Q5 is true if (as you assumed), the space of lotteries is the space of distributions over a finite set. (For a general convex set, you can get long-line phenomena.)
First, without proof, I’ll state the following generalization.
Theorem 1. Let ⪯ be a relation on a convex space L satisfying axioms A1, A2, A3, and the following additional continuity axiom. For all A,B1,B2,C∈L, the set
{p∈[0,1]∣A≺pB1+(1−p)B2≺C}
is open in [0,1]. Then, there exists a function u from L to the long line such that u(A)≤u(B) iff A⪯B.
The proof is not too different, but simpler, if we also assume A4. In particular, we no longer need the extra continuity axiom, and we get a stronger conclusion. Nate sketched part of the proof of this already, but I want to be clearer about what is stated and skip fewer steps. In particular, I’m not sure how Nate’s hypotheses rule out examples that require long-line-valued functions—maybe he’s assuming that the domain of the preference relation is a finite-dimensional simplex like I am, but none of his arguments use this explicitly.
Theorem 2. Let ⪯ be a relation on a finite-dimensional simplex L=ΔΩ satisfying axioms A1-A4. Then, there is a quasiconcave function u:L→R such that u(A)≤u(B) iff A⪯B.
First, I’ll set up some definitions and a lemma. For any lotteries A, B, let [A,B] denote the line segment
{pA+(1−p)B∣p∈[0,1]}.
We say that preferences are increasing along a line segment [A,B] if whenever p≤q, we have
(1−p)A+pB⪯(1−q)A+qB.
We will also use open and half-open interval notation in the corresponding way.
Lemma. Let ⪯ be a preference relation on a finite-dimensional simplex L=ΔΩ satisfying axioms A1-A4. Then, there are ⪯-minimal and -maximal elements in L.
Proof. First, we show that there is a minimal element. Axiom A4 states that for any mixture C=pA+(1−p)B, either C⪰A or C⪰B. By induction, it follows more generally that any convex combination C of finitely many elements (Ai)i∈I satisfies C⪰Ai for some i∈I. But every element is a convex combination of the vertices of L, so some vertex of L is ⪯-minimal.
The proof that there is a maximal element is more complex. Consider the family of sets
F={{B∈L∣B⪰A}∣A∈L}.
This is a prefilter, so since L is compact (L here carries the Euclidean metric), it has a cluster point B. Either B will be a maximal element, or we will find some other maximal element. In particular, take any A∈L. We are done if A is a maximal element; otherwise, pick A′≻A. By the construction of F, for every n∈N, we can pick some Cn⪰A′ within a distance of 1n from B. Now, if we show that B itself satisfies B⪰A, it will follow that B is maximal.
The idea is to pass from our sequence (Cn)n∈N, with limit B, to another sequence lying on a line segment with endpoint B. We can use axiom A4, which is a kind of convexity, to control the preference relation on convex combinations of our points Cn, so these are the points that we will construct along a line segment. Once we have this line segment, we can finish by using A3, which is a kind of continuity restricted to line segments, to control B itself.
Let S⊆L be the set of lotteries in the affine span of the set {Cn}n∈N. Then, if we take some index set I⊆N such that (Cn)n∈I is a maximal affinely independent tuple, it follows that {Cn}n∈I affinely generates S. Hence, the convex combination
D=∑n∈I1|I|Cn,
i.e. the barycenter of the simplex with vertices at (Cn)n∈I, is in the interior of the convex hull of {Cn}n∈I relative to S, so we can pick some r>0 such that the r-ball around D relative to S is contained in this simplex.
Now, we will see that every lottery E in the set (B,D] satisfies E⪰A′. For any ε>0, pick k so that Ck is in the ε-ball around B. Since the tangent vector v=B−Ck has length less than ε, the lottery
F=D+rε(B−Ck)
is in the r-ball around D, and it is in S, so it is in the simplex with vertices (Cn)n∈I. Then, F⪰A′ by A4, and Ck⪰A′ by hypothesis. So, applying A4 again,
A′⪯rr+εCk+εr+εF=rr+εB+εr+εD.
Using A4 one more time, it follows that every lottery
E∈[rr+εB+εr+εD,D]
satisfies E⪰A′, and hence every lottery E∈(B,D].
Now we can finish up. If B≺A then, using A3 and the fact that D⪰A′≻A, there would have to be some lottery in [B,D] that is ⪯-equivalent to A, but this would contradict what we just concluded. So, B⪰A, and so B is ⪯-maximal. □
Proof of Theorem 2. Let C be a ⪯-minimal and D a ⪯-maximal element of L. First, we will see that preferences are increasing on [C,D], and then we will use this fact to construct a function L→R and show that it has the desired properties. Suppose preferences we not increasing; then, there would be A,B∈[C,D] such that A is closer to C while B is closer to D, and A≻B. Then, B would be a convex combination of A and D, but B≺A⪯D by the maximality of D, contradicting A4.
Now we can construct our utility function u:L→R using A3; for each ∼-class [A], we have C⪯A⪯D, so there is some[1]p∈[0,1] such that
(1−p)C+pD∼A.
Then, let u(A′)=p for all A′∈[A]. Since preferences are increasing on [C,D], it is immediate that if u(A)≤u(B), then A⪯B. Conversely, if A⪯B, we have two cases. If A≺B, then B⋠A, so u(B)≰u(A), and so u(A)≤u(B). Finally, if A∼B, then u(A)=u(B) by construction.
Finally, since for all A,B∈L we have u(A)≤u(B) iff A⪯B, it follows immediately that u is quasiconcave by A4. □
Nate mentions using choice in his answer, but here at least the use of choice is removable. Since ⪯ is monotone on [C,D], the intersection of the ∼-class [A] with [C,D] is a subinterval of [C,D], so we can pick p based on the midpoint of that interval
Q5 is true if (as you assumed), the space of lotteries is the space of distributions over a finite set. (For a general convex set, you can get long-line phenomena.)
First, without proof, I’ll state the following generalization.
Theorem 1. Let ⪯ be a relation on a convex space L satisfying axioms A1, A2, A3, and the following additional continuity axiom. For all A,B1,B2,C∈L, the set
{p∈[0,1]∣A≺pB1+(1−p)B2≺C}is open in [0,1]. Then, there exists a function u from L to the long line such that u(A)≤u(B) iff A⪯B.
The proof is not too different, but simpler, if we also assume A4. In particular, we no longer need the extra continuity axiom, and we get a stronger conclusion. Nate sketched part of the proof of this already, but I want to be clearer about what is stated and skip fewer steps. In particular, I’m not sure how Nate’s hypotheses rule out examples that require long-line-valued functions—maybe he’s assuming that the domain of the preference relation is a finite-dimensional simplex like I am, but none of his arguments use this explicitly.
Theorem 2. Let ⪯ be a relation on a finite-dimensional simplex L=ΔΩ satisfying axioms A1-A4. Then, there is a quasiconcave function u:L→R such that u(A)≤u(B) iff A⪯B.
First, I’ll set up some definitions and a lemma. For any lotteries A, B, let [A,B] denote the line segment
{pA+(1−p)B∣p∈[0,1]}.We say that preferences are increasing along a line segment [A,B] if whenever p≤q, we have
(1−p)A+pB⪯(1−q)A+qB.We will also use open and half-open interval notation in the corresponding way.
Lemma. Let ⪯ be a preference relation on a finite-dimensional simplex L=ΔΩ satisfying axioms A1-A4. Then, there are ⪯-minimal and -maximal elements in L.
Proof. First, we show that there is a minimal element. Axiom A4 states that for any mixture C=pA+(1−p)B, either C⪰A or C⪰B. By induction, it follows more generally that any convex combination C of finitely many elements (Ai)i∈I satisfies C⪰Ai for some i∈I. But every element is a convex combination of the vertices of L, so some vertex of L is ⪯-minimal.
The proof that there is a maximal element is more complex. Consider the family of sets
F={{B∈L∣B⪰A}∣A∈L}.This is a prefilter, so since L is compact (L here carries the Euclidean metric), it has a cluster point B. Either B will be a maximal element, or we will find some other maximal element. In particular, take any A∈L. We are done if A is a maximal element; otherwise, pick A′≻A. By the construction of F, for every n∈N, we can pick some Cn⪰A′ within a distance of 1n from B. Now, if we show that B itself satisfies B⪰A, it will follow that B is maximal.
The idea is to pass from our sequence (Cn)n∈N, with limit B, to another sequence lying on a line segment with endpoint B. We can use axiom A4, which is a kind of convexity, to control the preference relation on convex combinations of our points Cn, so these are the points that we will construct along a line segment. Once we have this line segment, we can finish by using A3, which is a kind of continuity restricted to line segments, to control B itself.
Let S⊆L be the set of lotteries in the affine span of the set {Cn}n∈N. Then, if we take some index set I⊆N such that (Cn)n∈I is a maximal affinely independent tuple, it follows that {Cn}n∈I affinely generates S. Hence, the convex combination
D=∑n∈I1|I|Cn,i.e. the barycenter of the simplex with vertices at (Cn)n∈I, is in the interior of the convex hull of {Cn}n∈I relative to S, so we can pick some r>0 such that the r-ball around D relative to S is contained in this simplex.
Now, we will see that every lottery E in the set (B,D] satisfies E⪰A′. For any ε>0, pick k so that Ck is in the ε-ball around B. Since the tangent vector v=B−Ck has length less than ε, the lottery
F=D+rε(B−Ck)is in the r-ball around D, and it is in S, so it is in the simplex with vertices (Cn)n∈I. Then, F⪰A′ by A4, and Ck⪰A′ by hypothesis. So, applying A4 again,
A′⪯rr+εCk+εr+εF=rr+εB+εr+εD.Using A4 one more time, it follows that every lottery
E∈[rr+εB+εr+εD,D]satisfies E⪰A′, and hence every lottery E∈(B,D].
Now we can finish up. If B≺A then, using A3 and the fact that D⪰A′≻A, there would have to be some lottery in [B,D] that is ⪯-equivalent to A, but this would contradict what we just concluded. So, B⪰A, and so B is ⪯-maximal. □
Proof of Theorem 2. Let C be a ⪯-minimal and D a ⪯-maximal element of L. First, we will see that preferences are increasing on [C,D], and then we will use this fact to construct a function L→R and show that it has the desired properties. Suppose preferences we not increasing; then, there would be A,B∈[C,D] such that A is closer to C while B is closer to D, and A≻B. Then, B would be a convex combination of A and D, but B≺A⪯D by the maximality of D, contradicting A4.
Now we can construct our utility function u:L→R using A3; for each ∼-class [A], we have C⪯A⪯D, so there is some[1] p∈[0,1] such that
(1−p)C+pD∼A.Then, let u(A′)=p for all A′∈[A]. Since preferences are increasing on [C,D], it is immediate that if u(A)≤u(B), then A⪯B. Conversely, if A⪯B, we have two cases. If A≺B, then B⋠A, so u(B)≰u(A), and so u(A)≤u(B). Finally, if A∼B, then u(A)=u(B) by construction.
Finally, since for all A,B∈L we have u(A)≤u(B) iff A⪯B, it follows immediately that u is quasiconcave by A4. □
Nate mentions using choice in his answer, but here at least the use of choice is removable. Since ⪯ is monotone on [C,D], the intersection of the ∼-class [A] with [C,D] is a subinterval of [C,D], so we can pick p based on the midpoint of that interval