Solutions Manual Foundations of Mathematical Economics ( PDFDrive.com )

Solutions Manual
Foundations of Mathematical Economics
Michael Carter
November 15, 2002

Solutions for Foundations of Mathematical Economics
c
⃝2001 Michael Carter
All rights reserved
Chapter 1: Sets and Spaces
1.1
{1,3,5,7...}or {𝑛∈𝑁:𝑛is odd }
1.2 Every 𝑥∈𝐴also belongs to 𝐵. Every 𝑥∈𝐵also belongs to 𝐴. Hence 𝐴, 𝐵 have
precisely the same elements.
1.3 Examples 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.
Examples 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
1.6 The set of players is 𝑁={1,2,...,𝑛}. The strategy space of each player is the set
of feasible outputs
𝐴𝑖={𝑞𝑖∈ℜ
+:𝑞𝑖≤𝑄𝑖}
where 𝑞𝑖is 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/∈𝑆and 𝑥/∈𝑇,
or 𝑥∈𝑆𝑐and 𝑥∈𝑇𝑐. Consequently, 𝑥∈𝑆𝑐∩𝑇𝑐. Conversely, assume 𝑥∈𝑆𝑐∩𝑇𝑐.
This implies that 𝑥∈𝑆𝑐and 𝑥∈𝑇𝑐. Consequently 𝑥/∈𝑆and 𝑥/∈𝑇and therefore
𝑥/∈𝑆∪𝑇. This implies that 𝑥∈(𝑆∪𝑇)𝑐. The other identity is proved similarly.
1.9
𝑆∈𝒞
𝑆=𝑁
𝑆∈𝒞
𝑆=∅
1

Solutions for Foundations of Mathematical Economics
c
⃝2001 Michael Carter
All rights reserved
0-1 1 𝑥1
-1
1
𝑥2
Figure 1.1: The relation {(𝑥, 𝑦):𝑥2+𝑦2=1}
1.10 The sample space of a single coin toss is {𝐻, 𝑇 }. Thesetofpossibleoutcomesin
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 0is a feasible production plan. Therefore, 0∈𝑌. Also,
0∈ℜ
𝑛
+and therefore 0∈𝑌∩ℜ
𝑛
+.
To show that there is no other feasible production plan in ℜ𝑛
+, we assume the contrary.
Thatis,weassumethereissomefeasibleproductionplany∈ℜ
𝑛
+∖{0}. This implies
the existence of a plan producing a positive output with no inputs. This technological
infeasible, so that 𝑦/∈𝑌.
1.12 1. Let x∈𝑉(𝑦). This implies that (𝑦, −x)∈𝑌. Let x′≥x. Then (𝑦, −x′)≤
(𝑦, −x) and free disposability implies that (𝑦, −x′)∈𝑌. Therefore x′∈𝑉(𝑦).
2. Again assume x∈𝑉(𝑦). This implies that (𝑦, −x)∈𝑌. By free disposal,
(𝑦′,−x)∈𝑌for every 𝑦′≤𝑦, which implies that x∈𝑉(𝑦′). 𝑉(𝑦′)⊇𝑉(𝑦).
1.13 The domain of “<”is{1,2}=𝑋and 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, reflexive or symmetric.
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Solutions for Foundations of Mathematical Economics
c
⃝2001 Michael Carter
All rights reserved
1.16 The following table lists their respective properties.
<≤=
reflexive ×√√
transitive √√√
symmetric ×√√
asymmetric √××
anti-symmetric √√√
complete √√×
Note that the properties of symmetry and anti-symmetry are not mutually exclusive.
1.17 Let ∼be an equivalence relation of a set 𝑋∕=∅. That is, the relation ∼is reflexive,
symmetric and transitive. We first show that every 𝑥∈𝑋belongs to some equivalence
class. Let 𝑎be any element in 𝑋and let ∼(𝑎) be the class of elements equivalent to
𝑎,thatis
∼(𝑎)≡{𝑥∈𝑋:𝑥∼𝑎}
Since ∼is reflexive, 𝑎∼𝑎and so 𝑎∈∼(𝑎). Every 𝑎∈𝑋belongs to some equivalence
class and therefore
𝑋=
𝑎∈𝑋∼(𝑎)
Next, 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∼𝑎and by
symmetry 𝑎∼𝑥. Also 𝑥∼𝑏and so by transitivity 𝑎∼𝑏. Let 𝑦be any element
in ∼(𝑎)sothat𝑦∼𝑎. Again by transitivity 𝑦∼𝑏and 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 example, player 1 belongs to the coalitions
{1},{1,2}and so on.
1.19
𝑥≻𝑦=⇒𝑥≿𝑦and 𝑦∕≿𝑥
𝑦∼𝑧=⇒𝑦≿𝑧and 𝑧≿𝑦
Transitivity of ≿implies 𝑥≿𝑧. We need to show that 𝑧∕≿𝑥. Assume otherwise, that
is assume 𝑧≿𝑥This implies 𝑧∼𝑥and by transitivity 𝑦∼𝑥. But this implies that
𝑦≿𝑥which 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
3

Solutions for Foundations of Mathematical Economics
c
⃝2001 Michael Carter
All rights reserved
transitive Assume 𝑥≻𝑦and 𝑦≻𝑧.
𝑥≻𝑦=⇒𝑥≿𝑦and 𝑦∕≿𝑥
𝑦≻𝑧=⇒𝑦≿𝑧and 𝑧∕≿𝑦
Since ≿is transitive, we conclude that 𝑥≿𝑧.
It remains to show that 𝑧∕≿𝑥. Assume otherwise, that is assume 𝑧≿𝑥. We
know that 𝑥≿𝑦and transitivity implies that 𝑧≿𝑦, contrary to the assumption
that 𝑦≻𝑧. We conclude that 𝑧∕≿𝑥and
𝑥≿𝑧and 𝑧∕≿𝑥=⇒𝑥≻𝑧
This shows that ≻is transitive.
1.21 reflexive Since ≿is reflexive, 𝑥≿𝑥which implies 𝑥∼𝑥.
transitive Assume 𝑥∼𝑦and 𝑦∼𝑧. Now
𝑥∼𝑦⇐⇒ 𝑥≿𝑦and 𝑦≿𝑥
𝑦∼𝑧⇐⇒ 𝑦≿𝑧and 𝑧≿𝑦
Transitivity of ≿implies
𝑥≿𝑦and 𝑦≿𝑧=⇒𝑥≿𝑧
𝑧≿𝑦and 𝑦≿𝑥=⇒𝑧≿𝑥
Combining
𝑥≿𝑧and 𝑧≿𝑥=⇒𝑥∼𝑧
symmetric
𝑥∼𝑦⇐⇒ 𝑥≿𝑦and 𝑦≿𝑥
⇐⇒ 𝑦≿𝑥and 𝑥≿𝑦
⇐⇒ 𝑦∼𝑥
1.22 reflexive Every integer is a multiple of itself, that is 𝑚=1𝑚.
transitive Assume 𝑚=𝑘𝑛 and 𝑛=𝑙𝑝 where 𝑘, 𝑙 ∈𝑁. Then 𝑚=𝑘𝑙𝑝 so that 𝑚is a
multiple of 𝑝.
not symmetric If 𝑚=𝑘𝑛,𝑘∈𝑁,then𝑛=1
𝑘𝑚and 𝑘/∈𝑁. For example, 4 is a
multiple of 2 but 2 is not a multiple of 4.
1.23
[𝑎, 𝑏]={𝑎, 𝑦, 𝑏, 𝑧 }
(𝑎, 𝑏)={𝑦}
1.24
≿(𝑦)={𝑏, 𝑦, 𝑧 }
≻(𝑦)={𝑏, 𝑧 }
≾(𝑦)={𝑎, 𝑥, 𝑦 }
≺(𝑦)={𝑎, 𝑥 }
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