Order Embeddings with Irrational Codomain: Debreu Properties of Real Subsets

Transcription

1 Order (2006) 23: DOI /s y Order Embeddings with Irrational Codomain: Debreu Properties of Real Subsets M. J. Campión J. C. Candeal E. Induráin G. B. Mehta Received: 24 March 2006 / Accepted: 13 December 2006 / Published online: 27 January 2007 Springer Science + Business Media B.V Abstract The objective of this paper is to investigate the role of the set of irrational numbers as the codomain of order-preserving functions defined on topological totally preordered sets. We will show that although the set of irrational numbers does not satisfy the Debreu property it is still nonetheless true that any lower (respectively, upper) semicontinuous total preorder representable by a real-valued strictly isotone function (semicontinuous or not) also admits a representation by means of a lower (respectively, upper) semicontinuous strictly isotone function that takes values in the set of irrational numbers. These results are obtained by means of a direct construction. Moreover, they can be related to Cantor s characterization of the real line to obtain much more general results on the semicontinuous Debreu properties of a wide family of subsets of the real line. Key words total preorders semicontinuous strictly isotone functions irrational numbers Debreu properties Mathematics Subject Classifications (2000) Primary: 54 F 05 Secondary: 06 A 06 M. J. Campión E. Induráin (B) Departamento de Matemáticas, Campus Arrosadía, Universidad Pública de Navarra, Pamplona, Spain steiner@unavarra.es M. J. Campión mjesus.campion@unavarra.es J. C. Candeal Facultad de Ciencias Económicas y Empresariales, Departamento de Análisis Económico, Universidad de Zaragoza, c/ Doctor Cerrada 1 3, Zaragoza, Spain candeal@unizar.es G. B. Mehta Departament of Economics, University of Queensland, 4072 Brisbane, Queensland, Australia g.mehta@economics.uq.edu.au

2 344 Order (2006) 23: Introduction Let (X, ) and (Y, ) be totally preordered sets. Then a function f : X Y is said to be strictly isotone (oranorder monomorphism or order embedding) if x y is equivalent to f (x) f (y). We say that the function f is a representation of the preorder and the problem of proving the existence of such a function is said to be a representation problem. IfthesetX also has a topology τ X and the set Y has a topology τ Y then the problem of proving the existence of a continuous strictly isotone function f : (X,τ X ) (Y,τ Y ) is said to be a topological representation problem. Similarly, if the set X also has a differentiable or algebraic structure then the function f is also required to be a morphism with respect to these structures. There is now a very vast literature on the representation problem in mathematics and applied mathematics (see, e.g. [9, 18]). It is undeniable that the main emphasis in the literature has been on the case where the codomain (Y, ) is the set of real numbers with the natural ordering and the Euclidean topology. One consequence of this practice is that many total preorders, even ones with quite good properties, do not have real-valued representations. Also, from a purely mathematical point of view there is no reason or justification for studying only the very special case where R is the codomain. For further discussion of these ideas we refer the reader to [4] and[14]. Therefore, it is desirable to develop a general theory of representations of ordered structures in which the codomain is not necessarily the set of real numbers. One way to proceed is to consider codomains such as the long line or the lexicographic plane as in [14]. In this paper we follow a somewhat different approach by studying codomains of strictly isotone functions that are subsets of R but not necessarily order-isomorphic to R. (For initial results in this direction see [8] where order-monomorphisms are studied with codomains consisting of the rational numbers and the real algebraic numbers.) In this paper we continue the study of order-monomorphisms with codomains that are important subsets of the real numbers. As pointed out in [14], the most desirable codomains Y in studying continuity properties of strictly isotone functions are those for which the analogue of Debreu gap lemma holds, because in that case the existence of a strictly isotone function (which is not necessarily continuous) with such a codomain Y implies the existence of another strictly isotone function, that is continuous, and which also takes values in Y. We will prove that the set of irrational real numbers does not satisfy the Debreu property, so that, in a sense, it cannot be considered a good codomain for strictly isotone functions. Nevertheless, the set of irrational numbers R \ Q still has a good property if we consider semicontinuous representations instead of continuous ones. Thus, we shall also prove that a lower (respectively, upper) semicontinuous total preorder defined on a topological space (X,τ)admits a representation in the real line R if and only if it also admits a lower (respectively, upper) semicontinuous representation in the set R \ Q R of irrational numbers (that we will consider endowed with the inherited Euclidean topology). Unlike some general results concerning the Debreu property (see e.g. [14]) the proof we include here is given by means of a direct and explicit construction. In connection with the semicontinuous Debreu properties it is important to point out that the results achieved for the (at first glance) particular case of order-

3 Order (2006) 23: embeddings into the set of irrational numbers can be combined with other result (the Cantor characterization of the real line) to infer a much more general result, namely: All the subsets of the real line whose complement is countable and dense, do not satisfy the continuous Debreu property, but do satisfy both the upper and lower semicontinuous Debreu properties. In other words, the properties analyzed through a direct construction on the set of irrationals are shared by a wide family of real subsets. We also provide an example of a real subset that shares with the irrationals the same Debreu properties, but whose complement is uncountable. Finally, we provide an example to show that the semicontinuous Debreu properties cannot be taken for granted on every subset of the real line. We mean that, after showing that the irrationals (as well as many other real subsets) satisfy the semicontinuous Debreu properties, we could ask ourselves if such properties are satisfied by everyrealsubset.we prove that this is not the case, furnishing an example of a real subset that neither satisfies the upper semicontinuous Debreu property nor the lower semicontinuous one. For related ideas we refer the reader to [12]and[13]. 2 Preliminaries Let X be a nonempty set. Let be a total preorder (i.e.: a reflexive, transitive and complete binary relation) on X. (If is also antisymmetric, it is said to be a total order and the totally ordered structure (X, ) is said to be a chain). We denote x y instead of (y x). Alsox y will stand for (x y) (y x) for every x, y X. The total preorder is said to be dense in itself if for every x, y X with x y, there exist t X such that x t y. The total preorder is said to be representable if there exists a real-valued strictly isotone function f : X R such that x y f (x) f (y) (x, y X). This fact is characterized (see e.g. [9], p. 23) by equivalent conditions of orderseparability that must satisfy. Thus, the total preorder is said to be orderseparable in the sense of Debreu if there exists a countable subset D X such that for every x, y X with x y there exists an element d D with x d y. Such a subset D is said to be order-dense in (X, ). If X is endowed with a topology τ, the total preorder is said to be continuously representable if there exists a real-valued strictly isotone function f that is continuous with respect to the topology τ on X and the usual topology on the real line R. The total preorder is said to be τ-continuous if the sets U(x) ={y X, x y} and L(x) ={y X, y x} are τ-open, for every x X. In this case, the topology τ is said to be natural with respect to the preorder. (See[9], p. 19). The coarsest natural topology is the order topology θ whose subbasis is the collection {L(x) : x X} {U(x) : x X}. The total preorder defined on X is said to be τ-lower semicontinuous if the set U(x) ={y X, x y} is τ-open, for every x X. Ina similar way, is said to be τ-upper semicontinuous if the set L(x) ={y X, y x} is τ-open, for every x X. Now Debreu s open gap lemma is a powerful tool to obtain continuous representations of an order-separable totally preordered set (X, ) endowed with a natural topology τ.(see[11], or Ch. 3 in [9]). To explain this lemma let S be a subset of the

4 346 Order (2006) 23: real line R. Alacuna S corresponding to S is a nondegenerate interval of R that has both a lower bound and upper bound in S and that has no points in common with S. A maximal lacuna is said to be a Debreu gap. Debreu s open gap lemma states that if S is a subset of the extended real line R, then there exists a strictly increasing map, called a gap function, g : S R such that all the Debreu gaps of g(s) are open. Using Debreu s open gap lemma, the classical method to get a continuous realvalued strictly isotone function goes as follows: First, one constructs a strictly isotone function (which may not necessarily be continuous) f (see e.g. [6], Theorem 24 on p. 200, or else [9], Theorem on p. 14). Then Debreu s open gap lemma is applied to find a strictly increasing function g : f (X) R such that all the Debreu gaps of g( f (X)) are open. Consequently, the composition F = g f : X R is also a representation (X, ), but now F is continuous with respect to any given natural topology τ on X. If is a τ-upper (respectively τ-lower) semicontinuous total preorder on X, and it is also representable (or, equivalently, order-separable in the sense of Debreu), a similar argument allows us to get a τ-upper (respectively τ-lower) semicontinuous real-valued strictly isotone map that represents. The process is the same as above: Construct a strictly isotone function f representing (X, ) and apply Debreu s open gap lemma to find a strictly increasing function g : f (X) R such that all the Debreu gaps of g( f (X)) are open. The composition F = g f : X R becomes also a strictly isotone function that represents (X, ), but now F is τ-upper (respectively τ-lower) semicontinuous. An appealing proof of a variant of Debreu s gap lemma, due to Beardon, appears in [2]. In order to understand the significance of Beardon s version of the gap lemma it should be noted that the continuity of a strictly isotone function u : (X,τ, ) (R, ) depends upon the fact that the image u(x) R has no gaps of the form [a, b) which are half-closed half-open and no gaps of the form (a, b] which are halfopen half-closed. These are the bad gaps" which should be removed by the gap function g. The existence of a closed gap does not preclude the continuity of the strictly isotone function u. (See[9], pp ). Now let S be an arbitrary subset of R. Beardon proves that if S R then there is a strictly increasing function g : S R such that g(s) has no half-closed half-open or half-open half-closed gaps. It is important to point out that if a strictly isotone function u : (X,τ, ) (R, ) has no half-open half-closed gap (a, b],then it is upper semicontinuous. In the same way, if u has no half-closed half-open gap [a, b), then it is lower semicontinuous. In [3] the following question regarding the gap lemma appears: Does the Debreu gap lemma hold only for the real numbers R or can it be extended to more general totally preordered sets? This question is very important for representation theory because of the almost universal practice of using R as the codomain of any strictly isotone function. At this point, it is important to know some characterization of chains that are order-isomorphic to the reals. In this direction, the fundamental Cantor theorem on the linear continuum solves the question. A standard set-theoretical proof of this theorem may be found in pp of [15]. For a proof of this theorem that is based on concepts of utility theory we refer the reader to [5] where the Cantor theorem is proved by using Debreu s gap lemma.

5 Order (2006) 23: Lemma 2.1 (Cantor s fundamental theorem) Let (X, ) be a totally ordered set which satisfies the following properties: (a) (b) (X, ) is order-unbounded and has a countable order-dense subset. (X,θ) is a connected topological space (where θ stands for the order topology induced by on X). Then (X, ) is order-isomorphic to the real line R. Observe now that if the gap lemma of Debreu can be extended to totally ordered sets that are quite different from R with its usual Euclidean order, then it will be possible to begin the development of a theory of continuity of strictly isotone functions with other codomains. (For a complete account of the latest achievements about this question, consult [14]). Totally ordered sets where the analogue of the Debreu s gap lemma holds are said to have the Debreu property. In what follows, we shall say that a totally preordered set (X, ),hasthedebreu property if for an arbitrary subset S X there is a strictly increasing function g : S X such that g(s) has no half-open half-closed or halfclosed half-open gaps. Equivalently, a totally preordered set (X, ), endowed with a natural topology τ, has the Debreu property if for any strictly isotone function f defined on a totally preordered space (Y, Y ), and taking values in X there is another continuous strictly isotone function g : Y X, where we supposethat Y is endowed with the order topology. As an immediate consequence of the definition of the Debreu property, and the Debreu s gap lemma, we obtain the following result for subsets of the real line. Proposition 2.2 (a) Let X be a subset of the real line that contains an open interval (a, b) X R. Then (X, ) satisfies the Debreu property. (b) Let X be a nonempty subset of the real line that does not satisfy the Debreu property. Then its complement R \ XisdenseinR. Equivalently, X is totally disconnected. Wecan generalizethe conceptofdebreu property, in the following two directions: (a) (b) A totally preordered set (X, ), endowed with a natural topology τ, hasthe lower Debreu property if for any strictly isotone function f defined on a totally preordered space (Y, Y ), and taking values in X there is another lower semicontinuous strictly isotone function g : Y X, where we suppose that Y is endowed with the order topology, A totally preordered set (X, ), endowed with a natural topology τ, hasthe upper Debreu property if for any strictly isotone function f defined on a totally preordered space (Y, Y ), and taking values in X there is another upper semicontinuous strictly isotone function g : Y X, where we suppose that Y is endowed with the order topology.

6 348 Order (2006) 23: It is clear that if (X, ) has the Debreu property, then it also has both the lower and the upper Debreu properties. However the converse is not true, as we shall prove later. 3 Strictly Isotone Functions with Irrational Values Let (X, ) be a totally preordered set that is representable in (R, ) by means of a (not necessarily continuous) strictly isotone function u : X R. We wonder if it is possible to get another strictly isotone function v : X R such that v(x) is an irrational number (i.e.: v(x) / Q, where Q denotes the set of rational numbers) for every x X. The answer is positive, and it is a consequence of next proposition, in which we furnish a direct and explicit construction. Proposition 3.1 There is a strictly increasing map h : R R that only takes irrational values (i.e.: h(r) / Q for every r R). Proof Consider the decimal expansion of a real number r as the series a n N + 10, n where N is an integer and a n {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} (n N). Observe that there are some real numbers that admit two possible such expansions, for instance: 1 = (0)...= = n 9 = (9)... 10n In these situations, we shall only use the expansion in which the periodic digit is 9. (In the above example, we shall use (9)... disregarding the use of 1 as (0)...). Now, given we define r = N + h(r) = N + a n, (r R) 10n a n 10 n2 +n 2, (r R). (For instance h(1) = h( (9)...)= ). Observe that after the digit a n and just before the digit a n+1 asetofnzeroes has been inserted. By definition, the map h is strictly increasing. Notice also that, by construction, the decimal expansion of h(r) has no periods, so that it corresponds to an irrational number h(r) / Q. Remark 3.2 (a) As a consequence of Debreu s open gap lemma, the map h constructed in Proposition 3.1 has the property that the image h(r) is not all of R \ Q. Asa

7 Order (2006) 23: (b) matter of fact, observe that: There is no strictly increasing and surjective map H : R R \ Q. The reason is that such a map, if any, would be continuous because it has no gaps and the order and Euclidean induced topology on R \ Q coincide. Moreover, such a function would be plainly open. Therefore, R would be homeomorphic to R \ Q. But this is not true because the real line is connected but R \ Q is totally disconnected. The map h that appears in the proof of Proposition 3.1 is lower semicontinuous. Indeed, if (x n ) is a strictly increasing sequence of real numbers that has a limit l R, it is plain, by construction, that lim n + h(x n) = h(l). However, h fails to be upper semicontinuous: If we consider the strictly decreasing sequence (x n = n ) + whose limit as n goes to + is obviously 1,we observe that (c) But lim h(x n) = 1. n + h(1) = 9 10 n2 +n 2 = 1. As a matter of fact, we can also point out that h(r) contains half-open halfclosed gaps: For instance ( + ] 9 10 n2 +n 2, 1 is one of such gaps. Therefore h cannot be continuous. Suppose now that the decimal expansion of a real number r is given as the series r = N + a n 10 n. Assume now that in the case of real numbers that admit two possible such expansions we always choose the expansion in which the periodic digit is 0. (As an example, we would use 1 but not (9)...). Now, given we define where b n = a k r = N + j(r) = N + if n = k2 + k 2 a n, (r R) 10n b n 10 n (r R), for some k N ; b n = 9 otherwise.

8 350 Order (2006) 23: (For instance j(1) = ). Observe that after the digit a n and just before the digit a n+1 asetofndigits 9 has been inserted. By definition, the map j is strictly increasing. It is straightforward to see now that j : R R is upper semicontinuous but not lower semicontinuous and takes only irrational values, and j(r) contains in this case half-closed half-open gaps. For instance [ ) c n 1, 1 +, 10 n with c n = 0 if n = k2 +k 2 for some k N; c n = 9 otherwise, is one of such gaps. Therefore j cannot be continuous. Corollary 3.3 Let (X, ) a totally preordered set that admits a representation in R through a real-valued strictly isotone function u : X R. Then it also admits a representation v : X R which only takes irrational values (i.e.: v(x) Q = ). Proof Just consider v as the composition h u where h is the map given in Proposition 3.1, or, alternatively, take v = j u, where j is the map given in Remark 3.2 (c). Now we are ready to analyze the behaviour of the set of irrational numbers with respect to the Debreu property. Theorem 3.4 The set R \ Q of irrational numbers, endowed with its usual order and Euclidean topology, does not satisfy the Debreu property. However, it satisfies the lower Debreu property and the upper Debreu property. Proof Consider the real line R endowed with the usual total order, that is obviously continuous with respect to the usual Euclidean topology. Moreover, (R, ) is representable in (R \ Q, ) by means of the function h considered in Proposition 3.1 above (or, alternatively, by means of the function j introduced in Remark 3.2 (c)). Suppose that the set R \ Q of irrational numbers has the Debreu property with respect to its usual Euclidean topology inherited from R. This would imply the existence of another strictly isotone representation H : (R, ) (R \ Q, ), but now H being continuous. IfweconsiderthesetH(R) (R \ Q), endowed with the order topology, it is clear that the map H : R H(R) is bijective (because it is strictly isotone) and continuous (becausetheordertopologyon H(R) iscoarserthan the induced Euclidean topology). But, in adition, such map H is also open. Thisis because H maps a basic open subset (a, b) R into (H(a), H(b)), which is obviously a basic open subset in the order topology of H(R). Thus, H : (R, usual topology) (H(R), order topology) would be a topological homeomorphism. But, with the same arguments used in Remark 3.2 (a), this is plainly a contradiction,sincer is connected with respect to the usual Euclidean topology, and, however, H(R) is a subset of irrational numbers that has more than one element, so that it is not connected with respect to its order topology.

9 Order (2006) 23: Therefore, the set R \ Q of irrational numbers, endowed with its usual order and Euclidean topology, does not satisfy the Debreu property. 1 To see that R \ Q satisfies the lower Debreu property, let(y, Y ) beatotally preordered set endowed with its order topology. Let u : Y R \ Q be a strictly isotone map. In particular, considered as a function that takes values in the real line R, it is clear that u : Y R is a strictly isotone function (perhaps discontinuous!). Since, as a consequence of Debreu s gap lemma, the real line R, endowed with its usual order and Euclidean topology satisfies the Debreu property, there exists another strictly isotone function U : Y R that represents Y and is continuous with respecttothe ordertopology in Y andthe usualtopology in R. The composition v = h U, whereh is the map given in Proposition 3.1 is now a real-valued lower semicontinuous map that takes only irrational values. Since on R \ Q the induced Euclidean topology is the order topology, we conclude that R \ Q satisfies the lower Debreu property. To see that R \ Q satisfies the upper Debreu property, we can argue in an entirely analogous manner, now using, instead of h,themap j that appears in Remark 3.2 (c). Suppose now that we are interested in finding continuous or semicontinuous strictly isotone functions from a totally ordered set (Y, Y ) endowed with a topology τ Y (not necessarily a natural topology) that take only irrational values. The next result follows now as a consequence of Theorem 3.4. Corollary 3.5 Let (Y, Y ) be a totally preordered set endowed with a topology τ Y. Suppose also that Y is τ Y -lower semicontinuous (respectively, τ Y -upper semicontinuous) and order-separable in the sense of Debreu. Then there exists a lower semicontinuous (respectively, upper semicontinuous) strictly isotone function u : (Y, Y ) R that takes only irrational values. Proof Let θ Y be the order topology on Y associated to Y.Since Y is τ Y -lower (respectively, τ Y - upper semicontinuous), the identity map i : (Y,τ Y ) (Y,θ Y ) is lower semicontinuous (respectively, upper semicontinuous). Since Y is orderseparable in the sense of Debreu, using the Debreu s open gap lemma it follows that Y is representable through a continuous strictly isotone function u : (Y,θ Y ) R. Finally, the composition h u i : (Y,τ Y ) (R, usual Euclidean topology), (or, respectively, the composition j u i : (Y,τ Y ) (R, usual Euclidean topology)), where h is the map introduced in Proposition 3.1 (respectively, where j is the map introduced in Remark 3.2 (c)), is a lower semicontinuous (respectively, upper semicontinuous) strictly isotone function that also represents Y and takes only irrational values. Observe, however, that it is not true, in general, that any τ Y -continuous and order-separable total preorder Y defined on Y admits a continuous real-valued strictly isotone function taking only irrational values. Once more, the reason is that the set R \ Q does not satisfy the Debreu property. 1 An alternative proof of the fact that the irrationals do not have the Debreu property was communicated to the last author by Professor G. Herden (Essen, Germany) in October 2003.

10 352 Order (2006) 23: Definition Given a topological space (X,τ), the topology τ is said to satisfy the continuous representability property (CRP) if every τ-continuous total preorder defined on X admits a numerical representation by means of a continuous realvalued strictly isotone function. Also, the topology τ is said to satisfy the semicontinuous representability property (SRP) if every τ-upper (respectively τ-lower) semicontinuous total preorder defined on X admits a numerical representation by means of an upper (respectively lower) semicontinuous real-valued map. It is known that SRP implies CRP, but the converse is not true (See Proposition 4.4 in [7]). We can extend the definitions of CRP and SRP to the case of representations taking only irrational values, in the following sense: (a) The topology τ is said to satisfy the continuous representability property on irrationals (CRPI) if every τ-continuous total preorder defined on X admits a numerical representation by means of a continuous real-valued strictly isotone function that only takes irrational values. (b) The topology τ is said to satisfy the semicontinuous representability property on irrationals (SRPI) if every τ-upper (respectively τ-lower) semicontinuous total preorder defined on X admits a numerical representation by means of an upper (respectively lower) semicontinuous real-valued map that only takes irrational values. Now, we observe that it is not true that SRPI implies CRPI: It follows from the previous results that the usual Euclidean topology of R satisfies SRPI. However, it is straightforward to see that it does not satisfy CRPI, again because the set of irrationals fails to satisfy the Debreu property. Remark 3.6 In General Topology, a result in the context of metric spaces, known as the Alexandroff s lemma (see [1], p. 83), states that: << Let X be a nonempty set endowed with a complete metric d. LetY be a G δ subset (i.e.: a countable intersection of open subsets) of X. ThenY can be endowed with a complete metric d, perhaps different from d, such that d and d induce the same topology on Y. >> Since the set R \ Q of irrational numbers is a G δ subset of the real line R, which is a complete metric space with respect to the usual Euclidean topology, it follows that it can be given a complete metric, obviously different from the usual Euclidean one (that fails to be complete on R \ Q), such that this new metric also induces the usual Euclidean topology on R \ Q. As an immediate consequence of this fact and Theorem 3.4, we obtain an example of a totally ordered complete metric space that does not satisfy the Debreu property. 4 Debreu Properties of Certain Real Subsets Let us see now how we can use the results obtained in Section 3, as well as the explicit construction given there to get general results concerning the satisfaction of Debreu properties on certain families of real subsets. 2 2 Certain lines of enquiry pursued in this section have resulted from comments of an anonymous referee (although several proofs given in the present paper, based on direct constructions, are somewhat different from the referee s original ones).

11 Order (2006) 23: Lemma 4.1 (Cantor s characterization of the rationals) Let (X, ) be a totally ordered set which satisfies the following properties: (a) (b) (X, ) is order-unbounded and countable. (X, ) is dense-in-itself. Then (X, ) is order-isomorphic to the set Q of rational numbers, endowed with its usual order. Proof See e.g. [6], Theorem 23 on p Proposition 4.2 Let S be a real subset whose complement R \ S is countable and dense in R. Then, with respect to the usual Euclidean order, S is order-isomorphic to the set R \ Q of irrational numbers. Proof By Lemma 4.1, it follows that (R \ S, ) is order-isomorphic to (Q, ). Now we may apply the well-known argument of the classical proof of Lemma 2.1 (see [15]) in order to start with an order-isomorphism φ : (R \ S, ) (Q, ) and extend it to an order-isomorphism from (R, ) to (R, ). Then the restriction S : (S, ) (R \ Q, ) furnishes the desired isomorphism from S onto the irrationals. Corollary 4.3 Let S be a real subset whose complement R \ S is countable and dense in R. Then the following properties hold. (a) (b) (c) There exist order-monomorphisms α : (R, ) (S, ) and β : (R, ) (S, ) that are lower, respectively upper, semicontinuous with respect to the usual Euclidean topology on the real line R. (S, ) satisfies the lower and upper semicontinuous Debreu properties. (S, ) does not satisfy the Debreu property. Proof (a) Consider the composition α = i S φ h : R R where h : R R is the lower semicontinuous map that appears in the construction given in Proposition 3.1, so that, in particular, h(r) R \ Q ; φ : R \ Q S is an order-isomorphism (whose existence is guaranteed by Proposition 4.2), and i S : S R is the inclusion map. It is obvious that α is a lower semicontinuous order-monomorphism (simply, observe that the only Debreu gaps that α provokes in R are half-open half-closed, as those of the map h). Similarly, the composition β = i S φ j : R where j : R R \ Q is the upper semicontinuous map that appears in the construction given in Remark 3.2 (part iii) is also an upper semicontinuous order-monomorphism. (b) and (c) This is analogous to the proof of Theorem 3.4. The next Theorem 4.4. shows that the result in part iii) of Corollary 4.3 may still be improved, looking for a even bigger family of real subsets that do not satisfy the (continuous) Debreu property.

12 354 Order (2006) 23: Theorem 4.4 Let S be a real subset whose complement R \ SisdenseinR. Suppose that the real line R is order-embeddable in S (i.e.: there exists an order-monomorphism ψ : R S). Then (S, ) does not satisfy the Debreu property. Proof Suppose that the set S has the Debreu property with respect to its usual Euclidean topology inherited from R. This would imply the existence of a continuous order-embedding J : (R, ) (S, ). Again as in the proof of Theorem 3.4, if we consider the set J(R) S, endowed with the order topology, it is clear that the map J : R J(R) is bijective, continuous and open. Thus, J : (R, usual topology) (J(R), order topology) would be a topological homeomorphism. But this is a contradiction, since(r, ) is connected, and J(R) is not connected with respect to its order topology, because it has uncountably many points and its complement is dense in R. Remark 4.5 Theorem 4.4 completes the panorama of Proposition 2 in [14], that states that: << Given a dense-in-itself subchain (X, ) of the real line that contains a copy of the real line, it holds that (X, ) has the Debreu property if and only if the order topology on X that is induced by the Euclidean ordering is not totally disconnected >>. Notice that in Theorem 4.4 we do not ask the subchain to be dense-in-itself. Let us put an example of the situation considered in Theorem 4.4. To do so, we introduce some necessary facts. Definition (a) (b) An irrational number α R \ Q is said to be: Transcendental if it is not a root of a polynomials with rational coefficients, A Liouville number if, for every natural number n N, there exists a rational number p q Q with q 2 such that α p q < 1. q n It is well-known that Liouville s numbers are transcendental (see e.g. [10], pp ). However, not every transcendental number is a Liouville number. For instance π is a transcendental number but not a Liouville number, as proved in [17]. There are classical examples of Liouville numbers: For instance, if (a n ) {n 1, n N} is a sequence of digits such that a n {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} (n 1; n N) and, in addition it has a subsequence whose terms are all different from zero, then, as proved in [10], p. 105, the real number: z = a n 10 n! = 0.a 1 a 2 000a a is a Liouville number (actually, these were the first numbers that were proved to be transcendental by Liouville in 1844, see [16]), so that in particular it belongs to the set T of transcendental numbers. Also, if (b n ) {n 1, n N} is a sequence of numbers such that b n {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} (n 1; n N) and, in addition it has a subsequence whose terms are all different from 9, then the real number: w = 0.b 1 b 2 999b b = k=1 α k 10 k, where α k = b n if k = n! and α k = 9 otherwise, is also a Liouville number. Arguing as in Proposition 3.1 or Remark 3.2 (c) we can also obtain semicontinuous strictly isotone representations with values in the set T of transcendental numbers, using a suitable explicit construction. In addition, T does not have the Debreu

13 Order (2006) 23: property. The conventions here are as in the proof of Proposition 3.1. Now, given r = N + + a n (r R) we define H(r) = N + + a n 10 n (r R). As in Remark 10 n! 3.2 (b), this map H : R R is lower semicontinuous but not upper semicontinuous, strictly increasing, and takes values of the set of Liouville transcendental numbers. In the same way, but using now the conventions of Remark 3.2 (c), given r = N + + a n, (r R) we define J(r) = N + + b n 10 n (r R), where b 10 n n = a k if n = k! for some k N; b n = 9 otherwise. Now, the map J : R R is upper semicontinuous but not lower semicontinuous, strictly increasing, and takes values of the set of transcendental numbers. Example 4.6 Similarly to what happens for the set R \ Q of irrational numbers, the complement of the set T of transcendental numbers is countable (see e.g. [10], p. 103) and dense (because it contains the set Q of rational numbers). This set R \ T is also known as the set A of real algebraic numbers. Therefore, by Corollary 4.3, it follows that : <<The set T of transcendental numbers satisfies the semicontinuous Debreu properties, but not the (continuous) Debreu property>>. If we consider just the set of Liouville numbers L instead of the whole set T of transcendental numbers, we cannot make use of Corollary 4.3, since the Lebesgue measure and the Hausdorff dimension of the set L of Liouville numbers are equal to zero (see Ch. 2 in [19]),sothatitscomplementD = R \ L, calledthesetof diophantine numbers, is, in particular, dense in R and uncountable. But now, instead of Corollary 4.3 we can use Theorem 4.4 to conclude that: <<The set L of Liouville numbers does not satisfy the Debreu property>>. Notice, in addition, that by the above construction it follows that: << The set L of Liouville numbers satisfies both the lower and upper semicontinuous Debreu properties>>. In general, we find a difficulty if we try to apply Theorem 4.4 to concrete real subsets S. Notice that we must have at hand an order-embedding of the real line R into the given subset S. Although in the present paper we have considered two concrete cases, namely S = R \ Q and S = L, where we can explicitly exhibit the embedding, in other situations such kind of embeddings are not easy to be found, or even to be proved to exist. Let us give now a third example, that shows another situation where a direct construction can be given. Example 4.7 (Cantor ternary set) If we consider the closed interval [0, 1] in R, then we remove the open middle third ( 1 3, 2 ), and then we remove the open middle thirds 3 ( 1 9, 2 9 ) and ( 7 9, 8 ) of the two remaining intervals, and then we remove the open middle 9 thirds of the four remaining intervals; and so on, indefinitely, what remains is the Cantor ternary set C. It can be described as consisting precisely of those numbers of [0, 1] that have expansions in base 3 containing no 1s. To define an order-embedding from R into the set C we first map R into (0, 1) by means of the strictly increasing function μ : R R given by μ(x) = ex (x R). Then we interpret each number of e x +1 (0, 1) in base 2, using an expansion with 0 s and 1 s. Observe that there are some real

14 356 Order (2006) 23: numbers that admit two possible such expansions, for instance: (1)...= (1)... (in base 2). In these situations, we shall only use the expansion in which the periodic digit is 1. Finally, we define an order-embedding ρ : (0, 1) C such that ρ( + a n ) = ( + 2a n 2 n ) (where a 3 n n {0, 1}, n N). The composition ρ μ is the desired order-embedding from the real line R into the Cantor set C. Itis well-known that the Cantor set is an uncountable Borel set whose Lebesgue measure is equal to zero. However, its Hausdorff dimension is not zero but log 2, so that the log 3 analytical structure of the Cantor set C is different from that of the set L of Liouville numbers, which is also a Borel set. The complement R \ C of the Cantor ternary set is uncountable and dense in R. Thus, by Theorem 4.4, it follows that: << The Cantor set C does not satisfy the Debreu property>>. In addition, as in Proposition 3.1 and part (b) of Remark 3.2, it is straightforward to see that the composition ρ μ is lower semicontinuous. Suppose now that for real numbers in [0, 1] that admit two different expansions in base 2, we decide now to use only the expansion in which the periodic digit is 0. In this case, the composition ρ μ is upper semicontinuous, asin part (c) ofremark 3.2. Consequently, we also obtain that: << The Cantor set C satisfies both the lower and upper semicontinuous Debreu properties>>. To conclude our study, it is worthwhile to furnish an example of a real subset that does not satisfy any of the semicontinous Debreu properties. This example is important in order to say that the semicontinuous Debreu properties are non trivial, i.e., it is not true that every real subset satisfies both of them. Example 4.8 Let X ={ n 1 : n > 1, n N} { n+1 : n > 1, n N} R, endowed n n with the usual order and the discrete topology (indeed that is the induced Euclidean topology on X). Let us see that X fails to satisfy the lower Debreu property. To see this, let Y ={ n 1 : n > 1, n N} {2}. n Let us prove now that every strictly increasing map g : Y X has half-closed half-open gaps: For any n > 1, n N there must exist m(n) >n, m(n) N such that g( n 1 n ) = m(n) 1. Also, there must exist r > 1, r N such that g(2) = r+1. All this follows m(n) r from a cardinality argument: If for some k > 1, k N it holds that g( k 1 k ) = q+1 q, then, as there are infinitely many elements of Y that are bigger than k 1 and g is k increasing, we would obtain infinitely many terms of X that are bigger than q+1 q, but this is impossible (notice that only finitely many such elements are available). In the same way it is proved that g(2) cannot be of the form n 1 for any n > 1, n N. We n conclude that g gives raise to the Debreu gap [1, g(2)) that is half-closed half-open. To prove that X fails to satisfy the upper Debreu property we proceed in an entirely analogous manner, now taking Z ={0} { n+1 : n > 1, n N} and proving that n every strictly increasing map h : Z X has half-open half-closed gaps. Acknowledgements We want to express our gratitude to an anonymous referee, to whom we are indebted for the detailed reading of the manuscript and his/her very helpful suggestions and ideas.

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