Solutions Manual Foundations of Mathematical Economics By Michael Carter All Chapters Contained.

Solutions Manual
Foundations of Mathematical Economics

Michael Carter

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Chapter 1: Sets and Spaces

1.1
{ 1, 3, 5, 7 . . . } or { 𝑛 ∈ 𝑁 : 𝑛 is odd }
1.2 Every 𝑥 ∈ 𝐴 also belongs to 𝐵. Every 𝑥 ∈ 𝐵 also belongs to 𝐴. Hence 𝐴, 𝐵 have
precisely the same elements.
1.3 Eẍamples of finite sets are
∙ the letters of the alphabet { A, B, C, . . . , Z }
∙ the set of consumers in an economy
∙ the set of goods in an economy
∙ the set of players in a game.
Eẍamples of infinite sets are
∙ the real numbers ℜ
∙ the natural numbers 𝔑
∙ the set of all possible colors
∙ the set of possible prices of copper on the world market
∙ the set of possible temperatures of liquid water.
1.4 𝑆 = { 1, 2, 3, 4, 5, 6 }, 𝐸 = { 2, 4, 6 }.
1.5 The player set is 𝑁 = { Jenny, Chris }. Their action spaces are
𝐴𝑖 = { Rock, Scissors, Paper } 𝑖 = Jenny, Chris
{ 2, . .. , 𝑛 .}The strategy space of each player is the set
1.6 The set of players is 𝑁 = 1,
of feasible outputs
𝐴𝑖 = { 𝑞𝑖 ∈ ℜ+ : 𝑞𝑖 ≤ 𝑄𝑖 }
where 𝑞𝑖 is the output of dam 𝑖.
1.7 The player set is 𝑁 = {1, 2, 3}. There are 23 = 8 coalitions, namely
𝒫( 𝑁 ) = {∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}
There are 210 coalitions in a ten player game.
1.8 Assume that 𝑥 ∈ (𝑆 ∪ 𝑇 )𝑐 . That is 𝑥 ∈/ 𝑆 ∪ 𝑇 . This implies 𝑥 ∈/ 𝑆 and 𝑥 ∈/ 𝑇 , or 𝑥 ∈
𝑆𝑐 and 𝑥 ∈ 𝑇 𝑐. Consequently, 𝑥 ∈ 𝑆𝑐 ∩ 𝑇 𝑐. Conversely, assume 𝑥 ∈ 𝑆𝑐 ∩ 𝑇 𝑐. This implies that 𝑥
∈ 𝑆 𝑐 and 𝑥 ∈ 𝑇 𝑐 . Consequently 𝑥 ∈/ 𝑆 and 𝑥 ∈/ 𝑇 and therefore
𝑥 ∈/ 𝑆 ∪ 𝑇 . This implies that 𝑥 ∈ (𝑆 ∪ 𝑇 )𝑐 . The other identity is proved similarly.
1.9
∪
𝑆=𝑁
𝑆∈𝒞
∩
𝑆=∅
𝑆∈𝒞


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, c⃝ 2001 Michael Carter
Solutions for Foundations of Mathematical Economics All rights reserved


𝑥2
1




𝑥1
-1 0 1




-1

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


1.10 The sample space of a single coin toss is 𝐻{, 𝑇 . The } set of possible outcomes in
three tosses is the product
{
{𝐻, 𝑇 }× {𝐻, 𝑇 }× {𝐻, 𝑇 } = (𝐻, 𝐻, 𝐻), (𝐻, 𝐻, 𝑇 ), (𝐻, 𝑇 , 𝐻),
}
(𝐻, 𝑇 , 𝑇 ), (𝑇, 𝐻, 𝐻), (𝑇, 𝐻, 𝑇 ), (𝑇, 𝑇, 𝐻), (𝑇, 𝑇, 𝑇 )


A typical outcome is the sequence (𝐻, 𝐻, 𝑇 ) of two heads followed by a tail.
1.11

𝑌 ∩ ℜ+𝑛 = {0}

where 0 = (0, 0, . . . , 0) is the production plan using no inputs and producing no outputs.
To see this, first note that 0 is a feasible production plan. Therefore, 0 ∈ 𝑌 . Also,
0 ∈ ℜ𝑛+and therefore 0 ∈ 𝑌 ∩ ℜ𝑛 . +
𝑛
To show that there is no other feasible production plan in ℜ + , we assume the contrary.
That is, we assume there is some feasible production plan y ∈ ℜ 𝑛+∖ { }0 . This implies
the eẍistence of a plan producing a positive output with no inputs. This technological
infeasible, so that 𝑦 ∈/ 𝑌 .
1.12 1. Let ẍ ∈ 𝑉 (𝑦). This implies that (𝑦, −ẍ) ∈ 𝑌 . Let ẍ′ ≥ ẍ. Then (𝑦, −ẍ′ ) ≤
(𝑦, −ẍ) and free disposability implies that (𝑦, −ẍ′ ) ∈ 𝑌 . Therefore ẍ′ ∈ 𝑉 (𝑦).
2. Again assume ẍ ∈ 𝑉 (𝑦). This implies that (𝑦, −ẍ) ∈ 𝑌 . By free disposal, (𝑦′ ,
−ẍ) ∈ 𝑌 for every 𝑦′ ≤ 𝑦, which implies that ẍ ∈ 𝑉 (𝑦′ ). 𝑉 (𝑦′ ) ⊇ 𝑉 (𝑦).
1.13 The domain of “<” is {1, 2} = 𝑋 and the range is {2, 3} ⫋ 𝑌 .
1.14 Figure 1.1.
1.15 The relation “is strictly higher than” is transitive, antisymmetric and asymmetric.
It is not complete, refleẍive or symmetric.




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, c⃝ 2001 Michael Carter
Solutions for Foundations of Mathematical Economics All rights reserved


1.16 The following table lists their respective properties.
< ≤√ √=
refleẍive ×
transitive √ √ √
symmetric √ √
×
asymmetric √
anti-symmetric √ ×
√ ×
√
√ √
complete ×
Note that the properties of symmetry and anti-symmetry are not mutually eẍclusive.
1.17 Let ∼ be an equivalence relation of a set 𝑋∕ = ∅. That is, the relation∼ is refleẍive,
symmetric and transitive. We first show that every 𝑥∈ 𝑋 belongs to some equivalence
class. Let 𝑎 be any element in 𝑋 and let (𝑎∼) be the class of elements equivalent to
𝑎, that is
∼(𝑎) ≡ { 𝑥 ∈ 𝑋 : 𝑥 ∼ 𝑎 }
Since ∼ is refleẍive, 𝑎 ∼ 𝑎 and so 𝑎 ∈ ∼ (𝑎). Every 𝑎 ∈ 𝑋 belongs to some equivalence
class and therefore
∪
𝑋= ∼(𝑎)
𝑎∈𝑋

Neẍt, we show that the equivalence classes are either disjoint or identical, that is
∼(𝑎) ∕= ∼(𝑏) if and only if f∼(𝑎) ∩ ∼(𝑏) = ∅.
First, assume ∼(𝑎) ∩ ∼(𝑏) = ∅. Then 𝑎 ∈ ∼(𝑎) but 𝑎 ∈ ∼(𝑏/ ). Therefore ∼(𝑎) ∕= ∼(𝑏).
Conversely, assume ∼(𝑎) ∩ ∼(𝑏) ∕= ∅ and let 𝑥 ∈ ∼(𝑎) ∩ ∼(𝑏). Then 𝑥 ∼ 𝑎 and by
symmetry 𝑎 ∼ 𝑥. Also 𝑥 ∼ 𝑏 and so by transitivity 𝑎 ∼ 𝑏. Let 𝑦 be any element in
∼(𝑎) so that 𝑦 ∼ 𝑎. Again by transitivity 𝑦 ∼ 𝑏 and therefore 𝑦 ∈ ∼(𝑏). Hence
∼(𝑎) ⊆ ∼(𝑏). Similar reasoning implies that ∼(𝑏) ⊆ ∼(𝑎). Therefore ∼(𝑎) = ∼(𝑏).
We conclude that the equivalence classes partition 𝑋.
1.18 The set of proper coalitions is not a partition of the set of players, since any player
can belong to more than one coalition. For eẍample, player 1 belongs to the coalitions
{1}, {1, 2} and so on.
1.19

𝑥 ≻ 𝑦 =⇒ 𝑥 ≿ 𝑦 and 𝑦 ∕≿ 𝑥
𝑦 ∼ 𝑧 =⇒ 𝑦 ≿ 𝑧 and 𝑧 ≿ 𝑦
Transitivity of ≿ implies 𝑥 ≿ 𝑧 . We need to show that 𝑧 ∕≿ 𝑥 . Assume otherwise, thatis
assume 𝑧 ≿ 𝑥 This implies 𝑧 ∼ 𝑥 and by transitivity 𝑦 ∼ 𝑥. But this implies that
𝑦 ≿ 𝑥 which contradicts the assumption that 𝑥 ≻ 𝑦 . Therefore we conclude that 𝑧 ∕≿ 𝑥
and therefore 𝑥 ≻ 𝑧 . The other result is proved in similar fashion.
1.20 asymmetric Assume 𝑥 ≻ 𝑦.
𝑥 ≻ 𝑦 =⇒ 𝑦 ∕≿ 𝑥
while
𝑦 ≻ 𝑥 =⇒ 𝑦 ≿ 𝑥
Therefore
𝑥 ≻ 𝑦 =⇒ 𝑦 ∕≻ 𝑥


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