On the Preference Relations with Negatively Transitive Asymmetric Part. I (original) (raw)
On the Preference Relations with Negatively Transitive Asymmetric Part. I
Maria Viktorovna Droganova* Valentin Vankov Iliev †{ }^{\dagger}
Abstract
Given a linearly ordered set II, every surjective map p:A→Ip: A \rightarrow I endows the set AA with a structure of set of preferences by “replacing” the elements ι∈I\iota \in I with their inverse images p−1(ι)p^{-1}(\iota) considered as “balloons” (sets endowed with an equivalence relation), lifting the linear order on AA, and “agglutinating” this structure with the balloons. Every ballooning AA of a structure of linearly ordered set II is a set of preferences AA whose preference relation (not necessarily complete) is negatively transitive and every such structure on a given set AA can be obtained by ballooning of certain structure of a linearly ordered set II, intrinsically encoded in AA. In other words, the difference between linearity and negative transitivity is constituted of balloons. As a consequence of this characterization, under certain natural topological conditions on the set of preferences AA furnished with its interval topology, the existence of a continuous generalized utility function on AA is proved.
Introduction
The aim of this paper is twofold: To characterize the preference relations with negatively transitive asymmetric part and to prove the existence of a continuous generalized utility function on the corresponding set of preferences furnished with its interval topology with respect to which it is connected and separable. Since the authors can not find appropriate source(s) for citing, they allow themselves to gather here from a unique point of view material that can be found in various texts using varying terminology and notation at different levels of rigor and generality. Moreover, the systematic use of saturated subsets shortens and clarifies the exposition. In what follows, we use, in general, the terminology from [8][8].
Below we use the terms “preorder” as a synonym of “preference relation” and “preordered set” as a synonym of “set of preferences”.
- *Department of Economics, Duke University, Durham, NC 27708-0097, USA
†{ }^{\dagger} Institute of Mathematics and Informatics, Bulgarian Academy of Sciences, Sofia 1113, Bulgaria ↩︎
Any preference relation RR on a set AA is a disjoint union of its symmetric part E=IRE=I_{R} (an equivalence relation) and its asymmetric part F=PRF=P_{R} (an asymmetric and transitive relation). Moreover, FF is EE-saturated. Conversely, any ordered pair (E,F)(E, F) of such relations produces a reflexive and transitive relation RR by forming their union: R=E∪FR=E \cup F. The rule R↦(IR,PR)R \mapsto\left(I_{R}, P_{R}\right) establishes a one-to-one correspondence between the set of of all reflexive and transitive binary relations RR on AA and the set of all ordered pairs (E,F)(E, F), where EE is an equivalence relation on A,FA, F is an asymmetric transitive binary relation on AA, which is EE-saturated, and E∩F=∅E \cap F=\emptyset.
The nature of the connection between negative transitivity and completeness is the main subject of the present paper. In mathematical economics, the idea descends from Fishburn and his Theorem 2.1 in [6, Ch. 2, Sec. 2.2]. Under the assumptions of completeness of the preference relation RR and negative transitivity of its asymmetric part FF, he notes that E=EF(xEFy\mathcal{E}=E_{F}\left(x E_{F} y\right. means " xx and yy are not FF-comparable) is an equivalence relation and that the partition {F,E,F−1}\left\{F, \mathcal{E}, F^{-1}\right\} of A2A^{2} can be factorized with respect to E\mathcal{E}, thus producing a linearly ordered factor-set. In mathematics this idea is older and one can find it, for example, in [2, Ch. III, Sec. 1, Exercise 4], where N. Bourbaki factorizes a partially ordered set with respect to the transitive closure of the relation " xx and yy are not comparable or x=yx=y " and obtains a linearly ordered factor-set. The result is presented here, faintly generalized for a preorder RR, as Theorem 1.4.2. If we strengthen the condition of the later theorem by assuming that the asymmetric part FF of RR is negative transitive, we obtain our Theorem 2.1.1 which generalizes Fishburn’s theorem. More precisely, the negative transitivity allows us to show that FF is E\mathcal{E}-saturated (see 1.1) and the triple {F,E,F−1}\left\{F, E, F^{-1}\right\}, where EE is the symmetric part of RR, can be factored out with respect to the equivalence relation E\mathcal{E}. The triple of factors {F/E,E/E,F−1/E}\left\{F / \mathcal{E}, E / \mathcal{E}, F^{-1} / \mathcal{E}\right\} is a partition of the factor-set A′=A/EA^{\prime}=A / \mathcal{E}, we have E/E=DA′E / \mathcal{E}=D_{A^{\prime}} (the diagonal of A′2A^{\prime 2} ), F−1/E=(F/E)−1F^{-1} / \mathcal{E}=(F / \mathcal{E})^{-1}, and R′=DA′∪(F/E)R^{\prime}=D_{A^{\prime}} \cup(F / \mathcal{E}) is a linear order on A′A^{\prime}. Note that the E\mathcal{E}-equivalence classes are the fibres of the canonical surjective map c:A→A′c: A \rightarrow A^{\prime}, x↦x \mapsto the equivalence class of xx. Since E=E∪ER\mathcal{E}=E \cup E_{R}, every E\mathcal{E}-equivalence class is a disjoint union of EE-equivalence classes, that is, indifference curves, and the members of different indifference curves in a fixed E\mathcal{E}-equivalence class are not RR-comparable. In particular, every E\mathcal{E}-equivalence class is EE-saturated and hence inherits from EE its own equivalence relation. In case the later equivalence relation on every E\mathcal{E}-equivalence class is trivial, that is, every E\mathcal{E}-equivalence class coincides with an EE-indifference curve, or, what is the same, when E=E\mathcal{E}=E, then ER=∅E_{R}=\emptyset and we obtain the completeness of RR and Fishburn’s Theorem 2.1. Conversely, starting with a linearly ordered set II and with a family (Aι)i∈I\left(A_{\iota}\right)_{i \in I} of balloons, that is, sets AιA_{\iota} each endowed with an equivalence relation Eι,ι∈IE_{\iota}, \iota \in I, we construct in Theorem 2.2.1 a set AA - the coproduct of the family (Aι)i∈I\left(A_{\iota}\right)_{i \in I} with its natural projection p:A→Ip: A \rightarrow I and a preference relation RR on AA with negatively transitive asymmetric part F=PRF=P_{R}, such that the equivalence relation E=IRE=I_{R} induces on every balloon p−1(ι)=Aιp^{-1}(\iota)=A_{\iota} the equivalence relation EιE_{\iota}. Moreover, the equivalence relation associated with the partition (Aι)i∈I\left(A_{\iota}\right)_{i \in I} of AA is identical to E=EF\mathcal{E}=E_{F}, the linear order on II is the factor-relation of RR, and
the corresponding strict order on II is the factor-relation of FF with respect to E\mathcal{E}. Thus, every ballooning AA of a structure of linearly ordered set II has a structure of set of preferences AA whose preference relation has negatively transitive asymmetric part and every such structure on a given set AA can be obtained by ballooning of certain structure of a linearly ordered set II, intrinsically encoded in AA.
G. Debreu (not without influence from N. Bourbaki) studies in [4] the existence of a continuous utility representation u:A→Ru: A \rightarrow \mathbb{R} of a complete set of preferences AA. Lemma 2.3.1 reduces the problem to the existence of a strictly increasing and continuous map of the factor-set A/EA / E into the real line R\mathbb{R}. Under the condition of connectedness and separability on AA endowed with a “natural” topology (that is, topology which is finer than the interval topology on AA ), G. Debreu proves this existence by refereing to [4,(6.1)][4,(6.1)].
In his monograph [6], P. C. Fishburn proves (Theorem 2.2) that if the set AA is furnished with an asymmetric relation FF which is negatively transitive and if the factor-set A/EA / \mathcal{E} is countable, then there exists a utility function u:A→Ru: A \rightarrow \mathbb{R}. Further, in [6, Theorem 3.1] he presents a necessary and sufficient condition for existence of such utility function, which is, in fact, the Birkhoff’s criterium [1, Ch. III, Theorem 2] for the linearly ordered set A/EA / \mathcal{E} to be embedded in a order-preserving way into the real line.
In [7, Sec. 5], E. A. Ok argues that “…utility theory can be beneficially extended to cover incomplete preference relations” and, moreover, “Identifying a useful set of conditions on preorders that would lead to such a representation result [that is, functional representation of preorders] is an open problem worthy of investigation”.
Our setup uses the sufficient condition (connectedness and separability) from [3, Ch. IV, Sec. 2, Exercise 11 a] for the the existence of a homeomorphic embedding of factor-set A/EA / \mathcal{E} into the real line (see also Theorem A.2.24). If the preference relation on AA has negatively transitive asymmetric part, there is a close bond between the topologies of AA and A/EA / \mathcal{E} described in Proposition 1.6.1. Using this strong relationship, in case AA is connected and separable we prove the existence of a continuous generalized utility function on AA (Theorem 2.3.2).
The aim of the appendix is to give a proof of Theorem A. 2.24 which asserts that any linearly ordered set AA endowed with the interval topology, which is connected and separable possesses a strictly increasing homeomorphism onto one of the intervals in the real line with endpoints 0 and 1 . The text represents the slightly dressed up notes of the authors during reading of [2, Ch. III, Sec. 1], the problems from [3, Ch. IV, Sec. 2, Exercises 6, 7, 8, 9, 11], and other relevant literature.
Portions of this paper are parts of the senior thesis of the first author made under the supervision of the second one. We intend to devote Part 2 of the paper to some economic applications of our results.
1 Mathematical Background
Below, for the sake of completeness and non-ambiguous reading of this paper, we present the main statements which are used. Moreover, we show how saturatedness appears naturally in the context of factorization.
If the opposite is not stated explicitly, we assume that the occurrence of the variables x,y,z,…x, y, z, \ldots in a statement is bounded by the universal quantifier (“for all”).
1.1 Factorizations
Let AA be a non-empty set and let EE be an equivalence relation on AA. For x∈Ax \in A we denote by CxC_{x} the equivalence class with respect to EE (or, EE-equivalence class) with representative xx. The subset B⊂AB \subset A is said to be saturated with respect to EE, or EE-saturated, if x∈Bx \in B implies Cx⊂BC_{x} \subset B. The factor-set A/EA / E (that is, the set of all equivalent classes) is a partition of AA and c:A→A/E,x↦Cxc: A \rightarrow A / E, x \mapsto C_{x}, is the canonical surjective map. Sometimes we write xˉ=c(x)\bar{x}=c(x).
The relation RR is called left EE-saturated if xRyx R y and x′Exx^{\prime} E x imply x′Ryx^{\prime} R y, right EE-saturated if xRyx R y and y′Eyy^{\prime} E y imply xRy′x R y^{\prime}, and EE-saturated if xRy,x′Exx R y, x^{\prime} E x, and y′Eyy^{\prime} E y imply x′Ry′x^{\prime} R y^{\prime}.
Let EE be an equivalence relation on AA. The product E=E×E\mathcal{E}=E \times E is an equivalent relation on A×AA \times A and let c2:A×A→(A×A)/Ec_{2}: A \times A \rightarrow(A \times A) / \mathcal{E} be the corresponding canonical surjective map. The bijection (c(x),c(y))↦c2(x,y)(c(x), c(y)) \mapsto c_{2}(x, y) identifies the sets (A/E)×(A/E)(A / E) \times(A / E) and (A×A)/E(A \times A) / \mathcal{E} and then the canonical surjective map
c2:A×A→(A/E)×(A/E),c2(x,y)=(xˉ,yˉ)c_{2}: A \times A \rightarrow(A / E) \times(A / E), c_{2}(x, y)=(\bar{x}, \bar{y})
induces via the rule X↦c2(X),X⊂A×AX \mapsto c_{2}(X), X \subset A \times A, a canonical surjective map
c2:2A×A→2(A/E)×(A/E)c_{2}: 2^{A \times A} \rightarrow 2^{(A / E) \times(A / E)}
The map c2c_{2} from (1.1.1) establishes a bijection R↦RˉR \mapsto \bar{R}, where Rˉ=c2(R)\bar{R}=c_{2}(R), between the set of all EE-saturated binary relations RR on AA and the set of all binary relations Rˉ⊂(A/E)×(A/E)\bar{R} \subset(A / E) \times(A / E) on the factor-set A/EA / E. We have xRyx R y if and only if xˉRˉyˉ\bar{x} \bar{R} \bar{y}.
Thus, any EE-saturated binary relation RR on the set AA produces a relation Rˉ=c2(R)\bar{R}=c_{2}(R) on the factor-set A/EA / E which we call factor-relation of RR with respect to the equivalence relation EE. We have xRyx R y if and only if xˉRˉyˉ\bar{x} \bar{R} \bar{y}.
Below we present another natural factorization R′R^{\prime} of a binary relation RR on a set AA with respect to an equivalence relation EE. Let A′=A/EA^{\prime}=A / E be the factor-set and let A→A′,x↦xˉA \rightarrow A^{\prime}, x \mapsto \bar{x}, be the canonical surjective map. We define the binary relation R′R^{\prime} on A′A^{\prime} by the rule
xˉR′yˉ if for any x′∈xˉ there exists y′∈yˉ such that x′Ry′\bar{x} R^{\prime} \bar{y} \text { if for any } x^{\prime} \in \bar{x} \text { there exists } y^{\prime} \in \bar{y} \text { such that } x^{\prime} R y^{\prime}
and call R′R^{\prime} weak factor-relation of RR with respect to the equivalence relation EE, thus generalizing the notion of factor-relation introduced above. This is also a generalization of [2, Ch. III, Sec. 1, Exercise 2, a].
1.2 Derivative relations
Let RR be a binary relation on a non-empty set AA. We denote by R−1R^{-1} the inverse relation and by RcR^{c} the complementary relation. We set IR=R∩R−1I_{R}=R \cap R^{-1}, UR=R∪R−1,PR=R\IRU_{R}=R \cup R^{-1}, P_{R}=R \backslash I_{R}, and ER=URcE_{R}=U_{R}^{c}. The relation IRI_{R} is the symmetric part of RR and the relation PRP_{R} is the asymmetric part of RR. In case RR is reflexive and transitive (that is, a preorder) IRI_{R} is an equivalence relation and PRP_{R} is asymmetric and transitive. When, in addition, IR=DAI_{R}=D_{A}, where DAD_{A} is the diagonal of A2A^{2}, the preorder RR is a partial order. A balloon is a preordered set AA furnished with symmetric preorder RR. Thus, RR is an equivalence relation: R=IRR=I_{R}.
1.3 Saturatedness and Transitivity
Lemma 1.3.1 Let AA be a set, let RR be a reflexive binary relation on AA, and let EE be an equivalence relation on AA. If R⊂ER \subset E, then RR is weakly EE-saturated.
Proof: Let x′Exx^{\prime} E x and xRyx R y. Then x′Eyx^{\prime} E y and if we set y′=x′y^{\prime}=x^{\prime}, then yEy′y E y^{\prime} and x′Ry′x^{\prime} R y^{\prime} because RR is reflexive.
Proposition 1.3.2 Let RR be a binary relation on the set AA and let EE be an equivalence relation on AA. Let A′=A/EA^{\prime}=A / E be the factor-set and let c:A→A′c: A \rightarrow A^{\prime}, x↦xˉx \mapsto \bar{x}, be the canonical surjective map. Then the canonical map cc is increasing if and only if RR is weakly EE-saturated.
Under the condition that RR is weakly EE-saturated, the following statements hold:
(i) If RR is reflexive, then R′R^{\prime} is reflexive.
(ii) If RR is transitive, then R′R^{\prime} is transitive.
(iii) The relation R′R^{\prime} is antisymmetric if and only if RR satisfies the condition xRy,yRzx R y, y R z, and xEzx E z imply xEyx E y.
(iv) If RR is EE-saturated, then R′R^{\prime} coincides with the factor-relation Rˉ\bar{R}. If, in addition, IR⊂EI_{R} \subset E, then condition (1.3.1) holds.
(v) Let BB be a set endowed with a binary relation SS. For any increasing map p:A→Bp: A \rightarrow B which is constant on the members of the factor-set A′=A/EA^{\prime}=A / E there exists a unique increasing map p′:A′→Bp^{\prime}: A^{\prime} \rightarrow B such that the diagram
is commutative. If pp is surjective, then p′p^{\prime} is surjective. If, in addition,
E={(x,y)∈A2∣p(x)=p(y)}E=\left\{(x, y) \in A^{2} \mid p(x)=p(y)\right\}
then p′p^{\prime} is an increasing bijection.
Proof: Let us suppose that cc is an increasing map, let xRyx R y, and let x′Exx^{\prime} E x, that is, x′∈xˉx^{\prime} \in \bar{x}. We have xˉR′yˉ\bar{x} R^{\prime} \bar{y} which means there exists y′∈yˉy^{\prime} \in \bar{y} with x′Ry′x^{\prime} R y^{\prime} and hence RR is weakly EE-saturated. Conversely, if RR is weakly EE-saturated and if xRyx R y, then for any x′∈xˉx^{\prime} \in \bar{x} there exists y′∈yˉy^{\prime} \in \bar{y} with x′Ry′x^{\prime} R y^{\prime}, therefore xˉR′yˉ\bar{x} R^{\prime} \bar{y}.
The proofs of parts (i) and (ii) are immediate.
(iii) Let RR satisfies condition (1.3.1) and let xˉR′yˉ,yˉR′xˉ\bar{x} R^{\prime} \bar{y}, \bar{y} R^{\prime} \bar{x}. In other words, there exist y′∈yˉy^{\prime} \in \bar{y} and x′∈xˉx^{\prime} \in \bar{x}, such that xRy′x R y^{\prime} and y′Rx′y^{\prime} R x^{\prime}. Since xEx′x E x^{\prime}, we obtain xEy′x E y^{\prime}, that is, xˉ=yˉ\bar{x}=\bar{y}. Conversely, let R′R^{\prime} be antisymmetric and let us suppose that xRy,yRzx R y, y R z, and xEzx E z. In particular, xˉR′yˉ,yˉR′xˉ\bar{x} R^{\prime} \bar{y}, \bar{y} R^{\prime} \bar{x}, and this yields xˉ=yˉ\bar{x}=\bar{y}, that is, xEyx E y.
(iv) We have Rˉ⊂R′\bar{R} \subset R^{\prime}. Now, let xˉR′yˉ\bar{x} R^{\prime} \bar{y} and let x′∈xˉ,y′∈yˉx^{\prime} \in \bar{x}, y^{\prime} \in \bar{y}. We have that there exists y′′∈yˉy^{\prime \prime} \in \bar{y} with x′Ry′′x^{\prime} R y^{\prime \prime}. Since y′′Ey′y^{\prime \prime} E y^{\prime}, the saturatedness of RR yields x′Ry′x^{\prime} R y^{\prime}. Thus, we have xˉRˉyˉ\bar{x} \bar{R} \bar{y}.
Now, let us suppose, in addition, that IR⊂EI_{R} \subset E and let xRy,yRzx R y, y R z, and xEzx E z. Since RR is EE-saturated, we obtain yRxy R x, hence xIRyx I_{R} y and this, in turn, implies xEyx E y.
(v) For any map p:A→Bp: A \rightarrow B which is constant on the members of the factorset A′=A/EA^{\prime}=A / E there exists a unique map p′:A′→Bp^{\prime}: A^{\prime} \rightarrow B such that p=p′∘cp=p^{\prime} \circ c. Let xˉR′yˉ\bar{x} R^{\prime} \bar{y} and we can suppose that xRyx R y. Then p(x)Sp(y)p(x) S p(y) and this relation can be rewritten as p′(xˉ)Sp′(yˉ)p^{\prime}(\bar{x}) S p^{\prime}(\bar{y}). Therefore p′p^{\prime} is an increasing map. Since p(x)=p′(xˉ)p(x)=p^{\prime}(\bar{x}) the surjectivity of pp implies the surjectivity of p′p^{\prime}. Under the condition (1.3.2), if p(x)=p(y)p(x)=p(y), then xˉ=yˉ\bar{x}=\bar{y}, that is, p′p^{\prime} is a bijection.
Proposition 1.3.3 Let AA be a set and let FF be an asymmetric and negatively transitive relation on AA. Then E=EF\mathcal{E}=E_{F} is an equivalence relation and the relation FF is transitive and E\mathcal{E}-saturated.
Proof: The proofs that E\mathcal{E} is an equivalence relation and that the asymmetry and negative transitivity of FF imply transitivity of FF are straightforward. Since {E,F,F−1}\left\{\mathcal{E}, F, F^{-1}\right\} is a partition of A2A^{2}, we obtain that FF is E\mathcal{E}-saturated. Indeed, let x′Exx^{\prime} \mathcal{E} x and xFyx F y. If x′Eyx^{\prime} \mathcal{E} y (respectively, yFx′y F x^{\prime} ), then xEyx \mathcal{E} y (respectively, xFx′x F x^{\prime} ) an absurdity. Thus, x′Fyx^{\prime} F y. Similarly, let xFyx F y and yEy′y \mathcal{E} y^{\prime}. If xEy′x \mathcal{E} y^{\prime} (respectively, y′Fxy^{\prime} F x ), then xEyx \mathcal{E} y (respectively, y′Fyy^{\prime} F y ) - an absurdity. Thus, xFy′x F y^{\prime}.
1.4 Indifference and Completeness
Let AA be a set. A binary relation RR on AA is said to be indifference if RR is reflexive and symmetric. In case RR is an indifference on AA, its transitive closure R(t)R^{(t)} is an equivalence relation on AA and the R(t)R^{(t)}-equivalence classes are called RR-indifference curves. Thus, the factor-set A/R(t)A / R^{(t)} consists of all RR indifference curves. In case the relation RR is self-understood, we denote by xˉ\bar{x} the RR-indifference curve with representative x∈Ax \in A.
Lemma 1.4.1 Let the binary relation RR on AA be asymmetric and transitive.
(i) The relation ERE_{R} is an indifference.
(ii) If xˉ\bar{x} and yˉ\bar{y} are two different RR-indifference curves, then any two elements x′∈xˉx^{\prime} \in \bar{x} and y′∈yˉy^{\prime} \in \bar{y} are RR-comparable and if xRyx R y, then x′Ry′x^{\prime} R y^{\prime}.
Proof: (i) The proof is immediate.
(ii) Since xˉ≠yˉ\bar{x} \neq \bar{y}, any two elements x′∈xˉx^{\prime} \in \bar{x} and y′∈yˉy^{\prime} \in \bar{y} are RR-comparable. Let, for example, xRyx R y and let y′∈yˉy^{\prime} \in \bar{y}. Let y=b0,b1,…,bn=y′y=b_{0}, b_{1}, \ldots, b_{n}=y^{\prime} be the finite sequence such that biERbi+1,i=0,1,…,n−1b_{i} E_{R} b_{i+1}, i=0,1, \ldots, n-1. Note that bi∈yˉb_{i} \in \bar{y} for all i=0,1,…,n−1i=0,1, \ldots, n-1. We use induction with respect to ii to prove that xRbix R b_{i}. This statement is true for i=0i=0 and let us suppose that xRbi−1x R b_{i-1}. If biRxb_{i} R x, then the transitivity of RR imply biRbi−1b_{i} R b_{i-1} - a contradiction. Thus, xRbix R b_{i} for any i=0,1,…,ni=0,1, \ldots, n and, in particular, xRy′x R y^{\prime} for y′∈yˉy^{\prime} \in \bar{y}. Using the above argumentation for R−1R^{-1} (which is also reflexive and transitive) and taking into account that ER=ER−1E_{R}=E_{R^{-1}}, we obtain that xRyx R y and x′∈xˉx^{\prime} \in \bar{x} yield x′Ryx^{\prime} R y. Thus, if xRy,x′∈xˉx R y, x^{\prime} \in \bar{x}, and y′∈yˉy^{\prime} \in \bar{y}, then x′Ry′x^{\prime} R y^{\prime}.
Theorem 1.4.2 Let the binary relation RR on AA be reflexive and transitive, let F=PRF=P_{R} be its asymmetric part, and let EF(t)E_{F}^{(t)} be the transitive closure of the indifference EFE_{F}. One has:
(i) The relation RR is weakly EF(t)E_{F}^{(t)}-saturated.
(ii) The week factor-relation R′R^{\prime} on the factor-set A′=A/EF(t)A^{\prime}=A / E_{F}^{(t)} is reflexive, transitive, antisymmetric, and complete; A′A^{\prime} endowed with R′R^{\prime} is a linearly ordered set.
Proof: (i) Let xRyx R y and xEF(t)x′x E_{F}^{(t)} x^{\prime}.
Case 1. xˉ=yˉ\bar{x}=\bar{y}.
In this case yEF(t)x′y E_{F}^{(t)} x^{\prime} and it is enough to note that x′Rx′x^{\prime} R x^{\prime}.
Case 2. xˉ≠yˉ\bar{x} \neq \bar{y}.
Lemma 1.4.1, (ii), shows that xx and yy are RR-comparable and if xRyx R y, then, in particular, for any x′∈xˉx^{\prime} \in \bar{x} there exists y′∈yˉy^{\prime} \in \bar{y} such that x′Ry′x^{\prime} R y^{\prime}.
(ii) The weak factor-relation R′R^{\prime} on the factor-set A′A^{\prime} is reflexive and transitive because of Proposition 1.3.2, (i), (ii). In accord with Lemma 1.4.1, (ii), if xˉ≠yˉ\bar{x} \neq \bar{y}, then xx and yy are RR-comparable, say xRyx R y, and we obtain xˉR′yˉ\bar{x} R^{\prime} \bar{y}. Therefore the week factor-relation R′R^{\prime} on A′A^{\prime} is complete. Let xˉR′yˉ\bar{x} R^{\prime} \bar{y} and yˉR′xˉ\bar{y} R^{\prime} \bar{x}. If we suppose xˉ≠yˉ\bar{x} \neq \bar{y}, then Lemma 1.4.1, (ii), yields xRyx R y and yRxy R x which contradicts the asymmetry of RR. Thus, xˉ=yˉ\bar{x}=\bar{y} and R′R^{\prime} is an antisymmetric relation. In other words, R′R^{\prime} is a linear order on A′A^{\prime}.
1.5 Coproducts of preordered sets
Here we remind the definition of coproduct of a family (Aι)ι∈I\left(A_{\iota}\right)_{\iota \in I} of sets (see, for example, [2\left[2\right., Ch. II, Sec. 4, no8]\left.n^{o} 8\right] ) and endow this coproduct with a structure of preordered set, given such structures on the members AιA_{\iota} of the family and given a structure of partially ordered set on the index set II.
For any ι∈I\iota \in I there exists a bijection Aι→Aι×{ι},x↦(x,ι)A_{\iota} \rightarrow A_{\iota} \times\{\iota\}, x \mapsto(x, \iota). The set A=∪ι∈I(Aι×{ι})A=\cup_{\iota \in I}\left(A_{\iota} \times\{\iota\}\right) is called the coproduct of the family (Aι)ι∈I\left(A_{\iota}\right)_{\iota \in I}. Let U=∪ι∈IAιU=\cup_{\iota \in I} A_{\iota} be the union of this family. In case Aι∩Aκ=∅A_{\iota} \cap A_{\kappa}=\emptyset for any ι≠κ\iota \neq \kappa there exists a projection p:U→I,x↦ιp: U \rightarrow I, x \mapsto \iota, where x∈Aιx \in A_{\iota}. We have a bijection between the union UU and the coproduct AA, given by the rule
U→A,x↦(x,p(x))U \rightarrow A, x \mapsto(x, p(x))
When the family (Aι)ι∈I\left(A_{\iota}\right)_{\iota \in I} has pairwise disjoint members we identify the union UU and the coproduct AA via the above bijection and define the projection p:A→Ip: A \rightarrow I, x↦ιx \mapsto \iota for x∈Aιx \in A_{\iota}.
Below, when discussing the coproduct of a family of sets, we implicitly suppose without any loss of generality that the members of this family are pairwise disjoint. In other words, “coproduct” is a shorthand for “union of pairwise disjoint sets”.
Proposition 1.5.1 Let (Aι)ι∈I\left(A_{\iota}\right)_{\iota \in I} be a family of preordered sets, let (Rι)ι∈I\left(R_{\iota}\right)_{\iota \in I} be the family of the corresponding preorders, and let the index set II be partially ordered. Let A=∏ι∈IAιA=\prod_{\iota \in I} A_{\iota} be the coproduct of the family (Aι)ι∈I\left(A_{\iota}\right)_{\iota \in I}, let p:A→Ip: A \rightarrow I, x↦ιx \mapsto \iota for x∈Aιx \in A_{\iota}, be the natural projection, and let E\mathcal{E} be the equivalence relation associated with the partition (Aι)ι∈I\left(A_{\iota}\right)_{\iota \in I} of the coproduct AA.
(i) The coproduct AA of this family has a structure of preordered set with preorder RR defined by the rule
xRy if " p(x)<p(y)" or " p(x)=p(y) and xRp(x)y"x R y \text { if " } p(x)<p(y) " \text { or " } p(x)=p(y) \text { and } x R_{p(x)} y "
where << is the asymmetric part of the partial order ≤\leq on II.
(ii) One has
xIRyx I_{R} y if and only if " p(x)=p(y)p(x)=p(y) " and " xIRp(x)yx I_{R_{p(x)}} y ".
(iii) One has
xPRyx P_{R} y if and only if " p(x)<p(y)p(x)<p(y) " or " p(x)=p(y)p(x)=p(y) and xPRp(x)yx P_{R_{p(x)}} y ".
Let, in addition, (Aι)ι∈I\left(A_{\iota}\right)_{\iota \in I} be a family of balloons. Then one has:
(iv) xPRyx P_{R} y if and only if p(x)<p(y)p(x)<p(y).
(v) The asymmetric part PRP_{R} of the preorder RR is E\mathcal{E}-saturated.
Proof: (i) We have xRxx R x because xRιxx R_{\iota} x for ι=p(x)\iota=p(x). Now, let xRyx R y and yRzy R z. We obtain p(x)≤p(y)p(x) \leq p(y) and p(y)≤p(z)p(y) \leq p(z), hence p(x)≤p(z)p(x) \leq p(z). If p(x)<p(z)p(x)<p(z), then xRzx R z. Otherwise, there exists ι∈I\iota \in I such that ι=p(x)=p(y)=p(z),xRιy\iota=p(x)=p(y)=p(z), x R_{\iota} y, and yRιzy R_{\iota} z. Therefore xRιzx R_{\iota} z and this yields xRzx R z.
(ii) We have xRyx R y and yRxy R x if and only if there exists ι∈I\iota \in I such that p(x)=p(x)= p(y)=ι,xRιyp(y)=\iota, x R_{\iota} y, and yRιxy R_{\iota} x.
(iii) This is an immediate consequence of parts (i) and (ii).
(iv) We have PRp(x)=∅P_{R_{p(x)}}=\emptyset and part (iii) yields that xPRyx P_{R} y is equivalent to p(x)<p(y)p(x)<p(y).
(v) If x′Exx^{\prime} \mathcal{E} x and yEy′y \mathcal{E} y^{\prime} (that is, x′∈Ap(x)x^{\prime} \in A_{p(x)} and y′∈Ap(y)y^{\prime} \in A_{p(y)} ) and if xPRyx P_{R} y, then x′PRy′x^{\prime} P_{R} y^{\prime} because of part (iv).
Let (Aι)ι∈I\left(A_{\iota}\right)_{\iota \in I} be a family of preordered sets, whose index set II is partially ordered. The coproduct AA of this family, endowed with the preorder from Proposition 1.5.1, (i), is said to be the coproduct of the preordered sets (Aι)\left(A_{\iota}\right), ι∈I\iota \in I, or the coproduct of the family (Aι)ι∈I\left(A_{\iota}\right)_{\iota \in I} of preordered sets.
We note that definition from part (i) of the above proposition is a generalization of the definition from [2, Ch. 3, Sec. 1, Exercise 3, a].
1.6 Interval topology of coproduct of a family of balloons
Here we use freely the terminology, notation, and results from subsection 1.5 as well as from [3,Ch.I][3, \mathrm{Ch} . \mathrm{I}] ).
Let AA be a preordered set and let RR be its preorder. Let us set E=IRE=I_{R} and F=PRF=P_{R}. An open interval in AA is any subset of AA of the form (x,y)={z∈(x, y)=\{z \in A∣xFzA \mid x F z and zFy}z F y\}, or (←,x)={z∈A∣zFx}(\leftarrow, x)=\{z \in A \mid z F x\}, or (x,→)={z∈A∣xFz}(x, \rightarrow)=\{z \in A \mid x F z\}, where x,y∈Ax, y \in A. A closed interval in AA is any subset of AA of the form [x,y]=[x, y]= {z∈A∣xRz\{z \in A \mid x R z and zRy}z R y\}, or (←,x]={z∈A∣zRx}(\leftarrow, x]=\{z \in A \mid z R x\}, or [x,→)={z∈A∣xRz}[x, \rightarrow)=\{z \in A \mid x R z\}, where x,y∈Ax, y \in A. In order to emphasize that the corresponding interval is in AA, we write (x,y)=(x,y)A(x, y)=(x, y)_{A}, etc. The interval topology T0(A)\mathcal{T}_{0}(A) on AA is the topology generated by the set OA\mathcal{O}_{A} of all open intervals in AA, that is, U∈T0(A)(UU \in \mathcal{T}_{0}(A)(U is open) if UU is a union of finite intersections of open intervals in AA. In particular, the open (respectively, closed) intervals are open (respectively, closed) sets in the interval topology. When AA is a linearly ordered set, any finite intersection of open intervals is an open interval, hence in this case the set OA\mathcal{O}_{A} is a base of the topology of AA.
We remind that a topological space AA is said to be separable if there is a countable subset D⊂AD \subset A such that D∩U≠∅D \cap U \neq \emptyset for any non-empty open subset U⊂AU \subset A. In order to prove separability, it is enough to verify the inequality D∩U≠∅D \cap U \neq \emptyset for any non-empty member UU of a base of the topology of AA.
Proposition 1.6.1 Under the conditions of Proposition 1.5.1, let the index set II be linearly ordered, let (Aι)ι∈I\left(A_{\iota}\right)_{\iota \in I} be a family of balloons, and let AA and II be endowed with their interval topologies.
(i) The natural projection p:A→Ip: A \rightarrow I is a strictly increasing map which induces a pair of mutually inverse bijections
OA→OI,J↦p(J)\mathcal{O}_{A} \rightarrow \mathcal{O}_{I}, J \mapsto p(J)
and
OI→OA,K↦p−1(K)\mathcal{O}_{I} \rightarrow \mathcal{O}_{A}, K \mapsto p^{-1}(K)
(ii) The set OA\mathcal{O}_{A} is a base of the topology of AA.
(iii) p:A→Ip: A \rightarrow I is a continuous and open map.
Proof: Let F=PRF=P_{R} be the asymmetric part of the preorder RR on the coproduct A.
(i) According to Proposition 1.5.1, (iv), (v), the relations xFyx F y and p(x)<p(x)< p(y)p(y) are equivalent and hence any open interval JJ in AA is E\mathcal{E}-saturated. In particular, pp is a strictly increasing map. Moreover, pp maps any open interval JJ in AA into an open interval KK in II whose endpoint(s) is (are) the image(s) of the endpoint(s) of JJ. Since the sets Aι,ι∈IA_{\iota}, \iota \in I, are not empty, the natural projection pp is a surjective map and we have p(J)=Kp(J)=K. On the other hand, the inverse image p−1(K)p^{-1}(K) of an open interval KK in II is an open interval in AA with endpoint(s) chosen to be inverse image(s) of the endpoint(s) of KK. Thus, the existence of
the maps (1.6.1) and (1.6.2) is justified. Finally, we have p(p−1(K))=Kp\left(p^{-1}(K)\right)=K and the E\mathcal{E}-saturatedness of the open intervals JJ in AA yields J=p−1(p(J))J=p^{-1}(p(J)).
(ii) It is enough to show that a finite intersection of open intervals in AA is an open interval. For let Jλ∈OA,λ=1,…,sJ_{\lambda} \in \mathcal{O}_{A}, \lambda=1, \ldots, s and let Kλ∈OIK_{\lambda} \in \mathcal{O}_{I} be such that Jλ=p−1(Kλ)J_{\lambda}=p^{-1}\left(K_{\lambda}\right). Since II is a linearly ordered set, the intersection K=∩λ=1sKλK=\cap_{\lambda=1}^{s} K_{\lambda} is an open interval in II and we have
p−1(K)=p−1(∩λ=1sKλ)=∩λ=1sp−1(Kλ)=∩λ=1sJλp^{-1}(K)=p^{-1}\left(\cap_{\lambda=1}^{s} K_{\lambda}\right)=\cap_{\lambda=1}^{s} p^{-1}\left(K_{\lambda}\right)=\cap_{\lambda=1}^{s} J_{\lambda}
Now, part (i) yields the statement.
(iii) Apply part (i).
Corollary 1.6.2 (i) The topology of AA is the inverse image via pp of the topology of II.
(ii) The topological space AA is connected if and only if the topological space II is connected.
(iii) The topological space AA is separable if and only if the topological space II is separable and the countable dense subsets D⊂A,D′⊂ID \subset A, D^{\prime} \subset I can be chosen such that p(D)=D′p(D)=D^{\prime}.
Proof: (i) Proposition 1.6.1 implies immediately that T0(A)={p−1(V)∣V∈\mathcal{T}_{0}(A)=\left\{p^{-1}(V) \mid V \in\right. T0(I)\mathcal{T}_{0}(I).
(ii) According to Proposition 1.6.1, (iii), the surjective map p:A→Ip: A \rightarrow I is continuous, hence the connectedness of AA implies the connectedness of II (see [3, Ch. I, Sec. 11, n ∘{ }^{\circ}, Proposition 4]. Now, let II be a connected topological space and let us suppose that AA is not connected. In other words, there exist two open non-empty and disjoint subsets UU and U′U^{\prime} of AA, such that A=U∪U′A=U \cup U^{\prime}. Proposition 1.6.1, (ii), yields that U=∪λ∈LJλU=\cup_{\lambda \in L} J_{\lambda} and U′=∪μ∈MJμ′U^{\prime}=\cup_{\mu \in M} J_{\mu}^{\prime} for some non-empty families (Jλ)λ∈L,(Jμ′)μ∈M\left(J_{\lambda}\right)_{\lambda \in L},\left(J_{\mu}^{\prime}\right)_{\mu \in M} of open intervals in AA. Let (Kλ)λ∈L\left(K_{\lambda}\right)_{\lambda \in L}, (Kμ′)μ∈M\left(K_{\mu}^{\prime}\right)_{\mu \in M} be the families of open intervals in II, such that Jλ=p−1(Kλ)J_{\lambda}=p^{-1}\left(K_{\lambda}\right) and Jμ′=p−1(Kμ′)J_{\mu}^{\prime}=p^{-1}\left(K_{\mu}^{\prime}\right). We have p(U)=∪λ∈LKλ,p(U′)=∪μ∈MKμ′p(U)=\cup_{\lambda \in L} K_{\lambda}, p\left(U^{\prime}\right)=\cup_{\mu \in M} K_{\mu}^{\prime}, and I=p(U)∪I=p(U) \cup p(U′)p\left(U^{\prime}\right). Since II is a connected topological space, the non-empty open sets p(U)p(U) and p(U′)p\left(U^{\prime}\right) are not disjoint, that is, there exist indices λ∈L\lambda \in L and μ∈M\mu \in M with Kλ∩Kμ′≠∅K_{\lambda} \cap K_{\mu}^{\prime} \neq \emptyset. We have Jλ∩Jμ′=p−1(Kλ∩Kμ′)J_{\lambda} \cap J_{\mu}^{\prime}=p^{-1}\left(K_{\lambda} \cap K_{\mu}^{\prime}\right) and the surjectivity of pp implies Jλ∩Jμ′≠∅J_{\lambda} \cap J_{\mu}^{\prime} \neq \emptyset - a contradiction with U∩U′=∅U \cap U^{\prime}=\emptyset. Therefore AA is connected.
(iii) Let AA be separable and let DD be a countable subset of AA which meets every non-empty open interval JJ in AA. Then in accord with Proposition 1.6.1, (i), the countable image D′=p(D)D^{\prime}=p(D) meets any non-empty open interval KK in II.
Now, let II be a separable topological space and let D′⊂ID^{\prime} \subset I be a countable subset such that D′∩K≠∅D^{\prime} \cap K \neq \emptyset for any non-empty open interval KK in II. Using the axiom of choice, we fix an element d∈p−1(d′)d \in p^{-1}\left(d^{\prime}\right) for any d′∈D′d^{\prime} \in D^{\prime}. The subset D⊂AD \subset A of all d∈Ad \in A just chosen is countable and Proposition 1.6.1, (i), assures that DD meets every non-empty open interval JJ in AA. Thus, AA is separable and, moreover, D′=p(D)D^{\prime}=p(D).
2 Characterization of the negative transitivity
This section is the core of the paper. Here we give a complete characterization of the sets of preferences AA whose preference relation has negatively transitive asymmetric part. As a consequence of this characterization we prove the existence of a generalized continuous utility function on AA under the condition that the topological space AA is connected and separable.
2.1 The necessary condition
Theorem 2.1.1 Let AA be a set of preferences endowed with preference relation RR and let its asymmetric part F=PRF=P_{R} be negatively transitive. Then one has:
(i) The relation ERE_{R} is symmetric and E=EF\mathcal{E}=E_{F} is an equivalence relation on A.
(ii) E=E∪ER\mathcal{E}=E \cup E_{R}, where E=IRE=I_{R} is the symmetric part of RR.
(iii) The relation FF is E\mathcal{E}-saturated.
(iv) The relation EE is weakly E\mathcal{E}-saturated.
(v) The relation RR is weakly E\mathcal{E}-saturated.
Let A′=A/EA^{\prime}=A / \mathcal{E} be the factor-set and let R′R^{\prime} and Fˉ\bar{F} be the weak factor-relation of RR and the factor-relation of FF with respect to the equivalence relation E\mathcal{E}. Then one has:
(vi) R′=DA′∪FˉR^{\prime}=D_{A^{\prime}} \cup \bar{F} and DA′∩Fˉ=∅D_{A^{\prime}} \cap \bar{F}=\emptyset.
(vii) R′R^{\prime} is a reflexive, transitive, antisymmetric, and complete binary relation on A′A^{\prime} with asymmetric part Fˉ\bar{F} and the factor-set A′A^{\prime} endowed with R′R^{\prime} is a linearly ordered set.
(viii) The inverse images Aι=p−1(ι),ι∈A′A_{\iota}=p^{-1}(\iota), \iota \in A^{\prime}, of the canonical surjective map p:A→A′p: A \rightarrow A^{\prime} are EE-saturated and any AιA_{\iota} furnished with the equivalence relation EιE_{\iota} induced by EE is a balloon.
(ix) The set of preferences AA is equal to the coproduct ∐ι∈A′Aι\coprod_{\iota \in A^{\prime}} A_{\iota} of the family (Aι)ι∈A′\left(A_{\iota}\right)_{\iota \in A^{\prime}} of balloons.
Proof: (i) The relation ERE_{R} is symmetric and in accord with Proposition 1.3.3, E\mathcal{E} is an equivalence relation on AA.
(ii) We have E=(F∪F−1)c=Fc∩(Fc)−1=(Rc∪R−1)∩((Rc)−1∪R)=\mathcal{E}=\left(F \cup F^{-1}\right)^{c}=F^{c} \cap\left(F^{c}\right)^{-1}=\left(R^{c} \cup R^{-1}\right) \cap\left(\left(R^{c}\right)^{-1} \cup R\right)= IRc∪IR=ER∪IRI_{R^{c}} \cup I_{R}=E_{R} \cup I_{R}.
(iii) Since FF is asymmetric and negatively transitive, Proposition 1.3.3 yields that FF is E\mathcal{E}-saturated.
(iv) This is a consequence of Lemma 1.3.1.
(v) Part (iii) implies that FF is weakly E\mathcal{E}-saturated. Since R=E∪FR=E \cup F, we obtain that RR is weakly E\mathcal{E}-saturated.
(vi) Using parts (ii), (iii), and (iv), we have R′=E′∪F′=DA′∪FˉR^{\prime}=E^{\prime} \cup F^{\prime}=D_{A^{\prime}} \cup \bar{F}, the last equality because of Proposition 1.3.2, (iv). If DA′∩Fˉ≠∅D_{A^{\prime}} \cap \bar{F} \neq \emptyset, then there exists x∈Ax \in A with xˉFˉxˉ\bar{x} \bar{F} \bar{x}. In particular, xFxx F x and this contradicts E∩F=∅E \cap F=\emptyset. Therefore DA′∩Fˉ=∅D_{A^{\prime}} \cap \bar{F}=\emptyset.
(vii) The weak factor-relation R′R^{\prime} is reflexive and transitive because of Proposition 1.3.2, (i), (ii). Moreover, part (v) together with Proposition 1.3.2, (iii),
(iv), imply that R′R^{\prime} is an antisymmetric relation. Now, let xˉ≠yˉ\bar{x} \neq \bar{y}. Then xEFcyx E_{F}^{c} y and let us suppose, for example, that xFyx F y. Therefore xˉFˉyˉ\bar{x} \bar{F} \bar{y} and this implies xˉR′yˉ\bar{x} R^{\prime} \bar{y}. Thus, R′R^{\prime} is a complete relation. In accord with Proposition 1.3.3, FF is transitive and Proposition 1.3.2, (ii), assures that the factor-relation F′=FˉF^{\prime}=\bar{F} is transitive, too. If xˉFˉyˉ\bar{x} \bar{F} \bar{y} and yˉFˉxˉ\bar{y} \bar{F} \bar{x}, then xˉR′yˉ\bar{x} R^{\prime} \bar{y} and yˉR′xˉ\bar{y} R^{\prime} \bar{x}, and the antisymmetry of R′R^{\prime} implies xˉ=yˉ−\bar{x}=\bar{y}- a contradiction with part (vi). Therefore the relation Fˉ\bar{F} on A′A^{\prime} is asymmetric and the decomposition of R′R^{\prime} from part (vi) yields PR′=FˉP_{R^{\prime}}=\bar{F}.
(viii) It is enough to note that E⊂EE \subset \mathcal{E}.
(ix) The set of preferences AA is equal to the coproduct ∐ι∈A′Aι\coprod_{\iota \in A^{\prime}} A_{\iota} of the family (Aι)ι∈A′\left(A_{\iota}\right)_{\iota \in A^{\prime}} of sets. Since R=E∪FR=E \cup F and since the relation xFyx F y is equivalent to the inequality p(x)<p(y)p(x)<p(y) (part (vi)), we have xRyx R y if and only if " p(x)<p(y)p(x)<p(y) " or " p(x)=p(y)p(x)=p(y) and xEp(x)yx E_{p(x)} y ". In accord with Proposition 1.5.1, (i), AA is the coproduct of the family (Aι)ι∈A′\left(A_{\iota}\right)_{\iota \in A^{\prime}} of balloons.
2.2 The sufficient condition
According to Theorem 2.1.1, (ix), any set of preferences whose preference relation has negatively transitive asymmetric part can be identified with the coproduct of a family of balloons with linearly ordered index set. Theorem 2.2.1 below shows that starting with a family of balloons indexed with a linearly ordered set II, we can endow the coproduct AA of this family with a preference relation RR such that its asymmetric part FF is negatively transitive, the factor-order R′R^{\prime} on the factor-set A′=A/EFA^{\prime}=A / E_{F} is linear, and the linearly ordered set A′A^{\prime} is isomorphic to II.
Theorem 2.2.1 Let (Aι)ι∈I\left(A_{\iota}\right)_{\iota \in I} be a family of balloons, let (Eι)ι∈I\left(E_{\iota}\right)_{\iota \in I} be the family of the corresponding equivalence relations, and let the index set II be linearly ordered. Let E\mathcal{E} be the equivalence relation on the coproduct A=∐ι∈IAιA=\coprod_{\iota \in I} A_{\iota}, which corresponds to the partition (Aι)ι∈I\left(A_{\iota}\right)_{\iota \in I}, let p:A→I,x↦ιp: A \rightarrow I, x \mapsto \iota for x∈Aιx \in A_{\iota}, be the natural projection, and let RR be the preference relation on AA from Proposition 1.5.1, (i), defined by the rule
xRy if " p(x)<p(y)" or " p(x)=p(y) and xEp(x)y"x R y \text { if " } p(x)<p(y) " \text { or " } p(x)=p(y) \text { and } x E_{p(x)} y "
(i) The asymmetric part F=PRF=P_{R} of RR is negatively transitive and one has E=EF\mathcal{E}=E_{F}.
(ii) The relation RR is weakly E\mathcal{E}-saturated.
(iii) The factor-set A′=A/EA^{\prime}=A / \mathcal{E} endowed with the factor-relation R′R^{\prime} is isomorphic to the linearly ordered set II.
Proof: (i) Since Eι=IEιE_{\iota}=I_{E_{\iota}} for any ι∈I\iota \in I, using Proposition 1.5.1, (ii), we have xIRyx I_{R} y if and only if " p(x)=p(y)p(x)=p(y) and xEp(x)yx E_{p(x)} y " and this implies
xFyx F y if and only if p(x)<p(y)p(x)<p(y).
In particular, EF={(x,y)∈A∣p(x)=p(y)}=EE_{F}=\{(x, y) \in A \mid p(x)=p(y)\}=\mathcal{E}.
Now, let xFzx F z and let y∈Ay \in A. Then p(x)<p(z)p(x)<p(z) and the linearity of II yields p(x)<p(y)p(x)<p(y) or p(y)<p(z)p(y)<p(z), that is xFyx F y or yFzy F z. In other words, the asymmetric part FF of RR is negatively transitive.
(ii) Use Part (i) and Theorem 2.1.1, (v).
(iii) Since pp is a surjective map and since E={(x,y)∈A∣p(x)=p(y)}\mathcal{E}=\{(x, y) \in A \mid p(x)=p(y)\}, Proposition 1.3.2, (v), yields an increasing bijection p′:A′→Ip^{\prime}: A^{\prime} \rightarrow I of linearly ordered sets, that is, an isomorphism between A′A^{\prime} and II.
2.3 Existence of a continuous generalized utility function
Let R\mathbb{R} be the set of real numbers. Let AA be a set of preferences, let RR be its preference relation with asymmetric part FF, and let AA be endowed with its interval topology. By a generalized utility function on AA we mean a real function u:A→Ru: A \rightarrow \mathbb{R} such that: (a) u(x)<u(y)u(x)<u(y) if and only if xFyx F y; (b) u(x)=u(y)u(x)=u(y) if and only if xEyx \mathcal{E} y, where E=EF\mathcal{E}=E_{F} is the indifference " xx and yy are not FF comparable". The existence of a generalized utility function yields that E\mathcal{E} is an equivalence relation and that FF is negatively transitive. Conversely, in case FF is negatively transitive, Theorem 2.1.1, (i), (ii), assure that E\mathcal{E} is an equivalence relation and, moreover, under the condition that uu exists, (b) is equivalent to: (b′)u(x)=u(y)\left(\mathrm{b}^{\prime}\right) u(x)=u(y) if and only if " xRyx R y and yRxy R x " or " xx and yy are not RR comparable". When the preference relation RR is complete, condition ( b′\mathrm{b}^{\prime} ) has the form u(x)=u(y)u(x)=u(y) if and only if xRyx R y and yRxy R x, that is, uu is the ordinary utility function.
In the lemma below we use freely the terminology and notation from Theorem 2.1.1 and Theorem 2.2.1.
We denote by UU the set of the four intervals in R\mathbb{R} with endpoints 0 and 1 .
Lemma 2.3.1 Let AA be a set of preferences whose preference relation RR has a negatively transitive asymmetric part FF, let AA be the coproduct of the corresponding family Aι)ι∈IA_{\iota})_{\iota \in I} of balloons whose index set II, after the identification I=A/EI=A / \mathcal{E}, is furnished with the linear factor-order, and let p:A→I,x↦ιp: A \rightarrow I, x \mapsto \iota for x∈Aιx \in A_{\iota}, be the natural projection. Let AA and II be endowed with their interval topologies and let Λ∈U\Lambda \in U. The following two respective statements are then equivalent:
(i) There exists a surjective generalized (and continuous) utility function u:A→Λu: A \rightarrow \Lambda.
(ii) There exists a strictly increasing surjective (and continuous) map u′:I→u^{\prime}: I \rightarrow Λ\Lambda.
Parts (i), (ii) with continuous u,u′u, u^{\prime} imply:
(iii) There exists a homeomorphism u′:I→Λu^{\prime}: I \rightarrow \Lambda.
If, in addition, II is connected, then parts (i), (ii) with continuous u,u′u, u^{\prime} and part (iii) are equivalent.
Proof: (i) ⟺\Longleftrightarrow (ii) There exists a bijection between the maps u:A→Ru: A \rightarrow \mathbb{R} which are constant on the members of the factor-set I=A/EI=A / \mathcal{E} and the maps u′:I→Ru^{\prime}: I \rightarrow \mathbb{R}
such that the diagram
is commutative. The equivalence (2.2.1) shows that uu is strictly increasing if and only if u′u^{\prime} is strictly increasing. Moreover, since E={(x,y)∈A2∣u(x)=u(y)}\mathcal{E}=\left\{(x, y) \in A^{2} \mid u(x)=u(y)\right\}, Proposition 1.3.2, (v), implies that the map u′u^{\prime} is a strictly increasing bijection. Proposition 1.6.1, (iii), and the commutativity of the above diagram (u′=p∘u)\left(u^{\prime}=p \circ u\right) assure that uu is continuous if and only if u′u^{\prime} is continuous.
(ii) ⟹\Longrightarrow (iii) Let u′:I→Λu^{\prime}: I \rightarrow \Lambda be a strictly increasing surjective continuous map. Then u′u^{\prime} is an isomorphism of posets and maps any open interval KK in II onto the open interval u′(K)u^{\prime}(K) in Λ\Lambda. Thus, the bijection u′u^{\prime} is a continuous and open map, hence its inverse u′−1:Λ→Iu^{\prime-1}: \Lambda \rightarrow I is a continuous map, too.
(ii) ⟺\Longleftrightarrow (iii) Let uu and u′u^{\prime} be continuous. Under the condition of connectedness of II, this equivalence may be found in [3, Ch. IV, Sec. 2, Exercise 8 a].
Let Q\mathbb{Q} be the set of rational numbers.
Theorem 2.3.2 Let AA be a set of preferences whose preference relation has a negatively transitive asymmetric part. Let AA be connected and has a countable dense subset DD with respect to the interval topology on AA. Then there exists a continuous generalized utility function uu on AA which maps the set AA onto some interval Λ∈U\Lambda \in U and its subset DD onto the intersection Λ∩Q\Lambda \cap \mathbb{Q}.
Proof: Under the conditions of Lemma 2.3.1, Corollary 1.6.2, (ii), (iii), yields that the index set II is connected and separable with D′=p(D)D^{\prime}=p(D) as a countable dense subset. Now, in accord with [3, Ch. IV, Sec. 2, Exercise 11 a], there exist a interval Λ∈U\Lambda \in U and a map u′:I→Λu^{\prime}: I \rightarrow \Lambda which is an increasing homeomorphism of II onto Λ\Lambda and maps D′D^{\prime} onto the intersection Λ∩Q\Lambda \cap \mathbb{Q}. Composing u′u^{\prime} with the natural projection p:A→Ip: A \rightarrow I, we obtain a map u:A→Λu: A \rightarrow \Lambda which according to Lemma 2.3.1 is the desired continuous generalized utility function.
Remark 2.3.3 In the case of plane A=R2A=\mathbb{R}^{2} endowed with the lexicographical order and the interval topology, the connectedness of the topological space AA fails to be true because the connected components of AA are exactly the “vertical lines” La={(a,b)∈R2∣b∈R},a∈RL_{a}=\left\{(a, b) \in \mathbb{R}^{2} \mid b \in \mathbb{R}\right\}, a \in \mathbb{R}. Moreover, it is well known that on the (complete) set of preferences AA there is no utility function - see, for example, [7, Chapter B, Section 4, 4.2, Example 1]. Thus, we can not expect a general result without AA being connected.
A Appendix
A. 1 Partially Ordered sets: Miscellaneous Results
Let AA be a set. The set PO(A)\mathcal{P O}(A) consisting of all partial orders R⊂A2R \subset A^{2} on AA with the relation R⊂R′R \subset R^{\prime} (that is, inclusion of partial orders) is a poset. Any linear order RR on AA is a maximal element of the poset PO(A)\mathcal{P O}(A) because if (x,y)∉R(x, y) \notin R, then (y,x)∈R(y, x) \in R.
Lemma A.1.1 If RR is a non-linear partial order on the set AA and if a,b∈Aa, b \in A is a pair such that (a,b)∉R(a, b) \notin R and (b,a)∉R(b, a) \notin R, then there exists a partial order R′R^{\prime} which extends RR and contains (a,b)(a, b).
Proof: We set
R(a,b)={(x,y)∈A2∣(x,a)∈R and (b,y)∈R}R_{(a, b)}=\left\{(x, y) \in A^{2} \mid(x, a) \in R \text { and }(b, y) \in R\right\}
and R′=R∪R(a,b)R^{\prime}=R \cup R_{(a, b)}. Since (a,b)∈R(a,b)(a, b) \in R_{(a, b)} we have R′≠RR^{\prime} \neq R. The relation R′R^{\prime} is reflexive since RR is reflexive.
Now, let (x,y)∈R′(x, y) \in R^{\prime} and (y,x)∈R′(y, x) \in R^{\prime}. The relations (x,y)∈R(a,b)(x, y) \in R_{(a, b)} and (y,x)∈R(a,b)(y, x) \in R_{(a, b)} mean (x,a)∈R,(b,y)∈R,(y,a)∈R(x, a) \in R,(b, y) \in R,(y, a) \in R, and (b,x)∈R(b, x) \in R, which in turn implies (b,a)∈R(b, a) \in R - a contradiction. If, for example, (x,y)∈R(x, y) \in R and (y,x)∈R(a,b)(y, x) \in R_{(a, b)}, then (y,a)∈R(y, a) \in R and (b,x)∈R(b, x) \in R. We have (x,a)∈R(x, a) \in R and this yields (b,a)∈R(b, a) \in R - again a contradiction. Thus, the relation R′R^{\prime} is antisymmetric.
Let (x,y)∈R′(x, y) \in R^{\prime} and (y,z)∈R′(y, z) \in R^{\prime}. The case (x,y)∈R(a,b)(x, y) \in R_{(a, b)} and (y,z)∈R(a,b)(y, z) \in R_{(a, b)} is impossible because (b,y)∈R(b, y) \in R and (y,a)∈R(y, a) \in R imply (b,a)∈R(b, a) \in R : a contradiction. Now, let (x,y)∈R(x, y) \in R and (y,z)∈R(a,b)(y, z) \in R_{(a, b)}. Then (y,a)∈R,(b,z)∈R,(x,a)∈R(y, a) \in R,(b, z) \in R,(x, a) \in R, and this yields (x,z)∈R(a,b)(x, z) \in R_{(a, b)}. In the case (x,y)∈R(a,b)(x, y) \in R_{(a, b)} and (y,z)∈R(y, z) \in R we have (x,a)∈R,(b,y)∈R,(b,z)∈R(x, a) \in R,(b, y) \in R,(b, z) \in R, hence (x,z)∈R(x, z) \in R.
We obtain immediately
Corollary A.1.2 Any non-linear partial order on a set is not maximal.
Theorem A.1.3 (E. Szpilrajn) Let AA be set. Any partial order on AA can be extended to a linear order on AA.
Proof: Let RR be a partial order on AA and let us consider the non-empty set E(R)⊂PO(A)\mathcal{E}(R) \subset \mathcal{P O}(A) consisting of all partial orders that extend RR. The partial order in PO(A)\mathcal{P O}(A) induces a partial order on E(R)\mathcal{E}(R) and let C⊂E(R)C \subset \mathcal{E}(R) be a chain. The union U=∪S∈CS\mathrm{U}=\cup_{S \in C} S is a partial order that extends RR, that is, U∈E(R)U \in \mathcal{E}(R), and, moreover, UU is an upper bound of CC. Therefore E(R)\mathcal{E}(R) is an inductively ordered poset and Kuratowski-Zorn theorem yields that there exists a maximal element M∈E(R)M \in \mathcal{E}(R). If the order MM is not linear, then Lemma A.1.1 produces a contradiction.
Let AA and A′A^{\prime} be posets. Isomorphism of posets is a bijection f:A→A′f: A \rightarrow A^{\prime} such that the relations x≤yx \leq y and f(x)≤f(y)f(x) \leq f(y) are equivalent.
Proposition A.1.4 Let AA be a linearly ordered set, and let BB be a poset. Every strictly monotonic map f:A→Bf: A \rightarrow B is injective; if ff is strictly increasing, then ff is an isomorphism of AA and f(A)f(A).
Proof: Indeed, x≠yx \neq y implies x<yx<y or x>yx>y, hence f(x)<f(y)f(x)<f(y) or f(x)>f(y)f(x)>f(y), and in all cases f(x)≠f(y)f(x) \neq f(y). Finally, f(x)≤f(y)f(x) \leq f(y) implies x≤yx \leq y : otherwise we would have x>yx>y and then f(x)>f(y)f(x)>f(y) : a contradiction.
Proposition A.1.5 Let AA and BB be posets and let f:A→Bf: A \rightarrow B be a bijection. Then ff is an isomorphism of posets if and only if ff and its inverse f−1f^{-1} are increasing maps, and under this condition ff and f−1f^{-1} are strictly increasing maps.
Proof: The “only if” part is immediate. Now, let ff and its inverse f−1f^{-1} be increasing maps. The relation x≤yx \leq y in AA implies f(x)≤f(y)f(x) \leq f(y) in BB. The relation x′≤y′x^{\prime} \leq y^{\prime} in BB implies f−1(x′)≤f−1(y′)f^{-1}\left(x^{\prime}\right) \leq f^{-1}\left(y^{\prime}\right) in AA, and it is enough to set x′=f(x)x^{\prime}=f(x) and y′=f(y)y^{\prime}=f(y). Now, let ff be an isomorphism of posets. Then x<yx<y implies f(x)≤f(y)f(x) \leq f(y) and if f(x)=f(y)f(x)=f(y), then x=yx=y : a contradiction. Therefore f(x)<f(y)f(x)<f(y).
Lemma A.1.6 Let AA be a linearly preordered set. Then the intersection of any finite family of open intervals in AA is an open interval in AA.
Proof: Let ((aι,bι))ι∈I\left(\left(a_{\iota}, b_{\iota}\right)\right)_{\iota \in I} be a finite family of open intervals. Let aa be the greatest element of the finite family (aι)ι∈I\left(a_{\iota}\right)_{\iota \in I} and let bb be the least element of the finite family (bι)ι∈I\left(b_{\iota}\right)_{\iota \in I}. Note that a=←a=\leftarrow if aι=←a_{\iota}=\leftarrow for all ι∈I\iota \in I and b=→b=\rightarrow if bι=→b_{\iota}=\rightarrow for all ι∈I\iota \in I. Then we have
∩ι∈I(aι,bι)=(a,b)\cap_{\iota \in I}\left(a_{\iota}, b_{\iota}\right)=(a, b)
Theorem A.1.7 (G. Cantor) Let AA and BB be countable linearly ordered sets that have least elements and greatest elements.
(i) If BB is without gaps, then there exists a strictly increasing map f:A→Bf: A \rightarrow B.
(ii) If AA and BB are without gaps, then the map ff from part (i) is an isomorphism of posets.
Proof: (i) We set A={a1,a2,…,}A=\left\{a_{1}, a_{2}, \ldots,\right\} and B={b1,b2,…,}B=\left\{b_{1}, b_{2}, \ldots,\right\} and without any loss of generality we can suppose that a1,b1a_{1}, b_{1} are the least and a2,b2a_{2}, b_{2} are the greatest elements of AA and BB. Further, we set An={a1,a2,…,an},n≥2A_{n}=\left\{a_{1}, a_{2}, \ldots, a_{n}\right\}, n \geq 2. We define inductively a sequence of strictly increasing maps (fn:An→B)n≥2\left(f_{n}: A_{n} \rightarrow B\right)_{n \geq 2} such that for any nn the map fnf_{n} is an extension of fn−1f_{n-1}. First we define a strictly increasing map f2:A2→Bf_{2}: A_{2} \rightarrow B by f2(a1)=b1f_{2}\left(a_{1}\right)=b_{1} and f2(a2)=b2f_{2}\left(a_{2}\right)=b_{2}. Given n≥3n \geq 3, under the assumption that fn−1f_{n-1} is a strictly increasing map, we define fnf_{n} via the rule
fn(an)=bsf_{n}\left(a_{n}\right)=b_{s}
where s≥3s \geq 3 is the minimal index such that fnf_{n} is a strictly increasing map. It is enough to show that there exists an index s≥3s \geq 3 that satisfies (A.1.1) and has the
latter property. Let aia_{i} be the immediate predecessor and aja_{j} be the immediate successor of ana_{n} in the finite poset AnA_{n}. We have fn−1(ai)<fn−1(aj)f_{n-1}\left(a_{i}\right)<f_{n-1}\left(a_{j}\right) and since the open interval I=(fn−1(ai),fn−1(aj))I=\left(f_{n-1}\left(a_{i}\right), f_{n-1}\left(a_{j}\right)\right) is not empty, each bs∈Ib_{s} \in I works. The sequence (fn)n≥2\left(f_{n}\right)_{n \geq 2}, in turn, defines a strictly increasing map f:A→Bf: A \rightarrow B by the formula f(an)=fm(an)f\left(a_{n}\right)=f_{m}\left(a_{n}\right) for some m≥nm \geq n.
(ii) It is enough to prove that for any integer p≥2p \geq 2 we have bk∈Im(f)b_{k} \in \operatorname{Im}(f) for all k<pk<p. This statement is true for p=2p=2. Now, let p≥3p \geq 3, let us suppose that the statement is true for all integers <p<p, and let us choose n≥2n \geq 2 such that b1,…,bp−1∈fn(An)b_{1}, \ldots, b_{p-1} \in f_{n}\left(A_{n}\right). Let us suppose that bp∉fn(An)b_{p} \notin f_{n}\left(A_{n}\right), let bib_{i} be the immediate predecessor, bκb_{\kappa} be the immediate successor of bpb_{p} in the finite poset fn(An)f_{n}\left(A_{n}\right), and let fn(ai)=bi,fn(aj)=bκf_{n}\left(a_{i}\right)=b_{i}, f_{n}\left(a_{j}\right)=b_{\kappa}. We have An∩(ai,aj)=∅A_{n} \cap\left(a_{i}, a_{j}\right)=\emptyset and choose mm to be the minimal index such that am∈(ai,aj)a_{m} \in\left(a_{i}, a_{j}\right). Then m>n,Am∩(ai,aj)={am}m>n, A_{m} \cap\left(a_{i}, a_{j}\right)=\left\{a_{m}\right\}, and, in particular, aia_{i} is the immediate predecessor and aja_{j} is the immediate successor of ama_{m} in the finite poset AmA_{m}. Moreover, pp is the minimal index with bp∈(fn(ai),fn(aj))b_{p} \in\left(f_{n}\left(a_{i}\right), f_{n}\left(a_{j}\right)\right). In accord with the definition of the map ff from part (i), we have f(am)=fm(am)=bpf\left(a_{m}\right)=f_{m}\left(a_{m}\right)=b_{p}. Thus, our statement is true for p+1p+1 and the principle of mathematical induction yields it for all p≥2p \geq 2.
Corollary A.1.8 For any countable linearly ordered set AA there exists a strictly increasing map f:A→[0,1]Qf: A \rightarrow[0,1]_{\mathbb{Q}}. If, in addition, AA is without gaps, then:
(i) The map ff establishes an isomorphism between AA and the interval [0,1]Q[0,1]_{\mathbb{Q}} if and only if AA has a least element and a greatest element.
(ii) The map ff establishes an isomorphism between AA and the interval [0,1]Q[0,1]_{\mathbb{Q}} if and only if AA has least element but has no greatest element.
(iii) The map ff establishes an isomorphism between AA and the interval (0,1]Q(0,1]_{\mathbb{Q}} if and only if AA has a greatest element but has no least element.
(iv) The map ff establishes an isomorphism between AA and (0,1)Q(0,1)_{\mathbb{Q}} if and only if AA has no least element and no greatest element.
Proof: The existence of a strictly increasing map f:A→[0,1]Qf: A \rightarrow[0,1]_{\mathbb{Q}} is ensured by Theorem A.1.7, (i). The necessity part of the equivalences in (i) - (iv) is immediate.
(i) Note that the interval [0,1]Q[0,1]_{\mathbb{Q}} has least element 0 , greatest element 1 , has no gaps, and then use Theorem A.1.7, (ii).
(ii) We adjoint a greatest element γ\gamma to AA, thus obtaining A′=A⊕{γ}A^{\prime}=A \oplus\{\gamma\}, use part (i) for A′A^{\prime}, and then restrict the corresponding isomorphism on AA.
(iii) We adjoint a least element λ\lambda to AA, thus obtaining A′={λ}⊕AA^{\prime}=\{\lambda\} \oplus A, use part (i) for A′A^{\prime}, and then restrict the corresponding isomorphism on AA.
(iv) We adjoint a greatest element γ\gamma and a least element λ\lambda to AA, thus obtaining A′′={λ}⊕A⊕{γ}A^{\prime \prime}=\{\lambda\} \oplus A \oplus\{\gamma\}, use part (i) for A′′A^{\prime \prime}, and then restrict the corresponding isomorphism on AA.
Using Corollary A.1.8, (iv), twice we obtain
Corollary A.1.9 Every countable subset of the open interval (0,1)R(0,1)_{\mathbb{R}}, which is dense in this interval, is isomorphic to Q\mathbb{Q}.
Combining Theorem A.1.3 and Corollary A.1.8 we obtain
Corollary A.1.10 Given a countable partially ordered set AA, there exists a strictly increasing map of AA into Q\mathbb{Q}.
Corollary A.1.11 If AA is a countable linearly ordered set AA without gaps, then there exists a bounded from above subset BB of AA, which has no least upper bound in AA.
Proof: Let f:A→[0,1]Qf: A \rightarrow[0,1]_{\mathbb{Q}} be a strictly increasing map with image II which coincides with one of the four intervals in the rational line Q\mathbb{Q} with endpoints 0 and 1. The existence of ff is assured by Corollary A.1.8. Let J=I∩(0,22)RJ=I \cap\left(0, \frac{\sqrt{2}}{2}\right)_{\mathbb{R}} and let B=f−1(J)B=f^{-1}(J). Since the subset JJ of II has no least upper bound in II, the subset BB of AA has no least upper bound in AA.
A. 2 Interval Topology: Connectedness
Let AA be a preordered set with preorder RR and let AA be furnished with its interval topology T0(A)\mathcal{T}_{0}(A). A closed interval [x,y][x, y] with xRyx R y is said to be a gap in AA if [x,y]={x,y}[x, y]=\{x, y\}, that is, the open interval (x,y)(x, y) is empty.
We use the same terminology and notation for a partially ordered set AA.
Examples A.2.1 (1) The natural topology of the real line R\mathbb{R} coincides with its interval topology T0(R)\mathcal{T}_{0}(\mathbb{R}). The (countable) set of all open intervals with rational endpoints is a base of this topology and the trace of this set on Q⊂R\mathbb{Q} \subset \mathbb{R} is by definition the natural base of the interval top[ology T0(Q)\mathcal{T}_{0}(\mathbb{Q}) on the rational line Q\mathbb{Q}. In particular, the topology of Q\mathbb{Q} as a subspace of the real line R\mathbb{R} coincides with its interval topology T0(Q)\mathcal{T}_{0}(\mathbb{Q}).
(2) The natural topology of any interval II in the real line R\mathbb{R}, that is, the topology on II, considered as a subspace of R\mathbb{R}, coincides with its interval topology T0(I)\mathcal{T}_{0}(I) because the trace of any open interval in R\mathbb{R} on II is an open interval in II.
(3) The natural topology of any interval JJ in the rational line Q\mathbb{Q}, that is, the topology on JJ, considered as a subspace of Q\mathbb{Q}, coincides with its interval topology T0(J)\mathcal{T}_{0}(J) because the trace of any open interval in Q\mathbb{Q} on JJ is an open interval in JJ. In accord with Example (1), any open interval JJ in the rational line Q\mathbb{Q} with its natural topology is a subspace of the real line R\mathbb{R}, too.
(4) Given a natural number n∈Nn \in \mathbb{N}, we set A=[0,1]QA=[0,1]_{\mathbb{Q}},
Fn={(1i,1j)∈Q×Q∣i,j∈N,1≤j<i≤n}∪{(0,1i)∣i∈N,1≤i≤n}F_{n}=\left\{\left(\frac{1}{i}, \frac{1}{j}\right) \in \mathbb{Q} \times \mathbb{Q} \mid i, j \in \mathbb{N}, 1 \leq j<i \leq n\right\} \cup\left\{\left(0, \frac{1}{i}\right) \mid i \in \mathbb{N}, 1 \leq i \leq n\right\}
The set FnF_{n} is an asymmetric and transitive binary relation on AA and Rn=R_{n}= DA∪FnD_{A} \cup F_{n} is a reflexive and transitive binary relation on AA, that is, a partial order on AA, with asymmetric part FnF_{n}. We denote by AnA_{n} the set AA endowed with the partial order RnR_{n}. Every member of FnF_{n}, that is, an ordered pair of the
form (1i,1j),i,j∈N,1≤j<i≤n\left(\frac{1}{i}, \frac{1}{j}\right), i, j \in \mathbb{N}, 1 \leq j<i \leq n, or (0,1i),i∈N,1≤i≤n\left(0, \frac{1}{i}\right), i \in \mathbb{N}, 1 \leq i \leq n, defines an open interval in the interval topology T0(An)\mathcal{T}_{0}\left(A_{n}\right) :
(1i,1j)An={a∈A∣(1i,a)∈Fn and (a,1j)∈Fn}\left(\frac{1}{i}, \frac{1}{j}\right)_{A_{n}}=\left\{a \in A \mid\left(\frac{1}{i}, a\right) \in F_{n} \text { and }\left(a, \frac{1}{j}\right) \in F_{n}\right\}
and
(0,1i)An={a∈A∣(0,a)∈Fn and (a,1i)∈Fn}\left(0, \frac{1}{i}\right)_{A_{n}}=\left\{a \in A \mid(0, a) \in F_{n} \text { and }\left(a, \frac{1}{i}\right) \in F_{n}\right\}
respectively. Every element of the family Fn\mathbb{F}_{n} of open intervals thus obtained has the form
(1i,1j)An={1k∣ 1i<1k<1j and k∈N,1≤k≤n}\left(\frac{1}{i}, \frac{1}{j}\right)_{A_{n}}=\left\{\frac{1}{k} \left\lvert\, \frac{1}{i}<\frac{1}{k}<\frac{1}{j}\right.\right. \text { and } k \in \mathbb{N}, 1 \leq k \leq n\}
and
(0,1i)An={1k∣ 1k<1i and k∈N,1≤k≤n}\left(0, \frac{1}{i}\right)_{A_{n}}=\left\{\frac{1}{k} \left\lvert\, \frac{1}{k}<\frac{1}{i}\right.\right. \text { and } k \in \mathbb{N}, 1 \leq k \leq n\}
In particular, the closed intervals [0,1n]An\left[0, \frac{1}{n}\right]_{A_{n}} and [1j+1,1j]An,1≤j≤n−1\left[\frac{1}{j+1}, \frac{1}{j}\right]_{A_{n}}, 1 \leq j \leq n-1, are all gaps in the partially ordered set AnA_{n}.
Since the intersection of any finite number of members of the family Fn\mathbb{F}_{n} is again a member of Fn\mathbb{F}_{n}, we obtain that the finite family Fn\mathbb{F}_{n} is a base of the interval topology T0(An)\mathcal{T}_{0}\left(A_{n}\right). On the other hand, any union of members of Fn\mathbb{F}_{n} equals a finite union of disjoint members of Fn\mathbb{F}_{n}. Therefore the interval topology T0(An)\mathcal{T}_{0}\left(A_{n}\right) is a finite set consisting of finite unions of disjoint members of Fn\mathbb{F}_{n}. In particular, all open sets different from AnA_{n} are finite and all closed sets different from the empty set ∅\emptyset are infinite (and countable), which implies that AnA_{n} is a connected topological space.
Proposition A.2.2 Let AA and A′A^{\prime} be partially ordered sets endowed with interval topology.
(i) If f:A→A′f: A \rightarrow A^{\prime} is an isomorphism of posets, then ff is a homeomorphism.
(ii) Let AA be a linearly ordered set and f:A→A′f: A \rightarrow A^{\prime} be a strictly increasing map with image C⊂A′C \subset A^{\prime}. If the topology of CC induced from A′A^{\prime} coincides with its interval topology, then ff is a homeomorphism of AA onto the chain CC considered as a subspace of A′A^{\prime}.
Proof: (i) In accord with Proposition A.1.5, ff and its inverse f−1f^{-1} are strictly increasing maps. Therefore
f((x,y))=(f(x),f(y)), and f−1((x′,y′))=(f−1(x′),f−1(y′))f((x, y))=(f(x), f(y)), \text { and } f^{-1}\left(\left(x^{\prime}, y^{\prime}\right)\right)=\left(f^{-1}\left(x^{\prime}\right), f^{-1}\left(y^{\prime}\right)\right)
for any open interval (x,y)⊂A(x, y) \subset A and any open interval (x′,y′)⊂A′\left(x^{\prime}, y^{\prime}\right) \subset A^{\prime}. Thus, the map ff (respectively, f−1f^{-1} ) maps the sub(base) of the topological space AA (respectively, A′A^{\prime} ) onto the (sub)base of the topological space A′A^{\prime} (respectively, A)A) and therefore ff is a homeomorphism.
(ii) According to Proposition A.1.4, f:A→Cf: A \rightarrow C is an isomorphism of posets. If U′⊂A′U^{\prime} \subset A^{\prime} is open, then the trace U′∩CU^{\prime} \cap C is open in CC and hence f−1(U′)=f^{-1}\left(U^{\prime}\right)= f−1(U′∩C)f^{-1}\left(U^{\prime} \cap C\right) is an open set in AA because of part (i).
The map
f:R→(−1,1)R,x↦x1+∣x∣f: \mathbb{R} \rightarrow(-1,1)_{\mathbb{R}}, x \mapsto \frac{x}{1+|x|}
is a strictly increasing bijection and the map
g:(−1,1)R→(0,1)R,x↦2x+1x+3g:(-1,1)_{\mathbb{R}} \rightarrow(0,1)_{\mathbb{R}}, x \mapsto 2 \frac{x+1}{x+3}
is a strictly increasing bijection. Thus, the composition h=g∘fh=g \circ f,
h:R→(0,1)R,x↦x+1+∣x∣x+3+3∣x∣h: \mathbb{R} \rightarrow(0,1)_{\mathbb{R}}, x \mapsto \frac{x+1+|x|}{x+3+3|x|}
is a strictly increasing bijection and in accord with Proposition A.1.4, hh is an isomorphism of posets. The restriction
h∣Q:Q→(0,1)Q,x↦x+1+∣x∣x+3+3∣x∣h_{\mid \mathbb{Q}}: \mathbb{Q} \rightarrow(0,1)_{\mathbb{Q}}, x \mapsto \frac{x+1+|x|}{x+3+3|x|}
is a strictly increasing bijection and the same argument yields that it is an isomorphism of posets.
Proposition A.2.3 Both bijections hh and h∣Qh_{\mid \mathbb{Q}} are homeomorphisms of the corresponding sets endowed with the interval topology.
Proof: We apply Proposition A.2.2.
Remarks A.2.4 (1) We note that the base of the interval topology on the underlying set AA is invariant if we replace the poset AA with its dual Aop A^{\text {op }}. Therefore the dual poset structure produces the same interval topology T0(A)\mathcal{T}_{0}(A).
(2) If the subset B⊂AB \subset A is endowed with the induced linear order, then the corresponding interval topology T0(B)\mathcal{T}_{0}(B) is, in general, weaker than the topology of BB considered as a subspace of AA.
Proposition A.2.5 Let AA be a linearly ordered set endowed with the interval topology and let B⊂A,x,y∈AB \subset A, x, y \in A, and x<yx<y. If AA is a set without gaps and (x,y)⊂B(x, y) \subset B, then x∈Bˉx \in \bar{B} and y∈Bˉy \in \bar{B}, where Bˉ\bar{B} is the closure of BB in AA.
Proof: Let II be an open interval that contains xx (respectively, yy ). It is enough to consider intervals of the form I=(a,b)I=(a, b) with a<x<ba<x<b (respectively, a<a< y<b)y<b). We set m=min{y,b}m=\min \{y, b\} (respectively, M=max{x,a}M=\max \{x, a\} ). Since AA is a set without gaps, the open interval (x,m)(x, m) (respectively, (M,y)(M, y) ) is not empty and (x,m)⊂B∩I(x, m) \subset B \cap I (respectively, (M,y)⊂B∩I(M, y) \subset B \cap I ). In particular, B∩I≠∅B \cap I \neq \emptyset and hence x∈Bˉx \in \bar{B} (respectively, y∈Bˉy \in \bar{B} ).
Corollary A.2.6 If z=infBz=\inf B or z=supBz=\sup B, then z∈Bˉz \in \bar{B}.
Proof: Let z=infBz=\inf B (respectively, z=supBz=\sup B ) and let I=(a,b)I=(a, b) be an open interval that contains zz. Then there exists x∈Bx \in B with z<x<bz<x<b (respectively, a<x<za<x<z ) and, in particular, I∩B≠∅I \cap B \neq \emptyset. Thus, z∈Bˉz \in \bar{B}.
Proposition A.2.7 Let AA be a non-empty linearly ordered set equipped with the interval topology. The topological space AA is compact if and only if any subset of AA has a least upper bound and a greatest lower bound.
Proof: Let AA be a compact space and let BB be a non-empty subset of AA endowed with the induced linear order. Since BB is filtered to the right, we can form the section filter of BB, that is, the filter on BB of base
B={Bx∣Bx=[x,→)B,x∈B}\mathcal{B}=\left\{B_{x} \mid B_{x}=[x, \rightarrow)_{B}, x \in B\right\}
consisting of all closed right-unbounded intervals in BB. Let G\mathcal{G} be the filter generated by B\mathcal{B} when B\mathcal{B} is considered as a filter base on AA. Note that the filter G\mathcal{G} is finer than the filter on AA, generated by the filter base
A={Ax∣Ax=[x,→)A,x∈B}\mathcal{A}=\left\{A_{x} \mid A_{x}=[x, \rightarrow)_{A}, x \in B\right\}
Since AA is compact, the filter G\mathcal{G} has a cluster point a∈Aa \in A. Therefore aa is a cluster point of the filter base A\mathcal{A} which consists of closed subsets of AA; in particular a∈Axa \in A_{x} for all x∈Bx \in B, that is, aa is an upper bound of the set BB. If bb is an upper bound of BB and b<ab<a, then the open neighborhood (b,→)(b, \rightarrow) of the point aa does not contains elements of BB, which is a contradiction. Hence a≤ba \leq b for any upper bound bb of BB, that is, a=supABa=\sup _{A} B. The topological space AA is still compact if we change the structure of the linearly ordered set on AA with its dual AopA^{o p}. Then the latter statement means that any non-empty subset of AopA^{o p} has a least upper bound, that is, any non-empty subset of AA has a greatest lower bound. In particular, the whole set AA has a least element mm and a greatest element MM. If B=∅B=\emptyset, then supAB=m\sup _{A} B=m and infAB=M\inf _{A} B=M.
Conversely, let us suppose that any non-empty subset of AA has a least upper bound and a greatest lower bound and let F\mathcal{F} be a filter on AA. Let LL be the set of all greatest lower bounds infAC\inf _{A} C of members CC of the filter F\mathcal{F} and let a=supALa=\sup _{A} L. We will prove that aa is a cluster point of F\mathcal{F}. Let us suppose that there exists C∈FC \in \mathcal{F} such that supAC<a\sup _{A} C<a. Then there exists C′∈FC^{\prime} \in \mathcal{F} with supAC<infAC′\sup _{A} C<\inf _{A} C^{\prime}, hence
supAC∩C′≤supAC<infAC′≤infAC∩C′\sup _{A} C \cap C^{\prime} \leq \sup _{A} C<\inf _{A} C^{\prime} \leq \inf _{A} C \cap C^{\prime}
which is a contradiction because C∩C′∈FC \cap C^{\prime} \in \mathcal{F} and, in particular, C∩C′≠∅C \cap C^{\prime} \neq \emptyset. Thus, for any C∈FC \in \mathcal{F} we have a∈[infAC,supAC]a \in\left[\inf _{A} C, \sup _{A} C\right]. If infAC0=supAC0\inf _{A} C_{0}=\sup _{A} C_{0} for some C0∈FC_{0} \in \mathcal{F}, then C0={x},x∈AC_{0}=\{x\}, x \in A, and the filter F\mathcal{F} is the trivial ultrafilter U(x)U(x). Since x∈Cx \in C for all C∈FC \in \mathcal{F}, we have infAC≤x\inf _{A} C \leq x for all C∈FC \in \mathcal{F}. Therefore a=supAL≤xa=\sup _{A} L \leq x. On the other hand, {x}∈F\{x\} \in \mathcal{F}, so x=infA{x}≤ax=\inf _{A}\{x\} \leq a, hence x=ax=a. Since the ultrafilter F=U(a)\mathcal{F}=U(a) converges to aa, the point aa is a cluster point of F\mathcal{F}. Now, let us consider the case when infAC<supAC\inf _{A} C<\sup _{A} C for all C∈FC \in \mathcal{F}.
Let us suppose the existence of an element C∈FC \in \mathcal{F} and an open interval II that contains aa, such that C∩I=∅C \cap I=\emptyset. If I=(a′,→)I=\left(a^{\prime}, \rightarrow\right) (respectively, ParseError: KaTeX parse error: Expected '\right', got 'EOF' at end of input: … \prime}\right) ) then supAC≤a′<a\sup _{A} C \leq a^{\prime}<a (respectively, a<a′′≤infACa<a^{\prime \prime} \leq \inf _{A} C ), which in both cases is a contradiction. Now, suppose that I=(a′,a′′)I=\left(a^{\prime}, a^{\prime \prime}\right) where a′<a′′a^{\prime}<a^{\prime \prime}. Then C=C′∪C′′C=C^{\prime} \cup C^{\prime \prime}, where C′=C∩(←,a′]C^{\prime}=C \cap\left(\leftarrow, a^{\prime}\right] and C′′=C∩[a′′,→)C^{\prime \prime}=C \cap\left[a^{\prime \prime}, \rightarrow\right). Since a′<aa^{\prime}<a, there exists D∈FD \in \mathcal{F} such that a′<infDa^{\prime}<\inf D. Now, the inequalities infAD≤a<a′′\inf _{A} D \leq a<a^{\prime \prime} yield C∩D=C′′∩DC \cap D=C^{\prime \prime} \cap D and, in particular, infA(C∩D)≥a′′>a\inf _{A}(C \cap D) \geq a^{\prime \prime}>a. On the other hand, C∩D∈FC \cap D \in \mathcal{F} implies infA(C∩D)≤a\inf _{A}(C \cap D) \leq a which is a contradiction. Thus, aa is a cluster point of the filter F\mathcal{F}.
Proposition A.2.8 Let AA be a non-empty linearly ordered set endowed with the interval topology. If every closed interval [x,y],x,y∈A,x<y[x, y], x, y \in A, x<y, is a connected subset of AA, then the topological space AA is connected.
Proof: If A=X∪YA=X \cup Y where XX and YY are non-empty disjoint open subsets of AA and if x∈X,y∈Yx \in X, y \in Y with x<yx<y, then the closed interval I=[x,y]I=[x, y] is not connected. Indeed, I=(I∩X)∪(I∩Y)I=(I \cap X) \cup(I \cap Y), where the traces I∩X,I∩YI \cap X, I \cap Y are non-empty, disjoint, and open (with respect to the induced topology) sets of II.
Proposition A.2.9 Let AA be a non-empty linearly ordered set endowed with the interval topology and let BB be a non-empty subset of AA, bounded from above. Let UU be the set of all upper bounds of BB and LL be the set of all lower bounds of UU. Then one has:
(i) LL and UU are non-empty subsets of AA with A=L∪UA=L \cup U.
(ii) The following three statements are equivalent:
(a) z∈L∩Uz \in L \cap U;
(b) zz is the greatest element of LL;
© z=supBz=\sup B.
(iii) If the intersection L∩UL \cap U is empty, then LL and UU are open sets and AA is not connected.
Proof: (i) We have B⊂LB \subset L and, in particular, L≠∅L \neq \emptyset. Since BB is bounded from above, U≠∅U \neq \emptyset. If x∈Ax \in A then either x∈Ux \in U, or there exists b∈Bb \in B with x<bx<b and this implies x<yx<y for all y∈Uy \in U, hence x∈Lx \in L.
(ii) (a) ⟹\Longrightarrow (b) and ©. The members z∈L∩Uz \in L \cap U satisfy z∈L,x≤zz \in L, x \leq z for all x∈Lx \in L, and z∈U,z≤yz \in U, z \leq y for all y∈Uy \in U. In other words, zz is the greatest element of LL and the least element of UU.
(b) or © ⟹\Longrightarrow (a). Let zz be the greatest element of LL, that is, x≤zx \leq z for all x∈Lx \in L and z≤yz \leq y for all y∈Uy \in U. In particular, z∈Uz \in U. Finally, let z=supBz=\sup B, that is, z∈Uz \in U and z≤yz \leq y for all y∈Uy \in U. In particular, z∈Lz \in L.
(iii) Let L∩U=∅L \cap U=\emptyset. Part (i) implies that the subset BB has no least upper bound, that is, for any y∈Uy \in U there exists y′∈Uy^{\prime} \in U with y∈(y′,→)y \in\left(y^{\prime}, \rightarrow\right). Thus UU is an open set. Because of part (ii) the subset LL has no greatest element. Therefore for any x∈Lx \in L there exists x′∈Lx^{\prime} \in L with x∈(←,x′)x \in\left(\leftarrow, x^{\prime}\right). In other words, LL is an open set too and because of part (i), the topological space AA is not connected.
Proposition A.2.10 Let AA be a non-empty linearly ordered set endowed with the interval topology. If the topological space AA is connected, then:
(i) Every non-empty bounded from above subset B⊂AB \subset A has a least upper bound.
(ii) AA is a set without gaps.
Proof: (i) Let UU be the set of all upper bounds of BB and LL be the set of all lower bounds of UU. Proposition A.2.9, (iii), implies that the intersection L∩UL \cap U is not empty and then Proposition A.2.9, (ii), yields the result.
(ii) If AA is a set with gaps, then there exist x<yx<y such that the open interval (x,y)(x, y) is empty. Then
A=(←,x]∪[y,→)A=(\leftarrow, x] \cup[y, \rightarrow)
hence the set AA is not connected - a contradiction.
After passing to the dual structure AcpA^{c p}, part (i) of Proposition A.2.10 yields
Corollary A.2.11 Let AA be a non-empty linearly ordered set endowed with the interval topology. If the topological space AA is connected, then every non-empty bounded from below subset B⊂AB \subset A has a greatest lower bound.
Combining Proposition A.2.7, Proposition A.2.10, Corollary A.2.11, and taking into account Corollary 1.8.10, we obtain immediately
Proposition A.2.12 Let AA be a non-empty linearly ordered set endowed with the interval topology. If the topological space AA is connected, then every nonempty bounded and closed subset B⊂AB \subset A is compact.
Corollary A.2.13 Every bounded and closed interval is compact.
Proposition A.2.14 If the conditions (i), (ii) of Proposition A.2.10 hold, then every closed interval [x,y],x<y[x, y], x<y, is connected.
Proof: Suppose that conditions (i), (ii) hold and let I=[x,y],x<yI=[x, y], x<y, be a closed interval which is a disjoint union of two non-empty closed (with respect to the induced topology) sets XX and YY with y∈Yy \in Y. Since XX is bounded from above by yy, it possesses a least upper bound z=supAXz=\sup _{A} X and we have z≤yz \leq y. Moreover, there exists a x′∈Xx^{\prime} \in X and then the inequalities x≤x′,x′≤zx \leq x^{\prime}, x^{\prime} \leq z imply z∈Iz \in I. Since XX is closed in II, Corollary A.2.6 yields z∈Xz \in X. Both possibilities z=yz=y and z<yz<y lead to the contradiction z∈X∩Yz \in X \cap Y, the second one by using the inclusion (z,y)⊂Y(z, y) \subset Y, Proposition A.2.5, and the closeness of YY.
Corollary A.2.13 and Proposition A.2.14 imply
Corollary A.2.15 Let AA be a non-empty linearly ordered set endowed with the interval topology. If the topological space AA is connected, then AA is locally compact and locally connected.
Proposition A.2.16 If AA is a non-empty linearly ordered set endowed with the interval topology and if the two conditions (i) and (ii) from Proposition A.2.10 hold, then AA is connected.
Proof: Proposition A.2.14 and Proposition A.2.8 imply that the set AA is connected.
Propositions A.2.10 and A.2.16 yield immediately
Corollary A.2.17 let AA be a non-empty linearly ordered set endowed with the interval topology. The topological space AA is connected if and only if the two conditions (i) and (ii) from Proposition A.2.10 hold.
Proposition A.2.18 Let AA be a non-empty linearly ordered set endowed with the interval topology. If AA is connected, then a subset of AA is connected if and only if it is an interval (bounded or not bounded).
Proof: Proposition A.2.10 and Proposition A.2.14 yield that every bounded closed interval is connected. Now, let II be any interval and let x,y∈I,x<yx, y \in I, x<y. Then [x,y]⊂I[x, y] \subset I and the closed interval [x,y]I=[x,y][x, y]_{I}=[x, y] is a connected subset of II. According to Proposition A.2.8, the interval II is a connected subset of AA. Now, let B⊂AB \subset A be a connected subset of AA. If B=∅B=\emptyset or B={y}B=\{y\}, then B=(x,x)B=(x, x), for any x∈Ax \in A, or, B=[y,y]B=[y, y], respectively. Now, let x,y∈B,x<yx, y \in B, x<y and let z∈A,x<z<yz \in A, x<z<y. If z∉Bz \notin B, then J=(←,z]B=(←,z)BJ=(\leftarrow, z]_{B}=(\leftarrow, z)_{B} is a non-empty open and closed subset of BB. The connectedness of BB implies J=BJ=B and this equality contradicts y∉Jy \notin J. Thus, z∈Bz \in B and for any x,y∈B,x<yx, y \in B, x<y, we have [x,y]B=[x,y][x, y]_{B}=[x, y]. If BB is not bounded from below and from above we obtain B=A=(←,→)B=A=(\leftarrow, \rightarrow). If the subset BB is bounded from above, not bounded from below, and if z=supBz=\sup B, we obtain B=(←,z)B=(\leftarrow, z) or B=(←,z]B=(\leftarrow, z] in case z∉Bz \notin B or z∈Bz \in B, respectively. Dually, if BB is bounded from below, not bounded from above and if z=infBz=\inf B, we obtain B=(z,→)B=(z, \rightarrow) or B=[z,→)B=[z, \rightarrow) in case z∉Bz \notin B or z∈Bz \in B, respectively. Finally, if BB is bounded, x=infBx=\inf B, and y=supBy=\sup B, then x<yx<y ( BB has at least two points), and BB coincides with one of the four intervals with end-points xx and yy.
Lemma A.2.19 Let AA and BB be linearly ordered sets endowed with the interval topology, let AA be connected, and let f:A→Bf: A \rightarrow B be a continuous map with image C=f(A)C=f(A).
(i) For any x<yx<y in AA the image JJ of the closed interval [x,y][x, y] is an interval in CC, which contains the closed interval in CC with endpoints f(x)f(x) and f(y)f(y).
If, in addition, ff is an injective map, then:
(ii) The image JJ coincides with the closed interval in CC with endpoints f(x)f(x) and f(y)f(y).
(iii) The map ff is strictly monotonic on any three-element subset of AA.
Proof: (i) The image CC is a connected subset of BB and Proposition A.2.18 yields that JJ is an interval (bounded or not bounded) in CC. Moreover, f(x),f(y)∈Jf(x), f(y) \in J, so the closed interval in CC with endpoints f(x)f(x) and f(y)f(y) is a subset of JJ.
(ii) Let ff be an injective map and let z∈(x,y)z \in(x, y). First, let us suppose that f(x)<f(y)f(x)<f(y). If f(z)<f(x)f(z)<f(x), then in accord with part (i), there exists x′∈[z,y]x^{\prime} \in[z, y] such that f(x)=f(x′)f(x)=f\left(x^{\prime}\right) and the injectivity of ff contradicts the inequality x<x′x<x^{\prime}. The case f(y)<f(z)f(y)<f(z) can be treated similarly. Thus, the inequality f(x)<f(y)f(x)<f(y) implies f(x)<f(z)<f(y)f(x)<f(z)<f(y). Following the same way, the inequality f(y)<f(x)f(y)<f(x) implies f(y)<f(z)<f(x)f(y)<f(z)<f(x). In both cases we obtain that JJ coincides with the closed interval in CC with endpoints f(x)f(x) and f(y)f(y).
(iii) Let T={x,y,z}T=\{x, y, z\} be a three-element subset of AA, let x<z<yx<z<y, and let, for example, f(x)<f(y)f(x)<f(y). The inequality f(x)<f(y)<f(z)f(x)<f(y)<f(z) (respectively, f(z)<f(x)<f(y))f(z)<f(x)<f(y)) and part (ii) yield the existence of an element y′∈(x,z)y^{\prime} \in(x, z) (respectively, x′∈(z,y)x^{\prime} \in(z, y) ) such that f(y)=f(y′)f(y)=f\left(y^{\prime}\right) (respectively, f(x)=f(x′)f(x)=f\left(x^{\prime}\right) ) which produces contradiction to the injectivity of ff. Thus, f(x)<f(z)<f(y)f(x)<f(z)<f(y) and ff is strictly increasing on TT. By considering the dual order on YY, in the case f(y)<f(x)f(y)<f(x) we prove that ff is strictly decreasing on TT.
Corollary A.2.20 Let AA and BB be linearly ordered sets endowed with the interval topology, let AA be a connected topological space, and let f:A→Bf: A \rightarrow B be a map with image C=f(A)C=f(A). Then ff is a homeomorphism of AA onto the subspace CC if and only if ff is continuous and strictly monotonic.
Proof: We suppose that the set AA has at least two elements, the case of a singleton AA being clear. Let ff be a continuous and strictly monotonic map. After eventual change of the poset structure of BB with its dual Bop B^{\text {op }} (this leaves the interval topology on BB invariant), we can suppose that ff is a strictly increasing map. Proposition A.1.4 implies that ff is an isomorphism of posets of AA and CC. Therefore ff maps the open interval (x,y)(x, y) in AA onto the open interval (f(x),f(y))(f(x), f(y)) in CC. Thus, the bijection f:A→Cf: A \rightarrow C is a continuous and open map. In particular, its inverse f−1:C→Af^{-1}: C \rightarrow A is a continuous map, too.
Now, let ff be a homeomorphism of AA onto its image CC. In particular, ff is an injective and continuous map. Let us suppose that there exist two pairs x,y∈A,x<yx, y \in A, x<y, and x′,y′∈A,x′<y′x^{\prime}, y^{\prime} \in A, x^{\prime}<y^{\prime}, such that f(x)<f(y)f(x)<f(y) and f(y′)<f(x′)f\left(y^{\prime}\right)<f\left(x^{\prime}\right). In case the intersection {x,y}∩{x′,y′}\{x, y\} \cap\left\{x^{\prime}, y^{\prime}\right\} is a singleton, Lemma A.2.19, (iii), yields that ff is strictly monotonic on the triple {x,y}∪{x′,y′}\{x, y\} \cup\left\{x^{\prime}, y^{\prime}\right\} which is a contradiction. Otherwise, ff is strictly increasing on the triple {x,y,x′}\left\{x, y, x^{\prime}\right\} and strictly decreasing on the triple {x,x′,y′}\left\{x, x^{\prime}, y^{\prime}\right\}, which again is a contradiction because {x,y,x′}∩{x,x′,y′}={x,x′}\left\{x, y, x^{\prime}\right\} \cap\left\{x, x^{\prime}, y^{\prime}\right\}=\left\{x, x^{\prime}\right\}.
Corollary A.2.21 Let AA be a countable linearly ordered set endowed with the interval topology.
(i) There exists a strictly increasing homeomorphism of AA onto one of the intervals of the rational line Q\mathbb{Q} with endpoints 0 and 1 , endowed with its interval topology.
(ii) There exists a strictly increasing homeomorphism of AA onto a subspace of Q\mathbb{Q}.
Proof: (i) According to Corollary A.1.10, there exists a strictly increasing map ff of AA onto an interval II in Q\mathbb{Q} with endpoints 0 and 1 , which turns out to
be an poset isomorphism of AA onto II, because of Proposition A.1.4. Now, Proposition A.2.2 yields that ff is a homeomorphism of AA and II endowed with their interval topologies.
(ii) In accord with Examples A.2.1, (3), the map ff is a strictly increasing homeomorphism of AA onto a subspace of Q\mathbb{Q}.
Remark A.2.22 If AA is a countable partially ordered set endowed with the interval topology, then it is not necessarily true that it can be embedded as a subspace of the rational line Q\mathbb{Q}. Something more, AA can not be embedded as a subspace of any linearly ordered set LL endowed with its interval topology. Indeed, let AA be the countable partially ordered set with partial order ≤A\leq_{A} from Examples A.2.1, (4), and suppose that there exists a linearly ordered set LL with partial order ≤L\leq_{L}, and a strictly increasing homeomorphism ff of AA onto a subspace of LL. Using ff, we can identify AA and its image and can suppose that A⊂LA \subset L is a subspace of LL such that a<Aba<_{A} b implies a<Lba<_{L} b. In other words, the only inequalities in AA are
1i<A1j,i,j∈N,1≤j<i≤n0<A1i,i∈N,1≤i≤n\begin{aligned} & \frac{1}{i}<_{A} \frac{1}{j}, i, j \in \mathbb{N}, 1 \leq j<i \leq n \\ & 0<_{A} \frac{1}{i}, i \in \mathbb{N}, 1 \leq i \leq n \end{aligned}
and we have
1i<L1j,i,j∈N,1≤j<i≤n0<L1i,i∈N,1≤i≤n\begin{aligned} & \frac{1}{i}<_{L} \frac{1}{j}, i, j \in \mathbb{N}, 1 \leq j<i \leq n \\ & 0<_{L} \frac{1}{i}, i \in \mathbb{N}, 1 \leq i \leq n \end{aligned}
Let r1,r2,r3∈Ar_{1}, r_{2}, r_{3} \in A be three pairwise different rational numbers strictly greater than 12\frac{1}{2} and strictly less than 1 . We can suppose that r1<Lr2<Lr3r_{1}<_{L} r_{2}<_{L} r_{3}. then U=(r1,r3)L∩AU=\left(r_{1}, r_{3}\right)_{L} \cap A is a non-empty open set of A,r2∈UA, r_{2} \in U, and this contradicts the form of the open sets in AA, established in Examples A.2.1, (4).
Let UU be the set of the four intervals in R\mathbb{R} with endpoints 0,1 . For each I∈UI \in U we set IQ=I∩QI_{\mathbb{Q}}=I \cap \mathbb{Q}. Let AA be a linearly ordered set, let EE be the set of its extremal elements (the least and the greatest element of AA, if exist), and let CC be a proper countable subset of AA, which is order dense in AA. We can suppose that E⊂CE \subset C. Then the linearly ordered set AA has no gaps and hence CC itself has no gaps. Now, Corollary A.2.21 yields the existence of a strictly increasing homeomorphism ff of AA onto some interval IQI_{\mathbb{Q}} in the rational line Q\mathbb{Q} where I∈UI \in U. Moreover, f(E)⊂{0,1}f(E) \subset\{0,1\}. Let us set C′=C\EC^{\prime}=C \backslash E. Given a∈A\Ea \in A \backslash E, we have the decomposition
C′={((←,a)A∩C′)∪((a,→)A∩C′)∪{a} if a∈C′((←,a)A∩C′)∪((a,→)A∩C′) if a∈A\CC^{\prime}= \begin{cases}\left((\leftarrow, a)_{A} \cap C^{\prime}\right) \cup\left((a, \rightarrow)_{A} \cap C^{\prime}\right) \cup\{a\} & \text { if } a \in C^{\prime} \\ \left((\leftarrow, a)_{A} \cap C^{\prime}\right) \cup\left((a, \rightarrow)_{A} \cap C^{\prime}\right) & \text { if } a \in A \backslash C\end{cases}
and denote I(a)−=f((←,a)∩C′),I(a)+=f((a,→)∩C′)I(a)_{-}=f\left((\leftarrow, a) \cap C^{\prime}\right), I(a)_{+}=f\left((a, \rightarrow) \cap C^{\prime}\right). Then the pair left(I(a)−,I(a)+right)\left(I(a)_{-}, I(a)_{+}\right)left(I(a)−,I(a)+right)is a Dedekind cut. If a∈C′a \in C^{\prime}, then I(a)−=(0,f(a)),I(a)+=I(a)_{-}=(0, f(a)), I(a)_{+}= (f(a),1)(f(a), 1), and
I∘=I(a)−∪I(a)+∪{f(c)}I^{\circ}=I(a)_{-} \cup I(a)_{+} \cup\{f(c)\}
where I∘I^{\circ} is the open interval in Q\mathbb{Q} with endpoints 0 and 1 . Thus, for any a∈C′a \in C^{\prime} the Dedekind cut left(I(a)−,I(a)+right)\left(I(a)_{-}, I(a)_{+}\right)left(I(a)−,I(a)+right)represents the rational number f(a)∈I∘f(a) \in I^{\circ} and we identify them: f(a)=(I(a)−,I(a)+)f(a)=\left(I(a)_{-}, I(a)_{+}\right). If a∈A\Ca \in A \backslash C, then
I∘=I(a)−∪I(a)+I^{\circ}=I(a)_{-} \cup I(a)_{+}
and the Dedekind cut left(I(a)−,I(a)+right)\left(I(a)_{-}, I(a)_{+}\right)left(I(a)−,I(a)+right)represents an irrational number 1<α<11<\alpha<1 : α=(I(a)−,I(a)+)\alpha=\left(I(a)_{-}, I(a)_{+}\right). We set F(a)=αF(a)=\alpha for any a∈A\C,F(a)=f(a)a \in A \backslash C, F(a)=f(a) for any a∈Ca \in C, and obtain a map
F:A→IF: A \rightarrow I
which is an extension of ff on AA.
Lemma A.2.23 The map FF from (A.2.1) is strictly increasing and extends the homeomorphism f:C→IQf: C \rightarrow I_{\mathbb{Q}} on AA.
Proof: It is enough to show that FF is a strictly increasing map. Let a,b∈Aa, b \in A with a<ba<b. Since CC is order dense in AA, there exists an element c∈Cc \in C with a<c<ba<c<b. Then f©inI(a)+capI(b)−f© \in I(a)_{+} \cap I(b)_{-}fc◯inI(a)+capI(b)−and this implies F(a)<F(b)F(a)<F(b).
Theorem A.2.24 Let AA be a linearly ordered set endowed with the interval topology, let AA be a connected topological space, and let CC be a countable subset which is dense in AA. Then there exists a strictly increasing homeomorphism of AA onto one of the intervals I∈UI \in U, which maps CC onto the interval IQI_{\mathbb{Q}}.
Proof: We can suppose E⊂CE \subset C. In accord with Proposition A.2.10, (ii), AA has no gaps and under this condition Corollary A.1.11 guarantees that CC has a bounded from above subset BB which has no least upper bound in CC. In particular, Proposition A.2.10 yields C≠AC \neq A because AA is connected. Lemma A.2.23 implies that the map FF from (A.2.1) is strictly increasing and extends the homeomorphism f:C→IQf: C \rightarrow I_{\mathbb{Q}} on AA. Now, we will prove the surjectivity of FF. Let α\alpha, 1<α<11<\alpha<1, be an irrational number, represented by the Dedekind cut (I−,I+)\left(I_{-}, I_{+}\right) of the open interval I∘I^{\circ}, so
I∘=I−∪I+I^{\circ}=I_{-} \cup I_{+}
and let C−=f−1(I−),C+=f−1(I+)C_{-}=f^{-1}\left(I_{-}\right), C_{+}=f^{-1}\left(I_{+}\right). Then C−C_{-}C−and C+C_{+}C+are not empty,
C′=C−∪C+C^{\prime}=C_{-} \cup C_{+}
and C−∩C+=∅C_{-} \cap C_{+}=\emptyset. Let ainC−a \in C_{-}ainC−and b∈C+b \in C_{+}. Since f(a)inI−f(a) \in I_{-}f(a)inI−and f(b)∈I+f(b) \in I_{+}, we have f(a)<f(b)f(a)<f(b) and this inequality implies a<ba<b. Thus, C−C_{-}C−consists of
lower bounds of C+C_{+}C+and C+C_{+}C+consists of upper bounds of C−C_{-}. Let a−=supC−a_{-}=\sup C_{-} and a+=infC+a_{+}=\inf C_{+}. Then we have a−≤a+a_{-} \leq a_{+}. Since there are no gaps in AA and since CC is dense in AA, the assumption a−<a+a_{-}<a_{+}a−<a+implies existence of an element c∈(a−,a+)∩C′c \in\left(a_{-}, a_{+}\right) \cap C^{\prime}, which contradicts the equality (A.2.2). Therefore a=a−=a+a=a_{-}=a_{+}a=a−=a+and
C−⊂(←,a)∩C′,C+⊂(a,→)∩C′C_{-} \subset(\leftarrow, a) \cap C^{\prime}, C_{+} \subset(a, \rightarrow) \cap C^{\prime}
If a∈C′a \in C^{\prime}, then the equality (A.2.2) contradicts the equality (A.2.3). Thus, a∈A\Ca \in A \backslash C and we have
C′=C−∪C+⊂(←,a)∩C′∪(a,→)∩C′=C′C^{\prime}=C_{-} \cup C_{+} \subset(\leftarrow, a) \cap C^{\prime} \cup(a, \rightarrow) \cap C^{\prime}=C^{\prime}
The last inclusions imply
C−=(←,a)∩C′,C+=(a,→)∩C′C_{-}=(\leftarrow, a) \cap C^{\prime}, C_{+}=(a, \rightarrow) \cap C^{\prime}
and then I(a)−⊂I−,I(a)+⊂I+I(a)_{-} \subset I_{-}, I(a)_{+} \subset I_{+}, hence
F(a)=(I(a)−,I(a)+)=(I−,I+)=αF(a)=\left(I(a)_{-}, I(a)_{+}\right)=\left(I_{-}, I_{+}\right)=\alpha
and therefore the map FF is surjective. Thus, we have a strictly increasing bijection F:A→IF: A \rightarrow I which is a homeomorphism of AA onto the subspace II of the real line R\mathbb{R} because of Proposition A.2.2, (ii).
References
[1] G. Birkhoff, Latice Theory, American Mathematical Society, 1948.
[2] N. Bourbaki, Elements of Mathematics, Theory of Sets, Springer Verlag, Berlin, 2004.
[3] N. Bourbaki, Elements of Mathematics, General Topology, Ch. I - IV, Springer Verlag, Berlin, 1989.
[4] G. Debreu, Representation of a preference ordering by a numerical function, in: Decision Processes, John Wiley, 1954.
[5] S. Eilenberg, Ordered topological spaces, American Journal of Mathematics 63, No 1 (1941), 39-45.
[6] P. C. Fishburn, Utility Theory for Decision Making, Wiley, New York, 1970.
[7] E. A. Ok, Utility Representation of an Incomplete Preference Relation, Journal of Economic Theory 104, No 2 (2002), 429-449.
[8] E. A. Ok, Real Analysis with Economic Applications, Princeton University Press, 2007.