Solutions_manual_foundations_of_mathematical_economics[1]

Solutions Manual c




FoundationsofMathematicalEconomics
c c c c




Michael Carter
c c

, ⃝ c 2001 Michael Carter
c cc
c c




Solutions for Foundations of Mathematical Economics c c c c c All rightsreserved c c




Chapter 1: Sets and Spaces c c c c




1.1
{1,3,5,7. . . }or {�∈ � : � is odd}
c c c c c c c c c c c c c c c c




1.2 Every � ∈ � also belongs to �. Every � ∈ � also belongs to �. Hence �,�
c c c c c c c c c c c c c c




chaveprecisely the same elements.
c c c c




1.3 Examples of finite sets are c c c c




∙ the letters of the alphabet {A, B, C, . . . , Z} c c c c c c c c c c c c




∙ the set of consumers in an economy c c c c c c




∙ the set of goods in an economy c c c c c c




∙ the set of players in a c c c c c




c game.Examples of infinite sets are c c c c c




∙ the real numbers ℜ c c c




∙ the natural numbers � c c c




∙ the set of all possible colors c c c c c




∙ the set of possible prices of copper on the world market
c c c c c c c c c c




∙ the set of possible temperatures of liquid water. c c c c c c c




1.4 � = {1,2,3,4,5,6}, � = {2,4,6}.
c c c c c c c c c c c c c c c c c




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




�� = {Rock,Scissors,Paper} c c c c c c � = Jenny,Chris c c c




1.6 The set of players is � = 1,
{ 2 , . .. , � .}The strategy space of each player is the set of
c c c c c c c c c c c c c c c c c c c c




feasible outputs
c c




�� = {�� ∈ ℜ + : �� ≤ ��} c c c c c c c c c c




where ��is the output of dam �. c cc cc c c c c




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




� (�) = {∅ , {1}, {2}, {3}, {1, 2}, {1,3}, {2, 3}, {1, 2, 3}} c c c c c c c c c c c c c c c




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There are 2 c c c coalitions in a ten player game. c c c c c




1.8 Assume that �∈ (� ∪ �)�. That is �∈/ � ∪ �. This implies �∈/ �and �∈/ �, or �∈ ��and �∈ ��.
c c cc cc cc c c c c c cc cc cc cc cc c c c c cc cc cc cc cc cc cc cc cc c c c c c c c c c c




Consequently, �∈ ��∩ ��. Conversely, assume �∈ ��∩ ��. This implies that � ∈ ��and � ∈ ��.
c c c c c c c c c c c c c c c c cc cc cc c c cc cc c c c c




Consequently �∈/ �and �∈/ � and therefore
cc cc c cc cc cc c cc c c cc




�∈/ �∪ �. This implies that �∈ (�∪ �)�. The other identity is proved similarly.
c c c c c c cc c c c c c c c c c c c c




1.9
∪
�=� c c




�∈�
∩
� =∅ c c




�∈�


1

, ⃝ c 2001 Michael Carter
c cc
c c




Solutions for Foundations of Mathematical Economics c c c c c All rightsreserved c c




�2
1




�1
-1 0 1




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




1.10 The sample space of a single coin toss is �,�
c
{ . The}set of possible outcomes inthree
c c c c c c c c c c c c c c c c c c




tosses is the product
c c c c




{
{�,�} ×{�,�} ×{�,�}= (�,�,�),(�,�,�),(�,�,�),
}
c c c c c c c c c c c c c c c c c c c c




(�,�,�),(�,�, �),(�, �,�),(�, �,�), (�,�,�) c c c c c c c c c c c c c c c c c c




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




1.11

� ∩ℜ+� = {0} c c
c



c




where 0 = (0,0 , . . . ,0)is the production plan using no inputs and producing no outputs. To see
c c c c c c c c c c c c c c c c c c c




this, first note that 0 is a feasible production plan. Therefore, 0 ∈ �. Also,
c c c c c c c c c c c c c c c c




0 ∈ ℜ +and therefore 0 ∈ � ∩ℜ �. +
c
� c
c


c c c c c c
c




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




assume there is some feasible production plan y � 0 . This implies
c c
∈ ℜ +the
∖ { }existence of a plan c c c c c c c c c c c c c c c c c c c c c c c c c
c c
c
c c
c c c c




producing a positive output with no inputs. This technological infeasible, so that �∈/ �.
c c c c c c c c c c c c c c c c




1.12 1. Let x ∈ �(�). This implies that (�,− x) ∈ �. Let x′ ≥ x. Then (�,− x′ ) ≤
cc cc c c c cc cc cc cc c c c c cc cc c c c c cc c c




(�,− x) and free disposability implies that (�,− x′ ) ∈ �. Therefore x′ ∈ �(�).
c c c c c cc c c c c c c c c c c




2. Again assume x ∈ �(�). This implies that (�,− x) ∈ �. By free disposal, (�′ ,− x) ∈ � for
c c c c c c c c c c cc c c c c c c c c c c c c c ccc c c c c c c c c c c c




every �′ ≤ �, which implies that x ∈ �(�′ ). �(�′ ) ⊇ �(�).
c c c c c c cc c c c c cc c c c c




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




1.14 Figure 1.1. c




1.15 The relation “is strictly higher than” is transitive, antisymmetric and asymmetric.It
c c c c c c c c c c c




is not complete, reflexive or symmetric.
c c c c c c




2

, ⃝ c 2001 Michael Carter
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c c




Solutions for Foundations of Mathematical Economics c c c c c All rightsreserved c c




1.16 The following table lists their respective properties. c c c c c c




< ≤√ √= c c




× reflexive
√ √ √
c c




transitive c c




symmetric √ √
×
c c




√
c c




asymmetric
anti-symmetric √ × ×
√ √ c c
c c




√ √
×
c c




complete
Note that the properties of symmetry and anti-symmetry are not mutually exclusive.
c c c c c c c c c c c




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




and transitive. We first show that every ��belongs to some
c c
∈ equivalence class. Let � be c c c c c c c c c c c c c c c




any element in � and let (�) be the class of∼elements equivalent to
c c c c c c c c c c c c c c




�, that is
c c




∼(�) ≡{� ∈ � : � ∼ �} c c c c c c c c c c




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




equivalenceclass and therefore
c c c c




∪
�= ∼(�) c




�∈�

Next, we show that the equivalence classes are either disjoint or identical, that is
c c c c c c c c c c c c c c




∼(�) ∕= ∼(�) if and only if f∼(�) ∩∼ (�) = ∅ .
c c c c c c c c c c c




First, assume ∼(�) ∩∼ (�) = ∅ . Then �∈ ∼ (�) but ��
c /∈ ∼( c c c c c c c c c c cc ). Therefore ∼(�) ∕= ∼(�).
c c c c




Conversely, assume ∼ (�) ∩ ∼ (�) ∕= ∅ and let �∈ ∼ (�) ∩ ∼ (�). Then �∼ �and b
ysymmetry � ∼ �. cc c c c c c c c c c cc cc c c c c c c c c cc c c c c c cc c c c c c c




Also � ∼ �and so by transitivity � ∼ �. Let � be any element in ∼ (�) so that �∼ �. Again by
c c c c c c c c c c c ccc c c c c c cc cc cc cc cc c cc c cc




transitivity �∼ �and therefore �∈ ∼ (�). Hence
cc cc cc c cc cc cc cc c cc c




∼(�) ⊆ ∼ (�). Similar reasoning implies that ∼(�) ⊆ ∼ (�). Therefore ∼(�) = ∼(�).
c c c cc c cc c c c c c c c




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




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




belong to more than one coalition.For example, player 1 belongs to the coalitions
c c c c c c c c c c c c c c




{1}, {1,2}and so on. c c c c c




1.19

�≻� =⇒ �≿ � and � ∕≿ �
c c c c c c c c c c




� ∼ � =⇒ � ≿ � and � ≿ �
c c c c c c c c c c




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




assume � ≿ � This implies � ∼� and by transitivity � ∼�. But this implies that
c c c c c c c c c c c c c c c c c c c




� ≿ � which contradicts the assumption that �≻�. Therefore we conclude that � ∕≿ �
c c c c c c c c c c c c c c c c c




and therefore �≻�. The other result is proved in similar fashion.
c c c c c c c c c c c c




1.20 asymmetric Assume �≻�. c c c c




�≻� =⇒ � ∕≿ � c c c c c c




while

� ≻� =⇒ � ≿ � c c c c c c




Therefore
� ≻� =⇒ � ∕≻� c c c c c c




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