Foundations of Mathematical Economics – Michael Carter | Complete Solutions Manual

Solutions Manual
Foundations of Mathematical
Economics

Michael Carter

, c⃝ 2001 Michael Carter
xoxoxo xo x o




Solutions for Foundations of Mathematical Economic xo xo x o x o x o All rights reserved xo xo




s



Chapter 1: Sets and Spaces x o x o x o x o




1.1
{1, 3, 5, 7 . . . }or {𝑛 ∈𝑁 : 𝑛 is odd }
xo xo xo xo xo ox xo x o xo xo xo x o xo x o x o xo




1.2 Every 𝑥 ∈ 𝐴 also belongs to 𝐵. Every 𝑥 ∈ x o x o x o x o x o x o x o




𝐵 also belongs to 𝐴. Hence 𝐴, 𝐵 haveprecisely the same elements.
x o x o x o x o xo x o xo x o xo x o x o x o




1.3 Examples of finite sets are xo xo xo xo




∙ the letters of the alphabet {A, B, C, . . . , Z }
x o x o x o x o x o xo x o x o x o x o x o xo




∙ the set of consumers in an economy x o x o x o x o x o x o




∙ the set of goods in an economy x o x o x o x o x o x o




∙ the set of players in a game. xo xo xo xo xo xo xo




Examples of infinite sets are xo xo x o x o




∙ the real numbers ℜ xo xo xo




∙ the natural numbers 𝔑 x o xo xo




∙ the set of all possible colors xo xo xo xo xo




∙ the set of possible prices of copper on the world market
x o x o x o x o x o x o x o x o x o x o




∙ the set of possible temperatures of liquid water.
x o x o x o x o x o x o x o




1.4 𝑆 = {1,2, 3,4, 5,6 }, 𝐸 = {2,4,6 }.
xo x o xo o
x ox xo ox ox ox xo xo x o xo o
x ox ox xo




1.5 The player set is 𝑁 = {Jenny, Chris } . Their action spaces are
x o x o x o x o x o xo o
x xo xo xo x o x o x o




𝐴𝑖 = {Rock, Scissors, Paper }
x o xo o
x xo xo xo 𝑖 = Jenny, Chris
x o xo xo




1.6 The set of players is 𝑁 = { 1, 2 , .. . , 𝑛 } . The strategy space of each player is the s
x o x o x o x o x o x o x o xo xo x o xo x o x o x o x o x o x o x o x o




et of feasible outputs
xo x o xo




𝐴𝑖 = {𝑞𝑖 ∈ℜ+ : 𝑞𝑖 ≤𝑄𝑖 }
x o xo xo x o xo x o xo x o xo xo




where 𝑞𝑖 is the output of dam 𝑖.
x o xoxo xoxo x o x o x o x o




1.7 The player set is 𝑁 = {1, 2, 3}. There are 23 = 8 coalitions, namely
x o x o x o x o x o xo xo xo xo x o x o x o xo x o x o




𝒫(𝑁) = {∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}
ox x o x o xo xo xo xo xo xo xo xo xo xo xo xo




There are 210 coalitions in a ten player game.
xo xo x o x o x o xo x o xo




1.8 Assume that 𝑥 ∈(𝑆 ∪𝑇) . That is 𝑥 ∈/ 𝑆 ∪𝑇. This implies 𝑥 ∈/ 𝑆 and 𝑥 ∈/ 𝑇, or 𝑥 ∈ 𝑆𝑐 and 𝑥 ∈ 𝑇
xox o xoxo xoxo xoxo o
x x o o
x ox
𝑐
xoxoxo xoxo xoxo xoxo xoxo xo xo ox xoxoxo xoxo xoxo xoxo xoxo xoxo xoxo xoxo xoxo ox xo xo xo xo xo x o xo xo xo


𝑐
. Consequently, 𝑥 ∈ 𝑆𝑐 ∩ 𝑇 𝑐. Conversely, assume 𝑥 ∈ 𝑆𝑐 ∩ 𝑇 𝑐. This implies that 𝑥 ∈𝑆 𝑐 and 𝑥 ∈𝑇𝑐 . C
x o x o xo xo xo xo xo x o x o x o xo xo xo xo xo xo xoxo xoxo xoxo x o xo xoxo xoxo x o xo ox xoxoxo




onsequently 𝑥∈/ 𝑆 and 𝑥∈/ 𝑇 and therefore xoxo ox xoxo xoxo xoxo ox xoxo xoxo xoxo




𝑥 ∈/ 𝑆 ∪𝑇. This implies that 𝑥 ∈(𝑆 ∪𝑇)𝑐 . The other identity is proved similarly.
xo xo o
x ox xo x o xoxo x o xo o
x xo o
x ox xo x o x o x o x o xo




1.9
∪
𝑆 =𝑁 xo xo




𝑆∈𝒞
∩
𝑆 =∅
xo xo




𝑆∈𝒞


1

, c⃝ 2001 Michael Carter
xoxoxo xo x o




Solutions for Foundations of Mathematical Economic xo xo x o x o x o All rights reserved xo xo




s

𝑥2
1




𝑥1
-1 0 1




-1

Figure 1.1: The relation {(𝑥, 𝑦) : 𝑥2 + 𝑦2 = 1 } x o xo x o x o xo xo xo x o x o xo x o x o xo




1.10 The sample space of a single coin toss is {𝐻, 𝑇 . The
x o x o } set of possible outcomes inth x o x o x o x o x o x o x o xo xo x o xo x o x o x o x o x o xo




ree tosses is the product
xo xo xo x o




{
{𝐻, 𝑇} × {𝐻, 𝑇 } × {𝐻, 𝑇}= (𝐻, 𝐻, 𝐻), (𝐻, 𝐻, 𝑇), (𝐻, 𝑇, 𝐻),
xo ox o
x xo xo o
x xo ox o
x x o xo xo xo xo xo ox xo xo ox xo



}
(𝐻, 𝑇, 𝑇 ), (𝑇, 𝐻, 𝐻), (𝑇, 𝐻, 𝑇 ), (𝑇, 𝑇, 𝐻), (𝑇, 𝑇, 𝑇 ) xo ox xo xo xo xo xo xo xo xo xo xo xo xo xo xo xo xo




A typical outcome is the sequence (𝐻, 𝐻, 𝑇 ) of two heads followed by a tail.
x o x o x o x o x o x o xo xo xo x o x o x o x o x o x o x o




1.11

𝑌 ∩ℜ+𝑛 = {0}
x o xo
x o
xo




where 0 = (0,0 , . . . ,0) is the production plan using no inputs and producing no outputs. To
xo xo xo ox ox ox xo xo xo xo xo xo xo xo xo xo xo xo x o




see this, first note that 0 is a feasible production plan. Therefore, 0 ∈𝑌 . Also,
x o x o x o x o x o x o x o x o x o x o x o x o x o xo xo x o




0 ∈ℜ𝑛 +and therefore 0 ∈𝑌 ∩ℜ𝑛 . +
x o xo
x o
x o x o x o xo x o xo
xo




To show that there is no other feasible production plan in 𝑛 , we
xo xo ℜ +assume the contrary. That
xo xo xo xo xo xo xo xo xoxoxoxoxo xo xo xo xo xo xo xo



𝑛
is, we assume there is some feasible production plan y
xo xo
+∖{ implies
0∈ ℜ. This } the existe xo xo xo xo xo xo xo xoxoxoxoxoxoxoxo xoxoxoxoxoxo xoxo
xo xoxo
x o xo x xo
o xo xo




nce of a plan producing a positive output with no inputs. This technological infeasible, so
xo xo xo xo xo xo xo xo xo xo xo xo xo x o x




othat 𝑦∈/ 𝑌 . x o ox x o xo




1.12 1. Let x ∈𝑉 (𝑦 ). This implies that (𝑦, −x) ∈𝑌 . Let x′ ≥x. Then (𝑦, −x′ ) ≤
xoxo xoxo x o o
x xo xoxo xoxo xoxo xoxo xo xo xo xo xoxo xoxo xo xo xox o xoxo xo xo




(𝑦, −x) and free disposability implies that (𝑦, −x′ ) ∈𝑌 . Therefore x′ ∈𝑉 (𝑦 ).
xo x o x o x o x o xoxo x o xo xo xo xo xo x o xo xo xo




2. Again assume x ∈ 𝑉 (𝑦 ). This implies that (𝑦, −x) ∈ 𝑌 . By free disposal, (𝑦 ′ , −x
xox o xoxo xoxo xox o xo xo xoxoxoxo xox o xox o xox o xo xox o xo xo xoxoxoxo xox o xox o xo xo




) ∈𝑌 for every 𝑦 ′ ≤𝑦 , which implies that x ∈𝑉 (𝑦 ′ ). 𝑉 (𝑦 ′ ) ⊇𝑉 (𝑦 ).
xo o
x xox o x o xo xo o
x x o x o xoxo x o xo xo xo xoxo xo xo xo xo




1.13 The domain of “<” is {1,2}= 𝑋 and the range is {2,3}⫋ 𝑌 .xo x o xo xo x o ox o
x xo x o x o x o xo x o ox xo xo xo




1.14 Figure 1.1. xo




1.15 The relation “is strictly higher than” is transitive, antisymmetric and asymmetri
x o xo x o x o x o x o x o x o x o x o




c.It is not complete, reflexive or symmetric.
xo x o x o x o x o xo xo




2

, c⃝ 2001 Michael Carter
xoxoxo xo x o




Solutions for Foundations of Mathematical Economic xo xo x o x o x o All rights reserved xo xo




s
1.16 The following table lists their respective properties.
xo xo xo x o x o xo




< ≤ √ √= xox o




× reflexive
√ √ √
xox o



transitive xox o




symmetric √ √ xox o


×
√
xox o



asymmetric
anti-symmetric √ × √ √
×
xox o
xox o




√ √ x o x o


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




1.17 Let be∼ an equivalence relation of a set 𝑋 = . ∕That
xo ∅ is, the relation is reflexive,
xo ∼ symme xo xo xo xo xo xo xo xo x o x o xo xo xo xo xo xo




tric and transitive. We first show that every 𝑥 𝑋 belongs∈to some equivalence class. Let
xo xo xo xo xo xo xo xo xo xo xo xo xo xo x o xo




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




𝑎, that is
xo xo




∼(𝑎) ≡{𝑥 ∈𝑋 : 𝑥 ∼𝑎 } x o xo xo x o xo x o x o x o xo xo




Since ∼ is reflexive, 𝑎 ∼ 𝑎 and so 𝑎 ∈ ∼ (𝑎). Every 𝑎 ∈ xo xo xo xo xo xo x o xo




𝑋 belongs to some equivalenceclass and therefore x o x o x o x o xo x o x o



∪
𝑋 = ∼(𝑎) x o




𝑎∈𝑋

Next, we show that the equivalence classes are either disjoint or identical, that
xo x o x o x o x o x o x o x o x o x o x o xoxo




is x o




∼(𝑎) ∕= ∼(𝑏) if and only if f∼(𝑎) ∩∼(𝑏) = ∅.
xo xo x o x o x o x o x o xo o
x xo xo




First, assume ∼(𝑎) ∩∼(𝑏) = ∅. Then 𝑎 ∈∼(𝑎) but 𝑎 ∈ ∼(𝑏/
x o xo xo o
x xo xo xo x o xo xo x o xoxo ). Therefore ∼(𝑎) ∕= ∼(𝑏).
xo x o xo xo




Conversely, assume ∼(𝑎) ∩∼(𝑏) ∕= ∅and let 𝑥 ∈ ∼(𝑎) ∩∼(𝑏). Then 𝑥 ∼𝑎 and bysymmetry 𝑎 ∼
xoxo xoxo xo o
x xoxo xoxo xo xoxo xoxo xoxo xo xo xo xoxoxo xoxo xoxo xo xoxo xoxo xo x o x o xo




𝑥. Also 𝑥 ∼ 𝑏 and so by transitivity 𝑎 ∼ 𝑏. Let 𝑦 be any element in ∼(𝑎) so that 𝑦 ∼𝑎.
xoxoxo x o x o xo xo xo x o x o xo x o xo xoxoxo xo x o x o xo xo xoxo xoxo xoxo xoxo xoxo xo xoxoxo




Again by transitivity 𝑦 ∼𝑏 and therefore 𝑦 ∈ ∼(𝑏). Hence
xoxo xoxo xoxo xoxo xo xoxo xoxo xoxo xoxo xo xoxoxo




∼(𝑎) ⊆∼(𝑏). Similar reasoning implies that ∼(𝑏) ⊆∼(𝑎). Therefore ∼(𝑎) = ∼(𝑏).
xo o
x xo xoxo xo xoxo x o xo xo xo x o xo xo




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




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




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




{1}, {1, 2}and so on.
x o xo o
x x o x o




1.19

𝑥 ≻𝑦 =⇒ 𝑥 ≿ 𝑦 and 𝑦 ∕≿ 𝑥
xo xo x o x o xo xo x o x o x o xo




𝑦 ∼𝑧 =⇒ 𝑦 ≿ 𝑧 and 𝑧 ≿ 𝑦
x o xo x o x o x o xo x o x o x o xo




Transitivity of ≿ implies 𝑥 ≿ 𝑧 . We need to show that 𝑧 ∕≿ 𝑥 . Assume otherwise, thatis as
xo xo xo xo xo xo xo xo xo xo xo xo xo xo xo xo xo xo x o




sume 𝑧 ≿ 𝑥 This implies 𝑧 ∼𝑥 and by transitivity 𝑦 ∼𝑥. But this implies that
x o x o xo x o x o x o x o o
x x o x o x o x o x o xo x o x o x o x o




𝑦 ≿ 𝑥 which contradicts the assumption that 𝑥 ≻𝑦 . Therefore we conclude that 𝑧 ∕≿ 𝑥
x o xo x o x o x o x o x o x o x o xo xo x o x o x o x o x o xo




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




1.20 asymmetric Assume 𝑥 ≻𝑦. x o x o x o o
x




𝑥 ≻𝑦 =⇒ 𝑦 ∕≿ 𝑥
xo xo x o x o x o xo




while
𝑦 ≻𝑥 =⇒ 𝑦 ≿ 𝑥
x o xo x o x o x o xo




Therefore
𝑥 ≻𝑦 =⇒ 𝑦 ∕≻𝑥
x o xo x o x o x o xo




3