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• Solutions Manual

Foundations of Mathematical Economics

Michael Carter

November 15, 2002

• Solutions for Foundations of Mathematical Economics c⃝ 2001 Michael Carter

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 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 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 ∈ 𝑆𝑐 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∈𝒞 𝑆 = 𝑁

∩ 𝑆∈𝒞 𝑆 = ∅

1

• Solutions for Foundations of Mathematical Economics c⃝ 2001 Michael Carter

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 }. 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 ∩ ℜ𝑛+ = {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+. To show that there is no other feasible production plan in ℜ𝑛+, we assume the contrary. That is, we assume there is some feasible production plan y ∈ ℜ𝑛+ ∖ {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) ∈ 𝑌 for every 𝑦′ ≤ 𝑦, which implies that x ∈ 𝑉 (𝑦′). 𝑉 (𝑦′) ⊇ 𝑉 (𝑦). 1.13 The domain of “

• Solutions for Foundations of Mathematical Economics c⃝ 2001 Michael Carter

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 ∈ 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 reflexive, 𝑎 ∼ 𝑎 and so 𝑎 ∈ ∼(𝑎). Every 𝑎 ∈ 𝑋 belongs 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 ∼ 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 example, 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, that is 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

3

• Solutions for Foundations of Mathematical Economics c⃝ 2001 Michael Carter

transitive Assume 𝑥 ≻ 𝑦 and 𝑦 ≻ 𝑧. 𝑥 ≻ 𝑦 =⇒ 𝑥 ≿ 𝑦 and 𝑦 ∕≿ 𝑥 𝑦 ≻ 𝑧 =⇒ 𝑦 ≿ 𝑧 and 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 ≿ 𝑦 and transitivity implies that 𝑧 ≿ 𝑦, contrary to the assumption that 𝑦 ≻ 𝑧. We conclude that 𝑧 ∕≿ 𝑥 and

𝑥 ≿ 𝑧 and 𝑧 ∕≿ 𝑥 =⇒ 𝑥 ≻ 𝑧 This shows that ≻ is transitive.

1.21 reflexive Since ≿ is reflexive, 𝑥 ≿ 𝑥 which implies 𝑥 ∼ 𝑥. transitive Assume 𝑥 ∼ 𝑦 and 𝑦 ∼ 𝑧. Now

𝑥 ∼ 𝑦 ⇐⇒ 𝑥 ≿ 𝑦 and 𝑦 ≿ 𝑥

𝑦 ∼ 𝑧 ⇐⇒ 𝑦 ≿ 𝑧 and 𝑧 ≿ 𝑦 Transitivity of ≿ implies

𝑥 ≿ 𝑦 and 𝑦 ≿ 𝑧 =⇒ 𝑥 ≿ 𝑧

𝑧 ≿ 𝑦 and 𝑦 ≿ 𝑥 =⇒ 𝑧 ≿ 𝑥 Combining

𝑥 ≿ 𝑧 and 𝑧 ≿ 𝑥 =⇒ 𝑥 ∼ 𝑧

symmetric

𝑥 ∼ 𝑦 ⇐⇒ 𝑥 ≿ 𝑦 and 𝑦 ≿ 𝑥 ⇐⇒ 𝑦 ≿ 𝑥 and 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 𝑚 is a multiple of 𝑝.

not symmetric If 𝑚 = 𝑘𝑛, 𝑘 ∈ 𝑁 , then 𝑛 = 1𝑘𝑚 and 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 }

4

• Solutions for Foundations of Mathematical Economics c⃝ 2001 Michael Carter

1.25 Let 𝑋 be ordered by ≿. 𝑥 ∈ 𝑋 is a minimal element there is no element which strictly precedes it, that is there is no element 𝑦 ∈ 𝑋 such that 𝑦 ≺ 𝑥. 𝑥 ∈ 𝑋 is the first element if it precedes every other element, that is 𝑥 ≾ 𝑦 for all 𝑦 ∈ 𝑋 . 1.26 The maximal elements of 𝑋 are 𝑏 and 𝑧. The minimal element of 𝑋 is 𝑥. These are also best and worst elements respectively.

1.27 Assume that 𝑥 is a best element in 𝑋 ordered by ≿. That is, 𝑥 ≿ 𝑦 for all 𝑦 ∈ 𝑋 . This implies that there is no 𝑦 ∈ 𝑋 which strictly dominates 𝑥. Therefore, 𝑥 is maximal in 𝑋 . In Example 1.23, the numbers 5, 6, 7, 8, 9 are all maximal elements, but none of them is a best element.

1.28 Assume that the elements are denoted 𝑥1, 𝑥2, . . . , 𝑥𝑛. We can identify the maximal element by constructing another list using the following recursive algorithm

𝑎1 = 𝑥1

𝑎𝑖 =

{ 𝑥𝑖 if 𝑥𝑖 ≻ 𝑎𝑖−1 𝑎𝑖−1 otherwise

By construction, there is no 𝑥𝑖 which strictly succedes 𝑎𝑛. 𝑎𝑛 is a maximal element.

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