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1. April 25, 2016 Masaki Miyashita1
Microeconomics Homework 1: Answers
Comments
1. Some of you mixed “what you presuppose”, “what you wish to show” and “what is derived
from assumptions or definitions”. The typical proof should have the following structure:
(1) Suppose · · · . / Take any · · · .
(2) (By · · · , / Since · · · , · · · holds.)
(3) Therefore, WTS.
2. Care about quantifier. Is it “any” element or “particular” element?
3. In Problem 4, the presumption mentions about only one direction. Actually, it is assumed
that [u(x) > u(y) =⇒ x ≻ y], but not vice versa. You need to claim that [x ≿ y =⇒
u(x) ≥ u(y)] by using contraposition.
4. There are many solutions for Problem 5, but the simplest way is maybe to construct
in practice. Many of you constructed a utility function by induction about the number
of alternatives. On the way, you might partition X into equivalence classes. I left this
important theorem in Appendix, so please see if you are interested in. (Here, I constructed
by the alternative way, this is much more simple.)
Problem 1
Take any x, y, z ∈ X with x ≻ y ≿ z. By x ≻ y, x ≿ y. Then, by the transitivity of ≿,
x ≿ z. If z ≿ x, then y ≿ x by the transitivity of ≿. This is a contradiction to x ≻ y. Hence,
¬(z ≿ x). Therefore, x ≻ z. □
Problem 2
(i). (Irreflexivity). Take any x ∈ X. If x ≻ x, then both x ≿ x and ¬(x ≿ x) hold
simultaneously, which is a contradiction. Hence, ¬(x ≻ x).
(Transitivity). Take any x, y, z ∈ X with x ≻ y ≻ z. By x ≻ y, x ≿ y and ¬(y ≿ x). By
y ≻ z, y ≿ z and ¬(z ≿ y). Thus, by the transitivity of ≿, x ≿ z. If z ≿ x, then y ≿ z ≿ x
holds by the transitivity of ≿. This is a contradiction to ¬(y ≿ x). Hence, ¬(z ≿ x). Therefore,
x ≻ z. □
(ii). (Reflexivity). Take any x ∈ X. By the completeness of ≿, x ≿ x. Hence, x ∼ x.
(Transitivity). Take any x, y, z ∈ X with x ∼ y ∼ z. By x ∼ y, x ≿ y and y ≿ x. By y ∼ z,
y ≿ z and z ≿ y. Then, by the transitivity of ≿, x ≿ z and z ≿ x hold. Hence, x ∼ z.
(Symmetry). Take any x, y ∈ X with x ∼ y. Then, x ≿ y and y ≿ x. Hence, y ∼ x. □
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2. Problem 3
Take any x, y ∈ X.
First, suppose that x ≿ y. Since u is a utility function representing ≿, u(x) ≥ u(y). If
u(x) = u(y),
v(x) = f(u(x)) = f(u(y)) = v(y).
If u(x) > u(y), since f is strictly increasing,
v(x) = f(u(x)) > f(u(y)) = v(y).
In any case, v(x) ≥ v(y) holds.
Second, suppose that v(x) ≥ v(y). Since f is strictly increasing, u(x) > u(y). Then, since
u is a utility function representing ≿, x ≿ y holds. □
Problem 4
Take any x, y ∈ X.
First, suppose that x ≿ y. Let us show that u(x) ≥ u(y). If u(x) < u(y), then y ≻ x by the
assumption. This is a contradiction to x ≿ y. Hence, u(x) ≥ u(y) holds.
Second, suppose that u(x) ≥ u(y). If u(x) > u(y), then x ≻ y. If u(x) = u(y), then x ∼ y.
In any case, x ≿ y holds. □
Problem 5
For each x ∈ X, let L(x) = {y ∈ X : x ≿ y}. By the reflexibity of ≿,
x ∈ L(x), ∀x ∈ X. (1)
By the completeness of ≿,
x ∈ L(y) or y ∈ L(x), ∀x, y ∈ X. (2)
Define a real-valued function u : X → R by
u(x) = #L(x), ∀x ∈ X,
where #A expresses the number of elements in A. Because X is finite, u is well-defined.
Let us introduce two Lemmas.
Lemma 1. For any x, y ∈ X, x ∈ L(y) if and only if L(x) ⊂ L(y).
Proof. If part. Suppose that L(x) ⊂ L(y). By (1), x ∈ L(x). Hence, x ∈ L(y).
Only if part. Suppose that x ∈ L(y). Take any z ∈ L(x). Then, x ≿ z. By the assumption,
y ≿ x. By the transitivity of ≿, y ≿ z. Hence, z ∈ L(y). ■
Lemma 2. For any x, y ∈ X, #L(x) ≥ #L(y) implies L(y) ⊂ L(x).
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3. Proof. Show by contraposition. Suppose that L(y) ̸⊂ L(x). Then, there exists z ∈ L(y) L(x).
By z ∈ L(y) and Lemma 1, L(z) ⊂ L(y). By z /
∈ L(x) and (2), x ∈ L(z). Then, by Lemma 1,
L(x) ⊂ L(z). By the transitivity of ⊂, L(x) ⊂ L(y). Hence L(x) is a proper subset of L(y),
which implies #L(y) > #L(x). ■
Now, let us show that u is a utility function representing ≿. To see this, we use the result
of Problem 4. Take any x, y ∈ X.
First, suppose u(x) = u(y). Then, by Lemma 2, L(x) = L(y). This and (1) together imply
that x ∈ L(y) and y ∈ L(x). Hence, x ∼ y.
Second, suppose u(x) > u(y). Then, by Lemma 2, L(y) ⊂ L(x). From this and (1),
y ∈ L(x). Thus, x ≿ y. On the other hand, if x ∈ L(y), L(x) ⊂ L(y) by Lemma 1. Then
L(x) = L(y), but this is a contradiction to u(x) > u(y). So x /
∈ L(y), and thus ¬(y ≿ x).
Hence, x ≻ y.
Those arguments and Problem 4 together show that u is a utility function representing ≿.
Therefore, we have completed the proof. □
Appendix
A set can be partitioned into classes by a binary relation if and only if it is an equivalence
relation (reflexive, transitive and symmetric).
Theorem 1. Let ∼ be a binary relation on X. Define Ma ≡ {x ∈ X : a ∼ x} for all a ∈ X.
Then ∼ is an equivalence relation on X if and only if there exists ∆ ⊂ X satisfying
i) X =
∪
a∈∆
Ma;
ii) ∀a, b ∈ ∆ with a ̸= b, Ma ∩ Mb = ∅;
iii) ∀x ∈ X, [x ∈ Ma =⇒ Ma = Mx].
Proof. Step 1: If part. Let ∼ be a binary relation on X. Suppose that there exists ∆ ⊂ X
satisfying (i), (ii) and (iii).
(Reflexivity). Take any x ∈ X. By (i), there exists a ∈ ∆ such that x ∈ Ma. Then, by (iii),
Ma = Mx. Then x ∈ Xx, i.e., x ∼ x.
(Transitivity). Take any x, y, z ∈ X with x ∼ y and y ∼ z. Then,
y ∈ Mx and z ∈ My. (3)
By (i), there exist a, b, c ∈ ∆ such that
x ∈ Ma, y ∈ Mb and z ∈ Mc. (4)
Then, by (iii),
Ma = Mx, Mb = My and Mc = Mz. (5)
Then, by (3), (4) and (5), y ∈ Ma ∩ Mb and z ∈ Mb ∩ Mc, that is, Ma and Mb (Mb and Mc) are
not disjoint. Thus, by (ii), a = b = c. Hence, by (4) and (5), z ∈ Mx, i.e., x ∼ z.
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4. (Symmetry). Take any x, y ∈ X with x ∼ y. Then,
y ∈ Mx. (6)
By (i), there exist a, b ∈ ∆ such that
x ∈ Ma and y ∈ Mb. (7)
Then, by (iii),
Ma = Mx and Mb = My. (8)
Then, by (6), (7) and (8), y ∈ Ma ∩ Mb, that is, Ma and Mb are not disjoint. Thus, by (ii),
a = b. Hence, by (7) and (8), x ∈ Xy, i.e., y ∼ x.
Step 2: Only if part. Suppose that ∼ is an equivalence relation on X. We claim that for
any x, y ∈ X, Mx and My are either disjoint or identical. To see this, pick any x, y ∈ X.
Case 1: x ∼ y. Let us show that Mx = My. First, take any z ∈ Mx. Then, x ∼ z. By the
symmetry of ∼, z ∼ x. By the transitivity of ∼, z ∼ y. Again, by the symmetry of ∼, y ∼ z.
Hence, z ∈ My. Conversely, take any z ∈ My. Then, y ∼ z. By the transitivity of ∼, x ∼ z.
Hence, z ∈ Mx. Therefore, Mx = My.
Case 2: ¬(x ∼ y). Let us show that Mx ∩ My = ∅. Suppose, by contradiction, that there
exists z ∈ Mx ∩ My. Then, x ∼ z and y ∼ z. By the transitivity and symmetry of ∼, x ∼ y, a
contradiction. Therefore, Mx ∩ My = ∅.
Now, consider the family M of all Mx. Usually, a set doesn’t count exactly the same element
“twice”.2
Therefore, all sets in M are distinct, and thus, pairwise-disjoint. Then, assigning
appropriate indices to each set in M (pick any a ∈ M for all M ∈ M as a representative
element), we get the partition M = {Ma}a∈∆ of X. Clearly, ∆ satisfies (i), (ii) and (iii) by the
construction. □
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We never write {1, 2, 3, 5, 8, · · · } as {1, 1, 2, 3, 5, 8 · · · } even though 1 appears twice in the Fibonacci sequence.
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