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Solutions Manual Foundations of Mathematical Economics By Michael Carter
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Solutions ManualFoundations of Mathematical Economics
Michael Carter
, ⃝ c 2001 Michael Carte
XEXEXE XE XE
Solutions for Foundations of Mathematical Economi
XE XE XE XE XE r All rights reserve XE XE
cs d
Chapter 1: Sets and Spaces X E X E X E X E
1.1
{1, 3, 5, 7 . . . }or {� ∈ � : � is odd }
XE XE XE XE XE XE XE X E XE XE XE X E XE X E X E XE
1.2 Every � ∈ � also belongs to �. Every ∈ �
X E XE X E X E X E X E X E
� also belongs to �. Hence �, � haveprecisely the same elements.
X E X E X E X E XE X E XE X E EX XE X E XE
1.3 Examples of finite sets are XE XE XE XE
∙ the letters of the alphabet {A, B, C, . . . , Z }
X E XE XE X E X E E
X XE XE X E XE X E XE
∙ the set of consumers in an economy
X E X E X E X E X E X E
∙ the set of goods in an economy
X E X E X E XE X E X E
∙ the set of players in a gamXE XE XE XE XE XE
e.Examples of infinite sets are
EX XE XE XE XE
∙ the real numbers ℜ XE XE XE
∙ the natural numbers � XE XE XE
∙ the set of all possible colors
XE XE XE XE XE
∙ the set of possible prices of copper on the world market
X E X E X E X E X E X E X E X E X E X E
∙ the set of possible temperatures of liquid water.
X E X E X E X E X E X E X E
1.4 � = {1, 2, 3, 4, 5, 6 }, � = {2, 4, 6 }.
XE XE XE E
X XE XE XE XE XE XE XE XE XE E
X XE XE XE
1.5 The player set is � = {Jenny, Chris }. Their action spaces are
X E X E X E X E X E XE E
X XE XE XE X E X E X E
�� = {Rock, Scissors, Paper }
X E XE E
X XE XE XE � = Jenny, Chris
XE XE XE
1.6 The set of players is � {= 1, 2, . .. , }�
X E X E X E X E X E X E X E XE XE X E . The strategy space of each player is
XE X E X E X E X E X E X E X
Ethe set of feasible outputs
X E XE XE XE
�� = {�� ∈ ℜ + : �� ≤ �� }
XE XE E
X XE E
X XE XE XE E
X XE
where �� is the output of dam �.
XE XEXE XEXE XE X E XE XE
1.7 The player set is � = {1, 2, 3}. There are 23 = 8 coalitions, namely
X E XE X E X E X E XE XE XE XE X E XE XE XE X E XE
� (� ) = {∅ , {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}
XE X E X E XE XE XE XE XE XE XE XE XE XE XE XE
10
There are 2 XE XE X E coalitions in a ten player game. XE XE XE XE XE
1.8 Assume that � ∈ (� ∪ � )� . That is � ∈/ � ∪ � . This implies � ∈/ � and � ∈
XEX E XEXE XEXE XEXE XE XE E
X XE XEXEXE XEXE XEXE XEXE XEXE XE E
X XE XEXEXE XEXE XEXE XEXE XEXE XEXE XEXE XEXE
/ � , or � ∈ �� and � ∈ � �. Consequently, � ∈ �� ∩ � �. Conversely, assume � ∈ �� ∩
XEXE XE XE XE XE XE XE X E XE XE XE X E X E XE XE XE XE XE X E X E X E XE XE XE
� �. This implies that � ∈ � � and � ∈ � � . Consequently � ∈/ � and � ∈/ � and th
XE XE XE XEXE XEXE XEXE XE XE XEXE XEXE XE XE XE XEXEXE XEXE XE XEXE XEXE XEXE XE XEXE XEXE XEXE
erefore
�∈/ � ∪ � . This implies that � ∈ (� ∪ � )� . The other identity is proved similarly.
XE XE E
X XE XE XE XEXE XE XE E
X XE E
X XE XE X E XE X E XE XE
1.9
∪
� =� XE XE
�∈�
∩
� =∅ XE XE
�∈�
1
, ⃝ c 2001 Michael Carte
XEXEXE XE XE
Solutions for Foundations of Mathematical Economi
XE XE XE XE XE r All rights reserve XE XE
cs d
�2
1
�1
-1 0 1
-1
2 2
Figure 1.1: The relation {(�, �) : � + � = 1 }
X E XE X E X E E
X XE XE X E XE XE X E XE XE
1.10 The sample space of a single coin toss {is �, �} . The set of possible outcome
XE X E X E X E X E X E X E X E X E XE XE X E XE X E X E X E X E
s inthree tosses is the product
X E EX XE XE XE XE
{
{�, � }×{�, � }×{�, � }= (�, �, �), (�, �, � ), (�, � , �),
XE XE E
X XE XE E
X XE XE E
X X E XE XE XE XE XE XE XE XE X
E XE
}
(�, � , � ), (�, �, �), (�, �, � ), (�, �, �), (�, �, � ) XE XE XE XE XE XE XE XE XE XE XE XE XE XE XE XE XE XE
A typical outcome is the sequence (�, �, � ) of two heads followed by a tail.
X E X E X E X E X E X E XE XE XE X E X E X E X E X E X E X E
1.11
� ∩ ℜ+� = {0}
X E XE
X E
XE
where 0 = (0, 0, . . . , 0) is the production plan using no inputs and producing no output
XE XE XE XE XE XE XE XE XE XE XE XE XE XE XE XE XE
s. To see this, first note that 0 is a feasible production plan. Therefore, 0 ∈ �
XE X E X E X E X E X E X E X E X E X E X E X E X E X E X E XE X
. Also, E X E
� �
0 ∈ ℜ + and therefore 0 ∈ � ∩ℜ +.
XE XE
X E
X E X E XE XE X E XE
XE
To show that there is no other feasible production plan in ℜ �+ , we assume the contrary.
XE XE XE XE XE XE XE XE XE XE XEXEXEXEXE XE XE XE XE XE
That is, we assume there is some feasible production plan y∈ ℜ +∖�{ } 0 . This im
XE XE XE XE XE XE XE XE XE XE XE XEXEXEXEXEXEXEXE
XE X E XE XEXEXEXEXEXE
X E XEXE XEXE XE
plies the existence of a plan producing a positive output with no inputs. This technol
XE XE XE XE XE XE XE XE XE XE XE XE XE XE
ogical infeasible, so that � ∈/ � .
XE X E XE XE XE XE XE
1.12 1. Let x ∈ � (�). This implies that (�, − x) ∈ � . Let x′ ≥ x. Then (�, − x′ ) ≤
XEXE XEXE XE E
X XE XEXE XEXE XEXE XEXE XE XE E
X XE XEXE XEXE XE E
X XEX E XEXE XE XE
(�, − x) and free disposability implies that (�, − x′ ) ∈ � . Therefore x′ ∈ � (�).
XE XE XE XE XE XEXE X E XE XE E
X XE XE XE XE E
X XE
2. Again assume x ∈ � (�). This implies that (�, − x) ∈ � . By free disp
XEX E XEXE XEXE XEX E XE XE XEXEXEXE XEXE XEX E XEXE XE XEX E XE XE XEXEXEXE XEXE XEX E
osal, (�′ , − x) ∈ � for every �′ ≤ � , which implies that x ∈ � (�′ ). � (�′ ) ⊇ � (�).
XE XE XE E
X XEX E XE XE XE E
X XE XE XEXE XE XE E
X XE XEXE XE XE E
X XE
1.13 The domain of “<” is {1, 2}= � and the range is {2, 3}⫋ � .
XE XE XE XE XE XE E
X XE XE XE XE XE XE XE E
X XE XE
1.14 Figure 1.1. XE
1.15 The relation “is strictly higher than” is transitive, antisymmetric and asymm
XE XE XE XE XE X E XE XE XE XE
etric.It is not complete, reflexive or symmetric.
EX XE XE XE XE XE XE
2
, ⃝ c 2001 Michael Carte
XEXEXE XE XE
Solutions for Foundations of Mathematical Economi
XE XE XE XE XE r All rights reserve XE XE
cs d
1.16 The following table lists their respective properties.
XE XE XE XE XE XE
< ≤√ √= XEX E
× reflexive
√ √ √
XEX E
transitive XEX E
symmetric √ √ XEX E
×
√ XEX E
asymmetric
anti-symmetric √ ×√ × √ XEX E
XEX E
√ √ X E X E
complete ×
Note that the properties of symmetry and anti-symmetry are not mutually exclusive.
XE XE XE XE XE XE XE XE XE XE XE
∼ an equivalence relation of a set � =∕ . ∅ That is, the relation is∼ reflexive, sy
1.17 Let be XE XE XE XE XE XE XE XE XE X XEE X E XE XE XE XE XE XE
mmetric and transitive. We first show that every � �∈ belongs to some equivalence c
XE XE XE XE XE XE XE XE XE XE XE XE XE XE
lass. Let � be any element in � and let
X E ∼ (�) be the class of elements equivalent
XE XE XE XE XE XE XE XE XE XE XE XE XE XE XE
to
XE
�, that isXE XE
∼(�) ≡ {� ∈ � : � ∼ � }XE XE XE XE XE X E XE XE XE XE
Since ∼ is reflexive, �∼ � and so �∈ ∼ (�). Every �∈
XE XE XE XE XE XE X E XE
� belongs to some equivalenceclass and therefore
X E X E X E X E EX XE XE
∪
� = ∼(�) XE
�∈�
Next, we show that the equivalence classes are either disjoint or identical, t
XE X E X E X E X E X E X E X E X E X E X E XEXE
hat is X E
∼(�) ∕= ∼(�) if and only if f∼(�) ∩∼ (�) = ∅ .
XE XE XE X E XE XE XE XE E
X XE XE
First, assume ∼(�) ∩∼ (�) = ∅ . Then � ∈ ∼ (�) but � ∈ ∼(�/ ). Therefore ∼(�) ∕= ∼(�).
X E XE XE E
X XE XE XE XE XE E
X XE XEXE XE XE XE XE
Conversely, assume ∼(�) ∩∼ (�) ∕= ∅ and let � ∈ ∼(�) ∩ ∼ (�). Then � ∼ � and b
XEXE ysym XEXE XE E
X XEXE XEXE XE XEXE XEXE XEXE XE XE E
X XEXEXE XEXE XEXE XE XEXE XEXE XE
metry � ∼ �. Also � ∼ � and so by transitivity � ∼ �. Let � be any element
X E X E XE XEXEXE X E X E XE XE XE X E X E XE X E XE XEXEXE XE X E X E XE
in ∼(�) so that � ∼ �. Again by transitivity � ∼ � and therefore � ∈ ∼(�). Henc
XE XEXE XEXE XEXE XEXE XEXE XE XEXEXE XEXE XEXE XEXE XEXE XE XEXE XEXE XEXE XEXE XE XEXEXE
e
∼(�) ⊆ ∼ (�). Similar reasoning implies that ∼(�) ⊆ ∼ (�). Therefore ∼(�) = ∼(�).
XE E
X XE XEXE XE XEXE XE XE E
X XE XE XE XE
We conclude that the equivalence classes partition �.
XE XE XE XE XE XE XE
1.18 The set of proper coalitions is not a partition of the set of players, since any pl
XE XE XE XE XE XE XE XE XE XE XE XE XE XE XE XE
ayercan belong to more than one coalition. For example, player 1 belongs to the c
EX XE XE XE XE XE XE XE XE XE XE XE XE XE XE
oalitions
{1}, {1, 2}and so on.
XE XE E
X X E X E
1.19
� ≻� =⇒ � ≿ � and � ∕≿ �
XE XE X E X E XE XE X E X E XE XE
� ∼ � =⇒ � ≿ � and � ≿ �
X E XE X E X E XE XE X E X E X E XE
Transitivity of ≿ implies � ≿ � . We need to show that � ∕≿ �. Assume otherwise, th
XE XE XE XE XE XE XE XE XE XE XE XE XE XE XE XE XE
atis assume � ≿ � This implies � ∼� and by transitivity � ∼�. But this impli
EX X E X E XE XE XE X E X E XE E
X XE X E X E X E XE E
X X E X E X E
es that
X E
� ≿ � which contradicts the assumption that � ≻� . Therefore we conclude that � ∕≿ �
X E XE X E X E X E X E X E X E XE XE XE X E X E X E X E X E XE
and therefore � ≻� . The other result is proved in similar fashion.
XE X E XE E
X XE X E X E X E X E XE X E X E
1.20 asymmetric Assume � ≻�. X E X E XE E
X
Therefore
while
3
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