Solutions Manual Foundations of Mathematical Economics By Michael Carter

Solutions ManualM




M Foundations of Mathematical Economics
M M M




Michael Carter
M M

, ⃝ c 2001 Michael Carter
M MM M M




Solutions for Foundations of Mathematical M M M M All rights reserved M M




MEconomics



Chapter 1: Sets and Spaces M M M M




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




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




Mhaveprecisely the same elements.
M M M M




1.3 Examples of finite sets are M M M M




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




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




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




∙ the set of players in a M M M M M




M game.Examples of infinite setsM M M M




M are
∙ the real numbers ℜ M M M




∙ the natural numbers � M M M




∙ the set of all possible colors M M M M M




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




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




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




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




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




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




Mset of feasible outputs
M M M




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




where �� is the output of dam �.
M MM MM M M M M




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




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




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




�
1.8 Assume that � ∈ (� ∪ �) . That is � ∈/ � ∪ �. This implies � ∈/ � and � ∈/ �, or �∈
M M MM MM MM M M M M MMM MM MM MM MM M M M M MM MM MM MM MM MM MM MM MM M M M M




�� and �∈ ��. Consequently, �∈ �� ∩ ��. Conversely, assume �∈ �� ∩ ��. This implies
M M M M M M M M M M M M M M M M M M M M M M MM M




that � ∈ �� and � ∈ ��. Consequently �∈/ � and �∈/ � and therefore
M MM M M MM MM M M M M MM MM M MM MM MM M MM MM MM




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




1.9
∪
�= � M M




�∈�
∩
�=∅ M M




�∈�


1

, ⃝ c 2001 Michael Carter
M MM M M




Solutions for Foundations of Mathematical M M M M All rights reserved M M




MEconomics

�2
1




�1
-1 0 1




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




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




Minthree tosses is the product
M M M M M




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



}
(�, �, �), (�, �, �), (�, �, �), (�, �, �), (�, �, �) M M M M M M M M M M M M M M M M M M




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




1.11

� ∩ℜ+� = {0} M M
M
M




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




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




Also, M




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




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


�
That is, we assume there is some feasible production plan y ∈ ℜ +
M M M M ∖ { } 0 . This implies M M M M M M M M M M M MM M M M
M MM M M MM M M M M M M M




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




infeasible, so that �∈/ �.
M M M M M M M




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




(�, − x) and free disposability implies that (�, − x′ ) ∈ �. Therefore x′ ∈ �(�).
M M M M M MM M M M M M M M M M M




2. Again assume x ∈ �(�). This implies that (�, − x) ∈ �. By free disposal,
M M MM MM M M M M MMMM MM M M M M M M M M M MMMM MM M M




(�′ , − x) ∈ � for every �′ ≤ �, which implies that x ∈ �(�′ ). �(�′ ) ⊇ �(�).
M M M M M M M M M M M M MM M M M M MM M M M M




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




1.14 Figure 1.1. M




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




Masymmetric.It is not complete, reflexive or symmetric. M M M M M M M




2

, ⃝ c 2001 Michael Carter
M MM M M




Solutions for Foundations of Mathematical M M M M All rights reserved M M




MEconomics
1.16 The following table lists their respective properties.
M M M M M M




< ≤√ √= M M




× reflexive
√ √ √
M M



transitive M M




symmetric √ √
×
M M




√
M M




asymmetric
anti-symmetric √ × ×
√ √ M M
M M




√ √ M M


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




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




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




class. Let � be any element in � and let
M M ∼ (�) be the class of elements equivalent
M M M M M M M M M M M M M M M




Mto
�, that is
M M




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




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




Mequivalenceclass and therefore M M M



∪
�= ∼(�) M




�∈�

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




that is M M




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




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




Conversely, assume ∼(�) ∩∼ (�) ∕= ∅ and let � ∈ ∼(�) ∩∼ (�). Then � ∼ � and bysymmetry
MM MM M M MM MM M MM MM MM M M M MM M MM MM M MM MM M




M� ∼ �. Also � ∼ �and so by transitivity � ∼ �. Let � be any element in ∼(�) so
M M MMM M M M M M M M M M M MMM M M M M M MM MM M




that � ∼ �. Again by transitivity � ∼ � and therefore � ∈ ∼(�). Hence
M MM MM M MMM MM MM MM MM M MM MM MM MM M MMM




∼(�) ⊆ ∼ (�). Similar reasoning implies that ∼(�) ⊆ ∼ (�). Therefore ∼(�) = ∼(�).
M M M MM M MM M M M M M M M




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




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




playercan belong to more than one coalition. For example, player 1 belongs to the
M M M M M M M M M M M M M M M




Mcoalitions
{1}, {1, 2}and so on.
M M M M M




1.19

� ≻� =⇒ � ≿ � and � ∕≿ �
M M M M M M M M M M




� ∼ � =⇒ � ≿ � and � ≿ �
M M M M M M M M M M




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




Massume � ≿ � This implies � ∼� and by transitivity � ∼�. But this implies that
M M M M M M M M M M M M M M M M M M




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




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




1.20 asymmetric Assume � ≻�. M M M M




� ≻� =⇒ � ∕≿ � M M M M M M




while

� ≻� =⇒ � ≿ � M M M M M M




Therefore
3