Solutions Manual Foundations of Mathematical Economics By Michael Carter

Solutions Manual bh




bh Foundations of Mathematical Economics
bh bh bh




Michael Carter
bh bh

, ⃝ c 2001 Michael
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Solutions for Foundations of Mathematical bh bh b h b h b hCarter
All rights reserved bh bh




Economics
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Chapter 1: Sets and Spaces b h b h b h b h




1.1
{1, 3, 5, 7 . . . }or {� ∈ � : � is odd }
bh bh bh bh bh bh bh b h bh bh bh b h b h b h b h bh




1.2 Every �∈ � also belongs to �. Every∈ � b h b h b h b h b h b h b h � also belongs to �. Hence
b h b h b h b h b h




�, � haveprecisely the same elements.
b h bh b h b
h b h b h b h




1.3 Examples of finite sets are bh bh b h b h




∙ the letters of the alphabet {A, B, C, . . . , Z }
b h b h b h b h b h bh b h b h b h b h b h bh




∙ the set of consumers in an economy
b h b h b h b h b h b h




∙ the set of goods in an economy b h b h b h b h b h b h




∙ the set of players in a bh bh bh bh bh




bh game.Examples of infinite sets
b
h b h b h b h




b h are
∙ the real numbers ℜ bh bh bh




∙ the natural numbers � b h bh b h




∙ the set of all possible colors bh bh bh bh bh




∙ the set of possible prices of copper on the world market
b h b h b h b h b h b h b h b h b h b h




∙ the set of possible temperatures of liquid water.
b h b h b h b h b h b h b h




1.4 � = {1,2, 3,4, 5,6 }, � = {2, 4,6 }.
bh b h bh h
b bh bh bh bh bh bh bh b h bh h
b bh bh bh




1.5 The player set is � = {Jenny, Chris } . Their action spaces are
b h b h b h b h b h bh h
b bh bh bh b h b h b h




�� = {Rock, Scissors, Paper }
b h bh h
b bh bh bh � = Jenny, Chris
b h bh bh




1.6 The set of players is �{ = b h b h b h b h b h b h b h 1, 2 , . ..}, � bh bh b h . The strategy space of each player
bh b h b h b h b h b h




is the set of feasible outputs
b h b h b h bh b h b h




�� = {�� ∈ ℜ + : �� ≤ �� }
b h bh bh b h bh b h bh bh bh bh




where �� is the output of dam �.
b h bhb
h bhb
h b h b h b h b h




3
1.7 The player set is � = {1, 2, 3}. There are 2 = 8 coalitions, namely
b h b h b h b h b h bh bh bh bh b h b h b h bh b h b h




� (�) = {∅ , {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}
bh b h b h bh bh bh bh bh bh bh bh bh bh bh bh




10
There are 2 b h b h b h coalitions in a ten player game. b h b h b h b h b h




�
1.8 Assume that � ∈ (� ∪ �) . That is � ∈/ � ∪ �. This implies � ∈/ � and � ∈/ �, or �
bh b h bh bh bh bh bhb
h h
b b h h
b bh bh bh bh bh bh bh bh bhbh bh bh bh bh bh bh bh bh bh bh bh bh bh bh bh bh bh bh bh bh bhbh bh bh bh bh bh




∈ �� and � ∈ � �. Consequently, � ∈ �� ∩ � �. Conversely, assume � ∈ �� ∩ � �. This
bh bh bh b h bh bh bh b h b h bh bh bh bh bh b h b h b h bh bh bh bh bh bh bh




implies that � ∈ �� and � ∈ ��. Consequently �∈/ � and �∈/ � and therefore
bh bhbh bhbh b h h
b bhbh bhbh b h h
b bh bh bhbh bhbh bh bh bh bhbh bhbh bh bhbh bh bh bhbh




�∈/ � ∪ �. This implies that � ∈ (� ∪ �)�. The other identity is proved similarly.
bh bh h
b bh bh b h bhb
h b h bh h
b bh bh bh bh b h b h b h b h b h




1.9
∪
�=� bh bh




�∈�
∩
� =∅ bh bh




�∈�


1

, ⃝ c 2001 Michael
bh bhbh bh




Solutions for Foundations of Mathematical bh bh b h b h b hCarter
All rights reserved bh bh




Economics
b h




�2
1




�1
-1 0 1




-1
2 2
Figure 1.1: The relation {(�, �) : � + � = 1 }
b h bh b h b h bh bh bh b h b h bh b h b h bh




1.10 The sample space of a single coin toss{ is �,}� . The set of possible
b h b h b h b h b h b h b h b h b h bh bh b h bh b h b h b h




outcomes inthree tosses is the product
b h b h b
h b h b h b h b h




{
{�, �} ×{�, �} ×{�, �}= (�, �, �), (�, �, �), (�, �, �),
bh bh h
b bh bh h
b bh bh h
b b h bh bh bh bh bh bh bh bh bh bh



}
(�, �, �), (�, �, �), (�, �, �), (�, �, �), (�, �, �)
bh bh bh bh bh bh bh bh bh bh bh bh bh bh bh bh bh bh




A typical outcome is the sequence (�, �, �) of two heads followed by a tail.
b h b h b h b h b h b h bh bh bh b h b h b h b h b h b h b h




1.11

� ∩ℜ+� = {0} b h bh
b h
b h




where 0 = (0,0 , . . . ,0) is the production plan using no inputs and producing no outputs.
bh bh bh bh h bh
b bh bh bh bh bh bh bh bh bh bh bh




To see this, first note that 0 is a feasible production plan. Therefore, 0
bh b h b h b h b h b h b h b h b h b h b h b h b h b h




∈ �. Also, b h bh bh b h




0 ∈ ℜ �+ and therefore 0 ∈ � ∩ℜ+� .
b h bh
b h
b h b h b h bh b h bh
bh




To show that there is no other feasible production plan in ℜ �
bh bh
+ , we assume the contrary. bh bh bh bh bh bh bh bh bh bh bh bh bh bh bh bh bh bh




That is, we assume there is some feasible production plan y ∈ ℜ +� ∖ { 0} . This implies
bh bh bh bh bh bh bh bh bh bh bh bh bh bh bh bhbh bh bh bh
b h bh bh bhbh bh bhbbh h bh bh bh bh bh




the existence of a plan producing a positive output with no inputs. This technological
bh bh bh bh bh bh bh bh bh bh bh bh bh bh




infeasible, so that �∈/ �.
bh b h b h b h h
b b h bh




1.12 1. Let x ∈ �(�). This implies that (�, − x) ∈ �. Let x′ ≥ x. Then (�, − x′ ) ≤
bh bh bhb
h b h h
b bh bh bh bhb
h bhb
h bhb
h bh b h bh bh bh bh bhb
h bh bh bh b h bhb
h bh bh




(�, − x) and free disposability implies that (�, − x′ ) ∈ �. Therefore x′ ∈ �(�).
bh b h b h b h b h bhb
h b h bh bh bh bh bh b h bh bh bh




2. Again assume x ∈ �(�). This implies that (�, − x) ∈ �. By free
bh b h bh b h bh b h bh b h bh bh bh bh bh bh bh b h bh b h bh b h bh bh b h bh bh bh bh bh bh bh b h bh




disposal, (�′ , − x) ∈ � for every �′ ≤ �, which implies that x ∈ �(�′ ). �(�′ ) ⊇ �
bh bh bh bh h
b bh b h b h b h bh bh b h b h bhb
h b h bh bh bh bh bh bh bh bh




(�). bh




1.13 The domain of “<” is {1, 2}= � and the range is {2,3}⫋ �.
b h b h b h b h b h bh bh bh b h b h b h bh b h bh bh bh bh




1.14 Figure 1.1. bh




1.15 The relation “is strictly higher than” is transitive, antisymmetric and
b h b h b h b h b h b h b h b h b h




asymmetric.It is not complete, reflexive or symmetric.
b h b
h b h b h b h b h bh b h




2

, ⃝ c 2001 Michael
bh bhbh bh




Solutions for Foundations of Mathematical bh bh b h b h b hCarter
All rights reserved bh bh




Economics
b h




1.16 The following table lists their respective properties.
bh b h bh b h b h bh




< ≤√ bh b

√= h




reflexive ×
√ √ √
bh b h



transitive bh b h




symmetric √ √
×
bh b h




√
bh b h




asymmetric × ×
anti-symmetric √ √ bh b h
bh b h

√
√ √ b h b h


complete ×
Note that the properties of symmetry and anti-symmetry are not mutually exclusive.
b h b h b h bh b h b h bh b h bh b h b h




1.17 Let ∼be an equivalence relation of a set �∕ = .∅ That is, the relation∼ is reflexive,
bh bh bh bh bh bh bh bh bh b bh
h b h bh bh bh bh bh




symmetric and transitive. We first show that every �∈ � belongs to some equivalence
bh bh bh bh bh bh bh bh bh bh bh bh bh bh




class. Let � be any element in � ∼and let (�) be the class of elements
bh b h bh bh bh bh bh bh b h bh bh bh bh bh bh bh




equivalent to
bh bh




�, that is bh bh




∼(�) ≡{� ∈ � : � ∼ �} b h bh bh b h bh b h b h b h bh bh




Since ∼ is reflexive, �∼ �and so �∈ ∼ (�). Every �
∈ � belongs to some
bh bh bh bh bh bh b h b h b h b h b h




equivalenceclass and therefore
b h b
h b h b h



∪
�= ∼(�) b h




�∈�

Next, we show that the equivalence classes are either disjoint or
bh b h b h b h b h b h b h b h b h b h




identical, that is b h bh b h b h




∼(�) ∕= ∼(�) if and only if f∼(�) ∩∼ (�) = ∅ .
bh bh b h b h b h b h b h bh bh bh bh




First, assume ∼(�) ∩∼ (�) = ∅ . Then � ∈ ∼ (�) but ��
b h /∈ ∼( b h bh h
b bh bh bh b h bh bh b h bhb
h ). Therefore ∼(�) ∕= ∼(�).
bh b h bh bh




Conversely, assume ∼(�) ∩∼ (�) ∕= ∅ and let � ∈ ∼(�) ∩∼ (�). Then � ∼ � and bysymmetry
bhbh bh bh bh h
b bh bh bh bh bh bhbh bh bh bhbh bh bh bh bh bh bh bh bh bhbh bh bhbh bhbh bh




b� ∼ �. Also � ∼ � and so by transitivity � ∼ �. Let � be any element in
h b h bh bh bh bh b h b h bh bh bh b h b h bh b h bh bh bh bh bh b h b h bh bh bh




∼(�) so that � ∼ �. Again by transitivity � ∼ � and therefore � ∈ ∼(�). Hence
bh bh bh bh bh bhbh bhbh h
b bh bh bh bhbh bhbh bhbh bhbh bh bhbh bh bh bhbh bhbh bh bh bh bh




∼(�) ⊆ ∼ (�). Similar reasoning implies that ∼(�) ⊆ ∼ (�). Therefore ∼(�) = ∼(�).
bh h
b bh bhb
h bh bhb
h b h bh bh bh b h bh bh




We conclude that the equivalence classes partition �.
bh bh bh bh bh bh bh




1.18 The set of proper coalitions is not a partition of the set of players, since any
bh bh bh bh bh bh bh bh bh bh bh bh bh bh bh




playercan belong to more than one coalition. For example, player 1 belongs to the
bh b
h bh bh bh bh bh bh bh bh bh bh bh bh bh




coalitions
bh




{1}, {1, 2}and so on. b h bh h
b b h b h




1.19

� ≻� =⇒ � ≿ � and � ∕≿ �
bh bh b h b h bh bh b h b h b h bh




� ∼ � =⇒ � ≿ � and � ≿ �
b h bh b h b h b h bh b h b h b h bh




Transitivity of ≿ implies �≿ �. We need to show that � ∕≿ �. Assume otherwise,
bh bh bh bh bh bh bh bh bh bh bh bh bh bh bh bh




thatis assume � ≿ � This implies � ∼� and by transitivity � ∼�. But this
bh b
h b h b h b h bh b h b h b h b h bh b h b h b h b h b h bh b h b h



implies that
b h b h




� ≿ � which contradicts the assumption that � ≻�. Therefore we conclude that � ∕≿ �
b h bh b h b h b h b h b h b h b h bh bh b h b h b h b h b h bh




and therefore � ≻�. The other result is proved in similar fashion.
b h b h bh h
b bh b h b h b h b h b h b h b h




1.20 asymmetric Assume � ≻�. b h b h b h h
b




Therefore

while


3