Solutions Manual
Foundations of Mathematical Economics
Michael Carter
, c⃝ 2001 Michael Carte
COCOCO CO CO
Solutions for Foundations of Mathematical Econom
CO CO CO CO C O r All rights reserve CO CO
ics d
Chapter 1: Sets and Spaces C O C O C O C O
1.1
{1, 3, 5, 7 . . . }or {𝑛 ∈𝑁 : 𝑛 is odd }
CO CO CO CO CO CO CO CO CO CO CO C O CO CO C O CO
1.2 Every 𝑥 ∈ 𝐴 also belongs to 𝐵. Every ∈ 𝑥
C O C O C O C O C O C O C O
𝐵 also belongs to 𝐴. Hence 𝐴, 𝐵 haveprecisely the same elements.
C O CO C O C O CO C O CO C O O
C CO CO CO
1.3 Examples of finite sets are CO CO CO CO
∙ the letters of the alphabet {A, B, C, . . . , Z }
C O CO CO C O C O CO CO C O C O CO C O CO
∙ the set of consumers in an economy
C O CO C O CO C O CO
∙ the set of goods in an economy
C O CO C O CO C O CO
∙ the set of players in a ga
CO CO CO CO CO CO
me.Examples of infinite sets are
O
C CO CO CO CO
∙ the real numbers ℜ CO CO CO
∙ the natural numbers 𝔑 CO CO CO
∙ the set of all possible colors
CO CO CO CO CO
∙ the set of possible prices of copper on the world market
C O C O C O C O C O CO C O C O C O CO
∙ the set of possible temperatures of liquid water.
C O C O C O C O C O C O C O
1.4 𝑆 = {1, 2, 3, 4, 5, 6 }, 𝐸 = {2, 4, 6 }.
CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO
1.5 The player set is 𝑁 = {Jenny, Chris } . Their action spaces are
C O C O C O C O C O CO CO CO CO CO C O C O CO
𝐴𝑖 = {Rock, Scissors, Paper }
C O CO CO CO CO CO 𝑖 = Jenny, Chris
C O CO CO
1.6 The set of players is 𝑁 { = 1, 2 , . . . }, 𝑛
C O CO CO CO C O C O C O CO CO C O . The strategy space of each player i
CO C O CO C O CO CO CO
s the set of feasible outputs
C O C O CO CO CO
𝐴𝑖 = {𝑞𝑖 ∈ℜ+ : 𝑞𝑖 ≤𝑄𝑖 }
CO CO CO C O CO CO CO CO CO CO
where 𝑞𝑖 is the output of dam 𝑖.
CO COCO COCO CO CO CO CO
1.7 The player set is 𝑁 = {1, 2, 3}. There are 23 = 8 coalitions, namely
C O CO CO C O C O CO CO CO CO C O CO CO CO C O CO
𝒫(𝑁 ) = {∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}
CO C O C O CO CO CO CO CO CO CO CO CO CO CO CO
There are 210 coalitions in a ten player game.
CO CO C O CO CO CO CO CO
𝑐
1.8 Assume that 𝑥 ∈ (𝑆 ∪𝑇 ) . That is 𝑥 ∈/ 𝑆 ∪𝑇 . This implies 𝑥 ∈/ 𝑆 and 𝑥 ∈/
COC O COCO COCO COCO CO CO CO CO COCOCO COCO COCO COCO COCO CO CO CO COCOCO COCO COCO COCO COCO COCO COCO COCO C
OCO𝑇 , or 𝑥 ∈ 𝑆𝑐 and 𝑥 ∈ 𝑇 𝑐. Consequently, 𝑥 ∈ 𝑆𝑐 ∩ 𝑇 𝑐. Conversely, assume 𝑥 ∈ 𝑆𝑐 ∩ 𝑇 𝑐. This
CO CO CO CO CO CO C O CO CO CO C O C O CO CO CO CO CO C O C O C O CO CO CO CO CO CO C
implies that 𝑥 ∈ 𝑆 𝑐 and 𝑥 ∈ 𝑇 𝑐 . Consequently 𝑥 ∈/ 𝑆 and 𝑥 ∈/ 𝑇 and therefore
OCO COCO COCO CO CO COCO COCO CO CO CO COCOCO COCO CO COCO COCO COCO CO COCO COCO COCO
𝑥 ∈/ 𝑆 ∪𝑇 . This implies that 𝑥 ∈(𝑆 ∪𝑇 )𝑐 . The other identity is proved similarly.
CO CO CO CO CO CO COCO CO CO CO CO CO CO CO CO CO CO CO CO
1.9
∪
𝑆 =𝑁 CO CO
𝑆∈𝒞
∩
𝑆 =∅ CO CO
𝑆∈𝒞
1
, c⃝ 2001 Michael Carte
COCOCO CO CO
Solutions for Foundations of Mathematical Econom
CO CO CO CO C O r All rights reserve CO CO
ics d
𝑥2
1
𝑥1
-1 0 1
-1
Figure 1.1: The relation {(𝑥, 𝑦) : 𝑥2 + 𝑦2 = 1 }
C O CO C O C O CO CO CO CO CO CO C O CO CO
1.10 The sample space of a single coin toss{ is 𝐻, }𝑇 . The set of possible outcom
CO C O C O CO C O C O C O C O CO CO CO C O CO C O CO CO C O
es inthree tosses is the product
C O O
C CO CO CO CO
{
{𝐻, 𝑇 } × {𝐻, 𝑇 } × {𝐻, 𝑇 }= (𝐻, 𝐻, 𝐻), (𝐻, 𝐻, 𝑇 ), (𝐻, 𝑇 , 𝐻),
CO CO CO CO CO CO CO CO CO C O CO CO CO CO CO CO CO CO CO CO
}
(𝐻, 𝑇 , 𝑇 ), (𝑇, 𝐻, 𝐻), (𝑇, 𝐻, 𝑇 ), (𝑇, 𝑇, 𝐻), (𝑇, 𝑇, 𝑇 ) CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO
A typical outcome is the sequence (𝐻, 𝐻, 𝑇 ) of two heads followed by a tail.
C O C O C O C O C O C O CO CO CO C O C O C O C O C O C O C O
1.11
𝑌 ∩ℜ+𝑛 = {0}
C O CO
C O
CO
where 0 = (0, 0, . . . , 0) is the production plan using no inputs and producing no outp
CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO
uts. To see this, first note that 0 is a feasible production plan. Therefore, 0
CO C O C O C O C O C O C O C O C O C O C O C O C O C O C O
∈ 𝑌 . Also, CO CO C O
0 ∈ℜ𝑛 + and therefore 0 ∈𝑌 ∩ℜ𝑛 +
CO CO
C O
. C O C O CO CO C O CO
CO
To show that there is no other feasible production plan in ℜ + 𝑛 , we assume the contra
CO CO CO CO CO CO CO CO CO CO COCOCOCOCO CO CO CO CO CO
ry. That is, we assume there is some feasible production plan
CO CO CO CO∈ ℜ y+∖{ } 𝑛 0 . T CO CO CO CO CO CO CO COC OCOCOCOCOCOCOCOCO
CO C O COCOCOCOCOCO COCO COCO
his implies the existence of a plan producing a positive output with no inputs. This
CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO
technological infeasible, so that 𝑦 ∈/ 𝑌 . CO CO CO CO CO CO CO
1.12 1. Let x ∈𝑉 (𝑦 ). This implies that (𝑦, −x) ∈𝑌 . Let x′ ≥x. Then (𝑦, −x′ ) ≤
COCO COCO CO CO CO COCO COCO COCO COCO CO CO CO CO COCO COCO CO CO COC O COCO CO CO
(𝑦, −x) and free disposability implies that (𝑦, −x′ ) ∈𝑌 . Therefore x′ ∈𝑉 (𝑦 ).
CO CO CO CO CO COCO CO CO CO CO CO CO CO CO CO CO
2. Again assume x ∈ 𝑉 (𝑦 ). This implies that (𝑦, −x) ∈ 𝑌 . By free di
COC O COCO COCO COC O CO CO COCOCOCO COC O COC O COC O CO COC O CO CO COCOCOCO COC O COC O
sposal, (𝑦 ′ , −x) ∈𝑌 for every 𝑦 ′ ≤𝑦 , which implies that x ∈𝑉 (𝑦 ′ ). 𝑉 (𝑦 ′ ) ⊇𝑉 (𝑦 )
CO CO CO CO COC O CO CO CO CO CO CO COCO CO CO CO CO COCO CO CO CO CO
.
1.13 The domain of “<” is {1, 2}= 𝑋 and the range is {2, 3}⫋ 𝑌 .
CO CO CO CO CO CO CO CO C O CO CO CO CO CO CO CO CO
1.14 Figure 1.1. CO
1.15 The relation “is strictly higher than” is transitive, antisymmetric and asym
CO CO CO CO CO CO CO CO CO CO
metric.It is not complete, reflexive or symmetric.
O
C CO CO CO CO CO CO
2
, c⃝ 2001 Michael Carte
COCOCO CO CO
Solutions for Foundations of Mathematical Econom
CO CO CO CO C O r All rights reserve CO CO
ics d
1.16 The following table lists their respective properties.
CO CO CO CO CO CO
< ≤ √ COC O
√=
reflexive ×
√ √ √
COC O
transitive COC O
symmetric √ COC O √
×
√
COC O
asymmetric × ×
anti-symmetric √ √
COC O
COC O
√
√ √ C O C O
complete ×
Note that the properties of symmetry and anti-symmetry are not mutually exclusive.
CO CO CO CO CO CO CO CO CO CO CO
∼ an equivalence relation of a set 𝑋 ∕= . ∅ That is, the relation∼is reflexive,
1.17 Let be CO CO CO CO CO CO CO CO CO C CO
O C O CO CO CO CO CO CO
symmetric and transitive. We first show that every 𝑥∈ 𝑋 belongs to some equivalenc
CO CO CO CO CO CO CO CO CO CO CO CO CO
e class. Let 𝑎 be any element in 𝑋 and∼ let (𝑎) be the class of elements equiva
CO C O CO CO CO CO CO CO C O CO CO CO CO CO CO CO CO
lent to CO
𝑎, that is
CO CO
∼(𝑎) ≡ {𝑥 ∈ 𝑋 : 𝑥 ∼ 𝑎 }
CO CO CO CO CO C O C O CO CO CO
Since ∼ is reflexive, 𝑎 ∼ 𝑎 and so 𝑎 ∈ ∼ (𝑎). Every 𝑎 ∈
CO CO CO CO CO CO C O CO
𝑋 belongs to some equivalenceclass and therefore
C O C O CO CO O
C CO CO
∪
𝑋 = ∼(𝑎) CO
𝑎∈𝑋
Next, we show that the equivalence classes are either disjoint or identical,
CO C O C O C O C O C O C O C O C O C O C O CO
that is CO C O
∼(𝑎) ∕= ∼(𝑏) if and only if f∼(𝑎) ∩∼(𝑏) = ∅.
CO CO CO CO CO CO CO CO CO CO CO
First, assume ∼(𝑎) ∩∼(𝑏) = ∅. Then 𝑎 ∈∼(𝑎) but 𝑎 ∈ ∼(𝑏/
CO CO CO CO CO CO CO CO CO CO CO COCO ). Therefore ∼(𝑎) ∕= ∼(𝑏).
CO CO CO CO
Conversely, assume ∼(𝑎) ∩∼(𝑏) ∕= ∅ and let 𝑥 ∈ ∼(𝑎) ∩∼(𝑏). Then 𝑥 ∼ 𝑎 and bysy
COCO COCO CO CO COCO COCO CO COCO COCO COCO CO CO CO COCOCO COCO COCO CO COCO COCO CO
mmetry 𝑎 ∼ 𝑥. Also 𝑥 ∼ 𝑏 and so by transitivity 𝑎 ∼ 𝑏. Let 𝑦 be any eleme
C O C O CO COCOCO C O C O CO CO CO C O C O CO C O CO COCOCO CO C O C O CO
nt in ∼(𝑎) so that 𝑦 ∼ 𝑎. Again by transitivity 𝑦 ∼ 𝑏 and therefore 𝑦 ∈ ∼(𝑏). H
CO COCO COCO COCO COCO COCO CO COCOCO COCO COCO COCO COCO CO COCO COCO COCO COCO CO COCOCO
ence
∼(𝑎) ⊆∼(𝑏). Similar reasoning implies that ∼(𝑏) ⊆∼(𝑎). Therefore ∼(𝑎) = ∼(𝑏).
CO CO CO COCO CO COCO CO CO CO CO CO CO CO
We conclude that the equivalence classes partition 𝑋.
CO CO CO CO CO CO CO
1.18 The set of proper coalitions is not a partition of the set of players, since any
CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO
playercan belong to more than one coalition. For example, player 1 belongs to th
O
C CO CO CO CO CO CO CO CO CO CO CO CO CO
e coalitions
CO
{1}, {1, 2}and so on.
CO CO CO CO CO
1.19
𝑥 ≻𝑦 =⇒ 𝑥 ≿ 𝑦 and 𝑦 ∕≿ 𝑥
CO CO C O C O CO CO C O C O C O CO
𝑦 ∼𝑧 =⇒ 𝑦 ≿ 𝑧 and 𝑧 ≿ 𝑦
C O CO C O C O CO CO C O C O C O CO
Transitivity of ≿ implies 𝑥 ≿ 𝑧 . We need to show that 𝑧 ∕≿ 𝑥 . Assume otherwise, t
CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO
hatis assume 𝑧 ≿ 𝑥 This implies 𝑧 ∼𝑥 and by transitivity 𝑦 ∼𝑥. But this impli
O
C C O C O C O CO C O C O C O C O CO C O C O C O C O C O CO C O C O C O
es that
C O
𝑦 ≿ 𝑥 which contradicts the assumption that 𝑥 ≻𝑦 . Therefore we conclude that 𝑧 ∕≿ 𝑥
C O CO C O C O C O C O C O C O CO CO CO CO C O C O C O C O CO
and therefore 𝑥 ≻𝑧 . The other result is proved in similar fashion.
CO CO CO CO CO CO CO CO CO CO CO CO
1.20 asymmetric Assume 𝑥 ≻𝑦. C O C O CO CO
Therefore
while
3
Foundations of Mathematical Economics
Michael Carter
, c⃝ 2001 Michael Carte
COCOCO CO CO
Solutions for Foundations of Mathematical Econom
CO CO CO CO C O r All rights reserve CO CO
ics d
Chapter 1: Sets and Spaces C O C O C O C O
1.1
{1, 3, 5, 7 . . . }or {𝑛 ∈𝑁 : 𝑛 is odd }
CO CO CO CO CO CO CO CO CO CO CO C O CO CO C O CO
1.2 Every 𝑥 ∈ 𝐴 also belongs to 𝐵. Every ∈ 𝑥
C O C O C O C O C O C O C O
𝐵 also belongs to 𝐴. Hence 𝐴, 𝐵 haveprecisely the same elements.
C O CO C O C O CO C O CO C O O
C CO CO CO
1.3 Examples of finite sets are CO CO CO CO
∙ the letters of the alphabet {A, B, C, . . . , Z }
C O CO CO C O C O CO CO C O C O CO C O CO
∙ the set of consumers in an economy
C O CO C O CO C O CO
∙ the set of goods in an economy
C O CO C O CO C O CO
∙ the set of players in a ga
CO CO CO CO CO CO
me.Examples of infinite sets are
O
C CO CO CO CO
∙ the real numbers ℜ CO CO CO
∙ the natural numbers 𝔑 CO CO CO
∙ the set of all possible colors
CO CO CO CO CO
∙ the set of possible prices of copper on the world market
C O C O C O C O C O CO C O C O C O CO
∙ the set of possible temperatures of liquid water.
C O C O C O C O C O C O C O
1.4 𝑆 = {1, 2, 3, 4, 5, 6 }, 𝐸 = {2, 4, 6 }.
CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO
1.5 The player set is 𝑁 = {Jenny, Chris } . Their action spaces are
C O C O C O C O C O CO CO CO CO CO C O C O CO
𝐴𝑖 = {Rock, Scissors, Paper }
C O CO CO CO CO CO 𝑖 = Jenny, Chris
C O CO CO
1.6 The set of players is 𝑁 { = 1, 2 , . . . }, 𝑛
C O CO CO CO C O C O C O CO CO C O . The strategy space of each player i
CO C O CO C O CO CO CO
s the set of feasible outputs
C O C O CO CO CO
𝐴𝑖 = {𝑞𝑖 ∈ℜ+ : 𝑞𝑖 ≤𝑄𝑖 }
CO CO CO C O CO CO CO CO CO CO
where 𝑞𝑖 is the output of dam 𝑖.
CO COCO COCO CO CO CO CO
1.7 The player set is 𝑁 = {1, 2, 3}. There are 23 = 8 coalitions, namely
C O CO CO C O C O CO CO CO CO C O CO CO CO C O CO
𝒫(𝑁 ) = {∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}
CO C O C O CO CO CO CO CO CO CO CO CO CO CO CO
There are 210 coalitions in a ten player game.
CO CO C O CO CO CO CO CO
𝑐
1.8 Assume that 𝑥 ∈ (𝑆 ∪𝑇 ) . That is 𝑥 ∈/ 𝑆 ∪𝑇 . This implies 𝑥 ∈/ 𝑆 and 𝑥 ∈/
COC O COCO COCO COCO CO CO CO CO COCOCO COCO COCO COCO COCO CO CO CO COCOCO COCO COCO COCO COCO COCO COCO COCO C
OCO𝑇 , or 𝑥 ∈ 𝑆𝑐 and 𝑥 ∈ 𝑇 𝑐. Consequently, 𝑥 ∈ 𝑆𝑐 ∩ 𝑇 𝑐. Conversely, assume 𝑥 ∈ 𝑆𝑐 ∩ 𝑇 𝑐. This
CO CO CO CO CO CO C O CO CO CO C O C O CO CO CO CO CO C O C O C O CO CO CO CO CO CO C
implies that 𝑥 ∈ 𝑆 𝑐 and 𝑥 ∈ 𝑇 𝑐 . Consequently 𝑥 ∈/ 𝑆 and 𝑥 ∈/ 𝑇 and therefore
OCO COCO COCO CO CO COCO COCO CO CO CO COCOCO COCO CO COCO COCO COCO CO COCO COCO COCO
𝑥 ∈/ 𝑆 ∪𝑇 . This implies that 𝑥 ∈(𝑆 ∪𝑇 )𝑐 . The other identity is proved similarly.
CO CO CO CO CO CO COCO CO CO CO CO CO CO CO CO CO CO CO CO
1.9
∪
𝑆 =𝑁 CO CO
𝑆∈𝒞
∩
𝑆 =∅ CO CO
𝑆∈𝒞
1
, c⃝ 2001 Michael Carte
COCOCO CO CO
Solutions for Foundations of Mathematical Econom
CO CO CO CO C O r All rights reserve CO CO
ics d
𝑥2
1
𝑥1
-1 0 1
-1
Figure 1.1: The relation {(𝑥, 𝑦) : 𝑥2 + 𝑦2 = 1 }
C O CO C O C O CO CO CO CO CO CO C O CO CO
1.10 The sample space of a single coin toss{ is 𝐻, }𝑇 . The set of possible outcom
CO C O C O CO C O C O C O C O CO CO CO C O CO C O CO CO C O
es inthree tosses is the product
C O O
C CO CO CO CO
{
{𝐻, 𝑇 } × {𝐻, 𝑇 } × {𝐻, 𝑇 }= (𝐻, 𝐻, 𝐻), (𝐻, 𝐻, 𝑇 ), (𝐻, 𝑇 , 𝐻),
CO CO CO CO CO CO CO CO CO C O CO CO CO CO CO CO CO CO CO CO
}
(𝐻, 𝑇 , 𝑇 ), (𝑇, 𝐻, 𝐻), (𝑇, 𝐻, 𝑇 ), (𝑇, 𝑇, 𝐻), (𝑇, 𝑇, 𝑇 ) CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO
A typical outcome is the sequence (𝐻, 𝐻, 𝑇 ) of two heads followed by a tail.
C O C O C O C O C O C O CO CO CO C O C O C O C O C O C O C O
1.11
𝑌 ∩ℜ+𝑛 = {0}
C O CO
C O
CO
where 0 = (0, 0, . . . , 0) is the production plan using no inputs and producing no outp
CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO
uts. To see this, first note that 0 is a feasible production plan. Therefore, 0
CO C O C O C O C O C O C O C O C O C O C O C O C O C O C O
∈ 𝑌 . Also, CO CO C O
0 ∈ℜ𝑛 + and therefore 0 ∈𝑌 ∩ℜ𝑛 +
CO CO
C O
. C O C O CO CO C O CO
CO
To show that there is no other feasible production plan in ℜ + 𝑛 , we assume the contra
CO CO CO CO CO CO CO CO CO CO COCOCOCOCO CO CO CO CO CO
ry. That is, we assume there is some feasible production plan
CO CO CO CO∈ ℜ y+∖{ } 𝑛 0 . T CO CO CO CO CO CO CO COC OCOCOCOCOCOCOCOCO
CO C O COCOCOCOCOCO COCO COCO
his implies the existence of a plan producing a positive output with no inputs. This
CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO
technological infeasible, so that 𝑦 ∈/ 𝑌 . CO CO CO CO CO CO CO
1.12 1. Let x ∈𝑉 (𝑦 ). This implies that (𝑦, −x) ∈𝑌 . Let x′ ≥x. Then (𝑦, −x′ ) ≤
COCO COCO CO CO CO COCO COCO COCO COCO CO CO CO CO COCO COCO CO CO COC O COCO CO CO
(𝑦, −x) and free disposability implies that (𝑦, −x′ ) ∈𝑌 . Therefore x′ ∈𝑉 (𝑦 ).
CO CO CO CO CO COCO CO CO CO CO CO CO CO CO CO CO
2. Again assume x ∈ 𝑉 (𝑦 ). This implies that (𝑦, −x) ∈ 𝑌 . By free di
COC O COCO COCO COC O CO CO COCOCOCO COC O COC O COC O CO COC O CO CO COCOCOCO COC O COC O
sposal, (𝑦 ′ , −x) ∈𝑌 for every 𝑦 ′ ≤𝑦 , which implies that x ∈𝑉 (𝑦 ′ ). 𝑉 (𝑦 ′ ) ⊇𝑉 (𝑦 )
CO CO CO CO COC O CO CO CO CO CO CO COCO CO CO CO CO COCO CO CO CO CO
.
1.13 The domain of “<” is {1, 2}= 𝑋 and the range is {2, 3}⫋ 𝑌 .
CO CO CO CO CO CO CO CO C O CO CO CO CO CO CO CO CO
1.14 Figure 1.1. CO
1.15 The relation “is strictly higher than” is transitive, antisymmetric and asym
CO CO CO CO CO CO CO CO CO CO
metric.It is not complete, reflexive or symmetric.
O
C CO CO CO CO CO CO
2
, c⃝ 2001 Michael Carte
COCOCO CO CO
Solutions for Foundations of Mathematical Econom
CO CO CO CO C O r All rights reserve CO CO
ics d
1.16 The following table lists their respective properties.
CO CO CO CO CO CO
< ≤ √ COC O
√=
reflexive ×
√ √ √
COC O
transitive COC O
symmetric √ COC O √
×
√
COC O
asymmetric × ×
anti-symmetric √ √
COC O
COC O
√
√ √ C O C O
complete ×
Note that the properties of symmetry and anti-symmetry are not mutually exclusive.
CO CO CO CO CO CO CO CO CO CO CO
∼ an equivalence relation of a set 𝑋 ∕= . ∅ That is, the relation∼is reflexive,
1.17 Let be CO CO CO CO CO CO CO CO CO C CO
O C O CO CO CO CO CO CO
symmetric and transitive. We first show that every 𝑥∈ 𝑋 belongs to some equivalenc
CO CO CO CO CO CO CO CO CO CO CO CO CO
e class. Let 𝑎 be any element in 𝑋 and∼ let (𝑎) be the class of elements equiva
CO C O CO CO CO CO CO CO C O CO CO CO CO CO CO CO CO
lent to CO
𝑎, that is
CO CO
∼(𝑎) ≡ {𝑥 ∈ 𝑋 : 𝑥 ∼ 𝑎 }
CO CO CO CO CO C O C O CO CO CO
Since ∼ is reflexive, 𝑎 ∼ 𝑎 and so 𝑎 ∈ ∼ (𝑎). Every 𝑎 ∈
CO CO CO CO CO CO C O CO
𝑋 belongs to some equivalenceclass and therefore
C O C O CO CO O
C CO CO
∪
𝑋 = ∼(𝑎) CO
𝑎∈𝑋
Next, we show that the equivalence classes are either disjoint or identical,
CO C O C O C O C O C O C O C O C O C O C O CO
that is CO C O
∼(𝑎) ∕= ∼(𝑏) if and only if f∼(𝑎) ∩∼(𝑏) = ∅.
CO CO CO CO CO CO CO CO CO CO CO
First, assume ∼(𝑎) ∩∼(𝑏) = ∅. Then 𝑎 ∈∼(𝑎) but 𝑎 ∈ ∼(𝑏/
CO CO CO CO CO CO CO CO CO CO CO COCO ). Therefore ∼(𝑎) ∕= ∼(𝑏).
CO CO CO CO
Conversely, assume ∼(𝑎) ∩∼(𝑏) ∕= ∅ and let 𝑥 ∈ ∼(𝑎) ∩∼(𝑏). Then 𝑥 ∼ 𝑎 and bysy
COCO COCO CO CO COCO COCO CO COCO COCO COCO CO CO CO COCOCO COCO COCO CO COCO COCO CO
mmetry 𝑎 ∼ 𝑥. Also 𝑥 ∼ 𝑏 and so by transitivity 𝑎 ∼ 𝑏. Let 𝑦 be any eleme
C O C O CO COCOCO C O C O CO CO CO C O C O CO C O CO COCOCO CO C O C O CO
nt in ∼(𝑎) so that 𝑦 ∼ 𝑎. Again by transitivity 𝑦 ∼ 𝑏 and therefore 𝑦 ∈ ∼(𝑏). H
CO COCO COCO COCO COCO COCO CO COCOCO COCO COCO COCO COCO CO COCO COCO COCO COCO CO COCOCO
ence
∼(𝑎) ⊆∼(𝑏). Similar reasoning implies that ∼(𝑏) ⊆∼(𝑎). Therefore ∼(𝑎) = ∼(𝑏).
CO CO CO COCO CO COCO CO CO CO CO CO CO CO
We conclude that the equivalence classes partition 𝑋.
CO CO CO CO CO CO CO
1.18 The set of proper coalitions is not a partition of the set of players, since any
CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO
playercan belong to more than one coalition. For example, player 1 belongs to th
O
C CO CO CO CO CO CO CO CO CO CO CO CO CO
e coalitions
CO
{1}, {1, 2}and so on.
CO CO CO CO CO
1.19
𝑥 ≻𝑦 =⇒ 𝑥 ≿ 𝑦 and 𝑦 ∕≿ 𝑥
CO CO C O C O CO CO C O C O C O CO
𝑦 ∼𝑧 =⇒ 𝑦 ≿ 𝑧 and 𝑧 ≿ 𝑦
C O CO C O C O CO CO C O C O C O CO
Transitivity of ≿ implies 𝑥 ≿ 𝑧 . We need to show that 𝑧 ∕≿ 𝑥 . Assume otherwise, t
CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO CO
hatis assume 𝑧 ≿ 𝑥 This implies 𝑧 ∼𝑥 and by transitivity 𝑦 ∼𝑥. But this impli
O
C C O C O C O CO C O C O C O C O CO C O C O C O C O C O CO C O C O C O
es that
C O
𝑦 ≿ 𝑥 which contradicts the assumption that 𝑥 ≻𝑦 . Therefore we conclude that 𝑧 ∕≿ 𝑥
C O CO C O C O C O C O C O C O CO CO CO CO C O C O C O C O CO
and therefore 𝑥 ≻𝑧 . The other result is proved in similar fashion.
CO CO CO CO CO CO CO CO CO CO CO CO
1.20 asymmetric Assume 𝑥 ≻𝑦. C O C O CO CO
Therefore
while
3
