Zeitschr. f: math. Logik und Grundlagen d. Math Bd. 36. S. 233-236 (1990)

EFFECTIVE EXTENSIONS OF PARTIAL ORDERS

by DEV KUMAR ROY in Miami, Florida (U.S.A.)

1. The following result, due to E. SZPILRAJN [3], is well known: Every partial order has a lin- ear extension. SZPILRAJN shows that a partial order with incomparable elements x, y can al- ways be extended to one with x 5 y ; since this can be done again and again, he uses a maxi- mal principle to conclude that a linear extension must exist. The result can also be proved using the compactness theorem. Every finite suborder of a partial order is extendible to a lin- ear order, therefore the same is true for the entire partial order. SZPILRAJN’S proof is of course nonconstructive. If one looks for an effective version of the theorem, is it possible somehow to glue together all the linear models for the finite parts? The compactness proof does just that, but the passage from finite satisfiability to satisfiability is again nonconstructive. It turns out that the case of a countable partial order permits some effective extensions. The countable case is studied here, with varying degrees of effectiveness. It is seen that if a partial order is recursive, then its extension is completely effective, but if a partial order is enumer- able, then it may fail to have a recursive or even enumerable extension. Furthermore, it is possible for a partial order to be inherently enumerable and not recursive; by contrast a lin- ear order which is enumerable is always recursively isomorphic to one which is recursive.

Notat ion: 5 , S1, 5’ etc. will indicate partial order relations as defined below. Some- times it will be necessary to refer to the ordering among (integer) indices as well. The usual inequality sign (5, <, >, etc.) will be used for the indices.

2. We begin with some definitions.

Defini t ion 2 . 1 . A countable partial order (F, 5 ) consists of a countablefield F and an ordering 5 S F x F which is reflexive, antisymmetric, and transitive. If 5 linearly orders F, then (F, 5 ) is called a linear order. For x, y E F, x I y will mean x $ y and y $ x; x < y will mean x 5 y and x * y ; x 2 y (resp. x > y ) stands for y 5 x (resp. y < x) .

Definit ion 2 . 2 . A recursive partial order is a countable partial order with F E N (the set of natural numbers) and 5 a recursive subset of N X N .

By Definition 2.2, the field of a recursive partial order is a recursive set, since 5 is a recur- sive relation and x E F iff x 5 x.

The question now is: Given a recursive partial order, how effective is its linear extension? Clearly every recursive partial order is embeddable in a highly recursive object, namely the set of rational numbers; but what is being sought is very different: an effedtive extension of the actual ordering of the partial order.

Defini t ion 2 . 3 . Acountablepartialorder(F2,S,) isanextensionof(F,,S,) i f fF,SF, and 5, E 5,.

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234 D.K.ROY

Theorem 2 . 4 . Every recursive partial order ( F, 6 ) has a recursive linear extension ( F’, 6 ’) . Proof. F is a recursive set; let { x o , x l , ...} be an effective enumeration. F’ will be the same

as F, and at the end of every stage n z 1 , { x o , x l , ..., x,,} will be linearly ordered by 5‘. At stage 0, 5’ and 5 coincide; at succeeding stages various order relations are added to 6’. The following describes stage n 2 1 :

Case 1 : If (Vi 5 n - I) (x i g x , ) , then introduce x,, 5 ’xi for every i 5 n - 1. Case 2 : If (3 i 5 n - 1) (xi j x,,) and (Vi 5 n - 1) ( x , 4 x i ) , then introduce xi 6 ’x,, for

every i s n - 1 . Case 3 : If (3i j n - 1) ( x i s x,,) and (3 j 5 n - 1 ) ( x , 5 x j ) then let xk be the 5’-greatest

among {x i : i 5 n - 1 and xi 5 x,} and let xl be the s’-least among { x j : j 5 n - 1 and x,, 5 x i } . Introduce x,, 5 ‘ x , and xk 5 ‘x,,.

In every case, close the set { x o , ..., x,,} under S’-transitivity. (This is clearly an effective procedure.) This completes the construction. 5‘ is clearly recursive and linearly orders F = F I . The one thing to prove is that 5 5 5’. Suppose not. Then there are distinct xk, x , E F with xk 5 x f and xf 5’xk. Call a pair such as {xk, xi} a bad pair. Assume that { x k , xf} is a mini- mal bad pair in the sense that {x i , x j } bad implies (min{i,j} > min{k, 1 } or (min { i , j } = min {k , I } and max{i,j} 2 max {k, I } ) ) . Suppose k < 1 and xk 5 xI . (All other cases are similar.) At stage 1, whether case 2 or case 3 applies, x, Y x k . This contradicts the assump- tion that { x k , xr} is bad. The minimal character of {xk, x i } is used if case 3 applies.

3. There i s no way to improve this theorem; if the relation 5 is merely enumerable, the ex- tension may fail to be recursive or even recursively enumerable. Modifying Definition 2.2, call a partial order recursively enumerable if 5 is r.e.

Theorem 3 . 1 . There is an r.e. partial order with no r.e. linear extension.

Proof. Let W,, W,, . . . be an effective list of all r.e. subsets of N. Using some standard cod- ing between N and NZ, let V, = { ( x l , x2 ) : x E We}. Call the unordered set of first and second members F,. Clearly every r.e. partial order ( F , 5 ) is ( F e , V,) for some e . It is enough to prove that there is an r.e. partial order ( X , 5 ) such that no (Fe , V, ) is a linear extension of ( X , 5 ) . The construction of ( X , s ) proceeds as follows: Dovetail the enumerations of all (F,, V , ) , e = 0 , 1 , ... If and when (2e,2e + 1) or (2e + 1 , 2 e ) shows up in the enumeration of V,, violate the order relation between them in ( X , 6 ) (i.e., if (2e,2e + 1) is enumerated first in V,, then introduce 2e + 1s 2e in ( X , 5 ); if (2e + 1 , 2 e ) is enumerated first, then intro- duce 2e 6 2e + 1 in ( X , 5 ) . Close ( X , 5 ) under reflexivity by including n 5 n for all n E N . If 2e 12e + 1 in (F, , V,) , then no action will be taken for this pair; but then ( F,, V,) will not be a linear extension of ( X , 5 ) .

This theorem raises more questions than it answers. If enumerability is not sufficient to guarantee an effective linear extension, is it possible to take an isomorphic recursive partial order and use that to extend the partial order? How effectively can one find such an iso- morph? Are there partial orders which are inherently enumerable but not recursive? By con- trast, it is known that every recursively enumerable linear order is recursively isomorphic to a recursive linear order, though it needs to be shown that the notions of r.e. and recursive for linear orders coincide with the same notions for partial orders.

Starting with the last question, the literature defines a linear order to be recursively enu- merable (respectively, recursive) if its order type can be realized by an r.e. (respectively, re-

EFFECTIVE EXTENSIONS OF PARTIAL ORDERS 23 5

cursive) subset of the rational numbers. A general reference and bibliography may be found in 121 or [l, Chap. 161. It is also known that every r.e. linear order is recursively isomorphic to a recursive linear order.

Defini t ion 3 .2 . Partial orders (6 5 ) and (F‘ , 5’) are recursively isomorphic if there exists a partial recursive f: N + N with dom(j) 2 F, f ( F ) = F, f ( P ) E (F’)‘, and satisfying (Vx, y E F ) ( x 5 y - f ( x ) 5‘ f (y ) ) . Two partial orders are isomorphic if the requirement that f be partial recursive is dropped in the above. Using some standard effective 1-1 correspon- dence between Q and N, the term “partial order” as well as the notion of isomorphism (recur- sive or otherwise) will be extended to include situations where the field of a partial order is a set of rationals.

Remark 3 .3 . A recursively enumerable (resp. recursive) set A of rational numbers, with its effective order relation, can clearly be used to define, effectively, an r.e. (resp. recursive) partial order (F , 5 ) . The converse is just as easy. It follows that the notions of r.e. and recur- sive as they have been used in this paper coincide with the corresponding notions as defined in the literature for linear orders, in the following strong sense: A partial order (F , 5 ) which is linear is r.e. (resp. recursive) if and only if it is recursively isomorphic to an r.e. (resp. rec- ursive) set of rationals.

It follows then, from the statement immediately preceding Definition 3.2, that every r.e. partial order which is linear is recursively isomorphic to a recursive partial order. This will fail if the condition of linearity is dropped:

Proposit ion 3 .4 . There is an r.e. partial order which is not recursively isomorphic to any rec-

Proof. If every r.e. partial order were to be recursively isomorphic to a recursive partial or- der then the proof of Theorem 2.4 would go through with cases 1 , 2 , and 3 being determined by the recursive isomorph; this would yield a recursive linear extension for every r.e. partial order, contradicting Theorem 3.1.

ursive partial order.

4. Summarizing the latter part of the last section, every r.e. linear order is recursively isom- orphic to a recursive linear order but there are r.e. partial orders which are not recursively isomorphic to recursive partial orders. The final question is: Are there r.e. partial orders which are inherently r.e. in the sense that they have no isomorph at all which is recursive? In other words, does dropping the requirement that the isomorphism be recursive lead to the same answer? Information from a l7: set of natural numbers is coded into a partial order to settle this question.

Theorem 4 . 1 . There is an r.e. partial order (F , 5 ) which is not isomorphic to any recursive

Proof. Choose a @-.Z: set S of natural numbers. Let

partial order,

k E S ++= V i3jR (iJ, k ) ,

with R recursive. Define an r.e. partial order ( F , 5 ) as follows: Fix some effective 1-1 corre- spondence

h : {0, 0 .5 , 1, 1.5, 2,2.5, ...} u N2+N.

The partial order will be defined on the domain of h, for simplicity; clearly it can be defined

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236 D.K.ROY

in terms of N instead. With this understanding, let F = dom(h) and let 5 be the reflexive transitive closure of the following set of order relations:

0 5 0.5 5 1s 1.5 5 2 5 ...; for all integer k 2 1 and all i 2 0: k 2 ( k - 1, i ) ; for all integer k 2 0 and all i 2 0: k 6 (k, i ) iff 3jR(i,j, k ) .

(F, 5 ) is certainly r.e. Observe that k E S if and only if k 5 (k, i ) for every i. Therefore k E Sc if and only if every sequence of 2k + 3 members uo, a l , .. ., u 2 k + 2 chosen from the field of the partial order, with nothing below a. and nothing between successive ai, has the property that there is a member of the field below a 2 k + 2 which is incomparable with uZk . Allowing ni to de- note the exponent of the ith prime in the prime decomposition of n , and lh(n) to denote the number of primes which divide n , if (F' , 5') is a recursive partial order isomorphic to (F, 5 ) , then

k E S C o ( V n ) ( l h ( n ) = 2 k + 3

A ( V i lh(n) - 2) (ni <'ni+ I A i ( 3 x ) (x E F' A ni <'x A x <'ni+ A l ( 3 X ) (X E F' A X <'no)

+ ( ~ Y ) ( . Y E F ' A Y < ' n z k + z h n z k I Y ) ) .

A Tarski-Kuratowski reduction shows Sc to be If:, which contradicts the assumption that S is ZZ:-,Z:. It follows that no such (F' , 5') can exist.

References

[l] ROSENSTEIN, J., Linear Orderings. Academic Press, New York-London 1982. [2] ROY, D., Linear order types of nonrecursive presentability. This Zeitschrift 31 (1985), 495-501. [3] SZPILRAJN, E., Sur l'extension de l'ordre partial. Fundamenta Math. 16 (1930), 386-389.

D. K. Roy Florida International University College of A r t s and Sciences Department of Mathematics University Park Miami, Florida 33 199 U.S.A.

(Eingegangen am 2. Marz 1989)