Solutions Manual Foundations of Mathematical Economics By Michael Carter

Solutions Manual
Foundations of Mathematical Economics

Michael Carter

, c⃝ 2001 Michael Carter
DFDFDF DF D F




Solutions for Foundations of Mathematical Economic DF DF D F D F D F All rights reserved DF DF




s



Chapter 1: Sets and Spaces D F D F D F D F




1.1
{1, 3, 5, 7 . . . }or {𝑛 ∈𝑁 : 𝑛 is odd }
DF DF DF DF DF DF DF D F DF DF DF D F DF D F D F DF




1.2 Every 𝑥 ∈ 𝐴 also belongs to 𝐵. Every 𝑥∈ D F D F D F D F D F D F D F




𝐵 also belongs to 𝐴. Hence 𝐴, 𝐵 haveprecisely the same elements.
D F D F D F D F DF D F DF D F D
F D F D F D F




1.3 Examples of finite sets are DF DF DF DF




∙ the letters of the alphabet {A, B, C, . . . , Z }
D F D F D F D F D F DF D F D F D F DF D F DF




∙ the set of consumers in an economy D F D F D F D F D F D F




∙ the set of goods in an economy D F D F D F D F D F D F




∙ the set of players in a game DF DF DF DF DF DF




.Examples of infinite sets are
D
F DF DF D F D F




∙ the real numbers ℜ DF DF DF




∙ the natural numbers 𝔑 DF DF DF




∙ the set of all possible colors DF DF DF DF DF




∙ the set of possible prices of copper on the world market
D F D F D F D F D F D F D F D F D F D F




∙ the set of possible temperatures of liquid water.
D F D F D F D F D F D F D F




1.4 𝑆 = {1, 2, 3, 4, 5, 6 }, 𝐸 = {2, 4, 6 }.
DF D F DF F
D DF DF DF DF DF DF DF D F DF F
D DF DF DF




1.5 The player set is 𝑁 = {Jenny, Chris } . Their action spaces are
D F D F D F D F D F DF F
D DF DF DF D F D F D F




𝐴𝑖 = {Rock, Scissors, Paper }
D F DF F
D DF DF DF 𝑖 = Jenny, Chris
D F DF DF




1.6 The set of players is 𝑁 ={ 1, 2 , . . . , 𝑛} . The strategy space of each player is the
D F D F D F D F D F D F D F DF DF D F DF D F D F D F D F D F D F D F D




Fset of feasible outputsDF D F DF




𝐴𝑖 = {𝑞𝑖 ∈ℜ+ : 𝑞𝑖 ≤𝑄 𝑖 }
DF DF DF DF DF D F DF DF DF DF




where 𝑞𝑖 is the output of dam 𝑖. D F DFD
F DFD
F D F D F D F D F




1.7 The player set is 𝑁 = {1, 2, 3}. There are 23 = 8 coalitions, namely
D F D F D F D F D F DF DF DF DF D F D F D F DF D F D F




𝒫(𝑁 ) = {∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}
DF D F D F DF DF DF DF DF DF DF DF DF DF DF DF




There are 210 coalitions in a ten player game.
DF DF D F D F D F DF D F DF




1.8 Assume that 𝑥 ∈(𝑆 ∪𝑇 ) . That is 𝑥 ∈/ 𝑆 ∪𝑇 . This implies 𝑥 ∈/ 𝑆 and 𝑥 ∈/ 𝑇 , or 𝑥 ∈ 𝑆𝑐 and
DFD F DFDF DFDF DFDF F
D DF F
D DF
𝑐
DFDFDF DFDF DFDF DFDF DFDF DF DF DF DFDFDF DFDF DFDF DFDF DFDF DFDF DFDF DFDF DFDF DF DF DF DF DF DF D




F𝑥 ∈ 𝑇 𝑐. Consequently, 𝑥 ∈ 𝑆𝑐 ∩ 𝑇 𝑐. Conversely, assume 𝑥 ∈ 𝑆𝑐 ∩ 𝑇 𝑐. This implies that 𝑥 ∈𝑆 𝑐 and
DF DF DF D F D F DF DF DF DF DF D F D F D F DF DF DF DF DF DF DFDF DFDF DFDF D F DF DFDF D




𝑥 ∈𝑇 𝑐 . Consequently 𝑥∈/ 𝑆 and 𝑥∈/ 𝑇 and therefore
FDF D F DF DF DFDFDF DFDF D
F DFDF DFDF DFDF D
F DFDF DFD F DFDF




𝑥 ∈/ 𝑆 ∪𝑇 . This implies that 𝑥 ∈(𝑆 ∪𝑇 )𝑐 . The other identity is proved similarly.
DF DF F
D DF DF D F DFD
F D F DF F
D DF F
D DF DF D F D F D F D F DF




1.9
∪
𝑆 =𝑁 DF DF




𝑆∈𝒞
∩
𝑆 =∅ DF DF




𝑆∈𝒞


1

, c⃝ 2001 Michael Carter
DFDFDF DF D F




Solutions for Foundations of Mathematical Economic DF DF D F D F D F All rights reserved DF DF




s

𝑥2
1




𝑥1
-1 0 1




-1

Figure 1.1: The relation {(𝑥, 𝑦) : 𝑥2 + 𝑦2 = 1 }
D F DF D F D F DF DF DF D F D F DF D F D F DF




1.10 The sample space of a single coin toss is{𝐻, 𝑇 .}The set of possible outcomes int
D F D F D F D F D F D F D F D F D F DF DF D F DF D F D F D F D F D F D
F




hree tosses is the product DF DF DF D F




{
{𝐻, 𝑇 } × {𝐻, 𝑇 } × {𝐻, 𝑇 }= (𝐻, 𝐻, 𝐻), (𝐻, 𝐻, 𝑇 ), (𝐻, 𝑇, 𝐻),
DF DF F
D DF DF F
D DF DF F
D D F DF DF DF DF DF DF DF DF D
F DF



}
(𝐻, 𝑇, 𝑇 ), (𝑇, 𝐻, 𝐻), (𝑇, 𝐻, 𝑇 ), (𝑇, 𝑇, 𝐻), (𝑇, 𝑇, 𝑇 ) DF D
F DF DF DF DF DF DF DF DF DF DF DF DF DF DF DF DF




A typical outcome is the sequence (𝐻, 𝐻, 𝑇 ) of two heads followed by a tail.
D F D F D F D F D F D F DF DF DF D F D F D F D F D F D F D F




1.11

𝑌 ∩ℜ+𝑛 = {0}
D F DF
D F
DF




where 0 = (0, 0, . . . , 0) is the production plan using no inputs and producing no outputs. T
DF DF DF DF D
F DF DF DF DF DF DF DF DF DF DF DF DF DF




o see this, first note that 0 is a feasible production plan. Therefore, 0 ∈𝑌 . Also,
D F D F D F D F D F D F D F D F D F D F D F D F D F D F DF DF D F




0 ∈ℜ𝑛 +and therefore 0 ∈𝑌 ∩ℜ𝑛 . +
D F DF
D F
D F D F D F DF D F DF
DF




To show that there is no other feasible production plan in 𝑛 ,ℜwe
DF DF
+ assume the contrary. Tha
DF DF DF DF DF DF DF DF DFDFDFDFDF DF DF DF DF DF DF



𝑛
t is, we assume there is some feasible production plan y
DF DF DF ∈ ℜ 0 +∖.{ This
} implies the exi DF DF DF DF DF DF DF DFDFDFDFDFDFDFDF DFDFDFDFDFDF
DF D F DFDF DFDF
DF D F DF DF DF




stence of a plan producing a positive output with no inputs. This technological infeasible
DF DF DF DF DF DF DF DF DF DF DF DF DF




, so that 𝑦∈/ 𝑌 .
D F D F D F D
F D F DF




1.12 1. Let x ∈𝑉 (𝑦 ). This implies that (𝑦, −x) ∈𝑌 . Let x′ ≥x. Then (𝑦, −x′ ) ≤
DFDF DFD
F D F F
D DF DFDF DFD
F DFD
F DFD
F DF DF DF DF DFDF DFD
F DF DF DFD F DFD
F DF DF




(𝑦, −x) and free disposability implies that (𝑦, −x′ ) ∈𝑌 . Therefore x′ ∈𝑉 (𝑦 ).
DF D F D F D F D F DFD
F D F DF DF DF DF DF D F DF DF DF




2. Again assume x ∈ 𝑉 (𝑦 ). This implies that (𝑦, −x) ∈ 𝑌 . By free disposal, (𝑦 ′ ,
DFD F DFDF DFDF DFD F DF DF DFDFDFDF DFD F DFD F DFD F DF DFD F DF DF DFDFDFDF DFD F DFD F DF DF




−x) ∈𝑌 for every 𝑦 ′ ≤𝑦 , which implies that x ∈𝑉 (𝑦 ′ ). 𝑉 (𝑦 ′ ) ⊇𝑉 (𝑦 ).
DF F
D DFD F D F DF DF F
D D F D F DFD
F D F DF DF DF DFDF DF DF DF DF




1.13 The domain of “<” is {1, 2}= 𝑋 and the range is {2, 3}⫋ 𝑌 .
DF D F DF DF D F DF F
D DF D F D F D F DF D F DF DF DF DF




1.14 Figure 1.1. DF




1.15 The relation “is strictly higher than” is transitive, antisymmetric and asymmetr
D F DF D F D F D F D F D F D F D F D F




ic.It is not complete, reflexive or symmetric.
D
F D F D F D F D F DF DF




2

, c⃝ 2001 Michael Carter
DFDFDF DF D F




Solutions for Foundations of Mathematical Economic DF DF D F D F D F All rights reserved DF DF




s
1.16 The following table lists their respective properties.
DF DF DF D F D F DF




< ≤ √ √= DFD F




× reflexive
√ √ √
DFD F



transitive DFD F




symmetric √ √ DFD F


×
√
DFD F



asymmetric
anti-symmetric √ × √ √
×
DFD F
DFD F




√ √ D F D F


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




1.17 Let be∼ an equivalence relation of a set 𝑋 = .∕ That
DF ∅ is, the relation is reflexive,
DF ∼ symm
DF DF DF DF DF DF DF DF D DF F DF DF DF DF DF DF




etric and transitive. We first show that every 𝑥 𝑋 belongs
DF DF ∈ to some equivalence class. Le DF DF DF DF DF DF DF DF DF DF DF DF D F




t 𝑎 be any element in 𝑋 and let (𝑎) be the
DF DF DF ∼ class of elements equivalent to
DF DF DF D F DF DF DF DF DF DF DF DF DF




𝑎, that is DF DF




∼(𝑎) ≡{𝑥 ∈𝑋 : 𝑥 ∼𝑎 } D F DF DF D F DF D F D F D F DF DF




Since ∼ is reflexive, 𝑎 ∼ 𝑎 and so 𝑎 ∈ ∼ (𝑎). Every 𝑎 ∈ DF DF DF DF DF DF D F DF




𝑋 belongs to some equivalenceclass and therefore D F D F D F D F D
F D F D F



∪
𝑋 = ∼(𝑎) D F




𝑎∈𝑋

Next, we show that the equivalence classes are either disjoint or identical, tha
DF D F D F D F D F D F D F D F D F D F D F DFDF




t is D F




∼(𝑎) ∕= ∼(𝑏) if and only if f∼(𝑎) ∩∼(𝑏) = ∅.
DF DF D F D F D F D F D F DF F
D DF DF




First, assume ∼(𝑎) ∩∼(𝑏) = ∅. Then 𝑎 ∈∼(𝑎) but 𝑎 ∈ ∼(𝑏/
D F DF DF F
D DF DF DF D F DF DF D F DFD
F ). Therefore ∼(𝑎) ∕= ∼(𝑏).
DF D F DF DF




Conversely, assume ∼(𝑎) ∩∼(𝑏) ∕= ∅and let 𝑥 ∈ ∼(𝑎) ∩∼(𝑏). Then 𝑥 ∼𝑎 and bysymmetry 𝑎
DFDF DFDF DF F
D DFDF DFDF DF DFDF DFDF DFDF DF DF DF DFDFDF DFDF DFDF DF DFDF DFDF DF D F D




F∼ 𝑥. Also 𝑥 ∼ 𝑏 and so by transitivity 𝑎 ∼ 𝑏. Let 𝑦 be any element in ∼(𝑎) so that 𝑦
DF DFDFDF D F D F DF DF DF D F D F DF D F DF DFDFDF DF D F D F DF DF DFDF DFDF DFDF DFDF DF




∼𝑎. Again by transitivity 𝑦 ∼𝑏 and therefore 𝑦 ∈ ∼(𝑏). Hence
DF DF DFDFDF DFDF DFDF DFDF DFDF DF DFDF DFDF DFDF DFDF DF DFDFDF




∼(𝑎) ⊆∼(𝑏). Similar reasoning implies that ∼(𝑏) ⊆∼(𝑎). Therefore ∼(𝑎) = ∼(𝑏).
DF F
D DF DFD
F DF DFD
F D F DF DF DF D F DF DF




We conclude that the equivalence classes partition 𝑋.
DF DF DF DF DF DF DF




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




rcan belong to more than one coalition. For example, player 1 belongs to the coalitio
D
F DF DF DF DF DF DF DF DF DF DF DF DF DF DF




ns
{1}, {1, 2}and so on. D F DF F
D D F D F




1.19

𝑥 ≻𝑦 =⇒ 𝑥 ≿ 𝑦 and 𝑦 ∕≿ 𝑥
DF DF D F D F D F DF D F D F D F DF




𝑦 ∼𝑧 =⇒ 𝑦 ≿ 𝑧 and 𝑧 ≿ 𝑦
D F DF D F D F D F DF D F D F D F DF




Transitivity of ≿ implies 𝑥 ≿ 𝑧 . We need to show that 𝑧 ∕≿ 𝑥 . Assume otherwise, thatis a
DF DF DF DF DF DF DF DF DF DF DF DF DF DF DF DF DF D
F D F




ssume 𝑧 ≿ 𝑥 This implies 𝑧 ∼𝑥 and by transitivity 𝑦 ∼𝑥. But this implies that
D F D F DF D F D F D F D F F
D D F D F D F D F D F DF D F D F D F D F




𝑦 ≿ 𝑥 which contradicts the assumption that 𝑥 ≻𝑦 . Therefore we conclude that 𝑧 ∕≿ 𝑥
D F DF D F D F D F D F D F D F DF DF DF D F D F D F D F D F DF




and therefore 𝑥 ≻𝑧 . The other result is proved in similar fashion.
D F D F DF F
D DF D F D F D F D F D F D F D F




1.20 asymmetric Assume 𝑥 ≻𝑦. D F D F DF F
D




𝑥 ≻𝑦 =⇒ 𝑦 ∕≿ 𝑥
DF DF D F D F D F DF




while
𝑦 ≻𝑥 =⇒ 𝑦 ≿ 𝑥
D F DF D F D F D F DF




Therefore
𝑥 ≻𝑦 =⇒ 𝑦 ∕≻𝑥
D F DF D F D F D F DF




3