Foundations of Mathematical Economics
Michael Carter
,
, c⃝ 2001 Michael Carter
ujujuj uj u j
Solutions for Foundations of Mathematical Economics uj uj u j u j u j All rights reserved uj uj
Chapter 1: Sets and Spaces u j u j u j u j
1.1
{1, 3, 5, 7 . . . }or {𝑛 ∈𝑁 : 𝑛 is odd }
uj uj uj uj uj uj uj u j uj uj uj u j uj u j u j uj
1.2 Every 𝑥 ∈ 𝐴 also belongs to 𝐵. Every 𝑥 ∈ u j u j u j u j u j u j u j
𝐵 also belongs to 𝐴. Hence 𝐴,𝐵 haveprecisely the same elements.
u j u j u j u j u j u j uj u j u
j u j u j u j
1.3 Examples of finite sets are uj uj u j u j
∙ the letters of the alphabet {A, B, C, . . . , Z }
u j u j u j u j u j uj u j u j u j u j u j uj
∙ the set of consumers in an economy u j u j u j u j u j u j
∙ the set of goods in an economy u j u j u j u j u j u j
∙ the set of players in a game. uj uj uj uj uj uj u
j
Examples of infinite sets are u j u j u j u j
∙ the real numbers ℜ uj uj uj
∙ the natural numbers 𝔑 u j uj u j
∙ the set of all possible colors uj uj uj uj uj
∙ the set of possible prices of copper on the world market
u j u j u j u j u j u j u j u j u j u j
∙ the set of possible temperatures of liquid water.
u j u j u j u j u j u j u j
1.4 𝑆 = {1,2,3,4, 5,6 }, 𝐸 = {2,4,6 }.
uj u j uj j
u uj uj uj uj uj uj uj u j uj j
u uj uj uj
1.5 The player set is 𝑁 = {Jenny, Chris } . Their action spaces are
u j u j u j u j u j uj j
u uj uj uj u j u j u j
𝐴𝑖 = {Rock, Scissors,Paper }
u j uj j
u uj uj uj 𝑖 = Jenny, Chris
u j uj uj
1.6 The set of players is 𝑁 = {1, 2 , .. . , 𝑛 }. The strategy space of each player is the se
u j u j u j u j u j u j u j uj uj u j uj u j u j u j u j u j u j u j u j
t of feasible outputs
uj u j u j
𝐴𝑖 = {𝑞𝑖 ∈ℜ+ : 𝑞𝑖 ≤𝑄𝑖 }
u j uj uj u j uj u j uj u j uj uj
where 𝑞𝑖 is the output of dam 𝑖.u j uju
j uju
j u j u j u j u j
1.7 The player set is 𝑁 = {1, 2, 3 }. There are 23 = 8 coalitions, namely
u j u j u j u j u j uj uj uj uj u j u j u j uj u j u j
𝒫(𝑁) = {∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}
uj u j u j uj uj uj uj uj uj uj uj uj uj uj uj
There are 210 coalitions in a ten player game.
u j u j u j u j u j u j u j u j
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 𝑐. C
uju j ujuj ujuj uju
j j
u u j j
u j
u
𝑐
ujujuj ujuj ujuj ujuj ujuj uj uj uj ujujuj ujuj ujuj ujuj ujuj ujuj ujuj ujuj ujuj uj uj uj uj uj uj u j uj uj uj u j
onsequently, 𝑥 ∈ 𝑆𝑐 ∩ 𝑇 𝑐. Conversely, assume 𝑥 ∈ 𝑆𝑐 ∩ 𝑇 𝑐. This implies that 𝑥 ∈𝑆 𝑐 and 𝑥 ∈𝑇𝑐 . Conse
u j uj uj uj uj uj u j u j u j uj uj uj uj uj uj ujuj ujuj ujuj u j uj ujuj ujuj u j uj uj ujujuj
quently 𝑥∈/ 𝑆 and 𝑥∈/ 𝑇 and therefore ujuj j
u ujuj ujuj ujuj j
u ujuj uju j ujuj
𝑥 ∈/ 𝑆 ∪𝑇. This implies that 𝑥 ∈(𝑆 ∪𝑇)𝑐 . The other identity is proved similarly.
uj uj j
u uj uj u j uju
j u j uj j
u uj j
u uj uj u j u j u j u j u j
1.9
∪
𝑆 =𝑁 uj uj
𝑆∈𝒞
∩
𝑆 =∅ uj uj
𝑆∈𝒞
1
, c⃝ 2001 Michael Carter
ujujuj uj u j
Solutions for Foundations of Mathematical Economics uj uj u j u j u j All rights reserved uj uj
𝑥2
1
𝑥1
-1 0 1
-1
Figure 1.1: The relation {(𝑥, 𝑦) : 𝑥2 + 𝑦2 = 1 } u j uj u j u j uj uj uj u j u j uj u j u j uj
1.10 The sample space of a single coin toss is 𝐻{, 𝑇 . The
u j u j } set of possible outcomes inthr u j u j u j u j u j u j u j uj uj u j uj u j u j u j u j u j u
j
ee tosses is the product
u j u j u j u j
{
{𝐻, 𝑇} × {𝐻, 𝑇} × {𝐻, 𝑇}= (𝐻, 𝐻, 𝐻), (𝐻, 𝐻, 𝑇), (𝐻, 𝑇, 𝐻),
uj uj j
u uj uj j
u uj uj j
u u j uj uj uj uj uj uj uj uj uj uj
}
(𝐻, 𝑇, 𝑇 ), (𝑇, 𝐻, 𝐻), (𝑇, 𝐻, 𝑇 ), (𝑇, 𝑇, 𝐻), (𝑇, 𝑇, 𝑇 ) uj uj uj uj uj uj uj uj uj uj uj uj uj uj uj uj uj uj
A typical outcome is the sequence (𝐻, 𝐻, 𝑇) of two heads followed by a tail.
u j u j u j u j u j u j uj uj uj u j u j u j u j u j u j u j
1.11
𝑌 ∩ℜ+𝑛 = {0}
u j uj
u j
u j
where 0 = (0,0 , . . . ,0) is the production plan using no inputs and producing no outputs. To s
uj uj uj uj j uj
u uj uj uj uj uj uj uj uj uj uj uj uj u j
ee this, first note that 0 is a feasible production plan. Therefore, 0 ∈𝑌 . Also,
u j u j u j u j u j u j u j u j u j u j u j u j u j uj uj u j
0 ∈ℜ𝑛 +and therefore 0 ∈𝑌 ∩ℜ𝑛 . +
u j uj
u j
u j u j u j uj u j uj
uj
To show that there is no other feasible production plan in 𝑛 , weℜ assume
uj uj
+ the contrary. That is,
uj uj uj uj uj uj uj uj ujujujujuj uj uj uj uj uj uj uj uj
we assume there is some feasible production plan y 𝑛 0 . This
uj
+∖{ } the existence of
∈ ℜimplies uj uj uj uj uj uj uj ujujujujujujujuj ujujujujujuj ujuj ujuj uj uju j uj u j uj uj uj u
a plan producing a positive output with no inputs. This technological infeasible, so that 𝑦∈/
j uj uj uj uj uj uj uj uj uj uj uj u j u j u j uj
𝑌.
u j uj
1.12 1. Let x ∈𝑉 (𝑦 ). This implies that (𝑦,−x) ∈𝑌 . Let x′ ≥x. Then (𝑦, −x′ ) ≤
ujuj uju
j u j j
u uj ujuj uju
j uju
j uju
j uj u j uj uj ujuj uju
j uj uj uju j uju
j uj uj
(𝑦,−x) and free disposability implies that (𝑦, −x′ ) ∈𝑌 . Therefore x′ ∈𝑉 (𝑦 ).
uj u j u j u j u j uju
j u j uj uj uj uj uj u j uj uj uj
2. Again assume x ∈ 𝑉 (𝑦 ). This implies that (𝑦,−x) ∈ 𝑌 . By free disposal, (𝑦 ′ , −x) ∈
uju j uju j uju j uju j uj uj ujujujuj uju j uju j uju j uj uju j uj uj ujujujuj uju j uju j uj uj uj uj
𝑌 for every 𝑦 ′ ≤𝑦 , which implies that x ∈𝑉 (𝑦 ′ ). 𝑉 (𝑦 ′ ) ⊇𝑉 (𝑦 ).
uju j u j u j uj j
u u j u j uju
j u j uj uj uj ujuj uj uj uj uj
1.13 The domain of “<” is {1,2}= 𝑋 and the range is {2,3}⫋ 𝑌 . u j u j u j u j u j uj j
u uj u j u j u j uj u j uj uj uj uj
1.14 Figure 1.1. uj
1.15 The relation “is strictly higher than” is transitive, antisymmetric and asymmetric
u j u j u j u j u j u j u j u j u j u j
.It is not complete, reflexive or symmetric.
u
j u j u j u j u j uj u j
2