelling U. Then there exist i < j in co and an embedding f : Tt ?>

Tj such that for every x ? T?, U{x) <

lj(f{x)). -\

From Theorem 2.5 we derive the following result:

Corollary 2.6. Let { T? : i ? co} be an infinite collection of finite trees. Then

there is a number m ? co such that for every index i and every node x ? T? with

level t, [x) ?m, there exists an embedding f ofT? into some T? with j > i, such that

levels. (/(*)) > levels (x).




Proof. We may assume that sup, ht(T?) = co, since otherwise one could choose

m > htiTi) for all i and satisfy the claim vacuously.

Suppose no m G co satisfied the theorem. Then for every m, we would have an

index im and a node xm G T^ with levelr( }(xm) = m such that:

(1) Vembeddings /: T{im) ->

T? withy > im, levelTj(f(xm)) =

level^U )

Now the set {z"o, hJi,...} will be infinite, since each T? has finite height. Moreover, the index im satisfies Equation 1 not only for xm but also for all predecessors of xm.

Therefore we can choose im+\ > im for all m.

For each m, define the labelling lm on the tree T^im) by

/ N _ JO, iflevelr,w(x) < m,

ll, otherwise.

Thus lm{xm) = 1 for all m. However, for any embedding / :

T^ ??

T^ with

k > m, we have

levelj, ifixm)) =

levelrUm)(xm) = m < k.

This forces 4(/(xm)) = 0. Thus the sequence {T?0, T?] ,T?2,...} contradicts Theo

rem!^. H

The same result holds for all y above the level m :

Corollary 2.7. Let {T? : i e co} be as in Corollary 2.6. Then there is a number m G co such that for every index i and every node y G T? with levels, (y) > m, there

exists an embedding f ofT? into some T? with j > i, such that

\evQ\Tjifiy)) > levelr,-(y).

Proof. The conclusion follows for every y G T? with level^ iy) > m, simply by

finding the node x ^ y in Tz with level^ (x) = m and applying the embedding given

by Corollary 2.6 for that x. H

Finally, we combine the version for infinite trees with the version for embedding nodes at greater levels.

Corollary 2.8. Let { Tt : i G co} be any collection of trees. Then there exist an n

and an m with the property that for all indices i > n,for every finite subtree S ? T?, and for any node x G S with levels (x) > m, there is an embedding g\S?>TjofS into some T? with j > i, such that

levels (g{x)) > levels(x). Proof. Suppose the statement were false. Now if g is an embedding of S into

Tj, it is impossible to have levelTj(g(x)) < levels(x). Therefore, the negation of

the statement is as follows:

(V?) (Vm) (3i > n) (3 finite S ? T?) (3x G S)

levels (x) >m& (Vj > i) (V embeddings g : S ?> Tj)

[levels (g(x))=levels(x)]]. We apply this negation first with n = 0 and m= 0. yielding an index ?"o > 0 and

a node xo at level > 0 in some finite subtree So of Tio. Inductively, we apply the



116 RUSSELL MILLER

negation with n = i?c and m = k + 1 to get an index z'^+i > ik and a corresponding node x/c+i at level > k + 1 of a finite subtree Sk+\ of T? . From the negation, we see that every embedding of any Sk into any Tj with j > ik fixes the level of

Xk- In particular, the same holds for any embedding of Sk into any Sj with j > k.

However, we know that level^. (x?r) > k for all k. Thus the set { Sk : k ? co} contradicts Corollary 2.7. H

For computability-theoretic purposes, we note that if S and T are finite trees

(and we have strong indices for each. i.e. we know the number of nodes of each), then the statement

3 an embedding g:S?>T

is decidable, uniformly in S and T. From the decidability of this statement, we

conclude further that if S is finite with known strong index and T is any computable tree, then the question of embeddability of S into T is a Xi question: it asks whether

there exists a finite subtree of T into which S embeds. Therefore, if we know

that there exists an embedding of S into T, then we can effectively find such an

embedding, via an algorithm uniform in S and T.

?3. co-branching nodes with htx(T) = co. We first consider computable trees of

height co. The general theorem that no such tree is computably categorical will be

proven in the next section. In this section, to prepare for that proof, we prove that

a significant subclass of such trees cannot be computably categorical. We define the limit-supremum of a sequence (rii)ieco to be

lim sup (??/) = inf sup(?/). i J />/'

T will be a given computable tree under the partial order -<, with height co, which is co-branching at a node xq. (That is, xo has infinitely many immediate successors x\.X2,_) We assume further that limsupz ht(T[x/])

= co. This can

occur two ways: either infinitely many T[x?] have height co, or there exist trees T[x?] of arbitrarily large finite heights.

Since the universe of T is computable, we may take it to be co, pulling back

via a 1-1 computable function if necessary to make this so. We will construct a

computable tree T' isomorphic to T. such that there is no computable isomorphism between them.

The isomorphism / from T to T' will be a Aij function, the limit of a computable sequence of finite partial 1-1 functions fs, such that the domains Ds =

dom(fs) c

T form a strong array of finite sets. We will ensure that Ds ? Ds+\ for each s,

although fs+\ need not agree with fs on Ds. (If it did so for all s, then / would

be a computable isomorphism, which is precisely what we wish to avoid!) Also, we

will force range(/) = co, so that the universe of T' will be co. The ordering -<' on

T' will be given by lifting the ordering -< from T via /, thereby guaranteeing that

/ is an isomorphism. To make -<' computable, we force the approximations fs to

satisfy the following condition:

Condition 3.1. For all a. b ? range(/,). we have a, b ? range(/iS+i) and



THE COMPUTABLE DIMENSION OF TREES OF INFINITE HEIGHT 117

To ensure that T and T' are not computably isomorphic, we impose the require ments 9?e.

me: <pe total => (3x G Tf) [levels U) / levelr(^(;c))].

This will suffice to prove the proposition.

Proposition 3.2. Let T be a computable tree of height co containing an co-branching node xo with immediate successors x\, xi,..., such that

limsupht(r[xz]) = co.

i

Then T is not computably categorical.

Proof. As previously remarked, we may assume the universe of T to be co. A successor tree of xo is a tree of the form T[x?] with i > 1. {{x\,X2_} are all the immediate successors of xo, as stated above. This set need not be computable.)

Corollary 2.8, applied to the successor trees, provides m and n in co such that for

every finite subtree S ? T[x?] with /* > n and every node x G S with levels(x) > w, there is an embedding of S into some T[xj] with j > i which maps x to a node of

greater level. We fix these values of m and n for the rest of the proof. (Notice that therefore the proof is not uniform in T.)

Let Ts be the subtree of T with nodes {r, xq, x\, ..., xn} U

{0,1,2,..., s}, under

<, where r is the root of T.

For our purposes, the finite subtrees S will generally be of the form Ds [y], where

Ds D Ts is the domain of fs and y is an immediate successor of xq in Ds (although not necessarily in T). We will call Ds[y] a successor tree at stage s. Notice that it

may happen that two successor trees which are distinct at stage s acquire a common root at stage s + 1, e.g., if s H- 1 = x? for some i, and thus merge into a single successor tree at stage s + 1. A given successor tree at stage s, however, can only be

merged this way finitely often, since each of its nodes has finite level in T.

The following construction yields a computable tree T' which is isomorphic to T but satisfies every requirement 9e, proving that T is not computably categorical. The witness nodes we will be nodes in T[xo], and will be approximated at stage s by a node we_s. The successor tree at stage s

containing we_s will be denoted Se.s. This

is the successor tree which we use in order to satisfy requirement 9e. The sequence

(we.s)seco will converge to some we, and each successor tree in T will contain at most one we. The isomorphism / from T to T' will be approximated at stage s

by a finite map fs with domain Ds. If yeAf s{we,s)) converges to a node at the same level of Ts as the level of fs(we.s) in T's, then we redefine fs+\ and we,s+\ with

fs+\ iwe.s+i) =

fsiwe.s) at a higher level in T^+l. (The level of a node in T's is just the level of its preimage under fs in Ts.) Doing this requires us to redefine fs+\ on the entire successor tree containing we^s, in order to satisfy Condition 3.1, and we will appeal to Corollary 2.8 to ensure that the necessary embedding exists. Thus

fiwe) will be the witness required by 91 e.

Figure 3.3 gives an example of our basic strategy. Se.s is the successor tree

which we use to satisfy Me. We suppose that we have found at stage s that

ipeifsiwe.s)) = 6, which lies at level 2 in D,. This is bad, because fs(we,s)

lies at level 2 in D!s, so it appears that <pe might be an isomorphism from T' to

T. Se.s is the successor tree above the node 4 in Ds, and we use Corollary 2.8



11? RUSSELL MILLER

fA^e.s) fs (6)

/J+/(8) fs+l(6)

/,+/(10) /J+/(4)

/s+/U'o)

to find an embedding of Se,s upwards into the successor tree above the node 10

in Ds+\. (The embedding is indicated by the arrow to D's+{.) We use this em

bedding to make level/)/ {fs+$we.s+$) > level/)/(fs{we.s)). by defining fs+\ so

that /,+1(9) - /,(4). 71h(12) = fs(6), and /s+l{we.s+l)

- fs(we,s). We add new values to range(/,+i) for /.?+i(4), /J+i(6), /,-+i(8), and /,+i(10). Thus

level/)., {<Pe(fs+$we.s+$)) + levelD/+i(/J+i(^?S+i)).

Construction, fo is the identity map with dom(/o) = To. The witness nodes

we.Q and the successor trees Se$ are undefined for all e. We let Do = dom(/o)- (At

each stage s, Ds and Ts will both be subtrees of T, with Ts C Ds.) We immediately define the successor trees Tq[xj] with 1 < i < n: to be frozen.




At stage s + 1, we consider the successor trees of xo in Ds. For each successor tree

S (if any) of height > m which is not frozen and does not contain Se.s for any e < s, we choose the least e < s such that Se,s is undefined, let Se,s+\

= S and choose

we.s+\ to be the oleast node at the highest level of S. Thus level^ s.+1 (we.s+\ ) > m.

We then consider in turn each e for which Se_s was defined.

Step 1. If there is an i < e and a z ? 7^+1 such that xo -< z -< w^s and z -< we_s,

then we immediately make Sj.s+\ and WjmS+\ undefined for all j > e, and declare all

Sj.s with j > e frozen.

(This step ensures that if two successor trees Si,s and Se_s have acquired a common

root above xo, thus becoming the same successor tree, then we use the single new

successor tree to play against requirement M\ only.) Step 2. Otherwise, we consider f s{we_s), the potential witness for requirement

9le. If (pe.s{fs{we,s)) diverges, or converges to an element not in Ds, or if

\QVQ\DX<Pe.s{fs(we.s))) ^ levelDy(we.,),

then we define:

fs+i =

fs on Se,s

Se.s+\ =

{ y e Ds U Ts+i : (y A we.s+\) >- x0}.

(Here y A we,s+\ represents the infimum in Ds U Ts+\, which is a finite tree. Taking the infimum over all of T would not be computable.)

Thus S^y+i is just the same successor tree as Se.s, along with any new elements

that may have appeared in this successor tree at stage s.

Step 3. If \ev?lDs{(pe^(f s(weiS))) =

levels{we,s), then find the least stage t > s

with Ds ? Tt such that the following holds:

Condition 3.4. There exists a z ? Tt such that:

1. z is an immediate successor ofxo in Tt, and

2. Tt[z]nDs -0, and

3. There is an embedding g ofSe,s into Tt[z] with

levelTt{g(we.s)) > levels,(we.s).

Let Se.s+\ ?

Tt[z], with we,s+\ =

g(we.s). (By our choice of g, this forces

levelr,(^.j+i) > levels(we.s). Also, level^vfl (we.s+\) > level^v{we_s) > m.) For

every x ? Se,s, define fs+\{g{x)) =

fs{x), and define fs+\(x) to be the least

element which is not yet in range(/.s+i) U range(/?). Declare Se.s to be frozen, so that at no subsequent stage sf will any ww be defined in the successor tree

containing Se.s. Having executed Step 3 for e, we let iu/.J+i and SjmS+\ diverge and

freeze Sj^s for all j > e, and do not execute Steps 1, 2, or 3 for any j > e.

(We execute Step 3 if Me is not satisfied by fs(we.s). By Corollary 2.8, there must exist a successor tree T[xj] into which the required embedding g exists, be cause levelse.Awe.s) > ^ and Se.s ? T[x?] for some i > n. The successor trees

T[x\ ],..., T[xn] were all frozen right away at stage 0, so none of them contains Se.s.

Thus we have found a z such that fs is completely undefined on the successor tree

S c Tt containing z, and Se_s embeds into S via a map g. We use this embedding to



120 RUSSELL MILLER

satisfy 9le. as in the example of Figure 3.3.Freezing Se_s ensures that fs> will never

again be redefined on Se.s, so that linv f sf must exist.)

Having completed these three steps for each Se_s, we now define Ds+\ to be

(Ue Se.s+\) U Ds U Ts+\. For any y G A, such that f s+\[y) is not yet defined, take

fs+\iy) =

fsiy)- (This includes nodes on already-frozen successor trees, nodes

on successor trees of height < m, and nodes not on T[xo].) For each y G A+i, if fs+\(y) is not yet defined, take fs+\{y) to be the least integer not already in

range(7A+i) Thus Ds+\ = dom(/iS+i). This completes the construction.

We now prove that this construction really does yield a tree T' which is isomorphic to T but not computably isomorphic to it.

Lemma 3.5. For every e, the sequence we,s converges to a limit we.

Proof. Assume by induction that the Lemma holds for every i < e. Notice that

in our construction, once we,s and Se,s are defined, the only way they can become

undefined is in Step 1 (if a new node of T[xq] appears which is a predecessor ofw^s for some / < e) or Step 3 (if wis ^ wu+\ for some i < e). Once we reach a stage so such that w^s = w? for every i < e and s > so and every predecessor of every Wi

(i < e) has appeared in TSo, we know that once we,s is defined for some s > so, it

will stay defined at all subsequent stages, although its value may change. Also, we,s is only defined at stages s such that wu is also defined for all i < e.

By induction, for every i < e, (wi.s)s? converges to some w?. Pick a stage sq

such that wu = Wj and levels (i^) =

IgvgIt(wi) for all / < e and s > so. Now if

s > so and we,s is not defined, then no w?,s with j > e is defined either. But since

limsupz ht(7,[x/]) = co, there are infinitely many successor trees of height > m, so

a new one, S, with S D DSQ = 0, must appear at some stage s > so. It will not be

frozen, since w?iS =

wf for all i < e, so it will be chosen as Se^s, and one of its nodes

of maximal height will be we,s. Then weJ is defined for every t > s, since every

predecessor of every w? with i < e is already in Ts. Thus, by induction, for every e,

we.s is defined for all sufficiently large s.

Once it is defined at a stage beyond so, we,s will only be redefined at a subsequent

stage t + 1 if XQwdjXVeif tivoe.t))) =

levelrr(t?e?i) and Condition 3.4 holds. More

over, even when it is redefined, we will still have ft+\{weJ+\) =

ftiweJ). Since the

tree T has height co, we know that for all t,

levelTt((pe(ft(wej))) < \evdTi(peiftiwej))) < co.

But (levelr,(<A?(/tiwej))))tec? is a non-decreasing sequence, so it can only change value finitely often. Thus, once defined, we,s will only be redefined finitely often, so

it must converge. H

Lemma 3.6. For every x, lim^ f six) exists.

Proof. We know x G 7^ ? Ds = dom(/,) for all s > x. Furthermore, once

fs (x) is defined, the only way we can have fs{x) ^ fs+\ (x) is if x lies on a successor

tree Se,s for which we,s is redefined or undefined at stage s + 1. Once this happens, Se.s is declared frozen, and ft \Se.s

? fs+\ \Se.s for all t > s + 1. Thus, not only

does (fs(x))seco converge, but in fact it changes value at most once. H

We define the function / = lim^ fs




Lemma 3.7. The functions f s satisfy Condition 3.1. {Hence the relation -<' defined onT' =

range(/) by

a<'b ^ (Vi) [a,b 6 ranged)=>/,-'(a) -< /7'(*>)] ?s computable and gives a tree structure on co).

Proof. The construction makes it clear that range(/5) C range(/5+i) for all s.

Now fix a,b ? range(/j). lff~l{a) / f~lx{a), then f~l{a) must lie on a successor

tree S^ such that we,s / t^+i. Hence /j+iGK/T"1^))) =

fs{f7l(a)) = a, and

f~^{(a) ?

g{f7l{a)), where g is the upward embedding of Se.s into Se,s+\ used in

the construction. We consider four cases:

Case 1. Suppose f~l{b) ? Se.s as well. Then also f~+{{b) =

g{f~l{b)), and

since g is an embedding, we have

/;+? -< f:^(b) ?=? /r'(a) -< /r'(ft).

Case 2. Suppose f~ \b) e rfxo]-.^-{ *(>} Then/,-'(&) -L fil{a). By Part

2 of Condition 3.4. we know /^(?) T[xo] -

5e?+i -

{xo}> so also /^(?O

J

Case 3. Suppose /7'(?>) =< *o- Then ./T'O) -< /W). and f~+}(b) =

Case 4. If/r1^) ? x0, then /7'(?) _L //-'(a), and also /"? =

frl(b) ?

xo^/-11(fl),so/-11(6)?/,-+,1(a).

A similar analysis applies if f~[{a) =

f~+{{a) and f~l{b) ^ f~+x (b). H

Lemma 3.8. The tree {Tr.-<') is a computable tree isomorphic to T.

Proof. We defined every fs to be a 1-1 map, with range(/i) ? range(/,+i). By Lemma 3.6, then, / is also 1-1.

The range of / is co since at each of the (infinitely many) stages at which we

needed a new element for the range of fs, we took the smallest one available. If

f~l\{y) 7^ f7X{y) f?r some s, then y = fs{x) for some x on some Se,s which was

redefined at stage s + 1, and f~+x (y) ? Se.s+\. But Se,s can only be redefined finitely

often, since levels{ipe{f{we))) < co, so eventually f~x{y) will stabilize, forcing

y ? range(/). Moreover, dom(/)

= |J? Ds = T, so f is a bijection from T to T'. Since the

partial order -<' on T' is defined by lifting -< from T via /, we know that / is an

isomorphism. Computability of -<f follows from Lemma 3.7: given a, b ? T', find a stage s such that a, b ? range(/5). Then a <' b <=>

f^l{a) -< f~x{b). H

Lemma 3.9. For every e, either (pe{f{we)) diverges or

levelT>{f{we)) ̂ lQvelT{(pe{f{we))). Thus requirement 91 e is satisfied by the element f{we).

Proof. Let so be a stage such that for all s > so,we.s = we and fs{we.s) =

f{we). Since we.s is never redefined after stage so, we know that either (pe{f{we)) diverges, or levQlDs{(pe{f{we))) ^ levelsy (we) for all s > so. But since [js Ds = T, the

latter of these implies that levels{(pe{f{we))) ^ levels (i^). Now levels (i^) =



122 RUSSELL MILLER

lev?lT'{f{we)) since / is an isomorphism, so ipe maps the element f(we) of T' to an element at a different level in T. Thus Me is satisfied, and (pe is not an

isomorphism from T' to T. H

This completes the proof of Proposition 3.2. H

?4. Trees of height co.

4.1. Main Theorem. We now prove the desired result for trees of height co.

Theorem 4.1. No tree of height co is computably categorical. The theorem will be proved in subsection 4.5, after we have established the

necessary five propositions, covering five different types of tree. We use the notions of an extendible node and a side tree to define these cases. Recall (from page 113) that a node x G T is extendible if there exists an infinite path through T containing x. The set of all extendible nodes of T. if nonempty, forms a subtree of T, denoted

by rext. Text need not be computable, even though T is.

The side tree above a node x is denoted S[x], and is a subtree of T[x].

S[x] = {ye T[x] : (Vz g T) [x -< z <y^z i Text] }

(x itself may or may not be extendible.) Equivalently, consider the extendible immediate successors xi,X2,... ofx. The side tree S[x] is precisely T[x]-|J/ T\x(\.

Thus x itself is the only node of S[x] which can be extendible in T, and S[x] contains no infinite paths, although it can have height co if it is infinite-branching. S[x] is not necessarily computable.

4.2. Three cases using Proposition 3.2.

Proposition 4.2. Let T be a computable tree of height co, and suppose further that T has height co above some nonextendible node yo. Then T is not computably categorical.

Proof. Let T and yo be as in the proposition. We claim there exists an xo G T

with co-many immediate successors, such that htXQ(T) = co and T has finite height

above every x >^ xo. Indeed, consider the subtree

S = { x G T : htx(T)

= co Sex is nonextendible & x JL yo }

S contains a ^-least element (either y o or some predecessor of yo), so S is indeed a

subtree. However, S contains no infinite paths, so it must contain terminal nodes, all of which will lie above yo. We take xo to be one of these, (xq is terminal in

S, that is; T will have height co above xo.) Therefore, T has finite height above

every x y- xo, and moreover, this xo must be an co-branch point, since otherwise

one of its immediate successors in T would also be in S. Let X1.X2,... be the immediate successors of xo in T. Then sup, ht(T[x/]j

= co, because htXo(T) = co.

But ht(T[x/]) < co for all i > 1, since otherwise x? would lie in S. Therefore we

must have lim sup, ht(T[x?]) = co, and so Proposition 3.2 applies to T and T is not

computably categorical. H

Proposition 4.3. Suppose that the computable tree T of height co contains an

extendible node xo such that the side tree S[xo] has height co. Then T is not computably categorical.

Proof. If xo has an immediate successor in S[xo] above which T has height co, then we apply Proposition 4.2 to this node. If all immediate successors of xq in S[xq]




have finite height, then there must be infinitely many of them, say x\, X2,.... Then

limsupz>1 ht(T[xz]) = co, because sup^! ht(T[x/])

= co. Moreover, any immediate successor of xo in T either lies in S[xq] or is extendible. Hence Proposition 3.2

applies to xo itself. H

Proposition 4.4. Suppose that in the computable tree T of height co, there is a node

xo ? Text wzY/z infinitely many immediate successors in Texi. Then T is not computably categorical.

Proof. ht(T[y]) = co for every immediate successor^ of xo in Text, so Proposition

3.2 applies to xo. H

4.3. An isolated path.

Proposition 4.5. Let T be a computable tree of height co. Suppose there is a node

xo G T which is uniquely extendible, i.e., which lies on exactly one infinite path y

through T. If all side trees at nodes on y above xo have finite height, then T is not

computably categorical.

Proof. Let xo be a uniquely extendible node on an infinite path y through T, such that all side trees at nodes on y above xo have finite height.

Let xo -< x\ -< X2 -< ... be all the nodes of y above xo. We apply Corollary 2.4 to

the set of side trees S[x?] above nodes of y, yielding an n such that for every / > n

and every finite subtree S ? S[x?], there is some j > i for which S embeds into

S[xj]. Our diagonalization argument will take place entirely above xn. (Notice that

the sequence (x/)zGa; cannot necessarily be computed, and that the choice of n from

Corollary 2.4 is nonuniform.) We define Ts

? {r, xo, xi,..., xn} U {0,1,..., s},

a tree under -<. (As before, r

represents the root of T.) We computably approximate the sequence (xi)ie . For

each s, let

\Xn =

Xn^s -^ Xn + i%s ~> ' * ^ X[ssj

be the chain of maximal length in Ts[xn]. (If there is more than one such chain, take the first such in the dictionary order derived from <.) Since all side trees have

finite height, clearly x^s ?? x? for each i. Indeed, xu = xz for all s such that

{xn,..., xm} ? Ts, where m = max7</-{j + ht(5[x7-])). (However, ht^xy]) need

not be computable in j.) The requirements 9te are the same as in Proposition 3.2:

me\ (fie total => (3x G T') [leveljv(x) ^ levels(<pe(*))].

This time, however, we will say that 9e is satisfied at stage s only if the witness node

we,s is defined and ipe,s(f*s{we.s)) converges and lies at a level of Ts different from

levels (weiS). Instead of simply freezing nodes, as in the proof of Proposition 3.2, we must

freeze them with priority e. Thus, at each stage s, we define envelopes Ee.s for each

e, to provide negative restraints on redefining the isomorphism / on elements of

Ee.s- If x lies in the envelope Ee.s, then fs+\{x) =? f six) only if necessary for

the sake of a requirement 91 \ with i < e. Thus the envelopes will ensure that the

functions fs converge to a limit / with range co.

Construction, fo is the identity map on To, and the witness nodes we.o are

undefined for all e. We define Ee$ ? 0 for all e.



124 RUSSELL MILLER

At stage s + 1, we search for the least e < s + 1 such that one of the following holds:

1. we_s is undefined.

2. For each i with n < i < ls+\, the following holds:

Xi;s + l ^ We,s =>XLs + i ^ We-\j.

3. wemS is defined and (pe.s{fs(we.s))l and

levels (we,s) =

lQvolDs((pe(fs(we,s))).

(Such an e must exist, because Wj+u- is undefined.) Let WjmS+\ =

w^s and

?/.*+i -{^ A'+i : (3z ? ?,>) J ^ z }

for all i < e, and let Wj}S+\ be undefined and is7>+i = 0 for all j > e.

If Case 1 holds for e, we let we_s+\ to be the oleast node in Ds[xn] with

level/)v[^](t?;e5?+i) > e which does not lie in any E^s with i < e and such that

(3j) [xj.s -< we,s+\ &Xj.s -? we-i,s+i\.

We define Ee,s+\ =

Ds+\ =

Ds U Ts+\. (If no such node exists, then we,s+\ remains

undefined, with Ee,s+\ = 0 and Ds+\

= Ds U 7^+1.) If Case 2 holds, we let we_s+\ diverge with Ee.s+\

= 0 and Ds+\ = Ds U Ts+\.

(This is the case where we-\iS and we,s appear to lie in the same side tree along y, in which case we cannot embed one upwards without disturbing the other.)

Otherwise, Case 3 holds. We search for the least t > max(7) J satisfying either of the following two conditions. Let mt =

max{ k : Xkj ^ we,s } for each t.

Condition 4.6. There exists i < e such that xmtJ _^ wiA.

Condition 4.7. There exists an embedding g of Ds[xmtt] into Tt[xm?A] with

levelr,{g{we,s)) > lev?lDs{we.s).

If Condition 4.6 holds for t, then we make we,s+\ undefined, and set Ee_s+\ = 0

and Ds+\ =

Ds U Ts+\.

Otherwise, we use the embedding g given by Condition 4.7 to satisfy requirement 9le. Let i?v,+i =

g(we,s), and for all y ? Ds[xmtJ], define fs+\{g{y)) =

fs{y). For those y ? Ds[xmtJ]

- range(g), take f s+\ {y) to be the least element of co that is

not yet in range(/5+i) nor in range(/J. Let Ds+\ = DSU Tt, and let the envelope

Ee.s + \ = Ds + \.

(For the sake of clarity, we note that if xmtJ does not lie in Ds, then

Ds[xmtj] =

{y e Ds : xmtj <y}.

We do have we,s ? Ds[xm[J] by definition ofmt. If Ds[xmft] does not have a single root, then we consider each minimal element in it to have level 0.)

In all three cases, we then define fs+\{y) =

fs{y) for those y ? Ds on which

fs+\ is not yet defined. Also, for each y ? Ds+\ -

Ds on which f s+\ is not yet defined, choose the least element of co which is not yet in range(/i+i) to be fs+\ {y). This completes the construction.

(The idea of the construction is that each witness element we_s lies in the side tree above some xz. When we need to satisfy 91 e, we do so by embedding the side tree containing we_s into another side tree at a higher level. We define fs+\ so that




fs+$we.s+$ =

fsi^e.s)- Since levels{(pe(fsiwe.s))) is finite, we will only have to

repeat this process finitely often before reaching a stage s such that fs{we_s) will

satisfy 91 e permanently.) We first must prove that at each stage s at which we search for a t, we eventually

find one. This requires a lemma guaranteeing our ability to embed trees upwards in

T[xn].

Lemma 4.8. For every x? >z xn and every t, there is an embedding g of the tree

Tt[xj] into T[x/+i].

Proof. By the choice of n and Corollary 2.4, we know that every finite subtree

of every S[xj] with j > n embeds into some S[xk] with k > j. By induction,

then, every finite subtree of every such S[xj] embeds into infinitely many S[xk] with

k > j. Since there are only finitely many side trees S[xj0],..., S[xjn ] which intersect

the finite tree Tt, we can embed S[xyo] D Tt into some S[xko], then embed S[xj1 ] n Tt into some SJx^] with k\ > ko, and so on. The union of these embeddings is the

desired embedding g. H

Lemma 4.9. Fix any stage s, and take the corresponding e chosen in the construc

tion. Then there exists a t for which Condition 4.7 holds.

Proof. Since each sequence (xu)te converges to xz-, we know that mt converges

to a limit m as t ?> oo. Thus we.s G S[xm], and m > n. Moreover, there exists t

such that Ds ? Tt. By Lemma 4.8, there is an embedding g : Tt[xm] ?> T[xm+\], and then

levels v iwe.,s) < lev??Tt{we,s) < levels (g (i?^))

since level r, (x) < levelr(g(x)) for every x G Tt[xm]. H

Lemma 4.10. For every e, the sequence {we.s)se converges to a limit we, the se

quence (fs(we))seco converges to a limit f{we), and either (peifiwe))] or

levelsipeifiwe))) 7^ levelriwe)- iSince levels{we) = levels (/(i^)), this satis

fies 9 e.)

Proof. Assume by induction that there exists a stage so such that for all s > so and all i < e, the hypotheses of the theorem hold: wLs = Wi, fsi^i)

= f(wi), and

either ifi(f(wi))? or 9? is satisfied by f(wi) at stages. Moreover, assume x^s = xk for every k < 7 + 1 and every s > so, where j is maximal with x7 ^ we-\,ThGnms >

j -f 1 for every s > so, so xm^s -? Wi for all / < e and s > so. lfwe,So is undefined,

then at the first stage s after so at which ht{Ds) > ht(Ee-\,s) + levels(xn), we,s will

be defined. Moreover, it will never again become undefined, since Condition 4.6

will never again be satisfied and Case 2 will never apply. Now if there is no stage s > so such that ipe,s(,we,s)l and

level/), iwe.s) =

levdDs{<PeAfs(we.s))),

then neither we.s nor fsiwe.s) will ever be redefined after sq. Then 9e will be

satisfied by we = hms we.s, because levels[<pe(f s(iVe.s))) is finite. Thus the lemma

will be satisfied for e.

If there exist stages s > so such that (pe,sifsiwe.s))i and levels (iu^) =

levelsy i(pe.sifsiwe,s)))> then Condition 4.6 will not hold at those stages, so at each

such s we will find a t satisfying Condition 4.7 and follow the corresponding instruc

tions for that t. Thus, we.s+\ will be redefined, but with fs+$we,s+$ =

fs(we,s)



126 RUSSELL MILLER

Moreover, by our choice of g, we will have

levelAv+1(i?v,+i) > levels (we,s).

Now lQVQlDs{(pe^{fs{we.s))) niay increase as s increases, but only finitely often, since fs{we^s) is constant after so and levels{ipe{fs{we_s))) < co. Therefore, we

eventually reach a stage s\ with

levelDs[{(pe{f,S]{we,S]))) =

levelT((pe{fSl{we,S?))),

and for all s > s\ + 1, 9le will be satisfied by we,s. Therefore we,s will never again be redefined, and 9le will be satisfied by we

? lim^ we^s. H

Lemma 4.11. For every x ? T, the sequence {fs{x))sew converges to a limit. The limit function f

= X\ms fs has range co.

Proof. Fix x. The construction ensures that x ? Tx ? Dx C Ds = dom{fs) for

all s > x. If xn -ft x, then f s{x) =

f s+\{x) for all s for which f s{x) is defined.

Assume, therefore, that xn -< x. Let k = max{ / : X{ -< x }, so x ? S^]. Let

so be a stage such that for all s > so and for all / < k + 1, we have x^s = x?. Also, let h =

max{ i + ht(S[x?]) : i < k -\- 1 }. Then by Lemma 4.10, there exists a stage s\ > so such that for all s > s\ and for all / < h, we have u^ = w?.

Suppose s > s\ is a stage such that fs 2 fs+\, and take the corresponding index e. Then Condition 4.7 is satisfied for some t > s, yielding an embedding

g\ Ds[xm,,t] ??

Tt[xmiJ]. By the construction, we,s ^ we^s+\, so wemusthavee > h.

This forces levels (we;i ) > A, since each wi+\^s is at a level > z, so we,s ^ U/<A: ^[x?]

by choice of h. Hence Xk+\ ^ we,s, and m? > k + 1 by definition of mt (and since ? >

so). But then x^+i =

x/<+i.t ^ -^m,,*- Since x ? S[x^], we have Xk+\ -? x, so

xm/;/ y< x. Hence x fi Ds[xm?J], and so x ? dom(g). Therefore fs+\{x) =

fs{x) for all s > s\. We define /

= lim^ fs. To see that range(/)

= co, let y ? co. We assume inductively that {0,1,..., y

- 1} ? range(/). Therefore, if y ? range(/"), there would exist a stage at which

y would be the least available fresh element, and so there must be a stage so and an x ? T for which fSo{x)

= y. Moreover, then y ? range(/,) for all s > so. If there exists some stage s\ > so at which fSl_\(x) ^ fS] {x), say for the sake of

a requirement 91 e, then there must be an x' such that fS] {xf) = y. At each such

s\, we will have x' ? Ee^. Indeed, by taking s\ so large that all M\ with i < e are

satisfied at all stages s > s\, we may assume that x' ? Ee,s for all s > s\. But then

fs{x') =

fs+i{x'), soy = f{x') ? range(/). H

Thus / is a 1-1 A!> map from T to co, hence an isomorphism from T to the tree

{T', -<f), where T' = co and -<' is the ordering induced on T' by -< via f.

Lemma 4.12. The maps fs satisfy Condition 3.1. Thus -<f is computable.

Proof. The construction ensures that Ds ? Ds+\ for all s. For every x ?

Ds -

Ds[xn], we have fs{x) =

fs+\{x). Therefore, Condition 3.1 clearly holds if

either f~x{a) or f~l{b) is not in T[x?]. So take x, y ? Ds[xn], with a = fs{x),

b = fs{y), and let x' =

f~+x(a) and y' =

f~+x{b). We have four cases, depending on whether or not x = x' and y

? y'.

The first case, where x = x' and y = y', is trivial. Also, if x ^ x' and y ^ y',

then x and y must both lie in Ds[xmiJ], for which we find an embedding g into some




Tt[xmiJ]. In this case,

x =< y -^ g{x) ^ giy) <=> x' -< y'

since g{x) = x' and giy)

= y'. Thus Condition 3.1 is satisfied in these two cases.

Suppose x 7^ x' and y =

y'. Then x G Ds[xm/j]. If y G Ds[xmiJ] also, then x' ?

g{x) -< giy) =

y'. If not, then either y -< xmtJ (in which case y -< x and

y -< g{x) = x', since range(g) c T[xmtJ]) or y _L x,?7^ (in which case j> J_ x and

y _L g(x) = x', again because range(g) c T[xm/,,]).

The preceding paragraph shows that in the third case, not only

x -< y <=> x' -< y'

but also

x _L y <=> x' JL y'.

Hence by symmetry, the fourth case, with x = x' and y ^ y', is also satisfied. H

Thus (T/', -</) is a computable tree, isomorphic to T, which satisfies every require ment 9e. Hence T is not computably categorical, proving Proposition 4.5. H

4.4. No isolated paths. An extendible node which lies on more than one infinite

path is called multiply extendible, as opposed to the uniquely extendible nodes of Subsection 4.3. We now consider the case of a tree in which every extendible node is

multiply extendible. This implies that every extendible node lies on infinitely many infinite paths. (We also assume that the tree contains at least one extendible node!)

Proposition 4.13. Let T be a computable tree of height co such that Text is non

empty and finite-branching and every x G Text lies on infinitely many infinite paths through T. If all side trees in T have finite height, then T is not computably categorical.

Proof. We use the same requirements 9e as in Propositions 3.2 and 4.5. The idea of this construction is that for each e, we devote an entire level le of T to

satisfying 9e. By the assumptions of the Proposition, we know that there exists at least one extendible node at level le, and at most finitely many of them. Also, there

may exist any number of nonextendible nodes at level le. Since we cannot tell the extendible nodes from the nonextendible ones at any stage s, we consider all the nodes at level le,s at that stage, and denote them by v?s,v]s,..., v"eJ.

Now since the Proposition assumes that the side tree above each extendible node has finite height, and since there exist only finitely many extendible nodes at levels < le, there must exist a number de such that every node x at level le with ht x ( Ts ) > de must be extendible. We do not know de, but at each stage we focus on those nodes at

level le^s in Ds 2 Ts above which Ds has maximal height. Thus, we will eventually be

considering only extendible nodes and their successors. Above these nodes we look for upward embeddings to use to satisfy 9e. Since every extendible node x lies on

infinitely many infinite paths, and since Text is finite-branching, T[x] must contain a

subtree of type 2<co, and any finite tree can be embedded into 2< at arbitrarily high levels. Thus we can find upward embeddings of Ds[x] above x whenever needed, as

long as x is extendible. (For trees defined using the infimum function, Lemma 7.2 is required here.)

The notation is as in the previous proofs, except that there may be more than one

potential witness for a given requirement 9e at a given stage s. We denote these witnesses by w^s,w?s,..., wneeJ. Also, we will keep track of the original position of



128 RUSSELL MILLER

each of these witnesses. When wks is defined, we will set vk s = wks, but as wks is

embedded further up in the tree, vks stays fixed. The only stages at which vks will be

redefined are those at which a requirement of higher priority receives attention and

those at which vks acquires a new predecessor. For a given e and s, the elements

vks will be at the same level for all /c, and we will denote this level by le.s Let r be the root of T. We define Ts =

{r} U {0,1,..., s}, a tree under -<. Again, we will define envelopes Ee.s, in order to ensure that range(/)

= co.

The requirements 9e are as follows:

9e: ye total =>(3x G T') [levels(x) ^ levelr(^(x))].

9e receives attention at stage s if some witness node wks is embedded upwards at stage s, if w? s is newly defined at stage s, or if the height of the envelope Ee.s increases at stage s. When this happens, all actions previously taken for the sake of

requirements 9j with j > e are injured. However, this will only occur finitely often

for each e.

Construction, fo is the identity map on To, and the witness nodes wk0 and

their original positions vk0 are undefined for all e and k. Also undefined are ne.o

and le.o for all e, and all Ee# are empty. At stage s -f 1, we execute the following steps for each e < s, starting with e ? 0.

If a requirement 9e receives attention, then we do not execute the steps for any

j>e.

1. lfW?S is undefined, and there exists an element x of Ds with

level/)v(x) > max^J{level/)v(j)

: y G E^s }, i<e

then let le s+\ be its level, and let w? c,,,..., w"e'??\ be all the elements of Ds at

level le.s+\. Let vks+l

= wks+{

for each k. Requirement 9e has now received

attention. Let Ds+\ =

Ds U Ts+\. and set Ee.s+\ =

Ds+\. For each j > e we

set

Ej,s+\ ={yeDs: {3zeEj,s)y?z}.

2. If WgS is undefined, and there does not exist any element x at a sufficiently

high level to satisfy condition (1), then let we,s+\] also, and set

Ee,+\ ={yeDs: (3zg?w)^z}.

Then 9e has not received attention at this stage. 3. Otherwise, w?s,..., w^s are defined, as are the corresponding vk_s. Find the

least stage t > max(T)J such that one of the following holds:

(a) There exists m < ne.s and an embedding g : Ds[v s] ?> Tt[v s] such that

levels (g?J) > levelAf?J + s.

(b) There exists x G Tt with levelrr(a:) = k.s and h?x(Tt) > s, such that

either x ^ Ds or levels (x) < le^s. If (b) holds and (a) fails at stage t, let wks+x

= wks for all k < ne.s, and

let le.s+\ =

le.s- For each k, if level/),-ivks)

= le.s, let

vks+l =

vks; otherwise

let vks+x be the predecessor of vks at level le.s in Ds. If there exist elements




x ? Ds with level/)v(x) = le_s such that x ? {v?s+\,

- ^Vi}>

tnen define

those x's to be we^(,we^{,...,

with vk s+] = wk y+1 for each, and define

ne_s+\ to be the greatest superscript required. (If there are no such x. then

ne_s+\ =

ne_s.) Define

Ee.s+\ =

{yeD5\ (3z ? Ee,s) [y r< z]}.

If le+\.sl and ht(?V^+i) > /e+i.s, then we say that 9te has received attention at

stage s + 1, and for each j > e we set

?7,v+1 =

{ y ? ?, : (3z ? E,s) [y ? z]}.

Otherwise 91 e has not received attention.

If (a) holds at stage t, let m be the least index for which it holds, and let g be the corresponding embedding. If tpe^{fs{w s))], or if

level^., (/>?))) + levels ?J, then we proceed exactly as in the preceding paragraph. Otherwise, 9le receives

attention as follows. For every node y ? Ds[v s], define fs+\ {g{y)) =

f s{y) and define fs+\{y) to be the least element of co which is not already in

range(/,+i) U range(/,)- Let w s+x =

g{w s). We define /^+i = le_s.

For each k, let vkes+x be that predecessor of vks at level le_s in Ds. (Quite

possibly, this will be vks itself.) Also, if there are any x ? Ds at level le_s which

are not in { vks+] : k < ne,s }, then define those x's to be

we^'{ ,w^+s^{s, with

vks+l =

wks+[ for each, and define ne_s+\ to be the greatest super

script required. (If there are no such x, then ne_s+\ = ne_s.) Finally, let

Ds+\ = Ds U range(g) U Ts+\, and let Ee.s+\

= Ds+\, with E^s+\

= 0 for all

j>e.

4. If 91 e has received attention at stage s + 1, we make all ?/;.y+i, //.,+i, vk v+1 and

^y J+i undefined for all j > e, and skip all steps for all those j. Otherwise we

increment e by 1 and return to Step 1.

Once we have either given attention to a requirement or completed the steps with e = s, we define f s+\{y)

= fs(y) for those y ? Ds on which fs+\ is not yet defined.

Also, for each y ? Ds+\ -

Ds on which f s+\ is not yet defined, choose the least

element of co which is not yet in range(/J+i) to be /lV+i(^). This completes the

construction.

Lemma 4.14. For each s and each e < s, either 3 (a) or 3 (b) must hold for some t.

Proof. Suppose there exists an extendible node y among {v?s,..., v^s'}. Then

by the assumption of the proposition, there is a copy of 2<0J embedded into T[y], and any finite tree can be embedded into 2<0J with the root mapping to a node at an arbitrarily high level of 2<CD. Thus 3 (a) will eventually hold.

Otherwise, none of v?es,..., v"?s is extendible. Now some node x on level le_s of

T must be extendible. If x ? Ds, then we must have level/)y {x) < le_s. since no node at level le_s in Ds is extendible. Otherwise x ? Ds, and either way we will eventually reach a stage t at which 3 (b) holds of x. H

Lemma 4.15. For every e the following hold:

X\ms ht{Ee_s) exists and is finite.



130 RUSSELL MILLER

The sequence (le.s)seco converges to some le G oo.

For every k G co, either (wks)se and (vk.s)seco converge to elements wk andvk

in co, or there exists a stage t such that wks] andvks] for all s > t.

The requirement 9e receives attention at only finitely many stages, and is satis

fied.

Proof. We proceed by induction on e. Fix e, and assume so is a stage satisfying all of the following conditions for every s > so and every i < e:

1. 9j does not receive attention at stage s:

2. /,,. =

/,:

3. Every v G Text with levels (i0 = le satisfies level 7;fv)

= le, and hence is of the form vks for some k\

4. vks = vk and wks

= wk for all k such that vks G Text (Notice that each level of Text is finite, since the proposition assumes that Text is finitely branching.

Hence only finitely many vks lie in Text.); 5. ht(T,) >/,_!.

Condition 3 simply says that we have waited until all predecessors of each v G Text at level le have appeared in TVo. This is possible because Text is finite-branching.

Notice that this condition implies the same condition for all i < e.

Now le.s is never redefined in the construction, and it can only become undefined at stages at which some 9\ with i < e receives attention. Hence l6iS

? le.sQ+i

for all s > so, so le,s converges to a limit le =

4.,0+i. Also, after stage so in the

construction, vks can only be redefined to be a predecessor of itself, and that only when it has acquired a new predecessor. But by Condition 3, each vks acquires no

new predecessors in T after stage so, so each sequence {vks)seco converges to a limit

Similarly, wk_s is never undefined after stage so, although it may be redefined at

certain stages at which 9e receives attention. If vks ? Text, then htt,/c (T) is finite,

and the corresponding wks can only be embedded finitely often by step 3(a), since each embedding (at a stage s + 1) moves it up by at least s levels in Ds. Hence all those sequences {wk_s)seoj converge.

For each of the finitely many k with vks G Text, it is possible for 3 (a) to hold for k at infinitely many stages. However, we only actually apply the embedding g to redefine wks at stages s + 1 such that (pe.sifsiwks))i and levels(^^(/^(i?;^)))

=

levels iwks). By the construction, we always have fs+\(w^9+i)

= fs{w^s), even if

wes+\ 7^ we.s- At each stage s -f 1 at which wks is redefined, we have

level/)i+I(w?J+1) > levels (w?j + s.

If this happens sufficiently often, then we must reach a stage s\ at which level/)s. iwkSi ) > levels((pe.sl(fsl (we.$i)))> since T has height co, and after stage s\, we will never

redefine wks again, even if 3(a) does apply. Hence each of these sequences (wk_s)seco does converge to a limit wk.

Now there must be an element of Text on level le. and this element will be

designated at some stage s as vks for some k. We note first that since all side trees are finite and Text is finitely-branching, there is a d such that every nonextendible node x at any level < le satisfies htA (T) < d. (Also, assume d is sufficiently large




that lej = le.) Once we reach stages s > d, therefore, 3 (a) will never again hold

for any m with v s nonextendible, and 3 (b) will not hold for any nonextendible x.

Thus only the finitely many extendible nodes vks will satisfy either 3 (a) or 3 (b) at

any subsequent stage. But every extendible node v at level le in T already satifies

levels (v) = 4, by inductive hypothesis, so 3 (b) will never hold again. By Lemma

4.14, there must exist an m, with v extendible, which satisfies 3 (a) at infinitely many stages. (If there is more than one such, choose the least of them, just as we

did at each stage of the construction.) If ipe{fs(w s))] for the corresponding w , then w s is never redefined, and

fs+\ {w ) =

fs{w ) for all s, so ipe{f{w ))], where / = lim,. fs as defined below.

Hence 9e is satisfied, since ipe is not total. On the other hand, if (pe{fs{u)"'1s))l, then for every stage s at which

l?vela,(/,?,))) = levels ?J,

either there will be a subsequent stage s' at which 3 (a) applies to v s, and 9e receives attention and w s, is embedded at a greater level, or else

(W > s) [levels,?,) < levelz),(^v(/^?V)))] In the latter case, w s, will never again be redefined, leaving 9e satisfied by the

witness f{w ). In the former case, we again have

levelz)j/+1(^(/^+i?j/+1))) < levels?,,).

But

levelAv(^(/?))) < levelr(^(/?))) < co,

so eventually we reach a stage s with levels{(fe{f{w'?7))) < level^ (w s). After this

stage, w is never redefined, leaving

levelr(^(/?))) < levelr?) - levels(/?)).

Thus requirement 91 e is satisfied.

We note that since each sequence (wks )sE converges to wk, none of them changes value more than finitely often. Moreover, the stage d designated above has the

property that only finitely many elements wks are ever redefined after stage d,

namely those corresponding to extendible vk.

Moreover, since there are only finitely many stages s at which any of the elements

wks is redefined, we eventually reach a stage s\ after which none of them is ever

redefined. Now Ee.Sl is finite. Let ^2 be a stage such that

(Vv ? T) [{3z ? Ee,S}) [y ? z]=>y ?

TS2}.

That is, every predecessor of each of the (finitely many) elements x ? Ee.Sl appears in TS2. Then for all s > S2, we have Ee.s

= Ee.S2. Hence lmiv ht{Ee.s)

= ht{Ee.Sl).

Thus Me only receives attention finitely often.

This completes the induction. H

Lemma 4.16. For each x, the sequence (fs{x))seco converges. The limit function f

? lim^ fs has range co.



132 RUSSELL MILLER

Proof. We need to show that both lims fs (x) and limv f~l {y) exist for all x and

y in co.

First of all. we have x ? Ts ? Ds for every s > x. so fs{x)[ for all sufficiently

large s. Also, by the construction, we have range(/.,) ? range(/A+i) for every s.

Moreover, each time we need a new element for the range of f s+\, we take the least

available one, so clearly every y ? co lies in range(/,) for all sufficiently large s.

So suppose fs{x) 7^ fs+\{x) for some s. The only way this can occur in our construction is if 3 (a) holds for some e and m. and we execute an upwards

embedding g of Ds[v s] into T[v s] at stage s + 1 in order to satisfy 9e. If this

happens, then Ee.s+\ =

Ds+\ D range(g). so x ? Ee.s+\. Similarly, if f~[{y) /

/7+li Mfor some s>then f7+\ (y) G Ee.s+\

The only way we could then have ft(x) ^ ft+\ (x) or f^l{y) ^ f?~+} {y) for any t > s is if some 9? with i < e receives attention at stage t + 1. This could happen for the following reasons:

Case 1. Step 3 (a) applies to 9) for some i < e, and we execute the corresponding

upward embedding g. In this case, EiJ+\ =

Dt+\, so x ? Eu+\ and f~+x(y) =

g{x) ? Eu+\

Case 2. w?t] and wft+l[, for some i < e. However, although 9? does receive

attention in this case, the construction leaves EeJ C EeJ+\. Hence x ? EeJ+\, and

ft+x{x) =

ft{x). Similarly, f^{{y) =

fs~\y) ? EeJ+l.

Case 3. ht{EiA+\) > li+u for some /' < e. Again, the construction leaves EeJ ?

EeJ+u sox ? EeJ+i andft+i{x) =

ft{x) and f^{{y)

= f~\y) ? EeJ+l.

Thus, for every t > s. we have both x and f^x{y) in is,-., for some / < e.

Therefore, ft+\{x) ^ ft{x) and f~+x(y) ^ ft~l{y) each can occur only for the

sake of an upwards embedding on behalf of some 9? with i < e. By Lemma 4.15, this can only occur finitely often. Hence the sequences (fs{x))se and (f~l{y))se both converge, making /

= lim,v fs a A^-bijection from co to co. H

As usual, we lift the partial order -< from T to an order -<' on T', making / an

isomorphism from T to T'.

Lemma 4.17. The functions f s satisfy Condition 3.1. Hence -<' is computable.

Proof. We have already seen that range(/s) ? range(/?+i). Take a,b ?

range(/,). The only way for f~+x{b) ^ f7X{b) is if f~l{b) lies in some sub

tree Ds[v s] which is embedded upward via some g as part of Step 3 (a) for some e

at stage s + 1. If f~l{a) is also embedded upward at stage s + 1, then since g is a

homomorphism of trees, we have:

frlw*f-\b) ^ g(f7l(a))<g(f^(b)) ^ /-,1(a)x/.r+1,(6).

Otherwise, f~l{a) ? Ds[v s]. In this case:

f;x(a)^f:\b) ^ f;\a)^v s

^ f;lM)<f;!M).




The case f~+{{b)

= f~ lib) is simpler, since this implies f~Xib) ? Ds[v s]. Thus,

if f-\a) -< f-lib). we know that f~\a) =

/">), so

f-^(a) -< /">) and

conversely as well. H

Thus (T7. -<') is a computable tree, isomorphic to T via /, yet not computably

isomorphic to T, since every requirement 9e is satisfied. Therefore. T is not

computably categorical. This completes the proof of Proposition 4.13. H

4.5. Proof of the Theorem.

Proof of Theorem 4.1. We need only confirm that the preceding propositions cover all possible cases. First, if T contains no extendible nodes, then Proposition 4.2 applies to the root of T, since ht(T)

= co. If Text is nonempty and infinite

branching, then Proposition 4.4 covers this case. If Text is nonempty and finite

branching, then we ask whether there exist side trees of height co. If so. then

Proposition 4.3 gives the result. Otherwise, every side tree has finite height. If

every extendible node lies on infinitely many infinite paths, we apply Proposition 4.13. If there exists a node x G Text which lies on only finitely many infinite paths

through T, then by following those finitely many infinite paths upwards until they all diverge, we find a node xo G Text which fits Proposition 4.5. H

?5. Trees of height > co. Having established that no tree of height co is com

putably categorical, we now prove the same result for trees of height > co. Recall

that for trees T of height co, we considered two cases in which T contains an in

finite path, and used guessing procedures to find that (or those) paths. Now the

existence of a node x at level co simplifies matters considerably, since the prede cessors of x0J form a computable infinite chain in T. (Technically, this chain is not a path, since it is not a maximal chain, but it is still perfectly useful for our

purposes.) We will appeal again to Kruskal's Theorem to guarantee the existence

of the necessary embeddings upwards along this chain, and use them to satisfy the

requirements.

On the other hand, having ht(T) > co creates a different set of problems. Previ

ously, with every node in T sitting at a finite level, we knew that each requirement would only require finitely many upwards embeddings in order to be satisfied. Now, it is possible that the node ipe.s(fs{we.s)) lies at an infinite level in T, in which case we might have to redefine fs{we.s) to lie at higher levels in T' infinitely often,

thereby injuring the lower-priority requirements infinitely many times. (Also, this

would prevent fs{we) from converging, ruining the isomorphism from T to T'.) We avoid this difficulty by watching for predecessors of (pe,sifsiwe.s)) and using

their preimages (under cpe) as new witness nodes. Eventually we will find such a

predecessor sitting at a finite level of T, and for this one we will only need finitely many upwards embeddings.

Of course, the preimage under (pe of a predecessor of (pe.sifsiwe.s)) will not

necessarily be a predecessor of fsiwe.s)) in T'. However, if indeed it is not a

predecessor, then clearly ipe was not an isomorphism. We can check effectively whether or not this is the case, and if it is not a predecessor, then 9e is automatically satisfied.

Theorem 5.1. No computable tree of infinite height is computably categorical.



134 RUSSELL MILLER

Proof. Theorem 4.1 covers the case of a tree of height co, so assume that T is a

tree under -< with ht(T) > co. Then T contains a node xw at level co. The set S of

predecessors of x is a computable set. ordered in order type co.

Each of our requirements 9e guarantees that ipe is not an isomorphism from T'

to T, just as before, but the exact statement is slightly different:

9e\ (fe bijective => either (3x G T') [levels(x) ^ levels((pe(x))] or Ox,y G T') [x ̂ y and^(x) < (feiy)]

If the second clause of the conclusion applies, or if there exists an s for which (pe.s is not one-to-one, we will say that 9e is finitely satisfied, since each of these facts

will become evident at some finite stage of the construction. In contrast, we can

never be sure at any finite stage whether or not we have permanently satisfied the

first clause of the conclusion, or whether ipe is total or onto.

Let r = xo -< xi -< be all the predecessors of x in T. We apply Corollary

2.4 to the collection of trees {Si : i G co }, where

Si = T[x?]

- T[x7+i].

(Thus the tree Si has root x? and contains those nodes lying above xt but not above

xi+\.) Clearly Si is computable. In the construction below, we will write Su for

Si n Ds. Let n be the number given by the corollary, such that every finite subtree

of every Si with / > n embeds into some Sj with j > i.

Construction. f0 is the identity map on T0 = { x? : i < n } U {x^}, and the

witness nodes we.o and their traces ve.o are undefined for all e. For each s we define

Ts+\ =TsU{s}.

At stage s + 1, we say that a requirement 9e (e < s) is finitely satisfied if

there exist distinct numbers x < s and y < s in the domain of (pe.s such that

<Pe.s(x) =

(pe.siy), or such that x -<f y and (pe.s(x) -ft <pe.s{y), or such that x -fi' y and (fe.six) -< ipe.s(y)- (In any of these three cases, we know right away that (pe is

not an isomorphism.) Search for the least e < s + 1 such that 9e is not yet finitely satisfied and one of the following cases holds:

1. we.s is undefined; or

2. we?s is defined and levels iwe.s) < level/)vive.s) + 1 and <pe.s(fs(we.s))l and

level/), (w^) =

\evdDsi(Peifsiwe.s)))',

or

3. we.s is defined and <pe.s{f s{we.s))i and there exist nodes w G Ds and w' G

range (/j) such that w -< (pe.sifsiwe.s)) and (pe.s{w')i = w and level/) v (w)

=

1 +\ev?lDs{ve.s).

(Such an e must exist, because ws+\.s is undefined.) We say that 9e receives attention

at this stage. For all i < e, let wu+\ =

wu and v?.s+\ =

vu. For all j > e, let Wj,s+\

and Vj.s+i be undefined. We proceed according to which of the three cases above

held.

1. If we.s is undefined,, we search for the ^-least node w in Ds[xn] satisying the

following conditions: w -< x :

w -/< wu for every / < e such that 91 i is not yet finitely satisfied;




for every x < e, either w -fi x or xco < x: and

for every y < e, there exists x ? Ds such that fs (x) = y and either w -fc x

or x ^ x.

Let ve_s+\ =

we_s+\ = w. (If there is no such w, then leave we,s+\ undefined.)

Let Ds+] =DSU r,+].

2. Ifwe.s is defined and ipe.s{fs(we.s))i and

lwADs(We.s) =

levelDs(tPe(f?We.s)))r

then search for the least t > s such that there exists x < t with ve_s -< x -< xco

for which Ds[ve.s] embeds into Tt[x] via an embedding g such thatg(^.,y) = x

andg{x ) =

x0J. Since iv.v -< x. clearly

levelr,(g(we.s)) > level/),(^.5).

Fix this i, x, andg.

We use the embedding g to satisfy (for the time being) the first clause

of 9e. Let Ds+\ =

Ds U Tt, ve.s+\ =

ve,s, and we.s+\ =

g{we.s), and

for all y ? Ds[ve,s] -

Ds[xOJ], define fs+$g{y)) =

fs{y). For those y ?

Ds+\[ve.s] ?

T>s[xw] ?

range(g). take f s+\{y) to be the least element of co

that is not yet in range(/iV+i) nor in range(/.s). Notice that although we

have temporarily fulfilled 9e, we do not state that 9e is satisfied, since pos

sibly levelr((p?.(/.v(i?;?,..v+i))) > levels,{ipe{f,s{ive.s+$)). We will continue to

scrutinize 9e at subsequent stages.

3. Otherwise, we have the nodes w ? Ds and w1 ? range(/A.) given in Case

(3). Let Ds+\ = Ds U Ts+\. Since 9e is not finitely satisfied, we must have

f~l{w') -< we_s. Define we,s+\ =

f~x{w'), and let ve.s+\ =

ve,s.

In all three cases, we then define fs+\{y) =

fs{y) for those y ? Ds on which

fs+\ is not yet defined. Also, for each y ? Ds+\ ?

Ds on which fs+\ is not yet defined, choose the least element of co which is not yet in range(/,+i ) to be fs+\ (y). For each e such that we,s+\ is defined, let ve,s+\

= we_s+\ A x0J. This completes the

construction.

(This construction is most comparable to that of Proposition 4.5. in which we

assumed that T contained an isolated infinite path. Here the path may not be

isolated, but the node x0J allows us to identify it anyway. The twist which we must

add appears in Case (3) of the construction, in which we ensure that the lim.s we.s will lie at a finite level, or else that ipe fails to preserve the relation -</.)

We first must prove that at each stage s at which Case (2) applies, we do eventually find an embedding, This requires a lemma guaranteeing our ability to embed trees

upwards in T[xn].

Lemma 5.2. For every x? >z xn and every t, there is an embedding g of the tree

Ds[xi] into T[Xi+\] with g{x ) = x .

Proof. By the choice of n and Corollary 2.4, we know that every finite subtree

of every Sj with j > n embeds into some Sk with k > j. By induction, then, every finite subtree of every such Sj embeds into infinitely many Sk with k > j. We may also assume that in each such embedding, Xj is mapped to xk. Since there are only

finitely many side trees S?Q,..., S?n which intersect the finite tree Ds. we can embed

Sj0 H Ds into some S^0, then embed S?} D Ds into some S*., with k\ > ko, and so



136 RUSSELL MILLER

on. The union of these embeddings with the identity map on Ds[x ] respects the

order -< (since each x7/ is mapped to some other predecessor xk of x ), and is the

desired embedding g. -\

Having thus guaranteed that every stage will eventually terminate, we turn to the

question of convergence.

Lemma 5.3. For every e, either 9e is finitely satisfied at some stage s, or else:

the sequence (ve.s)S?co converges to a limit ve -< xco, andv? -< vefor every i < e

such that 9i is not finitely satisfied; the sequence (we.s)seco converges to a limit we with ve -< we\ and

the sequence {fs{we))se converges to a limit f{we).

Proof. We proceed by induction on e. Suppose that 9e is never finitely satisfied, and let so be a stage so large that for all s > so and all i < e, the hypotheses of the theorem hold. (In particular, assume that vu = Vi, w?.s = w?, and fsiwi)

= fiw?)

for all s > so.) Since xco has infinitely many predecessors, there must exist a

stage s\ > so at which ve.S] and we.S] are defined. Moreover, they will never again

become undefined, since no 9? with i < e will ever again receive attention. Indeed, Ve.s

= Ve^ for every s > s\, since Cases (2) and (3) both define ve,s+\

= ve.s, so we

may write ve =

ve.Sl. We may also assume that (pe.Sl ifS] (we,Sl )) converges, since if

there is no such s\, then 9e will never again receive attention and the theorem will

be satisfied.) Notice that ve.s+\ becomes undefined at any stage at which Vi.s+\ / vu for some

i < e. Now if s is the last stage at which ve.s is undefined, then ve.s+\ = we.s+\ ii wLs for every i < e. Hence ve.s+\ ~? Vi.s+\ for any such i. However, ve.s+\ -< x , forcing

vu+\ ~< ve.s+\ for each such i. The construction never allows veJ+\i / veJ[, so we

must have vt -< ve, as the theorem demands.

Notice that at any stage s + 1 > s\ at which 9e satisfies Case (2), it will receive

attention and the resulting embedding will guarantee fs+\iwe.s+\) =

fsiwe.s) Hence (peif s+\iwe.s+\))

= <Pe(fs{we,s)) for all such s. Also, at any stage at which

9e satisfies Case (3), either 9e will be finitely satisfied or

(feifs + \iWe.s + \)) =

tPeW)i =W^ ife (fs iwe.s ))

where w and w' are as given in Case (3). Thus, Lpeifs+\iwe.s+\)) ^ <fe{f si^e.s)) in T for every s + 1 > s\. Since T is a

tree, the infinite nonincreasing sequence (ipe if s iwe.s)))s>sx must converge to a limit, so there exists a stage ^2 after which this sequence is constant. The construction

then makes it clear that Case (3) will never apply after stage si- This implies that no more nodes w are found which satisfy the hypotheses of Case (3). Therefore, either

there is no predecessor w ofipeifs2iwe.s2)) in T with levels[w) ? 1 + level/-[ve), in

which case ipe[f Sliwe.S2)) must lie at a finite level of T, or else any such predecessor does not lie in the range of ipe, in which case (pe is not bijective. (Recall that we are

assuming here that 9e is not finitely satisfied.) In the latter case, neither Case (2) nor Case (3) will ever again apply to 9e, so we.s ?

we.S2 for every s > S2. In the former case, where (peifS2iwe.s2)) lies at a finite level of T, we know that

Case (3) will never again apply, so Case (2) will only apply finitely many more times.

(Each time Case (2) applies, we have level/),+1 iwe.s+\) > level/),(w^), but Case (2) is impossible when level/)y iwe.s) > 1 + levels{ve).) Hence we will reach a stage after




which 9e never again receives attention, and thus (we.s)s>S2 converges to a limit

we. We also note that ve ^ we,s ^ we.s+\ for every s > S2, so that ve -< we. as

the theorem claims. Furthermore, we already saw that fs+$we.s+$ =

fs{we.s) for

every s > s\, so clearly the sequence (fs{we))s>S2 converges. This completes the

proof of the lemma. H

Lemma 5.4. The functions f s converge to a limit f which is bijective.

Proof. Every x lies in Ds for all s > x, so fs (x) will be defined for all sufficiently

large s. Also, each time a fresh element y was needed for the range of f s, we chose

the least y available. Once f~x{y) is defined, y will remain in range(/5) at all

subsequent stages, since the embeddings in Case (2) always preserve the range. In Case (1) we ensured that ve -< x only if e < x or x ^ x. In the latter case,

fs (x) will never be redefined. In the former case, we may have f s+\ {x) ^ f s {x) at

stages s + 1 at which Case (2) applies to a requirement 9e with e < x. However, Lemma 5.3 shows that there are only finitely many such stages. Similarly, Case (1) and this lemma ensure that f~xx (y) will only be redefined finitely often.

Thus f = lim., fs has domain and range co. Injectivity follows from the injectivity

of each fs. H

Lemma 5.5. For every e, either 9 e is finitely satisfied at some stage s, or(pe{f{we))] or levels((/?e(/{we))) / levelT{we). {Since levels(we)

= levels (/(i^)), this guar antees that 9e is satisfied.)

Proof. Suppose Lpe{f{we)) converges. (We also continue to assume that 9e is

not finitely satisfied, so that tpe is one-to-one and maps -<' to -<.) Let to be so large that 9e never receives attention after stage to.

Let le ?

levels (^). If levels (we) > le -\- I, then let u be the predecessor of we at

level le + 1 in T. Now since ipe is assumed to be total, we know that there is some

w ? T such that <pe{f{u))l = w. If levels (w) ^ levels(w), then 9e holds, since

levelr (w) = l?vela (/(?)). Otherwise, levels (it;)

= levels (w) =4 + 1, and Case

(3) must apply to w at some stage, with w' = f{u). (Notice that w -< (pe{f{we))

since by assumption (pe maps the ordering -<f on Tf to ^ on T.) This contradicts

our choice of to, completing the proof of the lemma. H

Lemma 5.6. The maps fs satisfy Condition 3.1. Thus <' is computable.

Proof. The construction ensures that Ds ? Ds+\ for all s. For every x ?

Ds -

Ds[xn], we have fs{x) =

fs+\(x). Therefore, Condition 3.1 clearly holds if

either f~l{a) or f~l{b) is not in T[xn]. So take x, y ? Ds[xn], with a = fs{x).

b = fs{y), and let xf =

f~xx{a) and y' =

f~^{{b). We have four cases, depending

on whether or not x = x' and y =

y'.

The first case, where x = x' and y = y', is trivial. Also, if x ^ x' and y ^ y1',

then x and y must both lie in Ds[ve_s] -

Ds[x ], where 9e receives attention in

Case (2) at stage s + 1. In this case, we find an embedding g into some Tt[x] with

x? -< x -< xw. Thus

xd>y *=> g{x) r< g{y) <=^ x' ^ y'

since g{x) ?

x' and g{y) =

y'. Thus Condition 3.1 is satisfied in these two cases.

Suppose x 7^ x' andy =

y'. Thenx ? Ds[ve_s] as above. If y ? Ds[ve_s] ?

Ds[x ], then x' =

g{x) -< g{y) =

y'. If not, then either y -< ve_s (in which case y -< x and

y -< g{x) = xf, since range(g) c T[ve.s]) or y _L xmtA (in which case y _L x and



138 RUSSELL MILLER

y J_ g(x) = x', again because range(g) c T[x?h.t]), or x ^ y. In this last case

the consition is satisfied, since

x -< y 4=^> x -< x0J <=> g(x) -< g(x ) =

x0J ^=^ x' -< y =

y'.

The preceding paragraph shows that in the third case, not only

x -< y <=^> x' -< y'

but also

x _L y <=^> x1 _L y'.

Hence by symmetry, the fourth case, with x ? x' and y / y', is also satisfied. H

Thus (T;, -<') is a computable tree, isomorphic to T via the A2 function /. and

T' satisfies every requirement 9e. Hence T is not computably categorical, proving Theorem 5.1. H

Intuitively it can be difficult to see where the action occurs in the proof of Theorem

5.1, particularly in Lemma 5.3, which is the heart of the proof. Essentially the

argument for convergence ofwe,s comes down to the fact that each time 9e receives

attention in Case (3), we generate another element in a descending sequence in T, and the definition of tree guarantees that this sequence must be finite. (In fact, we

can say more: once the actual predecessors of (pe{fs<{we.s)) at all levels < le in T

have appeared, and once ip~l has converged on all of them, 91 e will never again receive attention in Case (3).) Thereafter, any more upwards embeddings of we.s

would cause Case (3) to apply again, which we know cannot occur, so Case (2) must never again apply either. We conclude that 9e must be satisfied, becasue if (pe really were an isomorphism, either (2) or (3) would apply at some subsequent stage. In

the architecture of this proof, therefore, it is the well-ordering of the predecessors of each element of T which drives the result home. (One could write a similar

proof for Proposition 4.5, but the one given was more straightforward and offered a better insight into the reasons why the Proposition held, perhaps at some cost in

elegance.)

?6. Effectively infinite dimension. Recall that the computable dimension of a com

putable structure is the number of computable isomorphism classes of computable

copies ofthat structure. Theorem 4.1 shows that every computable tree of infinite

height has computable dimension at least 2. A theorem of Goncharov from [8] states that if s? is a computable structure which has two computable copies that are

A2-isomorphic but not computably isomorphic, then s? has computable dimension co. The isomorphisms which we constructed in our proofs are all A?, so in fact the

these trees all have computable dimension co.

It is possible to strengthen this statement even further, by avoiding countably many isomorphism classes simultaneously. That is, given a uniformly presented list

{Ti : i G co} of computable copies of T. one can construct another computable copy T' of T which is not computably isomorphic to any Tz. Thus, the computable dimension of T is effectively infinite', one might even call it effectively uncountable.

for, although there are only countably many computable isomorphism classes, there

is no effective enumeration of them.




Proposition 6.1. Let T be a tree of infinite height, and let {Tj} be a computable

ifinite or infinite) sequence of computable trees isomorphic to T. Then there exists a

computable tree T! isomorphic to T. such that no T? is computably isomorphic to T'.

(The fact that the set {T{\ is allowed to be infinite gives rise to the term effectively uncountable. If we could only prove this proposition for finite sets {T?}, then we

would only say that the dimension was effectively infinite.)

Proof. For trees of height co. the construction proceeds exactly as in Propositions 3.2. 4.5, and 4.13, according to which of these propositions applies to T. We let To

play the role of T as a template for T', constructing T' to be isomorphic to To via a

A^-isomorphism / = lim, fs, with Ds =

domain(/5) c T0. The only difference is

that instead of checking whether level/) v iwe.s) = level/) v {(pe(fs{we,s))) at each stage

s, we have witness elements we.u ? To to ensure that (pe is not an isomorphism

from T' to Tz, and we check at each stage s whether

\QVdDsiWeJ.s) =

lev?lTLs(<Pe{fS{We,u))).

If it is, then we proceed to embed we.i.s further upwards in To, which pushes fs (we,u ) further up in T'. Eventually (peifsiwe.i.s)) reaches its final level in Tl-, and one last

upwards embedding guarantees that (pe is not an isomorphism from T' to T?. The case of a tree of height > co requires similar modifications. At stage s -f 1

ofthat construction, we say that a requirement 9ej is finitely satisfied if there exist

distinct numbers x < s and y < s in the domain of(pe.s such that ipe,s (x) =

ipe,s (y), or such that x -<' y and (fe.six) -fi? (fe.siy), or such that x -fi! y and <pe,s(x) -<z

(fe.s iy)- (Here -</ denotes the partial order on the tree T?, so each of these conditions ensures that ipe is not an isomorphism from T' to Tt.) Then we search for the least

pair {e, i) < s + 1 such that 9ej is not yet finitely satisfied and one of the following cases holds:

1. wej,s is undefined: or

2. level/),(we,^) < level/)v(?v/..y) + 1 and <pe.s(fs(we,i,s))l and

level/)s.(tcw..v) =

\evdTi fipeif siwe.i,s)))\

or

3. ipe.s{fs{we.i.s))l and there exist nodes w G Tu and w' G range(/,) such that w <i tfe.sifsiwe.i.s)) and (fe.s(w')l

= w and levelsv(w) = 1 + level/)vivej,s)

Corresponding adjustments through the rest of the proof guarantee that each 9te.i is satisfied, so that <pe is not an isomorphism from T' to T,. H

?7. Trees under the infimum function. As noted in the introduction, much work on trees uses the language with one binary function symbol A, the infimum function, rather than the language of a partial order <. Theorem 5.1 and Proposition 6.1

both remain true when this definition is used. In this section we give the details

necessary to convert their proofs from the language of -< to that of A.

Proposition 7.1. Theorem 5.1 and Proposition 6.1 hold for all computable trees of

infinite height in the language of A.

Proof. First we note that every tree under A is also a tree in the language of

partial orders, with -<< defined by

x -< y <^> x A y = x.



140 RUSSELL MILLER

The same holds when "tree" is replaced by "computable tree." (The converse is true

for trees of height < co. For a counterexample of height co + 1, let T have a single node at each finite level, but two nodes at level co. Also, the converse fails miserably for computable trees, even of finite height.)

Similarly, although every embedding of trees in the language of A (which we

will call a A-embedding, for short) is also a -<-embedding, the converse again fails.

However, every isomorphism of trees in either language is also an isomorphism in

the other language. This will allow the proofs of Theorem 5.1 and Proposition 6.1 to

go through for trees under A with only minor adjustments, which we now describe.

The basic argument is that the computable tree {T', -<') we built in each proof suffices to show that {T A) is not computably categorical. Since T' is ~<-isomorphic to T, it is also A-isomorphic to it. Moreover, any computable A-isomorphism between T and T' would also be a computable ^-isomorphism, which is impossible. The one subtle point is that one must also show that the function /\' on T' really is

computable. For <', we accomplished this by ensuring that

f7+M) -< /7+? *=* f;\a) < f;\b) for alla,b ? range(/5); this was the reason why we needed the -<-embeddings given in Section 2 by Kruskal's Theorem. For A7, we wish to ensure similarly that

f7l\(a) a /-1, (b) = f;\a) A f;\b) for all a, b ? range(/\s), so we need the embeddings in all the relevant lemmas and

propositions to be A-embeddings. Kruskal's Theorem was originally proven for trees in the language of A, and all the

theorems of Section 2 remain correct in this language. This gives us the existence

of all the embeddings required in the proofs, save one. The exception arises in

Proposition 4.13, where (in the second paragraph of the proof, and also in Lemma

4.14) we found the embeddings needed simply by noting that every finite tree can

be embedded into 2<0J at arbitrarily high levels. This is no longer true when we

switch to the language of A: for instance, a tree with three nodes at level 1 cannot

be embedded into 2<0J while preserving A. Therefore, we need the following lemma.

Lemma 7.2. Let T be a tree such that TQXt is nonempty and finite-branching and

contains no isolated paths. Then all but finitely many nodes x ? Text have the property that for every finite S ? T[x], there exists y y x with y ? TQX{ such that S A-embeds

into T[y].

Proof. Suppose the lemma failed, so the the set U of nodes where it fails (with the

root r of T adjoined) forms an infinite subtree of Text. Since 7eXt is finite-branching,

K?nig's Lemma provides an infinite path through U, which in turn yields an infinite

path through TQXt containing infinitely many nodes r -< uo -< u\ -< from U.

Now for each i, some finite S? ? T[u?] A-embeds into no T[y] with u? < y. In

particular, S? does not A-embed into any T[u?] with j > i. Such a sequence would

violate Lemma 2.3 for A-embeddings. H

It follows that, by considering only nodes at sufficiently high levels, we can guar antee the existence of A-preserving embeddings in Proposition 4.13, thus ensuring the computability of the function A' on the tree T' there constructed. So this

proposition also holds for trees defined using A. No other changes to the proofs of

Theorem 5.1 and Proposition 6.1 are required. H




We note that for trees of finite height, the situation is more complex. There exist

such trees which are computably categorical under the infimum function, yet are

not computably categorical as partial orders. The authors of [14] believe that they have a criterion for compLitable categoricity of finite-height trees under A, using a

proof very similar to their result for finite-height trees under ~<. but they leave the

criterion for A as a conjecture. See Section 6 of [14] for details.

references

[1] C. J. Ash. Categoricity in hyperarithmetical degrees. Annals of Pure and Applied Logic, vol. 34

(1987), pp. 1-14.

[2] J. N. Crossley. A. B. Manaster. and M. F. Moses. Recursive categoricity and recursive stability. Annals of Pure and Applied Logic, vol. 31 (1986), pp. 191-204.

[3] R. Downey. On presentations of algebraic structures. Complexity, logic, and vecuvsion theory

(A. Sorbi, editor). Dekker. New York. 1997. pp. 157-205.

[4] Y. L. Ershov and S. S. Goncharov, Constructive models, Kluwer Academic/Plenum Press. New

York. 2000.

[5] J. H. Gallier. What's so special about Kruskal's theorem and the ordinal Tq?. Annals of Pure and

Applied Logic, vol. 53 (1991). pp. 199-260.

[6] S. S. Goncharov. Autostability and computable families of constructivizations. Algebra and Logic, vol. 14 (1975). pp. 647-680 (Russian), pp. 392-409 (English translation).

[7]-. The problem of the number of non-self equivalent constructivizations. Algebra i Logika. vol. 19(1980).

[8]-. Nonequivalent constructivizations, Proceedings of the Mathematical Institute of Siberia

Branch of Academic Sciences. Nauka. Novosibirsk. 1982.

[9]-, Autostable models and algorithmic dimensions. Handbook of recursive mathematics (Yu. L.

Ershov et al., editors), vol. 1. Elsevier, Amsterdam, 1998. pp. 261-287.

[10] S. S. Goncharov and V. D. Dzgoev, Autostability of models. Algebra and Logic, vol. 19 (1980),

pp. 45-58 (Russian), pp. 28-37 (English translation).

[11] B. Khoussainov and R. A. Shore. Computable isomorphisms, degree spectra of relations, and

Scott families. Annals of Pure and Applied Logic, vol. 93 (1998), pp. 153-193.

[12]-. Effective model theory: The number of models and their complexity. Models and com

putability: Invited papers from Logic Colloquium '97 (S. B. Cooper and J. K. Truss, editors). London

Mathematical Society Lecture Notes Series, vol. 259. Cambridge University Press, Cambridge, 1999.

pp. 193-240.

[13] J. B. Kruskal, Well quasi-ordering, the tree theorem, and V?zsonyis conjecture, Transactions of the American Mathematical Society, vol. 95 (1960). pp. 210-225.

[14] S. Lempp, C. McCoy. R. G. Miller, and R. Solomon, Computable categoricity of trees of finite

height, this Journal, vol. 70 (2005). pp. 151-215.

[15] C. St. J. A. Nash-Williams. On well-quasi-ordering finite trees. Proceedings of the Cambridge

Philosophical Society, vol. 59 (1963). pp. 833-835.

[16] J. B. Remmel, Recursive isomorphism types of recursive Boolean algebras, this Journal, vol. 46

(1981). pp. 572-594.

[17]-. Recursively categorical linear orderings. Proceedings of the American Mathematical

Society, vol. 83 (1981). pp. 387-391.

[18] S. G. Simpson. Nonprov ability of certain combinatorial properties of finite trees. Harvey Friedman's

research on the foundations of mathematics (L. A. Harrington. M. D. Morley, A. Scedrov. and S. G.

Simpson, editors). North-Holland. Amsterdam. 1985. pp. 87-117.

[19] R. I. Soare. Recursively enumerable sets and degrees. Springer-Verlag. New York. 1987.

DEPARTMENT OF MATHEMATICS. QUEENS COLLEGE - C.U.N.Y.

65-30 KISSENA BLVD.

FLUSHING. NEW YORK 11367. USA

E-mail: [email protected]



the computable dimension of trees of infinite height

Documents