Solutions Manual for Foundations of Mathematical Economics – Michael Carter – Complete Step-by-Step Solutions

Solutions Manual
Foundations of Mathematical Economics

Michael Carter

, ⃝ c 2001 Michael Carte
MLKMLKMLK MLK M L K




Solutions for Foundations of Mathematical Econom MLK M LK M L K M L K M L K r All rights reserved MLK MLK




ics



Chapter 1: M L K M L K Sets and Spaces M L K M L K




1.1
{1, 3, 5, 7 . . . }or {� ∈ � : � is odd }
MLK MLK MLK MLK MLK MLK MLK M L K MLK MLK MLK M L K ML K M L K M L K MLK




1.2 Every �∈ � also belongs to �. Every∈ � M L K M L K M L K M L K M L K M L K M L K




� also belongs to �. Hence �, � haveprecisely the same elements.
M L K M L K M L K M L K M LK M L K MLK M L K LK
M M L K M L K M L K




1.3 Examples of finite sets are MLK MLK M L K M L K




∙ the letters of the alphabet {A, B, C, . . . , Z }
M L K M L K M L K M L K M L K LK
M M L K M L K M L K M L K M L K MLK




∙ the set of consumers in an economy
M L K M L K M L K M L K M L K M L K




∙ the set of goods in an economy
M L K M L K M L K M L K M L K M L K




∙ the set of players in a ga MLK MLK MLK MLK MLK MLK




me.Examples of infinite sets ar
LK
M M LK M LK M L K M L K




e
∙ the real numbers ℜ MLK MLK MLK




∙ the natural numbers � M L K MLK M L K




∙ the set of all possible colorsMLK MLK MLK MLK MLK




∙ the set of possible prices of copper on the world market
M L K M L K M L K M L K M L K M L K M L K M L K M L K M L K




∙ the set of possible temperatures of liquid water.
M L K M L K M L K M L K M L K M L K M L K




1.4 � = {1, 2, 3, 4, 5, 6 }, � = {2, 4, 6 }.
MLK M LK MLK LK
M MLK MLK MLK MLK MLK MLK M LK M L K MLK LK
M MLK MLK MLK




1.5 The player set is � = {Jenny, Chris } . Their action spaces are
M L K M L K M L K M L K M L K MLK LK
M MLK MLK M LK M L K M L K M L K




�� = {Rock, Scissors, Paper }
M L K MLK LK
M MLK MLK MLK � = Jenny, Chris
M L K MLK MLK




1.6 The set of players is �{ = 1, 2 , . .}. , �
M L K M L K M L K M L K M L K M L K M L K MLK MLK M L K . The strategy space of each playe
MLK M L K M L K M L K M L K M L K




r is the set of feasible outputs
M L K M L K M L K MLK M L K M L K




�� = {�� ∈ ℜ + : �� ≤ �� }
MLK M LK LK
M MLK LK
M M L K MLK MLK LK
M MLK




where �� is the output of dam �.
M L K MLKM
LK MLKM
LK M L K M L K M L K M L K




3
1.7 The player set is � = {1, 2, 3}. There are 2 = 8 coalitions, namely
M L K M L K M L K M L K M L K MLK MLK MLK M LK M L K M L K M L K MLK M L K M L K




� (� ) = {∅ , {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}
MLK M L K M L K MLK MLK MLK MLK MLK MLK MLK MLK MLK MLK MLK MLK




10
There are 2 M L K M L K M L K coalitions in a ten player game. M L K M L K M L K M L K M L K




�
1.8 Assume that � ∈ (� ∪ �) . That is � ∈/ � ∪ � . This implies � ∈/ � and � ∈/ �, o
MLKM L K MLKMLK MLKMLK MLKMLK MLK MLK LK
M MLK MLKMLKMLK MLKMLK MLKMLK MLKMLK MLKMLK MLK LK
M MLK MLKMLKMLK MLKMLK MLKMLK MLKMLK MLKMLK MLKMLK MLKMLK MLKMLK MLKMLK MLK MLK




r � ∈ �� and � ∈ � �. Consequently, � ∈ �� ∩ � �. Conversely, assume � ∈ �� ∩ � �. T
MLK MLK MLK MLK M L K MLK MLK MLK M L K M L K MLK MLK MLK MLK MLK M L K M L K M L K MLK MLK MLK MLK MLK MLK




his implies that � ∈ � � and � ∈ �� . Consequently �∈/ � and �∈/ � and therefore
MLKMLK MLKMLK MLKMLK M L K LK
M MLKMLK MLKMLK M L K LK
M MLK MLKMLKMLK MLKMLK LK
M MLKMLK MLKMLK MLKMLK LK
M MLKMLK MLKM LK MLKMLK




�∈/ � ∪ �. This implies that � ∈ (� ∪ �)� . The other identity is proved similarly.
MLK M LK LK
M MLK MLK M L K MLKM
LK M L K M LK LK
M M LK LK
M MLK MLK M L K M L K M L K M L K M LK




1.9
∪
� =� M LK MLK




�∈�
∩
� =∅ MLK MLK




�∈�


1

, ⃝ c 2001 Michael Carte
MLKMLKMLK MLK M L K




Solutions for Foundations of Mathematical Econom MLK M LK M L K M L K M L K r All rights reserved MLK MLK




ics

�2
1




�1
-1 0 1




-1
2 2
Figure 1.1: The relation {(�, �) : � + � M L K MLK M L K M L K LK
M MLK MLK M L K M LK M LK M L K = 1} M L K MLK




1.10 The sample space of a single coin toss
M L K { is �,}� . The set of possible outco
M L K M L K M L K M L K M L K M L K M L K M L K MLK MLK M L K MLK M L K M L K M L K M L K




mes inthree tosses is the product
M L K LK
M M L K M L K M L K M L K




{
{�, �} ×{�, � } ×{�, � }= (�, �, �), (�, �, � ), (�, �, �),
MLK MLK LK
M MLK MLK LK
M MLK MLK LK
M M L K MLK MLK MLK MLK MLK MLK MLK MLK LK MLK
M


}
(�, �, � ), (�, �, �), (�, �, � ), (�, �, �), (�, �, � ) MLK MLK MLK MLK MLK MLK MLK MLK MLK MLK MLK MLK MLK MLK MLK MLK MLK MLK




A typical outcome is the sequence (�, �, � ) of two heads followed by a tail.
M L K M L K M L K M L K M L K M L K MLK MLK MLK M L K M L K M L K M L K M L K M L K M L K




1.11

� ∩ℜ+� = {0}
M L K MLK
M L K
M L K




where 0 = (0, 0 , . . . , 0) is the production plan using no inputs and producing no outputs.
MLK MLK MLK MLK MLK MLK MLK MLK MLK MLK MLK MLK MLK M LK MLK MLK MLK MLK




To see this, first note that 0 is a feasible production plan. Therefore, 0 ∈
M L K M L K M L K M L K M L K M L K M L K M L K M L K M L K M L K M L K M L K M L K LK
M




� . Also, MLK M L K




0 ∈ ℜ �+ and therefore 0 ∈ � ∩+ℜ � .
M L K MLK
M L K
M L K M L K M L K MLK M L K MLK
M LK




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




That is, we assume there is some feasible production plan y ∈ ℜ + �∖ { } 0 . This impl
MLK MLK MLK MLK MLK MLK MLK MLK MLK MLK MLKMLKMLKMLKMLKMLKMLKMLK
MLK M L K MLKMLKMLKMLKMLKMLKMLK
M L K MLKMLK MLKMLK MLK




ies the existence of a plan producing a positive output with no inputs. This technologi
MLK MLK MLK MLK MLK MLK MLK MLK MLK MLK MLK MLK MLK MLK




cal infeasible, so that �∈/ � .
MLK M L K M L K M L K LK
M M L K MLK




1.12 1. Let x ∈ � (�). This implies that (�, − x) ∈ � . Let x′ ≥ x. Then (�, − x′ ) ≤
MLKMLK MLKM
LK ML K LK
M MLK MLKMLK MLKM
LK MLKM
LK MLKM
LK MLK M LK LK
M MLK MLKMLK MLKM
LK M LK LK
M MLKM L K MLKM
LK MLK MLK




(�, − x) and free disposability implies that (�, − x′ ) ∈ � . Therefore x′ ∈ � (�).
MLK M L K M L K M L K M L K MLKM
LK M L K MLK MLK LK
M MLK MLK M L K MLK LK
M MLK




2. Again assume x ∈ � (�). This implies that (�, − x) ∈ � . By free disp
MLKM L K MLKM LK MLKM LK MLKM L K MLK MLK MLKMLKMLKMLK MLKM L K MLKM L K MLKM L K MLK MLKM L K MLK MLK MLKMLKMLKMLK MLKM LK MLKM L K




osal, (� ′ , − x) ∈ � for every � ′ ≤ � , which implies that x ∈ � (� ′ ). � (� ′ ) ⊇ � (�).
MLK MLK MLK LK
M MLKM L K M L K M LK MLK LK
M M L K M L K MLKM
LK M L K MLK LK
M MLK MLKMLK MLK MLK LK
M MLK




1.13 The domain of “<” is {1, 2}= � and the range is {2, 3}⫋ � .
M L K M L K M LK M LK M L K MLK LK
M MLK M L K M L K M L K M LK M L K MLK LK
M MLK MLK




1.14 Figure 1.1. MLK




1.15 The relation “is strictly higher than” is transitive, antisymmetric and asym
M L K M L K M L K M L K M L K M L K M L K M L K M L K M L K




metric.It is not complete, reflexive or symmetric.
LK
M M L K M L K M L K M L K M LK M L K




2

, ⃝ c 2001 Michael Carte
MLKMLKMLK MLK M L K




Solutions for Foundations of Mathematical Econom MLK M LK M L K M L K M L K r All rights reserved MLK MLK




ics
1.16 The following table lists their respective properties.
M LK M LK M LK M L K M L K MLK




< ≤√ MLKM L K

√=
reflexive ×
√ √ √
MLKM L K



transitive MLKM L K




symmetric √ √
×
MLKM L K




√
MLKM L K




asymmetric × ×
anti-symmetric √ √ MLKM L K
MLKM L K

√
√ √ M L K M L K


complete ×
Note that the properties of symmetry and anti-symmetry are not mutually exclusive.
M L K M L K M L K MLK M L K M L K M LK M LK MLK M L K M L K




∼ an equivalence relation of a set � =∕ . ∅That is, the relation is∼ reflexive, sy
1.17 Let be MLK MLK MLK MLK MLK MLK MLK MLK MLK MLK L
M M KL K MLK MLK MLK MLK MLK MLK




mmetric and transitive. We first show that every � �∈ belongs to some equivalence cl
MLK MLK MLK MLK MLK MLK MLK MLK MLK MLK MLK MLK MLK MLK




ass. Let � be any element in � and let
M L K ∼ (�) be the class of elements equivalen
MLK MLK MLK MLK M LK MLK M L K M LK MLK M LK M LK M LK MLK MLK MLK




t to
MLK




�, that is MLK MLK




∼(�) ≡{� ∈ � : � ∼ � } M L K MLK MLK M L K MLK M L K M L K M L K MLK MLK




Since ∼ is reflexive, �∼ � and so �∈ ∼ (�). Every �∈ MLK MLK MLK MLK MLK MLK M L K M LK




� belongs to some equivalenceclass and therefore M L K M L K M L K M L K KM
L M L K M L K



∪
� = ∼(�) M L K




�∈�

Next, we show that the equivalence classes are either disjoint
M LK M L K M L K M L K M L K M L K M L K M L K M L K M L K or identic M L K




al, that is MLKM L K M L K




∼(�) ∕= ∼(�) if and only if f∼(�) ∩∼ (�) = ∅ .
MLK MLK M L K M L K M L K M L K M L K MLK LK
M MLK MLK




First, assume ∼(�) ∩∼ (�) = ∅ . Then � ∈ ∼ (�) but ��/∈ ∼(
M L K M LK MLK LK
M MLK MLK MLK M L K M LK LK
M M L K MLKM
LK ). Therefore ∼(�) ∕= ∼(�).
MLK M L K MLK MLK




Conversely, assume ∼(�) ∩∼ (�) ∕= ∅ and let � ∈ ∼ (�) ∩∼ (�). Then � ∼ � and bysymmet
MLKMLK MLKMLK MLK LK
M MLKMLK MLKMLK MLK MLKMLK MLKMLK MLKMLK MLK MLK LK
M MLKMLKMLK MLKMLK MLKMLK M
LK MLKMLK MLKMLK M LK




ry � ∼ �. Also � ∼ � and so by transitivity � ∼ �. Let � be any element
M L K M L K MLK MLKMLKMLK M L K M L K MLK MLK MLK M L K M L K MLK M L K MLK MLKMLKMLK MLK M L K M L K MLK MLK




in ∼(�) so that � ∼ �. Again by transitivity � ∼ � and therefore � ∈ ∼ (�). Hence
MLKMLK MLKMLK MLKMLK MLKMLK MLKMLK M
LK MLKMLKMLK MLKMLK MLKMLK MLKMLK MLKMLK MLK MLKMLK MLKMLK MLKMLK MLKMLK MLK MLKMLKMLK




∼(�) ⊆ ∼ (�). Similar reasoning implies that ∼(�) ⊆ ∼ (�). Therefore ∼(�) = ∼(�).
MLK LK
M MLK MLKM
LK MLK MLKM
LK M L K MLK LK
M MLK M L K MLK MLK




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




1.18 The set of proper coalitions is not a partition of the set of players, since any pl
MLK ML K M LK M LK M LK M LK MLK MLK MLK M LK MLK M LK M LK M LK MLK MLK




ayercan belong to more than one coalition. For example, player 1 belongs to the c
LK
M MLK MLK MLK MLK MLK MLK MLK MLK MLK MLK MLK MLK MLK MLK




oalitions
{1}, {1, 2}and so on. M L K MLK LK
M M L K M L K




1.19

� ≻� =⇒ � ≿ � and � ∕≿ �
MLK MLK M L K M L K MLK M LK M L K M L K M L K M LK




� ∼ � =⇒ � ≿ � and � ≿ �
M L K MLK M L K M L K M L K MLK M L K M L K M L K MLK




Transitivity of ≿ implies � ≿ � . We need to show that � ∕≿ � . Assume otherwise, tha
MLK MLK MLK MLK MLK MLK MLK MLK M LK MLK MLK MLK M LK MLK MLK MLK MLK




tis assume � ≿ � This implies � ∼� and by transitivity � ∼�. But this im
LK
M M L K M L K M L K M LK M L K M L K M L K M L K LK
M M L K M L K M L K M L K M L K LK
M M L K M L K M L K



plies that M L K




� ≿ � which contradicts the assumption that � ≻� . Therefore we conclude that � ∕≿ �
M L K MLK M L K M L K M L K M L K M L K M L K M L K MLK MLK M L K M L K M L K M L K M L K ML K




and therefore � ≻� . The other result is proved in similar fashion.
M L K M L K MLK LK
M MLK M L K M L K M L K M L K M L K M L K M L K




1.20 asymmetric Assume � ≻�. M L K M L K MLK LK
M




Therefore

while


3