Solutions Manual Foundations of Mathematical Economics By Michael Carter

Solutions Manual
Foundations of Mathematical Economics


Michael Carter

, c⃝ 2001 Michael Carter
vlvlvl vl vl


Solutions for Foundations of Mathematical Economic vl vl vl vl vl All rights reserved vl vl


s



Chapter 1: Sets and Spaces v l v l v l v l




1.1
{1, 3, 5, 7 . . . }or {𝑛 ∈𝑁 : 𝑛 is odd }
vl vl vl vl vl vl vl vl vl vl vl v l vl vl vl vl




1.2 Every 𝑥 ∈ 𝐴 also belongs to 𝐵. Every 𝑥∈ v l v l v l v l v l v l v l



𝐵 also belongs to 𝐴. Hence 𝐴, 𝐵 haveprecisely the same elements.
v l vl v l v l vl v l vl v l lv vl vl vl




1.3 Examples of finite sets are vl vl vl vl




∙ the letters of the alphabet {A, B, C, . . . , Z }
vl vl vl v l v l vl vl vl v l vl v l vl




∙ the set of consumers in an economy
v l vl v l vl v l v l




∙ the set of goods in an economy v l vl v l vl v l v l




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




.Examples of infinite sets are
lv vl vl vl vl




∙ the real numbers ℜ vl vl vl




∙ the natural numbers 𝔑 vl vl vl




∙ the set of all possible colors vl vl vl vl vl




∙ the set of possible prices of copper on the world market
v l v l v l v l v l vl v l v l v l vl




∙ the set of possible temperatures of liquid water.
v l v l v l v l v l v l v l




1.4 𝑆 = {1, 2, 3, 4, 5, 6 }, 𝐸 = {2, 4, 6 }.
vl vl vl vl vl vl vl vl vl vl vl vl vl vl vl vl vl




1.5 The player set is 𝑁 = {Jenny, Chris } . Their action spaces are
v l vl v l v l v l vl vl vl vl vl v l v l vl




𝐴𝑖 = {Rock, Scissors, Paper }
v l vl vl vl vl vl 𝑖 = Jenny, Chris
vl vl vl




1.6 The set of players is 𝑁 ={ 1, 2 , . . . , 𝑛 } . The strategy space of each player is the
v l vl vl vl v l v l v l vl vl v l vl v l vl v l vl v l vl v l v l


set of feasible outputs
vl vl vl




𝐴𝑖 = {𝑞𝑖 ∈ℜ+ : 𝑞𝑖 ≤𝑄 𝑖 }
vl vl vl v l vl vl vl vl vl vl




where 𝑞𝑖 is the output of dam 𝑖. vl vlvl vlvl vl vl vl vl




1.7 The player set is 𝑁 = {1, 2, 3 }. There are 23 = 8 coalitions, namely
v l vl vl v l v l vl vl vl vl v l vl vl vl vl vl




𝒫(𝑁 ) = {∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}
vl v l v l vl vl vl vl vl vl vl vl vl vl vl vl




There are 210 coalitions in a ten player game.
vl vl v l vl vl vl vl vl




1.8 Assume that 𝑥 ∈ (𝑆 ∪𝑇 ) . That is 𝑥 ∈/ 𝑆 ∪𝑇 . This implies 𝑥 ∈/ 𝑆 and 𝑥 ∈/ 𝑇 , or 𝑥 ∈ 𝑆𝑐 and 𝑥
vlv l vlvl vlvl vlvl vl vl vl vl
𝑐
vlvlvl vlvl vlvl vlvl vlvl vl vl vl vlvlvl vlvl vlvl vlvl vlvl vlvl vlvl vlvl vlvl vl vl vl vl vl vl v l vl


∈ 𝑇 𝑐. Consequently, 𝑥 ∈ 𝑆𝑐 ∩ 𝑇 𝑐. Conversely, assume 𝑥 ∈ 𝑆𝑐 ∩ 𝑇 𝑐. This implies that 𝑥 ∈ 𝑆 𝑐 and 𝑥
vl vl v l v l vl vl vl vl vl v l v l v l vl vl vl vl vl vl vlvl vlvl vlvl vl vl vlvl vlvl vl


∈ 𝑇𝑐 . Consequently 𝑥∈/ 𝑆 and 𝑥∈/ 𝑇 and therefore
vl vl vlvlvl vlvl vl vlvl vlvl vlvl vl vlvl vlvl vlvl



𝑥 ∈/ 𝑆 ∪𝑇 . This implies that 𝑥 ∈(𝑆 ∪𝑇 )𝑐 . The other identity is proved similarly.
vl vl vl vl vl vl vlvl vl vl vl vl vl vl vl vl vl vl vl vl




1.9
∪
𝑆 =𝑁 vl vl


𝑆∈𝒞
∩
𝑆 =∅ vl vl


𝑆∈𝒞


1

, c⃝ 2001 Michael Carter
vlvlvl vl vl


Solutions for Foundations of Mathematical Economic vl vl vl vl vl All rights reserved vl vl


s

𝑥2
1




𝑥1
-1 0 1




-1

Figure 1.1: The relation {(𝑥, 𝑦) : 𝑥2 + 𝑦2 = 1 }
v l vl v l v l vl vl vl vl vl vl v l vl vl




1.10 The sample space of a single coin toss is{𝐻, 𝑇 . }The set of possible outcomes int
v l v l v l vl vl v l v l v l vl vl vl v l vl v l vl vl v l v l lv


hree tosses is the product
vl vl vl vl



{
{𝐻, 𝑇 } × {𝐻, 𝑇 } × {𝐻, 𝑇 }= (𝐻, 𝐻, 𝐻), (𝐻, 𝐻, 𝑇 ), (𝐻, 𝑇, 𝐻),
vl vl vl vl vl vl vl vl vl v l vl vl vl vl vl vl vl vl vl vl

}
(𝐻, 𝑇, 𝑇 ), (𝑇, 𝐻, 𝐻), (𝑇, 𝐻, 𝑇 ), (𝑇, 𝑇, 𝐻), (𝑇, 𝑇, 𝑇 ) vl vl vl vl vl vl vl vl vl vl vl vl vl vl vl vl vl vl




A typical outcome is the sequence (𝐻, 𝐻, 𝑇 ) of two heads followed by a tail.
v l v l v l v l v l v l vl vl vl v l v l v l v l v l v l v l




1.11

𝑌 ∩ℜ+𝑛 = {0}
v l vl
v l
vl




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


see this, first note that 0 is a feasible production plan. Therefore, 0 ∈ 𝑌 . Also,
v l v l v l v l v l v l v l v l v l v l v l v l v l v l vl vl v l



0 ∈ℜ𝑛 +and therefore 0 ∈𝑌 ∩ℜ𝑛 . +
vl vl
v l
v l v l vl vl v l vl
vl




To show that there is no other feasible production plan in 𝑛 , ℜwe
vl vl
+ assume the contrary. That
vl vl vl vl vl vl vl vl vlvlvlvlvl vl vl vl vl vl vl v

𝑛
is, we assume there is some feasible production plan y
l vl vl ∖{ }implies the exist
∈0ℜ . +This vl vl vl vl vl vl vl vlvlvlvlvlvlvlvl vlvlvlvlvlvl vl vlvlv vlvl
l vl v l vl vl vl


ence of a plan producing a positive output with no inputs. This technological infeasible,
vl vl vl vl vl vl vl vl vl vl vl vl vl vl


so that 𝑦∈/ 𝑌 .
vl vl vl vl vl




1.12 1. Let x ∈𝑉 (𝑦 ). This implies that (𝑦, −x) ∈𝑌 . Let x′ ≥x. Then (𝑦, −x′ ) ≤
vlvl vlvl vl vl vl vlvl vlvl vlvl vlvl vl vl vl vl vlvl vlvl vl vl vlv l vlvl vl vl



(𝑦, −x) and free disposability implies that (𝑦, −x′ ) ∈𝑌 . Therefore x′ ∈𝑉 (𝑦 ).
vl vl vl vl vl vlvl vl vl vl vl vl vl vl vl vl vl




2. Again assume x ∈ 𝑉 (𝑦 ). This implies that (𝑦, −x) ∈ 𝑌 . By free disposal, (𝑦 ′ , −
vlv l vlvl vlvl vlv l vl vl vlvlvlvl vlv l vlv l vlv l vl vlv l vl vl vlvlvlvl vlv l vlv l vl vl



x) ∈𝑌 for every 𝑦 ′ ≤𝑦 , which implies that x ∈𝑉 (𝑦 ′ ). 𝑉 (𝑦 ′ ) ⊇𝑉 (𝑦 ).
vl vl vlv l vl vl vl vl vl vl vlvl vl vl vl vl vlvl vl vl vl vl




1.13 The domain of “<” is {1, 2}= 𝑋 and the range is {2, 3}⫋ 𝑌 .
vl vl vl vl vl vl vl vl v l vl vl vl vl vl vl vl vl




1.14 Figure 1.1. vl




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



c.It is not complete, reflexive or symmetric.
lv vl vl vl vl vl vl




2

, c⃝ 2001 Michael Carter
vlvlvl vl vl


Solutions for Foundations of Mathematical Economic vl vl vl vl vl All rights reserved vl vl


s
1.16 The following table lists their respective properties.
vl vl vl vl vl vl




< ≤ √ vlv l
√=
reflexive ×
√ √ √
vlv l

transitive vlv l




symmetric √ vlv l √
×
√
vlv l


asymmetric × ×
anti-symmetric √ √ vlv l
vlv l
√
√ √ v l v l

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




1.17 Let be∼ an equivalence relation of a set 𝑋 = . ∕ That
vl ∅ is, the relation is reflexive,
vl ∼ symmvl vl vl vl vl vl vl vl v lv l vl vl vl vl vl vl


etric and transitive. We first show that every 𝑥 𝑋 belongs
vl vl ∈ to some equivalence class. Let vl vl vl vl vl vl vl vl vl vl vl vl v l



𝑎 be any element in 𝑋 and let (𝑎) be the∼class of elements equivalent to
vl vl vl vl vl vl vl vl vl vl vl vl vl vl vl vl



𝑎, that isvl vl




∼(𝑎) ≡ {𝑥 ∈𝑋 : 𝑥 ∼ 𝑎 } vl vl vl vl vl v l v l vl vl vl




Since ∼ is reflexive, 𝑎 ∼ 𝑎 and so 𝑎 ∈ ∼ (𝑎). Every 𝑎 ∈ vl vl vl vl vl vl v l vl



𝑋 belongs to some equivalenceclass and therefore v l v l vl vl lv vl vl


∪
𝑋 = ∼(𝑎) vl


𝑎∈𝑋

Next, we show that the equivalence classes are either disjoint or identical, that
vl v l v l v l v l v l v l v l v l v l v l vlvl



is v l



∼(𝑎) ∕= ∼(𝑏) if and only if f∼(𝑎) ∩∼(𝑏) = ∅.
vl vl vl vl vl vl vl vl vl vl vl




First, assume ∼(𝑎) ∩∼(𝑏) = ∅. Then 𝑎 ∈∼(𝑎) but 𝑎 ∈ ∼(𝑏/
vl vl vl vl vl vl vl vl vl vl vl vlvl ). Therefore ∼(𝑎) ∕= ∼(𝑏).
vl vl vl vl




Conversely, assume ∼(𝑎) ∩∼(𝑏) ∕= ∅ and let 𝑥 ∈ ∼(𝑎) ∩∼(𝑏). Then 𝑥 ∼ 𝑎 and bysymmetry 𝑎
vlvl vlvl vl vl vlvl vlvl vl vlvl vlvl vlvl vl vl vl vlvlvl vlvl vlvl vl vlvl vlvl vl v l v l


∼ 𝑥. Also 𝑥 ∼ 𝑏 and so by transitivity 𝑎 ∼ 𝑏. Let 𝑦 be any element in ∼(𝑎) so that 𝑦
vl vlvlvl v l v l vl vl vl v l v l vl v l vl vlvlvl vl v l v l vl vl vlvl vlvl vlvl vlvl vlvl


∼ 𝑎. Again by transitivity 𝑦 ∼ 𝑏 and therefore 𝑦 ∈ ∼(𝑏). Hence
vl vlvlvl vlvl vlvl vlvl vlvl vl vlvl vlvl vlvl vlvl vl vlvlvl



∼(𝑎) ⊆∼(𝑏). Similar reasoning implies that ∼(𝑏) ⊆∼(𝑎). Therefore ∼(𝑎) = ∼(𝑏).
vl vl vl vlvl vl vlvl vl vl vl vl vl vl vl




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




1.18 The set of proper coalitions is not a partition of the set of players, since any player
vl vl vl vl vl vl vl vl vl vl vl vl vl vl vl vl lv



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



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




1.19

𝑥 ≻𝑦 =⇒ 𝑥 ≿ 𝑦 and 𝑦 ∕≿ 𝑥
vl vl v l v l vl vl v l v l vl vl




𝑦 ∼𝑧 =⇒ 𝑦 ≿ 𝑧 and 𝑧 ≿ 𝑦
v l vl v l v l vl vl v l v l v l vl




Transitivity of ≿ implies 𝑥 ≿ 𝑧 . We need to show that 𝑧 ∕≿ 𝑥 . Assume otherwise, thatis as
vl vl vl vl vl vl vl vl vl vl vl vl vl vl vl vl vl lv v l


sume 𝑧 ≿ 𝑥 This implies 𝑧 ∼𝑥 and by transitivity 𝑦 ∼𝑥. But this implies that
v l v l vl vl v l v l v l vl v l v l v l v l v l vl v l v l v l v l



𝑦 ≿ 𝑥 which contradicts the assumption that 𝑥 ≻𝑦 . Therefore we conclude that 𝑧 ∕≿ 𝑥
v l vl v l v l v l v l v l v l vl vl vl vl v l v l v l v l vl


and therefore 𝑥 ≻𝑧 . The other result is proved in similar fashion.
vl vl vl vl vl vl vl vl vl vl vl vl




1.20 asymmetric Assume 𝑥 ≻𝑦. v l v l vl vl




𝑥 ≻𝑦 =⇒ 𝑦 ∕≿ 𝑥
vl vl v l v l vl vl




while
𝑦 ≻𝑥 =⇒ 𝑦 ≿ 𝑥
v l vl v l v l v l vl




Therefore
𝑥 ≻ 𝑦 =⇒ 𝑦 ∕≻𝑥
vl vl v l v l v l vl




3