On a Possible Continuous Analogue of the Szpilrajn Theorem and its Strengthening by Dushnik and Miller

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1 Order (2006) 23: DOI /s On a Possible Continuous Analogue of the Szpilrajn Theorem and its Strengthening by Dushnik and Miller Gianni Bosi Gerhard Herden Received: 28 February 2005 / Accepted: 22 November 2006 / Published online: 16 January 2007 Springer Science + Business Media B.V Abstract The Szpilrajn theorem and its strengthening by Dushnik and Miller belong to the most quoted theorems in many fields of pure and applied mathematics as, for instance, order theory, mathematical logic, computer sciences, mathematical social sciences, mathematical economics, computability theory and fuzzy mathematics. The Szpilrajn theorem states that every partial order can be refined or extended to a total (linear) order. The theorem by Dushnik and Miller states, moreover, that every partial order is the intersection of its total (linear) refinements or extensions. Since in mathematical social sciences or, more general, in any theory that combines the concepts of topology and order one is mainly interested in continuous total orders or preorders in this paper some aspects of a possible continuous analogue of the Szpilrajn theorem and its strengthening by Dushnik and Miller will be discussed. In particular, necessary and sufficient conditions for a topological space to satisfy a possible continuous analogue of the Dushnik-Miller theorem will be presented. In addition, it will be proved that a continuous analogue of the Szpilrajn theorem does not hold in general. Further, necessary and in some cases necessary and sufficient conditions for a topological space to satisfy a possible continuous analogue of the Szpilrajn theorem will be presented. Key words weakly continuous binary relation Szpilrajn property κ-attracting set Mathematics Subject Classifications (2000) 54F05 91B16 06A05 G. Bosi Dipartimento di Matematica Applicata Bruno de Finetti, Università di Trieste, Piazzale Europa 1, Trieste, Italy G. Herden (B) Universität Duisburg-Essen, Campus Essen, Fachbereich 6 (Mathematik) Universitätsstraße 3, Essen, Germany gerhard.herden@uni-due.de

2 272 Order (2006) 23: Introduction Let (X, ) be an arbitrary non-empty partially ordered set. Then the classical Szpilrajn theorem [53] states that has a total (linear) refinement, i.e. there exists a total order on X such that x y whenever x y or, equivalently,. Ifa preorder on X is given one may apply the Szpilrajn theorem in order to get a total refinement of by a total preorder, which means that and < (as usual, we denote by the strict part of a preorder, and by < the strict part of a total preorder ). Indeed, in this case we choose the set X := {[x] x X} of indifference (equivalence) classes in order to get with the help of the Szpilrajn theorem a total order on X that refines. Then, by setting x y [x] [ y ] for all points x, y X, the desired linear refinement of has been defined. Despite of the fact that the usual term used in the literature is extension,weprefer to speak of a refinement of a preorder by a total order or preorder. Indeed, the concept of an extension of a preference relation was already used by Yi [57] in a different context and in addition this concept also is of some interest in this paper (cf. Proposition 4.3). Further, we have also adopted this terminology in a paper which was recently published in this journal and which concerns the case of orders and not of preorders like in the present paper (see Bosi and Herden [5]). Different proofs of the Szpilrajn theorem have been presented, for instance, by Erné [19, Satz and Korollar 10.14], Fishburn [23, pp ] and [24, pp ], Abian [1], Bondesen [4] and Fay [21] who simplified the proof of Bondesen somewhat by applying the Lemma of Teichmüller and Tuckey (cf., for instance, the book of Hajnal and Hamburger [28]). For further details the reader may consult the introduction in Bosi and Herden [5]. The original proof of Szpilrajn uses Zermelo s Well-Ordering theorem which is equivalent to the Lemma of Zorn. The strengthening of the Szpilrajn theorem by Dushnik and Miller [18] states, moreover, that every partial order on X is the intersection of its total (linear) refinements. Of course, also the Dushnik-Miller theorem immediately can be generalized to arbitrary preorders on X. The Szpilrajn theorem and its strengthening by Dushnik and Miller or the more recent generalizations by Morgado [40], Wu [56], Donaldson and Weymark [15], Bossert [6], Duggan [17], Lahiri [37] and others belong to the most quoted theorems in order theory, mathematical logic, computer sciences, mathematical social sciences, mathematical economics, computability theory, fuzzy mathematics and other fields in pure and applied mathematics. Their relevance in order theory is well known. In the context of algebraic structures we want to mention the papers by Kontolatou and Stabakis [36] and Martínez-Legaz and Singer [39] and the more recent paper by Fishburn [25] where possible generalizations of the Szpilrajn theorem to additive partial orders arediscussed.felgner and Truss [22] consider the Szpilrajn property or order-extension property, which means that every partial order can be extended or refined to a total order in order to discuss the question if the Szpilrajn property implies the prime ideal theorem which states that every Boolean algebra has a prime ideal. In Petri s concurrency theory one usually works with a discrete version of Dushnik s and Miller s strengthening of the Szpilrajn theorem. In their paper on modelling concurrency Janicki and Koutny [35, Theorem 2.9] characterize extension complete order structures and, therefore, generalize Dushnik s and Miller s strengthening of the Szpilrajn theorem. Stehr [51] applies the Szpilrajn theorem in

3 Order (2006) 23: his characterization of global orientability. In addition, the Szpilrajn theorem and generalizations of it are of importance in social choice theory (cf., for instance, the paper by Nehring and Puppe [43] on a unifying structure of abstract choice theory or Sholomov [48] who uses an analogue of the Szpilrajn theorem in order to characterize ordinal relations and Sprumont [49] in his interesting study of quasiorders). Diaye [14, Lemma 1 and Proposition 1 of Section 3] applies the Szpilrajn theorem in order to prove a nice result on the representation of acyclic relations. Weymark [55] applies Dushnik s and Miller s strengthening of the Szpilrajn theorem in order to prove a generalization of Moulin s Pareto Extension theorem [41]. In the paper of Suck [52] the Szpilrajn theorem is used in order to prove a theorem on the applicability of conjoint structures. Roubens and Vincke [45] presentand discuss the Szpilrajn theorem in their book on preference modelling. In computability theory Roy [46] proved, for instance, the recursive version of the Szpilrajn theorem which states that every recursive partial order has a recursive total refinement or extension. Downey [16, Section 6] considers the effective content of the Szpilrajn theorem in order to approach the theory of dimension for partially ordered sets (cf. also Rozen [47]). Finally, we still want to mention the papers by Zadeh [58], Hashimoto [29], Wang [54], Ovchinnikov [44], Blanchard [2, 3] and Höhle and Blanchard [33]where the properties of fuzzy relations are discussed. In all these papers fuzzy versions of theorems of Szpilrajn type are proved. In case that X is endowed with some topology t one is mainly interested in continuous total orders or preorders instead of only total orders or preorders. A total order or preorder on (X, t) is said to be continuous if for every point x X both sets d(x) := {y X y x} and i(x) := {z X x z} are closed subsets of X. In any theory that combines the concepts of topology and order, therefore, the questions arise if there exist continuous analogues of the Szpilrajn theorem and its strengthening by Dushnik and Miller. In order to be as general as possible we do not restrict our discussion of a possible continuous analogue of the Szpilrajn theorem to orders or preorders but also consider arbitrary binary relations R on X. Sincethe intersection of preorders also is a preorder, of course, a continuous analogue of the Dushnik-Miller theorem cannot hold for arbitrary binary relations on X but only for preorders on X. Surely, one cannot expect that every binary relation R on X can be refined to a continuous total order or preorder. In the next section (cf. Lemma 2.1) we, therefore, must determine at first a necessary condition for the existence of a continuous total refinement of a binary relation R on (X, t). In the remainder of this paper binary relations that satisfy such a condition are said to be weakly continuous. Then three natural problems have to be discussed: (P1) (P2) (P3) Let R be a weakly continuous binary relation on (X, t). Determine necessary and sufficient conditions for R to have a continuous total refinement, which means that there exists some continuous total preorder on (X, t) such that R and R S := {(x, y) R (y, x) R} <. Determine necessary and sufficient conditions for a topology t on X to have the Szpilrajn property, which means that every weakly continuous binary relation R on X has a continuous total refinement. Determine necessary and sufficient conditions for a topology t on X to have the Dushnik-Miller property, which means that every weakly continuous preorder on X is the intersection of its continuous total refinements.

4 274 Order (2006) 23: Problem (P1) already has been discussed extensively in Herden and Pallack [32]. Therefore we concentrate in this paper solely on Problem (P2) and Problem (P3). The semicontinuous analogue of Problem (P1) has been discussed in some degree by Jaffray [34] and more recently by Bossert, Sprumont and Suzumura [7] and Herden and Mehta [30]. We recall that Theorem 1 in the paper by Bossert, Sprumont and Suzumura was not enunciated in a correct way (see Bosi and Herden [5]). In mathematical utility theory problems of Szpilrajn-type seem to be of particular importance. Indeed, problems of Szpilrajn-type include the general continuous representation problem. In order to somewhat discuss this problem we consider the views of Chipman [10, cf. pp and pp ] that there are strong reasons for not choosing the real line as the codomain of a utility function on a preordered set because utility is inherently of lexicographic nature. Indeed, in a sense, the nonexistence of a real valued utility function reflects in some degree the properties of a given binary relation, but is also partly consequence of a probably injudicious choice of the codomain. In the opinion of the authors the choice of an appropriate codomain of a (continuous) utility function depends on the particular situation to be considered. In a very general sense, the problem concerning the existence of a continuous utility representation for a binary relation may be described as follows. Let R be a weakly continuous binary relation on a topological space (X, t). Further, assume that there exists some topological space (Y, t ) that is endowed with a continuous total preorder. If there exists some continuous strictly increasing or, equivalently, order preserving function f : (X, R, t) (Y,, t ),then f maybereferredtoas a continuous utility representation of the binary relation R on (X, t). One easily verifies that the problem of the existence of (Y,, t ) is equivalent to Problem (P1). Furthermore, a solution of Problem (P2) provides necessary and sufficient conditions for a topology t on X to immediately decide if for any weakly continuous binary relation R on (X, t) there exists a totally preordered topological space (Y,, t ) such that a continuous order preserving representation f : (X, R, t) (Y,, t ) is possible. In addition, a solution of Problem (P3) provides necessary and sufficient conditions for a topology t on X to immediately decide if any weakly continuous preorder on (X, t) is the intersection of its continuous total refinements on (X, t). Therefore, continuous Szpilrajn-type theorems are also within the main stream of efforts that have been made in order to clarify the difference between preference and utility (cf. Herden and Pallack [32]). Theorem 3.2 characterizes the structure of topological spaces (X, t) that have the Dushnik-Miller property. On the other hand, at present the authors are not able to completely solve Problem (P2) in a satisfactory way. Nevertheless, in this paper conditions that are necessary and in some relevant cases also sufficient for a topological space (X, t) to have the Szpilrajn property will be presented. The main tool for proving these results is the concept of a κ-attracting subset of a topological space (X, t). This concept will be introduced in Section 6 of this paper. With its help in Section 7 and Section 8 of this paper the Szpilrajn property of topological spaces that are limit point compact or compact or connected canbestudied.werestrict our considerations on limit point compact or compact or connected topological spaces since compact or connected topological spaces are of particular interest in mathematical utility theory. In addition, in many cases the assumption (X, t) to be compact can be replaced by the weaker assumption (X, t) to be limit point compact.

5 Order (2006) 23: We want to finish this introduction by shortly discussing some general aspects of the problems that are considered in this paper. Therefore, we choose a class S of structures and classes R 1, R 2 of (binary) relations. The particular selection of S as well as the particular selections of R 1 and R 2 may vary. In this paper S consists of topologies on an arbitrarily chosen non-empty set X and R 1 and R 2 are particular sets of binary relations on X. In addition, we assume that for n = 1 and n = 2 and the pair (1, 2) incidence relations I n S R n and I 1,2 R 1 R 2 are given. Then the following four problems are fundamental in any theory that studies (topological) structures that are endowed with some binary relation. (R1) Let (s, R) S R 1 be given. Determine necessary and sufficient conditions that guarantee that (s, R) I 1. (R2) Determine all structures s S such that (s, R) I 1 for every R R 1. (R3) Let (s, R) I 1 be given. Determine necessary and sufficient conditions that guarantee the existence of some T R 2 such that (R, T) I 1,2 and(s, T) I 2. (R4) Determine all structures s S such that for all R R 1 for which (s, R) I 1 there exists some T R 2 such that (R, T) I 1,2 and (s, T) I 2. The reader will have no difficulties in order to verify that the problems (P1), (P2) and (P3) can be subsumed under the problems (R3) and (R4). Moreover, the authors are convinced that every problem that asks for the interrelations between topology and order, actually, can be subsumed under one of the problems (R1), (R2), (R3) or (R4). Indeed, in the opinion of the authors, who follow the spirit of Nachbin [42], any theory of topology and order has to characterize topologies that satisfy compatibility criteria with respect to binary relations. This means that any theory of topology and order studies the structure of topologies with the help of their behaviour with respect to binary relations. A theory that studies, in contrast, the structure of lattices or, more generally, partially ordered sets with the help of topologies as, for instance, the Scott topology or the Lawson topology (cf. [26, 27]) is of different nature. 2 Weakly Continuous Binary Relations Let (X, R, t) be some R-space, i.e. a topological space that is, in addition, endowed with an arbitrary binary relation R on X. A real-valued function f on X is said to be increasing if xry implies that f (x) f (y). If, in addition, f (x) < f (y) whenever xr S y then f is said to be strictly increasing or order preserving. Lemma 2.1 and Lemma 2.2 in Herden and Pallack [32] imply the following lemma which motivates our concept of a weakly continuous binary relation R on (X, t). Lemma 2.1 The following assertions hold: (a) (b) In order for R to have a continuous total refinement it is necessary that for every pair (x, y) R S there exists some continuous increasing real-valued function f xy on X such that f xy (x) < f xy (y). In order for a total preorder on X to be continuous it is necessary and sufficient that for every pair (x, y) < there exists some continuous increasing real-valued function f xy on X such that f xy (x) < f xy (y).

6 276 Order (2006) 23: Because of Lemma 2.1 R is said to be weakly continuous if for every pair (x, y) R S there exists a continuous increasing real-valued function f xy on X such that f xy (x) < f xy (y). In order to illustrate the concept of a weakly continuous binary relation R on (X, t) the reader may recall from the introduction that a total preorder on (X, t) is continuous if and only if for every point x X both sets d(x) and i(x) are closed subsets of X or, equivalently, if and only if the order topology t on X that is induced by is coarser than t. Let, therefore, R be an arbitrary binary relation on X. Then a subset S of X is said to be decreasing if y S and xry imply that x S. By duality one defines the concept of an increasing subset T of X. For every point x X we set d(x) := {S S is a decreasing subset of X that contains x}, i(x) := {T T is an increasing subset of X that contains x}, l(x) := {S S is a decreasing subset of d(x) that does not contain x}, r(x) := {T T is an increasing subset of i(x) that does not contain x}. Then { } {l(x) x X} {r(x) x X} is a sub-basis of a topology t R on X. Clearly, t coincides with t R if R = is a preorder on X. In this case we also have that d(x) := {y X y x} and i(x) := {z X x z} (see the definition of d(x) and i(x) presented in the introduction). Now the authors suggest to define an arbitrary binary relation R on X as being continuous if R is acyclic, t R is coarser than t and if for every point x X both sets d(x) and i(x) are closed subsets of X. The following lemma, the proof of which is implicitly in the proof of Proposition 2.8 in Herden and Pallack [32], justifies our concept of a weakly continuous binary relation R on (X, t). Lemma 2.2 Let R be a continuous binary relation on (X, t). Then R is weakly continuous. The converse of Lemma 2.2 does not hold. Indeed, in Example 2.9 in Herden and Pallack [32] a topological space (X, t) has been constructed that is endowed with some weakly continuous preorder such that t is not coarser than t. In addition, there exist points x, y X suchthat neither d(x) nori(y) is a closedsubsetof X. Let t prod := t t be the product topology on X X that is induced by t. Thena binary relation R on X is said to be closed if R is a closed subset of (X X, t prod ). It follows from the proof of Proposition 1 in Nachbin [42] that for a closed preorder on (X, t) and every point x X both sets d(x) and i(x) are closed subsets of X.Itis well known that the converse does not hold. Now the following lemma shows that our approach is not as general as it seems to be at first sight (cf. Herden and Pallack [32, Proposition 2.11]). Lemma 2.3 Every weakly continuous binary relation R on (X, t) has a weakly continuous refinement by a closed preorder.

7 Order (2006) 23: Proof Let R be a weakly continuous binary relation on X and let t nat be the natural topology on the real line. Then we denote by F(R) the set of all continuous increasing functions f : (X, R, t) (R,, t nat ) and set := {(x, y) X X f F(R) (f (x) f (y))}. Clearly, is a weakly continuous preorder on X such that R and R S. Hence, it suffices to verify that is closed. Let, therefore, (u,v) be a pair of X X that is not contained in. Then there exists some function f F(R) such that f (u) > f (v). The continuity of f guarantees the existence of neighbourhoods U of u and V of v such that f (s) > f (t) for every pair (s, t) U V. Thismeansthat (U V) = and the desired conclusion follows. Remark 2.4 Lemma 2.3 implies that in order to solve Problem (P2) we may restrict attention to weakly continuous preorders on (X, t). In addition, it follows that in order to discuss Problem (P2) we may assume in the remainder of this paper that for every point x X both sets d(x) and i(x) are closed subsets of X. In order to complete Lemma 2.3 we want to finish this section by describing situations in which a closed preorder on (X, t) already is weakly continuous (cf. Herden and Pallack [32, Proposition 2.12]). Proposition 2.5 Let (X, t) be a compact (Hausdorff-)space or a locally compact second countable (Hausdorff-)space that is endowed with some closed preorder. Then is weakly continuous. Proof If (X, t) is a compact (Hausdorff-)space then the desired conclusion follows with the help of the arguments that are applied in the proofs of Propositions 1, 4 and 5 and Theorem 4 in Nachbin [42]. The reader may notice that the proofs of these results do not depend on the assumption to be an order. If (X, t) is a locally compact second countable (Hausdorff-)space one may consult, for instance, the first part of the proof of Lemma in Bridges and Mehta [8]. This part of the proof is essentially based upon the compact version of Nachbin s Lifting theorem (cf. Nachbin [42, Theorem 6]). Also the proof of this theorem does not depend on the assumption to be an order. A weaker version of this theorem was proved by Levin [38]. 3 A Continuous Analogue of the Dushnik-Miller Theorem Let (X, t) be an arbitrarily chosen topological space. The reader may recall from the introduction that t is said to have the Dushnik-Miller property if every weakly continuous preorder on (X, t) is the intersection of its continuous total refinements. Let, in addition, A be an arbitrary subset of X. We recall, moreover, that a point x X is a limit point of A if U \{x} A = for every open neighborhood U of x. In order to characterize all topological spaces (X, t) that have the Dushnik- Miller property we consider the set C(X) := C(X, R) of all continuous real-valued functions f on X and choose the weak topology σ(x, C(X)) on X.Hereσ(X, C(X))

8 278 Order (2006) 23: is the coarsest topology on X for which all functions f C(X) are continuous. Denote by (X,σ(X, C(X)) ) the quotient space of (X,σ(C(X))) that is induced by the equivalence relation x y f C(X) (f (x) = f (y)). Then we consider the Hausdorff-space (X,σ(X, C(X)) ) that is associated with (X,σ(C(X))). Itis well known that (X,σ(X, C(X)) ) is a completely regular Hausdorff-space (cf., for instance, Cigler andreichel [11, Satz 10, page 101]). Hence, the following lemma, which is an immediate consequence of the definition of a weakly continuous preorder on (X, t), reduces the study of arbitrary topological spaces that have the Dushnik- Miller property to the study of completely regular Hausdorff-spaces that have the Szpilrajn property. Lemma 3.1 In order that t has the Dushnik-Miller property it is necessary and sufficient that σ(x, C(X)) has the Dushnik-Miller property. With the help of Lemma 3.1 we are already able to completely characterize all topological spaces (X, t) that have the Dushnik-Miller property. Theorem 3.2 In order that (X, t) has the Dushnik-Miller property it is necessary and sufficient that σ(x, C(X)) is the discrete topology on X. Proof The sufficiency part of the theorem follows with the help of Lemma 3.1 from the generalization of the original theorem of Dushnik and Miller to arbitrary preorders that already has been mentioned in the introduction. Therefore, we may concentrate on the necessity part of the theorem. In order to verify the necessity part of the theorem it suffices to prove that the Dushnik-Miller property implies that (X,σ(X, C(X)) ) does not contain any limit points. Let us assume, in contrast, that x is a limit point of (X,σ(X, C(X)) ).Sinceσ(X, C(X)) is a completely regular Hausdorff-topology on X there exists for every point y X \{x} a continuous function f y : (X,σ(X, C(X)) ) ([0, 1], t nat ) such that f y (y) = 0 and f y (x) = 1. This means that the preorder on X that is defined by := {(ν, ν) ν X} {(y, x) y X \{x}} is a weakly continuous preorder on (X,σ(X, C(X)) ).Since(X,σ(X, C(X)) ) has the Dushnik-Miller property it has, in particular, the Szpilrajn property. Hence, there exists a continuous total preorder on (X,σ(X, C(X)) ) such that and <. The definition of implies that x is the greatest element of. Nowwe choose the preorder on X that is given by := \{(y, x) y X \{x}}. The continuity of allows us to conclude that, actually, is a weakly continuous preorder on (X,σ(X, C(X)) ).Sincexis a limit point of (X,σ(X, C(X)) ) that, in addition, is the greatest element of it follows with the help of the continuity of that is the unique total refinement of by a continuous total preorder on (X,σ(X, C(X)) ). Hence, the existence of a limit point x contradicts the Dushnik- Miller property of (X,σ(X, C(X)) ). This conclusion finishes the proof of the theorem.

9 Order (2006) 23: Remark 3.3 Theorem 3.2 has the consequence that the existence of a useful continuous analogue of the Dushnik-Miller theorem can be excluded. 4 Fundamental Aspects of a Continuous Analogue of the Szpilrajn Theorem Let, as in the previous section, (X, t) be an arbitrarily chosen topological space. Of course, the considerations of the previous section also apply for the Szpilrajn property. This means, in particular, that the following lemma holds. Lemma 4.1 In order that t has the Szpilrajn property it is necessary and sufficient that σ(x, C(X)) has the Szpilrajn property. The following necessity result even reduces the study of topological spaces that have the Szpilrajn property to the study of normal Hausdorff-spaces that have the Szpilrajn property. Proposition 4.2 In order that t has the Szpilrajn property it is necessary that (X,σ(X, C(X)) ) is a normal Hausdorff-space. Proof Let A, B be two disjoint closed subsets of (X,σ(X, C(X)) ).Thenwemay assume without loss of generality that X \ (A B) =. Otherwise, both sets A and B are open and closed and nothing remains to be shown. Let, therefore, some point z X \ (A B) be arbitrarily chosen. Then we define a preorder on X by setting := {(ν, ν) ν X } {(x, z) x A} {(z, y) y B} {(x, y) x A and y B}. Since (X,σ(X, C(X)) ) is a completely regular Hausdorff-space there exist for any two points x A and y B continuous functions f x : (X,σ(X, C(X)) ) ([0, 1], t nat ) and f y : (X,σ(X, C(X)) ) ([0, 1], t nat ) such that f x (x) = 0 and f x (B {z}) ={1} and f y (A {z}) ={0} and f y (y) = 1. These properties of f x and f y, respectively, imply that both functions f x and f y are increasing and that f x (x) < f x (z) f x (y) and f y (x) f y (z) < f y (y) for all points x A and y B. Hence,it follows that is a weakly continuous preorder on (X,σ(X, C(X)) ). Let now be some continuous total preorder on (X,σ(X, C(X)) ) such that and <. Then l(z) := {u X u < z} and r(z) := {v X z <v} are disjoint open subsets of X that contain A and B, respectively. Proposition 4.2 implies, in particular, that every completely regular Hausdorffspace that is not normal (cf., for instance, the famous Niemytzki plane (Steen and Seebach [50, Example 82]) does not have the Szpilrajn property. It has been mentioned in the introduction that Yi [57] studied topological spaces (X, t) that have the extension property, i.e. every total preorder C that is defined on a closed subset C of X and is continuous with respect to the relativized topology t C on C can be extended to a continuous total preorder on X.

10 280 Order (2006) 23: Obviously, a topological space (X, t) that has the extension property must be normal. Indeed, let A, B be two disjoint closed subsets of X. Then one only has to extend the t A B -continuous total preorder A B on A B that is defined by A B := {(ν, ν) ν A B} {(x, x ) x A and x A} {(y, y ) y B and y B} {(x, y) x A and y B} to a continuous total preorder on (X, t) in order to then apply the same argument as in the proof of Proposition 4.2. Because of Proposition 4.2 the following proposition, therefore, describes in some degree the connection between the Szpilrajn property and the extension property. Proposition 4.3 Let (X, t) be a normal space. Then, if t has the Szpilrajn property, (X, t) also has the extension property. Proof Let C be a closed subset of X that is equipped with some t C-continuous total preorder C. Then for any two points x < C y C there exists a continuous increasing function f xy : (C, C, t C) (R,, t nat ) such that f xy (x) < f xy (y).the normality of (X, t) allows us to apply the classical Tietze-Urysohn Extension theorem in order to conclude that for any two points x < C y C there exists some continuous function g xy : (X, t) (R,, t nat ) that extends f xy. This means, in particular, that the relation := {(ν, ν) ν X} C is a weakly continuous preorder on (X, t). Since t has the Szpilrajn property the desired conclusion, therefore, follows. We want to finish this section by presenting some examples of topological spaces (X, t) that have the Szpilrajn property. We recall that a topological space (X, t) is said to be hereditarily Lindelöf if for every collection {O i } i I of open subsets of X there exists a countable subset J of I such that O i = O j. Let now H be the i I j J class of all topological spaces (X, t) for which the product space (X X, t prod ) := (X X, t t) is hereditarily Lindelöf. Clearly, H includes the class of second countable topological spaces. The proof of Theorem 2.15 in Herden and Pallack [32] implies that every topological space (X, t) H has the Szpilrajn property. Since the Szpilrajn theorem implies that every set X that is endowed with the discrete topology has the Szpilrajn property there also exist topological spaces (X, t) that have the Szpilrajn property and are not contained in H. Nontrivial topological spaces (X, t) that have the Szpilrajn property and are not necessarily contained in H are given by the following examples that will be used in the next section. Let κ be an arbitrary infinite regular cardinal number and let X κ be the set of all ordinal numbers α κ. For every ordinal number α<κwe set I α := {β X κ α<β}.then b := {{α} α X κ \{κ}} {I α α X κ \{κ}} is a basis for a topology t on X κ. Clearly, (X κ, t) H if and only if κ is a countable cardinal number, i.e. (X κ, t) is second countable. Now the following proposition holds. Proposition 4.4 (X κ, t) has the Szpilrajn property. Proof Since the proof of the proposition does not depend on the cardinality of κ but only on the regularity of κ we may assume without loss of generality κ to be the first uncountable cardinal number. Indeed, in Example 5.3 only the cases κ to be the

11 Order (2006) 23: first infinite or the first uncountable cardinal number will be considered. Let be a weakly continuous preorder on (X κ, t) such that for every point x X κ both sets d(x) and i(x) are closed subsets of X κ. Because of Remark 2.4 it suffices to verify that can be refined to a continuous total preorder on (X κ, t). Therefore, we set [κ] :=d(κ) i(κ). Then we choose for every point y X κ \[κ] the set d(y) if κ d(y). Otherwise, we choose the set i(y).inthiswayx κ \[κ] has been partitioned into two disjoint subsets Y and Z such that [κ] d(y) = for every point y Y and [κ] i(z) = for every point z Z. In the remainder of the proof both sets Y and Z will be considered separately. Hence, we may assume without loss of generality that none of the sets Y or Z is empty. With the help of the partition of X κ \[κ] into the sets Y and Z we, therefore, may refine to a preorder on X κ that is defined by := {(y,ν) y Y and ν [κ]} {(ν, z) z Z and ν [κ]} {(y, z) y Y and z Z }. The construction of Y and Z, respectively, allows us to immediately conclude that, actually, is a weakly continuous preorder on (X κ, t) that refines. LetY := Y [κ] and Z := Z [κ]. We want to show that there exists a continuous total preorder d(κ) on (Y, t Y ) that refines the restriction Y of to Y. Then it follows by duality that there exists a continuous total preorder i(κ) on (Z, t Z ) that refines the restriction Z of to Z.SinceX κ = Y Z we, therefore, may conclude that the transitive closure of d(κ) i(κ) is the desired continuous total refinement of. In order to finish the proof of the proposition it, thus, remains to prove the existence of d(κ). Therefore, we consider in the remainder of the proof for every point y Y the set d(y) with respect to. Nevertheless, the reader may notice that for every point y Y the sets d(y) that have been chosen with respect to coincide with the sets d(y) that have been chosen with respect to. Letx 0, x 1,..., x α,... be an arbitrary well ordering of Y. In order to avoid artificial considerations we may assume without loss of generality that α runs through all countable ordinal numbers. Then we construct d(κ) by transfinite induction on all countable ordinal numbers α. α = 0: Inthiscasewesetd (0) = d(x 0 ). It follows from the definition of Y that d (0) is a countable set. Then, as it has been demonstrated in the introduction, we may apply the original Szpilrajn theorem in order to guarantee the existence of a total preorder 0 on d (0) that refines d (0). α>0 is not a limit ordinal. We may assume that d (α 1) is a countable set and that d (0) d (α 1). Letγ α be the minimum of all ordinal numbers β such that x β d (α 1). Then we set d (α) := d (α 1) d(x γα ) and use the Szpilrajn theorem in order to verify the existence of a total preorder γα on d(x γα ) \ d (α 1) that refines d(xγα )\d (α 1). Now α is defined by α 1 γα {(y, z) y d (α 1) and z d (α) \ d (α 1)}. Obviously, α is a total refinement of d (α). In addition, since d(x γα ) is a countable set also d (α) is a countable set. α>0 is a limit ordinal. In this case we set d (α) := d (β) and α := β. β<α β<α Clearly, d (α) is a countable set and α a total preorder on d (α) that refines d (α). Finally, we set d(κ) := α {(y,ν) y Y and ν [κ]}. Sinceκ is the first α<κ uncountable cardinal number, i.e. an uncountable regular cardinal number, there exists no countable ordinal number α such that κ d (α). Hence, the construction of d(κ) implies with the help of the definition of t that d(κ) is a continuous total

12 282 Order (2006) 23: preorder on (Y, t Y ) that refines d(κ). This last conclusion finishes the proof of the proposition. Remark 4.5 The topological spaces (X κ, t) are not so artificial as it seems at first sight. Indeed, they appear in a natural way as representatives of topological spaces (X, t) that have the strong Szpilrajn property. A topological space (X, t) has the strong Szpilrajn property if every weakly continuous order on (X, t) can be refined to a continuous total order on (X, t). Let us denote for every well ordered chain (C,<) by κ C the length of (C,<),i.e.κ C is the uniquely determined ordinal number that is order isomorphic to (C,<). Then it has been proved in Bosi and Herden [5] thatin order that a topological space (X, t) has the strong Szpilrajn property it is necessary and sufficient that t is the discrete topology on X or that there exists some point x X such that the following conditions are satisfied: Szp 1: Szp 2: t X \{x} is the discrete topology on X \{x}. There exists a continuous well ordering on (X, t) the greatest element of which is x. 5 Subspaces, Quotients, Sums and Products of Topological Spaces that have the Szpilrajn Property In this section we study the question if subspaces, quotients, sums and products of topological spaces that have the Szpilrajn property also have the Szpilrajn property. Let, therefore, (X, t) be an arbitrarily chosen topological space. Because of Proposition 4.2 we assume throughout this section (X, t) to be a normal Hausdorffspace. Then the following propositions hold. Proposition 5.1 Let t have the Szpilrajn property and let C be a closed subset of X. Then the relativized topology t C has the Szpilrajn property. Proof The proof of Proposition 5.1 is based upon the same arguments as the proof of Proposition 4.3. Hence, we may omit the details of the proof for the sake of brevity. At present the authors do not know if Proposition 5.1 can be generalized to arbitrary subsets of X. On the other hand, every topological space (X, t) can be extended to a topological space (Y, t ) that has the Szpilrajn property. Indeed, let y be some point that is not contained in X.ThenwesetY := X {y} and t := t {Y}. The definition of t implies that every continuous function f : (Y, t ) (R,, t nat ) is constant. Hence, = for every weakly continuous preorder on (Y, t ),which immediately implies that (Y, t ) has the Szpilrajn property. But if X = then (Y, t ) is not a normal Hausdorff-space. Proposition 5.2 Let t have the Szpilrajn property and let P be an equivalence relation on X. Then the quotient topology t P has the Szpilrajn property. Proof Let P be an equivalence relation on X. Because of the proof of Lemma 2.3 we may consider without loss of generality a weakly continuous preorder P on

13 Order (2006) 23: the quotient space (X P, t P ) for which there exists a family F of continuous realvalued functions on (X P, t P ) such that for any two equivalence classes [x] X P and [y] X P the equivalence [x] P [y] f F ( f ([x]) f ([y])) holds. We must verify that P has a continuous total refinement. Let, therefore, p : X X P be the canonical projection. Since a subset A of X P is open, respectively, closed, if and only if p 1 (A) is an open, respectively, closed, subset of X we may conclude that P induces the weakly continuous preorder on (X, t) that is defined by setting x y f F ( f (p(x)) f (p(y))) for any two points x X and y X. We abbreviate this argument by (*). Since t has the Szpilrajn property has a continuous total refinement. Let us denote for every point x X by [x] I the corresponding indifference (equivalence) class of. Since it follows that for every point x X the inclusion [x] [x] I holds. Hence, the relation P on X P that is defined by [x] P [y] x [x] y [y] (x y) is independent of any particular chosen pair (x, y) [x] [y] of representatives. With the help of (*) we, therefore, may conclude that P is a continuous total refinement of P. Example 5.3 Let (X, t) and (Y, t ) be two topological spaces that have the Szpilrajn property. Then the following example shows that the topological sum (X Y, t t ), in general, does not have the Szpilrajn property. Let ℵ 0 be the first infinite cardinal number and let ℵ 1 be the first uncountable cardinal number. Proposition 4.4 implies that both spaces (X ℵ0, t) and (X ℵ1, t) have the Szpilrajn property. We show that (X ℵ0 X ℵ1, t t) does not have the Szpilrajn property. Therefore, we choose the closed subset S := {ℵ 0, ℵ 1 } of X ℵ0 X ℵ1.Since(X ℵ0 X ℵ1, t t) is a completely regular space there exists for every point x (X ℵ0 X ℵ1 ) \ S a continuous function f x : (X ℵ0 X ℵ1, t t) ([0, 1], t nat ) such that f x (x) = 0 and f x (S) ={1}. Hence, the preorder on (X ℵ0 X ℵ1, t t) that is defined by := {(x, x) x X ℵ0 X ℵ1 } {(ℵ 0, ℵ 1 ), (ℵ 1, ℵ 0 )} {(x, ℵ 0 ) x (X ℵ0 X ℵ1 ) \ S} {(x, ℵ 1 ) x (X ℵ0 X ℵ1 ) \ S} is weakly continuous. Let us assume, in contrast, that (X ℵ0 X ℵ1, t t) has the Szpilrajn property. Then there exists a continuous total preorder on (X ℵ0 X ℵ1, t t) such that and <. In the remainder of this example for every point x X ℵ0 X ℵ1 the closed decreasing set d(x) is always defined with respect to the total preorder. Let now x (X ℵ0 X ℵ1 ) \ S be arbitrarily chosen. Then the definition of implies with the help of the definition of the topology t on X ℵ1 that d(x) S = and, therefore, that d(x) contains at most countably many points y X ℵ1 \{ℵ 1 }. The uncountability of X ℵ1, thus, implies the existence of an uncountable well ordered chain d(x 0 ) d(x 1 )... d(x α ) of closed decreasing subsets of (X ℵ0 X ℵ1 ) \ S such that d(x α ) = (X ℵ0 X ℵ1 ) \ S.Since,on α the other hand, X ℵ0 is a countable set there exists some countable ordinal number α such that X ℵ0 \{ℵ 0 } d(x α ).Butd(x α ) is a closed subset of X ℵ0 X ℵ1.Hence, the definition of the topology t on X ℵ0 implies that ℵ 0 d(x α ), which means that d(x α ) S =. This contradiction proves that (X ℵ0 X ℵ1, t t) cannot have the Szpilrajn property. The reader may notice that the subspace (X ℵ0 X ℵ1 ) \{ℵ 0 } endowed with the relativized topology (t t) (X ℵ0 X ℵ1 ) \{ℵ 0 } is homeomorphic to (X ℵ1, t).hence, there exist topological spaces (X, t) that have the Szpilrajn property and that can be

14 284 Order (2006) 23: extended by only one point to topological spaces (Y, t ) that do not have the Szpilrajn property. Example 5.4 With the help of the same arguments as have been applied in Example 5.3 it also follows that the topological product (X ℵ0 X ℵ1, t t) of the topological spaces (X ℵ0, t) and (X ℵ1, t) does not have the Szpilrajn property. This means that the topological product (X Y, t t ) of topological spaces (X, t) and (Y, t ) that have the Szpilrajn property, in general, does not have the Szpilrajn property. 6 κ-attracting Sets and the Szpilrajn Property Let (X, t) be an arbitrarily chosen topological space and let A be a subset of X. Then A denotes the topological closure of A. In addition, for every set T we denote by T the cardinality of T and by cf( T ) the cofinality of T. AsinSection5 we may assume without loss of generality that throughout this section (X, t) is a normal Hausdorff-space. It already has been mentioned in the introduction that in the following sections conditions that are necessary and in some relevant cases also sufficient for (X, t) to have the Szpilrajn property will be presented. The main tool for proving these results is the concept of a κ-attracting subset of (X, t).withitshelp the Szpilrajn property of topological spaces that are limit point compact or compact or connected can be studied. Definition 6.1 Let 0 <κ< X be a cardinal number. Then a subset S of X is said to be κ-attracting if it satisfies the following conditions: A1: S is a closed subset of X. A2: There exist some regular cardinal number γ such that κ<γ X and a family {Z α } α<γ of (non-empty) subsets Z α of X \ S such that O S = for every open subset O of X such that {α<κ O Z α = } κ and every open subset O of X such that {α<γ O Z α = } γ. The following lemma is fundamental for the results of the following sections. Lemma 6.2 Let (X, t) have the Szpilrajn property and let 0 <κ< X be a cardinal number. Then there exists no κ-attracting subset S of X. Proof Let us assume, in contrast, that there exists a κ -attracting subset S of X.Then there exists some regular cardinal number γ such that κ<γ X and a family {Z α } α<γ of (non-empty) subsets Z α of X \ S such that Ō S = for every open subset O of X such that {α <κ O Z α = } κ and every open subset O of X such that {α <γ O Z α = } γ.since(x, t) is a completely regular space and S is a closed subset of X there exists for every point x X \ S a continuous function f x : (X, t) ([0, 1], t nat ) such that f x (x) = 0 and f x (S) ={1}. This means that the preorder on (X, t) that is defined by setting := {(x, x) x X} {(u,ν) u S and ν S} {(x, u) x X \ S and u S} is weakly continuous. Our assumption that (X, t) has the Szpilrajn property implies the existence of a continuous total refinement of. AsinExample5.3inthe

15 Order (2006) 23: remainder of the proof of the lemma for every point x X the decreasing sets d(x) and l(x), respectively, are defined with respect to the continuous total refinement of. The continuity of implies that for every point x X the sets l(x) are open subsets of X. In addition, the definition of implies that ( ) l(x) = X \ S. x X\S Then we show that there exists a well ordered strictly increasing chain l(x 0 ) l(x 1 )... l(x α ) the length of which is γ such that α<γl(x α ) = X \ S. Indeed, let us assume that such a chain does not exist. Then we may conclude with the help of ( ) that there exists some point x X \ S such that {α<γ l(x) Z α = } γ.this means that l(x) S =. The inclusion l(x) d(x), thus, implies that d(x) S =, which contradicts the definition of.sinceγ is a regular cardinal number it follows, in addition, that there must exist an ordinal number λ<γsuch that {α<κ l(x λ ) Z α = } κ. As it already has been shown we now may conclude that d(x λ ) S =, which contradicts the definition of and, therefore, finishes the proof of the lemma. In order to present a first application of Lemma 6.2 we generalize Example 5.3. Let, therefore, x X be some arbitrarily chosen point. Then we denote by S(x) the setofallsubsetst of X \{x} such that x T and define the order o(x) of x by { o(x) := 0, if S(x) = min{ T T S(x)}, if S(x) =. Let, for the moment, a point x X said to be regular (cf. Proposition 6.3 for a justification of this definition) if o(x) >0 and there exists some set T S(x) that can be chosen in such a way that T = o(x) and x P for every subset P of T the cardinality of which coincides with the cardinality of T. The topological space that is considered in Example 5.3 contains exactly two regular points the orders of which are ℵ 0 and ℵ 1, respectively. This means that the result in Example 5.3 is an immediate consequence of the following proposition. Proposition 6.3 Let (X, t) have the Szpilrajn property. Then there exists a uniquely determined regular cardinal number κ such that o(x) = κ for all regular points x X. Proof The proof of the proposition is divided into two steps. 1. Let x X be an arbitrarily chosen regular point. Then o(x) is a regular cardinal number. Indeed, let some T S(x) be chosen in such a way that the properties that have been described above are satisfied. Then we consider a well ordering y 0 < y 1 <... < y α <... <... of T. Let now Ɣ be a subset of {α α<o(x)} such that Ɣ = cf(o(x)) and sup γ = o(x). ThenwesetK := {y γ T γ Ɣ}. Since γ Ɣ T \ U < o(x) for every neighbourhood U of x it follows that U K = for every neighbourhood U of x. Hence,x K. The definition of o(x), thus,implies that cf(o(x)) = o(x), which means that o(x) is a regular cardinal number. 2. Let us assume in the second step, in contrast, that there exist regular points x, y X such that o(x) <o(y). We show that this assumption contradicts Lemma 6.2.

16 286 Order (2006) 23: Therefore, we choose sets T S(x) and T S(y) that satisfy with respect to x and y, respectively, the properties that have been described above. Now we consider some well ordering z 0 < z 1 <... < z α <... <... < o(y) of T T such that T ={z α α<o(x)}. ThenwesetS := {x, y}, κ := o(x), γ := o(y) and {Z α } α<o(y) := {{z α }} α<o(y).sincet ={z α α<o(x)} we may conclude that S is a κ-attracting subset of X. This conclusion contradicts Lemma 6.2, which still was to be shown. 7 Limit Point Compact Topological Spaces and Compact Topological Spaces that have the Szpilrajn Property Let (X, t) be an arbitrarily chosen topological space. As in the previous sections we assume throughout this section (X, t) to be a normal Hausdorff-space. The reader may recall from Section 2 of this paper that for every (continuous) total preorder on (X, t) the induced order topology is denoted by t. Let now A be an arbitrary subset of X. Then we denote by Lim(A) the set of all limit points of A. We recall that (X, t) is said to be limit point compact (weakly countably compact)iflim( A) = for every infinite subset A of X. In order to characterize limit point compact topological spaces that have the Szpilrajn property we first prove the following proposition which independently of the aim of this section surely is of its own interest. Proposition 7.1 Let ω be an ordinal number and let t be the order topology on ω.then in order that (ω, t) has the Szpilrajn property it is necessary and sufficient that ω is a countable ordinal number. Proof Let ω be a countable ordinal number. Then (ω, t) is a second countable topological space. Hence, (ω, t) has the Szpilrajn property (cf. Section 4). Let, on the other hand, ω be an uncountable ordinal number. Then we set X := {α <ω α ℵ 1 }. Because of Proposition 5.1 it suffices to verify that (X, t X) does not have the Szpilrajn property. Setting S := {λ X λ is a limit ordinal }, κ := ℵ 0, γ := ℵ 1 and {Z α } α<ℵ1 := {{α} α X is not a limit ordinal }, we may conclude that S is an ℵ 0 - attracting subset of X. Lemma 6.2, therefore, implies that (X, t X) does not have the Szpilrajn property. The following proposition generalizes Proposition 7.1. Proposition 7.2 Let (X, t) be limit point compact. Then in order that (X, t) has the Szpilrajn property it is necessary that Lim(A) A = for every uncountable subset AofX. Proof Let us assume, in contrast, that there exists an uncountable subset A of X such that Lim(A) A =. Then there exists an uncountable family {O x } x A of open subsets of X such that x O x and O x A ={x} for every point x A. Sinceℵ 1 A we may assume without loss of generality that A = ℵ 1. Because of Zermelo s Well-Ordering theorem we, thus, may assume that the family {O x } x A is given in the

17 Order (2006) 23: form {O xα } α<ℵ1. Therefore, we set S := X \ x AO x, κ := ℵ 0, γ := ℵ 1 and {Z α } α<ℵ1 := {O xα } α<ℵ1 in order to immediately conclude that S is an ℵ 0 -attracting subset of X. Hence, it follows from Lemma 6.2 that (X, t) does not have the Szpilrajn property. We recall that (X, t) satisfies ccc (countable chain condition) if every collection of pairwise disjoint non-empty open subsets of X is countable. This means that a totally preordered set (X, ) satisfies ccc if the order topology t that is induced by satisfies ccc. In addition, (X, t) is said to be a GRP-space (GRP means that (X, t) has the General Representation Property) if for every weakly continuous binary relation R on (X, t) there exists some continuous order preserving function f : (X, R, t) (R,, t nat ). The considerations in the introduction also motivate the importance of GRP-spaces for the theoretical foundation of mathematical utility theory. With the help of this definition it follows that the following three corollaries are consequences of Proposition 7.2. Corollary 7.3 Let (X, t) be limit point compact. Then in order that (X, t) has the Szpilrajn property it is necessary that (X, t) satisfies ccc. In order to present the next corollaries we recall that a Souslin Line is an unbordered order dense and Dedekind complete chain (X, ) that satisfies ccc but is not separable or, equivalently, is not order-isomorphic to the real line. Order dense and Dedekind complete means that t is a connected topology on X. In addition, a chain (X, ) is said to be a Souslin chain if it only has countably many jumps and satisfies ccc but is not order-embeddable into the real line. The Souslin Hypothesis (SH) states that there do not exist Souslin lines or, equivalently, Souslin chains. Now the following corollaries hold in ZFC+SH (Zermelo-F raenkel + Axiom of Choice + Souslin H ypothesis). Corollary 7.4 follows with the help of the Debreu Open Gap Lemma (cf. Debreu [12, 13]) from Corollary 7.3. Corollary 7.5 follows from Corollary 7.4 and the fact that second countable topological spaces have the Szpilrajn property. Corollary 7.4 (ZFC+SH) Let (X, t) be a limit point compact topological space. Then in order that (X, t) has the Szpilrajn property it is necessary and sufficient that (X, t) is a GRP-space. Corollary 7.5 (ZFC+SH) The following assertions are valid: (a) (b) Let (X, ) be a Dedekind-complete totally preordered set that only has countably many jumps. Then in order that (X, t ) has the Szpilrajn property it is necessary and sufficient that there exists a continuous order-embedding f : (X,, t ) (R,, t nat ). Let (X, ) be a totally preordered set and let t be connected. Then in order that (X, t ) has the Szpilrajn property it is necessary and sufficient that there exists a real interval I such that the quotient (X, ) is order-isomorphic to (I, ). In Section 9 of this paper we discuss the problem if the order topology of a Souslin line has the Szpilrajn property.