In the context of examining if the individualistic assumptions used in economics can be used in the aggregation of individual preferences ([7, Definition 5, Theorem 2], Arrow proved a key lemma that extends the famous Szpilrajn’s Theorem.
Proof. To prove necessity, let (G,+,≽) be an ordered group and let ≽ be normal and Δ-consistent. Suppose that x and y be two ≽¯-incomparable elements of G.
We put(1)≽∗=Δ∪≽∪a,b, a,b∈G, a≠b∣there are two non-negative integers p,q,not both zero,such that pa-b≽qx-y, x,y∈Y and x⊒y.We show that (G,+,≽∗) is an ordered group with ≽∗ being a normal, reflexive, and Δ-consistent extension of ≽ satisfying x ≻∗y.
First of all, we note that p is never zero, because otherwise we should have 0≽q(x-y) for some q∈N∗, whence by normality of ≽ we have y≽x against hypothesis.
To show that (G,+,≽∗) is an ordered group, we show that ≽∗ is contradictory and homogenous.
To prove that ≽∗ is contradictory, we show that any two of the three relations a ≻∗ b, a=b, and b ≻∗ a cannot be satisfied simultaneously. We prove the case of a ≻∗ b and b ≻∗ a; the other cases are obvious. We have four cases to consider.
Case 1. a ≻ b and b≻a. This is impossible because of the homogeneity of the relation ≽.
Case 2. a ≻ b as well as p(b-a)≽q(x-y) for some p,q∈N\{0}. By adding p times the a-b≻0 in the second inequality, one obtains p(a-b)+p(b-a)≽q(x-y); that is to say, 0≽q(x-y). But then, by normality we are led to y≽x, a contradiction.
Case 3. p ( a - b ) ≽ q ( x - y ) and b≻a for some p,q∈N\{0}. This is impossible as in the preceding case.
Case 4. In this case we have p1(a-b)≽q1(x-y), p2(b-a)≽q2(x-y), and a≠b for some p1,p2,q1,q2∈N\{0}. By adding p2 times the first, p1 times the second inequality, we have p1p2(a-b)+p1p2(b-a)≽(q1p2+q2p1)(x-y); that is, 0≽(q1p2+q2p1)(x-y). If q1p2+q2p1≠0, by normality we have y≽x, which is impossible. On the other hand, if q1p2+q2p1=0, then since p1,p2≠o we have q1=q2=0. It follows that p1(a-b)≽0 and p2(b-a)≽0. Therefore, by normality we have a≽b and b≽a. Since ≽ is contradictory, we conclude that a=b, an absurdity. Therefore, ≽∗ is contradictory.
To prove homogeneity for ≽∗, let a,b,c,d∈G such that a ≽∗ b and c ≽∗ d. We have four cases to consider: (a) a≽b and c≽d; (b) a≽b and p(c-d)≽q(x-y) for some p,q∈N∗; (c) p(a-b)≽q(x-y) for some p,q∈N∗ and c≽d; (d) p1(a-b)≽q1(x-y) and p2(c-d)≽q2(x-y) for some p1,p2,,q1,q2∈N∗. We only prove the fourth case, as the proof of the others is similar. In this case, we have that p1p2[(a+c)-(b+d)]≽(p2q1+p1q2)(x-y) and p1,p1≠0. Thus, a+c ≽∗ b+d.
We proceed now to prove that ≽∗ is Δ-consistent. Indeed, suppose to the contrary that there are a,b∈G such that a≽∗¯b and b≻a. Thus, there exists z0,z1,…,zn∈G such that (2)a=z0 ≽∗z1…zn-1 ≽∗zn=band b≻a. Therefore, there exist nonnegative integers pi,qi, i∈{1,2,…,n}, pi≠0, such that (3)p1a-z1≽q1x-y,p2z1-z2≽q2x-y,…,pnzn-1-b≽qnx-y.
By adding p2p3…pn times the first, p1p3…pn, times the second,…,p1p2…pn-1, times the n-th and by using the fact that p1p2…pn(b-a)≻0, we have (4)0=p1p2…pna-a≽q1p2…pm+p1q2p3…pm+…+p1p2…pn-1qnx-y.We have that p1p2…pn is nonzero. On the other hand, q1p2…pm+p1q2p3…pm+…+p1p2…pm-1qm=0, because, otherwise, 0≽(q1p2…pm+p1q2p3…pm+…+p1p2…pm-1qm)(x-y) implies that y≽x, an absurdity. It follows that qi=0, i∈{1,…,n}. But then, (5)a≽z1≽…≽zn-1≽zn=b≻a,a contradiction to Δ-consistency of ≽. It follows that ≽∗ is Δ-consistent.
To prove normality, let na ≽∗ 0 for some integer n and some a∈G. Then, na≽0 or there exist nonnegative integers p,q such that p(na-0)≽q(x-y). If na≽0, then by the normality of ≽ we have a≽0 which implies that a ≽∗ 0. Otherwise, pn(a-0)≽q(x-y) which implies a ≽∗0 as well.
It remains to prove that ≽∗ is an extension of ≽. Clearly, ≽⊆≽∗. On the other hand, if a≻b for some a,b∈G, then a-b≻0(x-y) which implies that a ≻∗b. It follows that ≻⊆≻∗.
Finally, since x-y≽x-y and x≠y, we conclude that x ≻∗y.
Suppose that G=G,+,≽i∣i∈I denotes the set of ordered groups such that, for each i∈I, ≽i is a normal, reflexive, and Δ-consistent extension of ≽ satisfying x ≻i y. Since (G,+,≽∗)∈G this set is nonempty. Let ≽~=(≽j)j∈I be a chain in ≽i∣i∈I, and let ≽^=⋃j∈I≽j. It is easy to check that (G,+,≽^) is an ordered group such that ≽^ is a normal, reflexive, and Δ-consistent extension of ≽ satisfying x≻^y.
By Zorn’s lemma, (≽i)i∈I possesses an element, say ⊵, that is maximal with respect to set inclusion. It follows that (G,+,⊵) is an ordered group such that ⊵ is a normal, reflexive, and Δ-consistent extension of ≽ satisfying x⊳y. Let ⊵¯ be the transitive closure of ⊵. Then, (G,+,⊵¯) is an ordered group such that ⊵¯ is a normal, reflexive, transitive, and Δ-consistent extension of ≽ satisfying x⊳¯y. To prove it, we show only the normality for ⊵¯. All the other conditions are easily verified from the fact that they are also satisfied by ⊵. Indeed, let na ⊵¯ 0 for some integer n and some a∈G. Therefore, there exist a1,…,an∈G such that (6)na⊵a1⊵…⊵an-1⊵an⊵0.Since an-1⊵an and an⊵0, by the homogeneity of ⊵ we conclude that an-1+an⊵an+0. But then, by the cancellativity of ⊵ we have that an-1⊵0 and, by an induction argument based on this logic, we obtain na⊵0 which implies that a⊵0. To prove that (G,+,⊵¯) is a linearly ordered group, it remains to show that ⊵¯ is complete and antisymmetric. Since antisymmetry of ⊵¯ is an immediate consequence of the Δ-consistency of ⊵, we prove the completeness of ⊵¯. Suppose to the contrary that there exist a,b∈X such that (a,b)∉⊵¯ and (b,a)∉⊵¯. We define (7)⊵∗= ⊵∪a,b, a,b∈G, a≠b∣there exists p,q∈N∗, not both zero,such that pa-b⊵qx-y, x,y∈Y and x⊒y.Then, as in the case of ≽∗ above, we can prove that (G,+,⊵∗) is an ordered group such that ⊵∗ is a normal, reflexive, and Δ-consistent extension of ≽ satisfying x ⊳∗y, a contradiction of maximality of ⊵. Therefore, ⊵¯ is complete.
To complete the necessity part we show that ⊵¯/Y=⊒. Evidently, ⊒⊆⊵¯/Y. To prove the converse, let a(⊵¯/Y)b for some a,b∈G. Suppose to the contrary that (a,b)∉⊒. Since ⊒ is complete, b⊒a holds which implies b ≻∗a (a≠b since a,b∈Y). Since ⊳ is a linear order extension of ≻∗, we have that b ⊳¯ a. But then, a≠b and antisymmetry of ⊵ imply that (a,b)∉⊵¯/Y, a contradiction. The last contradiction shows that ⊵¯/Y=⊒.
Conversely, suppose that (G,+,≽) is an ordered group and ≽ has a linear order extension ⊵. Suppose to the contrary that ≽ is non-Δ-consistent. Then, I(≽¯)⊈Δ. Thus, there exists a,b∈X, such that (a,b)∈I(≽¯)⊆I(⊵¯)=I(⊵) and a≠b, which contradicts with the fact that ⊵ is antisymmetric. It remains to prove that ≽ is normal. Suppose to the contrary that na≽0 for some integer n and a∈G and a⋡0. Then, as in the case of ⊵ above, the homogeneity and cancellativity of ≽ implies that (a,0)∉≽¯. On the other hands, na≽0 implies that na⊵0. It follows that (0,a)∉≽¯, because otherwise (0,a)∈≽¯⊆⊵¯ jointly to na⊵0 implies that (a,o)∈I(⊵¯)=I(⊵) which concludes that a=0 (⊵ is antisymmetric), a contradiction to a⋡0. Then, as in the case of ≽∗ above, for Y={0,a} (0 and a are incomparable with respect to ≽¯) and ⊒={(0,a)} there exists a homogeneous extension ≽∗∗ of ≽ such that 0 ≻∗∗a. It follows that 0 ≻∗∗na. On the other hand, na≻0 (na≽0 and a⋡0) implies na ≻∗∗0, a contradiction to the contradictory of ≻∗∗. The last contradiction shows that ≽ is normal.
The following result generalizes the classical Dushnik-Miller’s type extension theorem for an ordered group (see [2, Theorem 2]).
The following corollary is an immediate consequence of the necessity part of Theorem 8 for ≽ being a partial order and {+}=∅.