Midterm exam: Advanced Microeconomics, with Solutions SS 2012
1. Define a relation on R2+ by letting (x1 , x2 ) (y1 , y2 ) if and only if
x1 > y1 or both x1 = y1 and x2 > y2 . Show that is reflexive, complete,
and transitive, but has no utility representation.
Solution: Let (x1 , x2 ) and (y1 , y2 ) be two elements of R2+ . If we have
x1 > x2 or x2 > x1 , the elements are comparable. If not, then x1 =
y1 . Since the usual ordering of real numbers is complete, either x2 >
y2 or y2 > x2 . So the elemets are comparable in this case too. Since
both elements can coincide or be distinct, we have shown that is both
reflexive and complete. Now let (x1 , x2 ) , (y1 , y2 ), and (z1 , z2 ) be elements
of R2+ such that (x1 , x2 ) (y1 , y2 ) (z1 , z2 ). In order to show that is
transitive, we have to show that (x1 , x2 ) (z1 , z2 ). Two elements can
only be indifferent if they coincide. So we can restrict ourselves to the
interesting case (x1 , x2 ) (y1 , y2 ) (z1 , z2 ). If x1 > y1 then also x1 > z1 ,
hence (x1 , x2 ) (z1 , z2 ). If x1 = y1 then x2 > y2 . Since y1 > z1 , this
means either x1 > z1 or x1 = z1 and x2 > z2 . In both cases (x1 , x2 ) (z1 , z2 ).
We now show that has no utility representation by contradiction. Suppose u : R2+ → R is a utility representation for
. Now for every x ∈ R+ ,
(x, 1) (x, 0) and hence u(x, 0), u(x, 1) is a nonempty open interval and contains a rational number.
So let f : R+ → Q be a function
such that f(x) ∈ u(x, 0), u(x, 1) for all x ∈ R+ . We show that f is an
injection from the uncountable set of nonnegative real numbers to the
countable set of rational numbers, which is impossible. Let x > y. Then
(x, 1) (x, 0) (y, 1) (y, 0). Hence the intervals u(x, 0), u(x, 1)
and u(y, 0), u(y, 1) are disjoint, so f(x) 6= f(y), which shows that f is
injective.
2. Let F be a non-empty finite set and be an acyclic binary relation on F.
Show that there exists x∗ ∈ F such that x x∗ holds for no x ∈ F.
Solution: We prove the result by contradiction. Suppose there is no such
x∗ . Then we can find for each x ∈ X an x 0 ∈ X with x 0 x. Therefore, we
can construct a sequence (x1 , x2 , . . .) of elements of X such that xn+1 xn
for all indices n. If there exists n and m such that xn = xn+m , then
xn xn+1 . . . xn+m = xn , which violates acyclicity. So all terms
of the sequence are distinct, which shows that we have found infinitely
many different elements in the finite set F. This is impossible.
3. Let X ⊆ Rl be convex. Let be a reflexive, transitive, and complete
relation on X. Suppose that the set {y ∈ X : y x} is convex for all x ∈ X.
Show that the set {y ∈ X : y x} is convex for all x ∈ X.
Solution: We prove this again by contradiction. Suppose there exists x ∈
X such that the set {y ∈ X : y x} is not convex. Then there are elements
z and z 0 and a number α ∈ (0, 1) such that both z and z 0 are in the
set {y ∈ X : y x}, but αz + (1 − α)z 0 is not. So by completeness
αz+(1−α)z 0 ≺ x. By transitivity, z αz+(1−α)z 0 and z 0 αz+(1−α)z 0 .
Hense both z and z 0 are in the set {y ∈ X : y αz + (1 − α)z 0 }. Since
this set is by assumption convex, it must contain the convex combination
αz + (1 − α)z 0 , which implies αz + (1 − α)z 0 αz + (1 − α)z 0 . This is
impossible.
4. Give an example of a set X and a reflexive and complete relation ⊆ X×X
such that the asymmetric part is transitive, but is not transitive. Explain why your example has these properties. Hint: There is an example
in which X has three elements.
Solution: Exercise 6 on the first problem set provides one possible solution. Here is anoher one: Let X = {a, b, c} and
= (a, a), (b, b), (c, c), (a, b), (b, a), (a, c), (c, a), (b, c) .
The relation is reflexive, since for each x ∈ X, the pair (x, x) is in .
It is complete, since for all x and y in X, either
(x, y)
or (y, x) is in .
The asymmetric part is transitive, since = (b, c) and there are no
combinations possible that violate transitivity. But is not transitive,
since c a b but b c.
Here is how one can come up with the example: Requiring to be reflexive means simply that for all x ∈ X, the pair(x, x) is in . Completeness
has similar straightforward consequence. Now we want to be not transitive but to be transitive. The less pairs there are in , the harder it is
to find combinations that violate transitivity. The more pairs there are in
that are not in , the easier it is for to violate transitivity without doing so. So we make all pairs of elements but one indifferent and this
gives us our example (or something essentially the same).