Solutions Manual Foundations of Mathematical Economics 2009-11-17¢  Solutions for Foundations of Mathematical

  • View
    15

  • Download
    2

Embed Size (px)

Text of Solutions Manual Foundations of Mathematical Economics 2009-11-17¢  Solutions for...

  • Solutions Manual

    Foundations of Mathematical Economics

    Michael Carter

    November 15, 2002

  • Solutions for Foundations of Mathematical Economics c⃝ 2001 Michael Carter

    All rights reserved

    Chapter 1: Sets and Spaces

    1.1

    { 1, 3, 5, 7 . . .} or {𝑛 ∈ 𝑁 : 𝑛 is odd } 1.2 Every π‘₯ ∈ 𝐴 also belongs to 𝐡. Every π‘₯ ∈ 𝐡 also belongs to 𝐴. Hence 𝐴,𝐡 have precisely the same elements.

    1.3 Examples of finite sets are

    βˆ™ the letters of the alphabet {A, B, C, . . . , Z } βˆ™ the set of consumers in an economy βˆ™ the set of goods in an economy βˆ™ the set of players in a game.

    Examples of infinite sets are

    βˆ™ the real numbers β„œ βˆ™ the natural numbers 𝔑 βˆ™ the set of all possible colors βˆ™ the set of possible prices of copper on the world market βˆ™ 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 = { π‘žπ‘– ∈ β„œ+ : π‘žπ‘– ≀ 𝑄𝑖 } where π‘žπ‘– is the output of dam 𝑖.

    1.7 The player set is 𝑁 = {1, 2, 3}. There are 23 = 8 coalitions, namely 𝒫(𝑁) = {βˆ…, {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 π‘₯ ∈ (𝑆 βˆͺ 𝑇 )𝑐. That is π‘₯ /∈ 𝑆 βˆͺ 𝑇 . This implies π‘₯ /∈ 𝑆 and π‘₯ /∈ 𝑇 , or π‘₯ ∈ 𝑆𝑐 and π‘₯ ∈ 𝑇 𝑐. Consequently, π‘₯ ∈ 𝑆𝑐 ∩ 𝑇 𝑐. Conversely, assume π‘₯ ∈ 𝑆𝑐 ∩ 𝑇 𝑐. This implies that π‘₯ ∈ 𝑆𝑐 and π‘₯ ∈ 𝑇 𝑐. Consequently π‘₯ /∈ 𝑆 and π‘₯ /∈ 𝑇 and therefore π‘₯ /∈ 𝑆 βˆͺ 𝑇 . This implies that π‘₯ ∈ (𝑆 βˆͺ 𝑇 )𝑐. The other identity is proved similarly. 1.9 βˆͺ

    π‘†βˆˆπ’ž 𝑆 = 𝑁

    ∩ π‘†βˆˆπ’ž 𝑆 = βˆ…

    1

  • Solutions for Foundations of Mathematical Economics c⃝ 2001 Michael Carter

    All rights reserved

    0-1 1 π‘₯1

    -1

    1 π‘₯2

    Figure 1.1: The relation { (π‘₯, 𝑦) : π‘₯2 + 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 } Γ— {𝐻,𝑇 } Γ— {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

    π‘Œ ∩ β„œπ‘›+ = {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 ∈ π‘Œ . Also, 0 ∈ β„œπ‘›+ and therefore 0 ∈ π‘Œ ∩ β„œπ‘›+. To show that there is no other feasible production plan in β„œπ‘›+, we assume the contrary. That is, we assume there is some feasible production plan y ∈ β„œπ‘›+ βˆ– {0}. This implies the existence of a plan producing a positive output with no inputs. This technological infeasible, so that 𝑦 /∈ π‘Œ . 1.12 1. Let x ∈ 𝑉 (𝑦). This implies that (𝑦,βˆ’x) ∈ π‘Œ . Let xβ€² β‰₯ x. Then (𝑦,βˆ’xβ€²) ≀

    (𝑦,βˆ’x) and free disposability implies that (𝑦,βˆ’xβ€²) ∈ π‘Œ . Therefore xβ€² ∈ 𝑉 (𝑦). 2. Again assume x ∈ 𝑉 (𝑦). This implies that (𝑦,βˆ’x) ∈ π‘Œ . By free disposal,

    (𝑦′,βˆ’x) ∈ π‘Œ for every 𝑦′ ≀ 𝑦, which implies that x ∈ 𝑉 (𝑦′). 𝑉 (𝑦′) βŠ‡ 𝑉 (𝑦). 1.13 The domain of β€œ

  • Solutions for Foundations of Mathematical Economics c⃝ 2001 Michael Carter

    All rights reserved

    1.16 The following table lists their respective properties.

    < ≀ = reflexive Γ— √ √ transitive

    √ √ √ symmetric Γ— √ √ asymmetric

    √ Γ— Γ— anti-symmetric

    √ √ √ complete

    √ √ Γ— Note that the properties of symmetry and anti-symmetry are not mutually exclusive.

    1.17 Let ∼ be an equivalence relation of a set 𝑋 βˆ•= βˆ…. That is, the relation ∼ is reflexive, symmetric and transitive. We first show that every π‘₯ ∈ 𝑋 belongs to some equivalence class. Let π‘Ž be any element in 𝑋 and let ∼ (π‘Ž) be the class of elements equivalent to π‘Ž, that is

    ∼(π‘Ž) ≑ { π‘₯ ∈ 𝑋 : π‘₯ ∼ π‘Ž } Since ∼ is reflexive, π‘Ž ∼ π‘Ž and so π‘Ž ∈ ∼(π‘Ž). Every π‘Ž ∈ 𝑋 belongs to some equivalence class and therefore

    𝑋 = βˆͺ π‘Žβˆˆπ‘‹ ∼(π‘Ž)

    Next, we show that the equivalence classes are either disjoint or identical, that is ∼(π‘Ž) βˆ•= ∼(𝑏) if and only if f∼(π‘Ž) ∩ ∼(𝑏) = βˆ…. First, assume ∼(π‘Ž) ∩ ∼(𝑏) = βˆ…. Then π‘Ž ∈ ∼(π‘Ž) but π‘Ž /∈ ∼(𝑏). Therefore ∼(π‘Ž) βˆ•= ∼(𝑏). Conversely, assume ∼(π‘Ž) ∩ ∼(𝑏) βˆ•= βˆ… and let π‘₯ ∈ ∼(π‘Ž) ∩ ∼(𝑏). Then π‘₯ ∼ π‘Ž and by symmetry π‘Ž ∼ π‘₯. Also π‘₯ ∼ 𝑏 and so by transitivity π‘Ž ∼ 𝑏. Let 𝑦 be any element in ∼(π‘Ž) so that 𝑦 ∼ π‘Ž. Again by transitivity 𝑦 ∼ 𝑏 and therefore 𝑦 ∈ ∼(𝑏). Hence ∼(π‘Ž) βŠ† ∼(𝑏). Similar reasoning implies that ∼(𝑏) βŠ† ∼(π‘Ž). Therefore ∼(π‘Ž) = ∼(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

    π‘₯ ≻ 𝑦 =β‡’ π‘₯ β‰Ώ 𝑦 and 𝑦 βˆ•β‰Ώ π‘₯ 𝑦 ∼ 𝑧 =β‡’ 𝑦 β‰Ώ 𝑧 and 𝑧 β‰Ώ 𝑦

    Transitivity of β‰Ώ implies π‘₯ β‰Ώ 𝑧. We need to show that 𝑧 βˆ•β‰Ώ π‘₯. Assume otherwise, that is assume 𝑧 β‰Ώ π‘₯ This implies 𝑧 ∼ π‘₯ and by transitivity 𝑦 ∼ π‘₯. But this implies that 𝑦 β‰Ώ π‘₯ which contradicts the assumption that π‘₯ ≻ 𝑦. Therefore we conclude that 𝑧 βˆ•β‰Ώ π‘₯ and therefore π‘₯ ≻ 𝑧. The other result is proved in similar fashion. 1.20 asymmetric Assume π‘₯ ≻ 𝑦.

    π‘₯ ≻ 𝑦 =β‡’ 𝑦 βˆ•β‰Ώ π‘₯ while

    𝑦 ≻ π‘₯ =β‡’ 𝑦 β‰Ώ π‘₯ Therefore

    π‘₯ ≻ 𝑦 =β‡’ 𝑦 βˆ•β‰» π‘₯

    3

  • Solutions for Foundations of Mathematical Economics c⃝ 2001 Michael Carter

    All rights reserved

    transitive Assume π‘₯ ≻ 𝑦 and 𝑦 ≻ 𝑧. π‘₯ ≻ 𝑦 =β‡’ π‘₯ β‰Ώ 𝑦 and 𝑦 βˆ•β‰Ώ π‘₯ 𝑦 ≻ 𝑧 =β‡’ 𝑦 β‰Ώ 𝑧 and 𝑧 βˆ•β‰Ώ 𝑦

    Since β‰Ώ is transitive, we conclude that π‘₯ β‰Ώ 𝑧. It remains to show that 𝑧 βˆ•β‰Ώ π‘₯. Assume otherwise, that is assume 𝑧 β‰Ώ π‘₯. We know that π‘₯ β‰Ώ 𝑦 and transitivity implies that 𝑧 β‰Ώ 𝑦, contrary to the assumption that 𝑦 ≻ 𝑧. We conclude that 𝑧 βˆ•β‰Ώ π‘₯ and

    π‘₯ β‰Ώ 𝑧 and 𝑧 βˆ•β‰Ώ π‘₯ =β‡’ π‘₯ ≻ 𝑧 This shows that ≻ is transitive.

    1.21 reflexive Since β‰Ώ is reflexive, π‘₯ β‰Ώ π‘₯ which implies π‘₯ ∼ π‘₯. transitive Assume π‘₯ ∼ 𝑦 and 𝑦 ∼ 𝑧. Now

    π‘₯ ∼ 𝑦 ⇐⇒ π‘₯ β‰Ώ 𝑦 and 𝑦 β‰Ώ π‘₯

    𝑦 ∼ 𝑧 ⇐⇒ 𝑦 β‰Ώ 𝑧 and 𝑧 β‰Ώ 𝑦 Transitivity of β‰Ώ implies

    π‘₯ β‰Ώ 𝑦 and 𝑦 β‰Ώ 𝑧 =β‡’ π‘₯ β‰Ώ 𝑧

    𝑧 β‰Ώ 𝑦 and 𝑦 β‰Ώ π‘₯ =β‡’ 𝑧 β‰Ώ π‘₯ Combining

    π‘₯ β‰Ώ 𝑧 and 𝑧 β‰Ώ π‘₯ =β‡’ π‘₯ ∼ 𝑧

    symmetric

    π‘₯ ∼ 𝑦 ⇐⇒ π‘₯ β‰Ώ 𝑦 and 𝑦 β‰Ώ π‘₯ ⇐⇒ 𝑦 β‰Ώ π‘₯ and π‘₯ β‰Ώ 𝑦 ⇐⇒ 𝑦 ∼ π‘₯

    1.22 reflexive Every integer is a multiple of itself, that is π‘š = 1π‘š.

    transitive Assume π‘š = π‘˜π‘› and 𝑛 = 𝑙𝑝 where π‘˜, 𝑙 ∈ 𝑁 . Then π‘š = π‘˜π‘™π‘ so that π‘š is a multiple of 𝑝.

    not symmetric If π‘š = π‘˜π‘›, π‘˜ ∈ 𝑁 , then 𝑛 = 1π‘˜π‘š and π‘˜ /∈ 𝑁 . For example, 4 is a multiple of 2 but 2 is not a multiple of 4.

    1.23

    [π‘Ž, 𝑏] = { π‘Ž, 𝑦, 𝑏, 𝑧 } (π‘Ž, 𝑏) = { 𝑦 }

    1.24

    β‰Ώ (𝑦) = {𝑏, 𝑦, 𝑧 } ≻ (𝑦) = {𝑏, 𝑧 } β‰Ύ (𝑦) = {π‘Ž, π‘₯, 𝑦 } β‰Ί (𝑦) = {π‘Ž, π‘₯ }

    4

  • Solutions for Foundations of Mathematical Economics c⃝ 2001 Michael Carter

    All rights reserved

    1.25 Let 𝑋 be ordered by β‰Ώ. π‘₯ ∈ 𝑋 is a minimal element there is no element which strictly precedes it, that is there is no element 𝑦 ∈ 𝑋 such that 𝑦 β‰Ί π‘₯. π‘₯ ∈ 𝑋 is the first element if it precedes every other element, that is π‘₯ β‰Ύ 𝑦 for all 𝑦 ∈ 𝑋 . 1.26 The maximal elements of 𝑋 are 𝑏 and 𝑧. The minimal element of 𝑋 is π‘₯. These are also best and worst elements respectively.

    1.27 Assume that π‘₯ is a best element in 𝑋 ordered by β‰Ώ. That is, π‘₯ β‰Ώ 𝑦 for all 𝑦 ∈ 𝑋 . This implies that there is no 𝑦 ∈ 𝑋 which strictly dominates π‘₯. Therefore, π‘₯ 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 π‘₯1, π‘₯2, . . . , π‘₯𝑛. We can identify the maximal element by constructing another list using the following recursive algorithm

    π‘Ž1 = π‘₯1

    π‘Žπ‘– =

    { π‘₯𝑖 if π‘₯𝑖 ≻ π‘Žπ‘–βˆ’1 π‘Žπ‘–βˆ’1 otherwise

    By construction, there is no π‘₯𝑖 which strictly succedes π‘Žπ‘›. π‘Žπ‘› is a maximal element.