Solution Manual for Foundations of Mathematical Economics By Michael Carter A+ verified, ISBN: 978-0262531924

Solutions Manual
Foundations of Mathematical Economics


Michael Carter

, c⃝ 2001 Michael Carter
rqrqrq rq r q




Solutions for Foundations of Mathematical Economics rq rq r q r q r q All rights reserved rq rq




Chapter 1: Sets and Spaces r q r q r q r q




1.1
{1, 3, 5, 7 . . . }or {𝑛 ∈𝑁 : 𝑛 is odd }
rq rq rq rq rq qr rq r q rq rq rq r q rq r q r q rq




1.2 Every 𝑥 ∈ 𝐴 also belongs to 𝐵. Every 𝑥 ∈ r q r q r q r q r q r q r q




𝐵 also belongs to 𝐴. Hence 𝐴, 𝐵 haveprecisely the same elements.
r q r q r q r q r q r q qr r q q
r r q r q r q




1.3 Examples of finite sets are rq rq r q r q




∙ the letters of the alphabet {A, B, C, . . . , Z }
r q r q r q r q r q rq r q r q r q r q r q q
r




∙ the set of consumers in an economy r q r q r q r q r q r q




∙ the set of goods in an economy r q r q r q r q r q r q




∙ the set of players in a game. rq rq rq rq rq rq q
r




Examples of infinite sets are r q rq r q r q




∙ the real numbers ℜ rq rq rq




∙ the natural numbers 𝔑 r q rq r q




∙ the set of all possible colors rq rq rq rq rq




∙ the set of possible prices of copper on the world market
r q r q r q r q r q r q r q r q r q r q




∙ the set of possible temperatures of liquid water.
r q r q r q r q r q r q r q




1.4 𝑆 = {1,2,3, 4,5,6 }, 𝐸 = {2,4,6 }.
rq r q rq q
r qr qr qr qr qr rq rq r q rq q
r qr qr rq




1.5 The player set is 𝑁 = {Jenny, Chris } . Their action spaces are
r q r q r q r q r q rq q
r rq rq rq r q r q r q




𝐴𝑖 = {Rock, Scissors, Paper }
r q rq q
r rq rq rq 𝑖 = Jenny, Chris
r q rq rq




1.6 The set of players is 𝑁 = { 1, 2 , .. ., 𝑛 } . The strategy space of each player is the s
r q r q r q r q r q r q r q rq rq r q rq r q r q r q r q r q r q r q r q




et of feasible outputs
rq r q r q




𝐴𝑖 = {𝑞𝑖 ∈ℜ+ : 𝑞𝑖 ≤𝑄𝑖 }
r q rq rq r q rq r q rq r q rq rq




where 𝑞𝑖 is the output of dam 𝑖.
r q rqrq rqrq r q r q r q r q




1.7 The player set is 𝑁 = {1, 2, 3}. There are 23 = 8 coalitions, namely
r q r q r q r q r q rq qr qr rq r q r q r q rq r q r q




𝒫(𝑁) = {∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}
qr r q r q rq rq rq rq rq rq rq rq rq rq rq rq




There are 210 coalitions in a ten player game.
r q r q r q r q r q r q r q r q




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.
rqr q rqrq rqrq rqrq q
r r q q
r qr
𝑐
rqrqrq rqrq rqrq rqrq rqrq rq rq qr rqrqrq rqrq rqrq rqrq rqrq rqrq rqrq rqrq rqrq qr rq rq rq rq rq r q rq rq rq




r qConsequently, 𝑥 ∈ 𝑆𝑐 ∩ 𝑇 𝑐. Conversely, assume 𝑥 ∈ 𝑆𝑐 ∩ 𝑇 𝑐. This implies that 𝑥 ∈𝑆 𝑐 and 𝑥 ∈𝑇𝑐 . Co
r q rq rq rq rq rq r q r q r q rq rq rq rq rq rq rqrq rqrq rqrq r q rq rqrq rqrq r q rq qr rqrqrq




nsequently 𝑥∈/ 𝑆 and 𝑥∈/ 𝑇 and therefore rqrq qr rqrq rqrq rqrq qr rqrq rqrq rqrq




𝑥 ∈/ 𝑆 ∪𝑇. This implies that 𝑥 ∈(𝑆 ∪𝑇)𝑐 . The other identity is proved similarly.
rq rq q
r qr rq r q rqrq r q rq q
r rq q
r qr rq r q r q r q r q r q




1.9
∪
𝑆 =𝑁 rq rq




𝑆∈𝒞
∩
𝑆 =∅ rq rq




𝑆∈𝒞


1

, c⃝ 2001 Michael Carter
rqrqrq rq r q




Solutions for Foundations of Mathematical Economics rq rq r q r q r q All rights reserved rq rq




𝑥2
1




𝑥1
-1 0 1




-1

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




1.10 The sample space of a single coin toss is 𝐻
r q r q { , 𝑇 . The
} set of possible outcomes inthr r q r q r q r q r q r q r q rq rq r q rq r q r q r q r q r q q
r




ee tosses is the product
r q r q r q r q




{
{𝐻, 𝑇} × {𝐻, 𝑇} × {𝐻, 𝑇}= (𝐻, 𝐻, 𝐻), (𝐻, 𝐻, 𝑇), (𝐻, 𝑇, 𝐻),
rq qr q
r rq qr q
r rq qr q
r r q rq rq rq rq rq qr qr rq qr rq



}
(𝐻, 𝑇, 𝑇 ), (𝑇, 𝐻, 𝐻), (𝑇, 𝐻, 𝑇 ), (𝑇, 𝑇, 𝐻), (𝑇, 𝑇, 𝑇 ) rq qr rq rq rq rq rq rq rq rq rq rq rq rq rq rq rq rq




A typical outcome is the sequence (𝐻, 𝐻, 𝑇) of two heads followed by a tail.
r q r q r q r q r q r q rq rq qr r q r q r q r q r q r q r q




1.11

𝑌 ∩ℜ+𝑛 = {0}r q rq
r q
rq




where 0 = (0,0 , . . . ,0) is the production plan using no inputs and producing no outputs. To
rq rq rq qr qr qr rq rq rq rq rq rq rq rq rq rq rq rq r q




see this, first note that 0 is a feasible production plan. Therefore, 0 ∈𝑌 . Also,
r q r q r q r q r q r q r q r q r q r q r q r q r q rq rq r q




0 ∈ℜ𝑛 +and therefore 0 ∈𝑌 ∩ℜ𝑛 . +
r q rq
r q
r q r q r q rq r q rq
rq




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



𝑛
s, we assume there is some feasible production plan y
rq rq
+∖{implies
0 .∈ ℜThis } the existenc
rq rq rq rq rq rq rq rqrqrqrqrqrqrqrq rqrqrqrqrqrq rqrq rqrqrq r q rqrq r q rq rq




e of a plan producing a positive output with no inputs. This technological infeasible, so th
rq rq rq rq rq rq rq rq rq rq rq rq rq r q r q




at 𝑦∈/ 𝑌 . r q qr r q rq




1.12 1. Let x ∈𝑉 (𝑦 ). This implies that (𝑦, −x) ∈𝑌 . Let x′ ≥x. Then (𝑦, −x′ ) ≤
rqrq rqrq r q q
r rq rqrq rqrq rqrq rqrq qr r q rq rq rqrq rqrq rq rq rqr q rqrq qr rq




(𝑦,−x) and free disposability implies that (𝑦,−x′ ) ∈𝑌 . Therefore x′ ∈𝑉 (𝑦 ).
qr r q r q r q r q rqrq r q qr rq rq rq rq r q rq rq rq




2. Again assume x ∈ 𝑉 (𝑦 ). This implies that (𝑦,−x) ∈ 𝑌 . By free disposal, (𝑦 ′ ,−x)
rqr q rqr q rqr q rqr q rq rq rqrqrqrq rqr q rqr q rqr q qr rqr q rq rq rqrqrqrq rqr q rqr q rq qr rq




∈𝑌 for every 𝑦 ′ ≤𝑦 , which implies that x ∈𝑉 (𝑦 ′ ). 𝑉 (𝑦 ′ ) ⊇𝑉 (𝑦 ).
q
r rqr q r q r q rq q
r r q r q rqrq r q rq rq rq rqrq rq rq rq rq




1.13 The domain of “<” is {1,2}= 𝑋 and the range is {2,3}⫋ 𝑌 .r q r q r q r q r q qr q
r rq r q r q r q r q r q qr rq rq rq




1.14 Figure 1.1. rq




1.15 The relation “is strictly higher than” is transitive, antisymmetric and asymmetri
r q r q r q r q r q r q r q r q r q r q




c.It is not complete, reflexive or symmetric.
q
r r q r q r q r q rq r q




2

, c⃝ 2001 Michael Carter
rqrqrq rq r q




Solutions for Foundations of Mathematical Economics rq rq r q r q r q All rights reserved rq rq




1.16 The following table lists their respective properties. r q r q r q r q r q rq




< ≤ √ √= rqr q




× reflexive
√ √ √
rqr q



transitive rqr q




symmetric √ √ rqr q


×
√
rqr q



asymmetric
anti-symmetric √ × √ √
×
rqr q
rqr q




√ √ r q r q


complete ×
Note that the properties of symmetry and anti-symmetry are not mutually exclusive.
r q r q r q rq r q r q rq rq rq r q r q




1.17 Let be ∼ an equivalence relation of a set 𝑋 = . ∕That
∅ is, the relation is reflexive,
rq ∼ rq symme rq rq rq rq rq rq rq rq r q r q rq rq rq rq rq rq




tric and transitive. We first show that every 𝑥 𝑋 belongs ∈to some equivalence class. Let 𝑎
rq rq rq rq rq rq rq rq rq rq rq rq rq rq r q rq r




be any element in 𝑋 and let (𝑎) be the class
q rq ∼ of elements equivalent to
rq rq rq r q rq rq rq rq rq rq rq rq rq




𝑎, that is rq rq




∼(𝑎) ≡{𝑥 ∈𝑋 : 𝑥 ∼𝑎 } r q rq rq r q rq r q r q r q rq rq




Since ∼ is reflexive, 𝑎 ∼ 𝑎 and so 𝑎 ∈ ∼ (𝑎). Every 𝑎 ∈ rq rq rq rq rq rq r q r q




𝑋 belongs to some equivalenceclass and therefore r q r q r q r q q
r r q r q



∪
𝑋 = ∼(𝑎) r q




𝑎∈𝑋

Next, we show that the equivalence classes are either disjoint or identical, that
rq r q r q r q r q r q r q r q r q r q r q rqr q r




is q




∼(𝑎) ∕= ∼(𝑏) if and only if f∼(𝑎) ∩∼(𝑏) = ∅.
rq rq r q r q r q r q r q rq q
r rq rq




First, assume ∼(𝑎) ∩∼(𝑏) = ∅. Then 𝑎 ∈∼(𝑎) but 𝑎 ∈ ∼(𝑏/
r q r q rq q
r rq rq rq r q rq rq r q rqrq ). Therefore ∼(𝑎) ∕= ∼(𝑏).
rq r q rq rq




Conversely, assume ∼(𝑎) ∩∼(𝑏) ∕= ∅and let 𝑥 ∈ ∼(𝑎) ∩∼(𝑏). Then 𝑥 ∼𝑎 and bysymmetry 𝑎 ∼ 𝑥
rqrq rqrq rq q
r rqrq rqrq rq rqrq rqrq rqrq rq rq rq rqrqrq rqrq rqrq rq rqrq rqrq rq r q r q rq




. Also 𝑥 ∼ 𝑏 and so by transitivity 𝑎 ∼ 𝑏. Let 𝑦 be any element in ∼(𝑎) so that 𝑦 ∼𝑎. A
rqrqrq r q r q rq rq rq r q r q rq r q rq rqrqrq rq r q r q rq rq rqrq rqrq rqrq rqrq rqrq rq rqrqrq




gain by transitivity 𝑦 ∼𝑏 and therefore 𝑦 ∈ ∼(𝑏). Hence
rqrq rqrq rqrq rqrq rq rqrq rqrq rqrq rqrq rq rqrqrq




∼(𝑎) ⊆∼(𝑏). Similar reasoning implies that ∼(𝑏) ⊆∼(𝑎). Therefore ∼(𝑎) = ∼(𝑏).
rq q
r rq rqrq rq rqrq r q rq rq rq r q rq rq




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




1.18 The set of proper coalitions is not a partition of the set of players, since any playerc
rq rq rq rq rq rq rq rq rq rq rq rq rq rq rq rq q
r




an belong to more than one coalition. For example, player 1 belongs to the coalitions
rq rq rq rq rq rq rq rq rq rq rq rq rq rq




{1}, {1, 2}and so on. r q rq q
r r q r q




1.19

𝑥 ≻𝑦 =⇒ 𝑥 ≿ 𝑦 and 𝑦 ∕≿ 𝑥
rq rq r q r q rq rq r q r q r q rq




𝑦 ∼𝑧 =⇒ 𝑦 ≿ 𝑧 and 𝑧 ≿ 𝑦
r q rq r q r q r q rq r q r q r q rq




Transitivity of ≿ implies 𝑥 ≿ 𝑧 . We need to show that 𝑧 ∕≿ 𝑥 . Assume otherwise, thatis ass
rq rq rq rq rq rq rq rq rq rq rq rq rq rq rq rq rq q
r r q




ume 𝑧 ≿ 𝑥 This implies 𝑧 ∼𝑥 and by transitivity 𝑦 ∼𝑥. But this implies that
r q r q rq r q r q r q r q q
r r q r q r q r q r q rq r q r q r q r q




𝑦 ≿ 𝑥 which contradicts the assumption that 𝑥 ≻𝑦 . Therefore we conclude that 𝑧 ∕≿ 𝑥
r q rq r q r q r q r q r q r q r q rq rq r q r q r q r q r q rq




and therefore 𝑥 ≻𝑧 . The other result is proved in similar fashion.
r q r q rq q
r rq r q r q r q r q r q r q r q




1.20 asymmetric Assume 𝑥 ≻𝑦. r q r q r q q
r




𝑥 ≻𝑦 =⇒ 𝑦 ∕≿ 𝑥
rq rq r q r q r q rq




while
𝑦 ≻𝑥 =⇒ 𝑦 ≿ 𝑥
r q rq r q r q r q rq




Therefore
𝑥 ≻𝑦 =⇒ 𝑦 ∕≻𝑥
r q rq r q r q r q rq




3