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Solutions Manual Foundations of Mathematical Economics Michael Carter November 15, 2002 ⃝c 2001 Michael Carter Solutions for Foundations of Mathematical Economics All rights reserved 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 𝐴 ={Rock,Scissors,Paper} 𝑖=Jenny,Chris 𝑖 1.6 The setofplayersis 𝑁 ={1,2,...,𝑛}. The strategyspace ofeachplayeris the set of feasible outputs 𝐴 ={𝑞 ∈ℜ :𝑞 ≤𝑄 } 𝑖 𝑖 + 𝑖 𝑖 where 𝑞 is the output of dam 𝑖. 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 ∈ 𝑇𝑐. Consequently, 𝑥 ∈ 𝑆𝑐 ∩𝑇𝑐. Conversely, assume 𝑥 ∈ 𝑆𝑐∩𝑇𝑐. This implies that 𝑥 ∈ 𝑆𝑐 and 𝑥 ∈ 𝑇𝑐. 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 =𝑁 𝑆∈𝒞 ∩ 𝑆 =∅ 𝑆∈𝒞 1 ⃝c 2001 Michael Carter Solutions for Foundations of Mathematical Economics All rights reserved 𝑥 2 1 𝑥 1 -1 0 1 -1 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} + where0=(0,0,...,0)istheproductionplanusingnoinputsandproducingnooutputs. To see this, first note that 0 is a feasible production plan. Therefore, 0 ∈ 𝑌. Also, 0∈ℜ𝑛 and therefore 0∈𝑌 ∩ℜ𝑛. + + Toshowthatthereisnootherfeasibleproductionplaninℜ𝑛, weassumethecontrary. + 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 “<” is {1,2}=𝑋 and the range is {2,3}⫋𝑌. 1.14 Figure 1.1. 1.15 Therelation“isstrictlyhigherthan”istransitive,antisymmetricandasymmetric. It is not complete, reflexive or symmetric. 2 ⃝c 2001 Michael Carter Solutions for Foundations of Mathematical Economics 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∼beanequivalencerelationofaset𝑋 ∕=∅. Thatis,therelation∼isreflexive, symmetricandtransitive. We firstshowthatevery𝑥∈𝑋 belongsto someequivalence 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) 𝑎∈𝑋 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 Thesetofpropercoalitionsisnotapartitionofthesetofplayers,sinceanyplayer can belong to more than one coalition. For example, player1 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 whichcontradicts the assumption that 𝑥≻𝑦. Thereforewe 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 ⃝c 2001 Michael Carter Solutions for Foundations of Mathematical Economics All rights reserved 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 = 1𝑚 and 𝑘 ∈/ 𝑁. For example, 4 is a 𝑘 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 ⃝c 2001 Michael Carter Solutions for Foundations of Mathematical Economics All rights reserved 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 orderedby ≿. That is, 𝑥≿𝑦 for all 𝑦 ∈𝑋. Thisimpliesthatthereisno𝑦 ∈𝑋 whichstrictlydominates𝑥. Therefore,𝑥ismaximal 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 Assumethattheelementsaredenoted𝑥 ,𝑥 ,...,𝑥 . Wecanidentifythemaximal 1 2 𝑛 element by constructing another list using the following recursive algorithm 𝑎 =𝑥 1 1 { 𝑥 if 𝑥 ≻𝑎 𝑎 = 𝑖 𝑖 𝑖−1 𝑖 𝑎 otherwise 𝑖−1 By construction, there is no 𝑥 which strictly succedes 𝑎 . 𝑎 is a maximal element. 𝑖 𝑛 𝑛 1.29 𝑥∗ is maximal ⇐⇒ there does not exist 𝑥≻𝑥∗ that is ≻(𝑥∗)={𝑥:𝑥≻𝑥∗}=∅ 𝑥∗ is best ⇐⇒ 𝑥∗ ≿𝑥 for every 𝑥∈𝑋 ⇐⇒ 𝑥≾𝑥∗ for every 𝑥∈𝑋 That is, every 𝑥∈𝑋 belongs to ≾(𝑥∗) or ≾(𝑥∗)=𝑋. 1.30 Let 𝐴 be a nonempty set of a set 𝑋 ordered by ≿. 𝑥 ∈ 𝑋 is a lower bound for 𝐴 if it precedes every element in 𝐴, that is 𝑥 ≾ 𝑎 for all 𝑎 ∈ 𝐴. It is a greatest lower bound if it dominates every lower bound, that is 𝑥≿𝑦 for every lower bound 𝑦 of 𝐴. 1.31 Any multiple of 60 is an upper bound for 𝐴. Thus, the set of upper bounds of 𝐴 is {60,120,240,...}. The least upper bound of 𝐴 is 60. The only lower bound is 1, hence it is the greatest lower bound. 1.32 The least upper bounds of interval [𝑎,𝑏] are 𝑏 and 𝑧. The least upper bound of (𝑎,𝑏) is 𝑦. 1.33 𝑥 is an upper bound of 𝐴 ⇐⇒ 𝑥≿𝑎 for every 𝑎∈𝐴 ⇐⇒ 𝑎≾𝑥 for every 𝑎∈𝐴 ⇐⇒ 𝐴⊆≾(𝑥) Similarly 𝑥 is a lower bound of 𝐴 ⇐⇒ 𝑥≾𝑎 for every 𝑎∈𝐴 ⇐⇒ 𝑎≿𝑥 for every 𝑎∈𝐴 ⇐⇒ 𝐴⊆≿(𝑥) 5 ⃝c 2001 Michael Carter Solutions for Foundations of Mathematical Economics All rights reserved 1.34 For every 𝑥∈ℜ2, 𝑥≻𝑦 if 𝑥 >𝑦 or 𝑥 =𝑦 and 𝑥 >𝑦 1 1 1 1 2 2 Since all elements 𝑥∈ℜ2 are comparable, ≻ is complete; it is a total order. 1.35 Assume ≿ is complete for every 𝑖. Then for every 𝑥,𝑦 ∈ 𝑋 and for all 𝑖 = 𝑖 1,2,...,𝑛, either 𝑥 ≿ 𝑦 or 𝑦 ≿ 𝑥 or both. Either 𝑖 𝑖 𝑖 𝑖 𝑖 𝑖 𝑥 ∼ 𝑦 for all 𝑖 Then define 𝑥∼𝑦. 𝑖 𝑖 𝑖 𝑥 ∕∼ 𝑦 for some 𝑖 Let 𝑘 be the first individual with a strict preference, that is 𝑘 = 𝑖 𝑖 𝑖 min (𝑥 ∕∼𝑦 ). (Completeness of ≿ ensures that 𝑘 is defined). Then define 𝑖 𝑖 𝑖 𝑖 𝑥≻𝑦 if 𝑥 ≻ 𝑦 𝑘 𝑖 𝑘 𝑦 ≻𝑥 otherwise 1.36 Let 𝑆, 𝑇 and 𝑈 be subsets of a finite set 𝑋. Set inclusion ⊆ is reflexive since 𝑆 ⊆𝑆. transitive since 𝑆 ⊆𝑇 and 𝑇 ⊆𝑈 implies 𝑆 ⊆𝑈. anti-symmetric since 𝑆 ⊆𝑇 and 𝑇 ⊆𝑆 implies 𝑆 =𝑇 Therefore ⊆ is a partial order. 1.37 Assume 𝑥 and 𝑦 are both least upper bounds of 𝐴. That is 𝑥 ≿ 𝑎 for all 𝑎 ∈ 𝐴 and 𝑦 ≿ 𝑎 for all 𝑎 ∈ 𝐴. Further, if 𝑥 is a least upper bound, 𝑦 ≿ 𝑥. If 𝑦 is a least upper bound, 𝑥≿𝑦. By anti-symmetry, 𝑥=𝑦. 1.38 𝑥∼𝑦 =⇒ 𝑥≿𝑦 and 𝑦 ≿𝑥 which implies that 𝑥=𝑦 by antisymmetry. Each equivalence class ∼(𝑥)={𝑦 ∈𝑋 :𝑦 ∼𝑥} comprises just a single element 𝑥. 1.39 max𝒫(𝑋)=𝑋 and min𝒫(𝑋)=∅. 1.40 The subset {2,4,8} forms a chain. More generally, the set of integer powers of a given number {𝑛,𝑛2,𝑛3,...} forms a chain. 1.41 Assume 𝑥 and 𝑦 are maximal elements of the chain 𝐴. Then 𝑥 ≿ 𝑎 for all 𝑎 ∈ 𝐴 and in particular 𝑥 ≿ 𝑦. Similarly, 𝑦 ≿ 𝑎 for all 𝑎 ∈ 𝐴 and in particular 𝑦 ≿ 𝑥. Since ≿ is anti-symmetric, 𝑥=𝑦. 1.42 1. By assumption, forevery𝑡∈𝑇∖𝑊,≺(𝑡)is anonempty finite chain. Hence, it has a unique maximal element, 𝑝(𝑡). 2. Let 𝑡 be any node. Either 𝑡 is an initial node or 𝑡 has a unique predecessor 𝑝(𝑡). Either 𝑝(𝑡) is an initial node, or it has a unique predecessor 𝑝(𝑝(𝑡)). Continuing in this way, we trace out a unique path from 𝑡 back to an initial node. We can be sure of eventually reaching an initial node since 𝑇 is finite. 1.43 (1,2)∨(3,1)=(3,2) and (1,2)∧(3,2)=(1,2) 6 ⃝c 2001 Michael Carter Solutions for Foundations of Mathematical Economics All rights reserved 1.44 1. 𝑥∨𝑦 isanupperboundfor{𝑥,𝑦},thatisx∨y ≿𝑥andx∨y≿𝑦. Similarly, 𝑥∨𝑦 is a lower bound for {𝑥,𝑦}. 2. Assume 𝑥 ≿ 𝑦. Then 𝑥 is an upper bound for {𝑥,𝑦}, that is 𝑥 ≿ 𝑥∨𝑦. If 𝑏 is any upper bound for {𝑥,𝑦}, then 𝑏 ≿ 𝑥. Therefore, 𝑥 is the least upper bound for {𝑥,𝑦}. Similarly, 𝑦 is a lower bound for {𝑥,𝑦}, and is greater than any other lower bound. Conversely,assume 𝑥∨𝑦 =𝑥. Then 𝑥 is an upper bound for {𝑥,𝑦}, that is 𝑥≿𝑦. 3. Using the preceding equivalence 𝑥≿𝑥∧𝑦 =⇒ 𝑥∨(𝑥∧𝑦)=𝑥 𝑥∨𝑦 ≿𝑥 =⇒ (𝑥∨𝑦)∧𝑥=𝑥 1.45 A chain 𝑋 is a complete partially ordered set. For every 𝑥,𝑦 ∈ 𝑋 with 𝑥 ∕= 𝑦, either 𝑥≻𝑦 or 𝑦 ≻𝑥. Therefore, define the meet and join by { 𝑦 if 𝑥≻𝑦 𝑥∧𝑦 = 𝑥 if 𝑦 ≻𝑥 { 𝑥 if 𝑥≻𝑦 𝑥∨𝑦 = 𝑦 if 𝑦 ≻𝑥 𝑋 is a lattice with these operations. 1.46 Assume𝑋 and𝑋 arelattices, andlet𝑋 =𝑋 ×𝑋 . Consideranytwoelements 1 2 1 2 x = (𝑥 ,𝑥 ) and y = (𝑦 ,𝑦 ) in 𝑋. Since 𝑋 and 𝑋 are lattices, 𝑏 = 𝑥 ∨𝑦 ∈ 𝑋 1 2 1 2 1 2 1 1 1 1 and 𝑏 = 𝑥 ∨𝑦 ∈ 𝑋 , so that b = (𝑏 ,𝑏 ) = (𝑥 ∨𝑦 ,𝑥 ∨𝑦 ) ∈ 𝑋. Furthermore 2 2 2 2 1 2 1 1 2 2 b ≿ x and b ≿ y in the natural product order, so that b is an upper bound for the {x,y}. Every upper bound bˆ = (ˆ𝑏 ,ˆ𝑏 ) of {x,y} must have 𝑏 ≿ 𝑥 and 𝑏 ≿ 𝑦 , 1 2 𝑖 𝑖 𝑖 𝑖 𝑖 𝑖 so that bˆ ≿ b. Therefore, b is the least upper bound of {x,y}, that is b = x∨y. Similarly, x∧y=(𝑥 ∧𝑦 ,𝑥 ∧𝑦 ). 1 1 2 2 1.47 Let 𝑆 be a subset of 𝑋 and let 𝑆∗ ={𝑥∈𝑋 :𝑥≿𝑠 for every 𝑠∈𝑆} be the set of upper bounds of𝑆. Then 𝑥∗ ∈𝑆∗ ∕=∅. By assumption, 𝑆∗ has a greatest lower bound 𝑏. Since every 𝑠 ∈ 𝑆 is a lower bound of 𝑆∗, 𝑏 ≿ 𝑠 for every 𝑠 ∈ 𝑆. Therefore 𝑏 is an upper bound of 𝑆. Furthermore, 𝑏 is the least upper bound of 𝑆, since 𝑏≾𝑥 for every 𝑥∈𝑆∗. This establishes that every subset of 𝑋 also has a least upper bound. In particular, every pair of elements has a least upper and a greatest lower bound. Consequently 𝑋 is a complete lattice. 1.48 Without loss of generality, we will prove the closed interval case. Let [𝑎,𝑏] be an interval in a lattice 𝐿. Recall that 𝑎 = inf[𝑎,𝑏] and 𝑏 = sup[𝑎,𝑏]. Choose any 𝑥,𝑦 in [𝑎,𝑏]⊆𝐿. Since 𝐿 is a lattice, 𝑥∨𝑦 ∈𝐿 and 𝑥∨𝑦 =sup{𝑥,𝑦}≾𝑏 Therefore 𝑥∨𝑦 ∈ [𝑎,𝑏]. Similarly, 𝑥∧𝑦 ∈ [𝑎,𝑏]. [𝑎,𝑏] is a lattice. Similarly, for any subset𝑆 ⊆[𝑎,𝑏]⊆𝐿,sup𝑆 ∈𝐿if𝐿iscomplete. Also,sup𝑆 ≾𝑏=sup[𝑎,𝑏]. Therefore sup𝑆 ∈[𝑎,𝑏]. Similarly inf𝑆 ∈[𝑎,𝑏] so that [𝑎,𝑏] is complete. 7 ⃝c 2001 Michael Carter Solutions for Foundations of Mathematical Economics All rights reserved 1.49 1. The strong set order ≿ is 𝑆 antisymmetric Let 𝑆 ,𝑆 ⊆ 𝑋 with 𝑆 ≿ 𝑆 and 𝑆 ≿ 𝑆 . Choose 𝑥 ∈ 𝑆 1 2 1 𝑆 2 2 𝑆 1 1 1 and 𝑥 ∈𝑆 . Since 𝑆 ≿ 𝑆 , 𝑥 ∨𝑥 ∈𝑆 and 𝑥 ∧𝑥 ∈𝑆 . On the other 2 2 1 𝑆 2 1 2 1 1 2 2 hand, since 𝑆 ≿𝑆 , 𝑥 =(𝑥 ∨(𝑥 ∧𝑥 )∈𝑆 and𝑥 =𝑥 ∧(𝑥 ∨𝑥 )∈𝑆 2 1 1 1 1 2 2 2 2 1 2 1 (Exercise 1.44. Therefore 𝑆 =𝑆 and ≿ is antisymmetric. 1 2 𝑆 transitive Let 𝑆 ,𝑆 ,𝑆 ⊆ 𝑋 with 𝑆 ≿ 𝑆 and 𝑆 ≿ 𝑆 . Choose 𝑥 ∈ 𝑆 , 1 2 3 1 𝑆 2 2 𝑆 3 1 1 𝑥 ∈ 𝑆 and 𝑥 ∈ 𝑆 . Since 𝑆 ≿ 𝑆 and 𝑆 ≿ 𝑆 , 𝑥 ∨𝑥 and 𝑥 ∧𝑥 2 2 3 3 1 𝑆 2 2 𝑆 3 1 2 2 3 are in 𝑆 . Therefore 𝑦 =𝑥 ∨(𝑥 ∧𝑥 )∈𝑆 which implies 2 2 1 2 3 2 ( ) 𝑥 ∨𝑥 =𝑥 ∨ (𝑥 ∧𝑥 )∨𝑥 1 3 1 2 3 3 ( ) = 𝑥 ∨(𝑥 ∧𝑥 ) ∨𝑥 1 2 3 3 =𝑦 ∨𝑥 ∈𝑆 2 3 3 since 𝑆 ≿ 𝑆 . Similarly 𝑧 =(𝑥 ∨𝑥 )∧𝑥 ∈𝑆 and 2 𝑆 3 2 1 2 3 2 ( ) 𝑥 ∧𝑥 = 𝑥 ∧(𝑥 ∨𝑥 ) ∧𝑥 1 3 1 1 2 3 ( ) =𝑥 ∧ (𝑥 ∨𝑥 )∧𝑥 1 1 2 3 =𝑥 ∧𝑧 ∈𝑆 1 2 1 Therefore, 𝑆 ≿ 𝑆 . 1 𝑆 3 2. 𝑆 ≿ 𝑆 if and only if, for every 𝑥 ,𝑥 ∈ 𝑆, 𝑥 ∨𝑥 ∈ 𝑆 and 𝑥 ∧𝑥 ∈ 𝑆, which 𝑆 1 2 1 2 1 2 is the case if and only if 𝑆 is a sublattice. 3. Let 𝐿(𝑋) denote the set of all sublattices of 𝑋. We have shown that ≿ is 𝑆 reflexive, transitive and antisymmetric on 𝐿(𝑋). Hence, it is a partial order on 𝐿(𝑋). 1.50 Assume 𝑆 ≿ 𝑆 . For any 𝑥 ∈𝑆 and 𝑥 ∈ 𝑆 , 𝑥 ∨𝑥 ∈ 𝑆 and 𝑥 ∧𝑥 ∈ 𝑆 . 1 𝑆 2 1 1 2 2 1 2 1 1 2 2 Therefore sup𝑆 ≿𝑥 ∨𝑥 ≿𝑥 for every 𝑥 ∈𝑆 1 1 2 2 2 2 which implies that sup𝑆 ≿sup𝑆 . Similarly 1 2 inf𝑆 ≾𝑥 ∧𝑥 ≾𝑥 for every 𝑥 ∈𝑆 2 1 2 1 1 1 which implies that inf𝑆 ≾ inf𝑆 . Note that completeness ensures the existence of 2 1 sup𝑆 and inf𝑆 respectively. 1.51 An argument analogous to the preceding exercise establishes =⇒ . (Complete- ness is not required, since for any interval 𝑎=inf[𝑎,𝑏] and 𝑏=sup[𝑎,𝑏]). To establish the converse, assume that 𝑆 = [𝑎 ,𝑏 ] and 𝑆 = [𝑎 ,𝑏 ]. Consider any 1 1 1 2 2 2 𝑥 ∈𝑆 and 𝑥 ∈𝑆 . There are two cases. 1 1 2 2 Case 1. 𝑥 ≿𝑥 Since 𝑋 is a chain, 𝑥 ∨𝑥 =𝑥 ∈𝑆 . 𝑥 ∧𝑥 =𝑥 ∈𝑆 . 1 2 1 2 1 1 1 2 2 2 Case 2. 𝑥 ≺𝑥 Since 𝑋 is a chain, 𝑥 ∨𝑥 = 𝑥 . Now 𝑎 ≾ 𝑥 ≺ 𝑥 ≾ 𝑏 ≾ 𝑏 . 1 2 1 2 2 1 1 2 2 2 Therefore, 𝑥 = 𝑥 ∨𝑥 ∈ 𝑆 . Similarly 𝑎 ≾ 𝑎 ≾ 𝑥 ≺ 𝑥 ≾ 𝑏 . Therefore 2 1 2 1 2 1 1 2 2 𝑥 ∧𝑥 =𝑥 ∈𝑆 . 1 2 1 2 We have shown that 𝑆 ≿ 𝑆 in both cases. 1 𝑆 2 1.52 Assume that ≿ is a complete relation on 𝑋. This means that for every 𝑥,𝑦 ∈𝑋, either 𝑥≿𝑦 or 𝑦 ≿𝑥. In particular, letting 𝑥=𝑦, 𝑥≿𝑥 for 𝑥∈𝑋. ≿ is reflexive. 8 ⃝c 2001 Michael Carter Solutions for Foundations of Mathematical Economics All rights reserved 1.53 Anti-symmetryimpliesthateachindifferenceclasscontainsasingleelement. Ifthe consumer’spreferencerelationwasanti-symmetric,therewouldbe nobasketsofgoods between which the consumer was indifferent. Each indifference curve which consist a single point. 1.54 We previously showed (Exercise 1.27) that every best element is maximal. To provethe converse,assumethat𝑥ismaximalinthe weaklyorderedset𝑋. Wehaveto show that 𝑥≿𝑦 for all 𝑦 ∈𝑋. Assume otherwise, that is assume there is some 𝑦 ∈𝑋 for which 𝑥 ∕≿ 𝑦. Since ≿ is complete, this implies that 𝑦 ≻ 𝑥 which contradicts the assumption that 𝑥 is maximal. Hence we conclude that 𝑥 ≿ 𝑦 for 𝑦 ∈ 𝑋 and 𝑥 is a best element. 1.55 False. Achainhasatmostonemaximalelement(Exercise1.41). Here,uniqueness is ensured by anti-symmetry. A weakly ordered set in which the order is not antisymmetric may have multiple maximal and best elements. For example, 𝑎 and 𝑏 are both best elements in the weakly ordered set {𝑎∼𝑏≻𝑐}. 1.56 1. For every 𝑥∈𝑋, either 𝑥≿ 𝑦 =⇒ 𝑥∈≿(𝑦) or 𝑦 ≿𝑥 =⇒ 𝑥∈≾(𝑦) since ≿ is complete. Consequently, ≿(𝑦)∪≺(𝑦) = 𝑋 If 𝑥 ∈ ≿(𝑦)∩≾(𝑦), then 𝑥 ≿ 𝑦 and 𝑦 ≿𝑥 so that 𝑥∼𝑦 and 𝑥∈𝐼 . 𝑦 2. For every 𝑥∈𝑋, either 𝑥≿𝑦 =⇒ 𝑥∈≿(𝑦) or 𝑦 ≻𝑥 =⇒ 𝑥∈≺(𝑦) since ≿ is complete. Consequently, ≿(𝑦)∪≺(𝑦)=𝑋 and ≿(𝑦)∩≺(𝑦)=∅. 3. For every 𝑦 ∈ 𝑋, ≻(𝑦) and 𝐼 partition ≿(𝑦) and therefore ≻(𝑦), 𝐼 and ≺(𝑦) 𝑦 𝑦 partition 𝑋. 1.57 Assume 𝑥≿𝑦 and𝑧 ∈≿(𝑥). Then 𝑧 ≿𝑥≿𝑦 by transitivity. Therefore𝑧 ∈≿(𝑦). This shows that ≿(𝑥)⊆≿(𝑦). Similarly, assume 𝑥 ≻ 𝑦 and 𝑧 ∈ ≻(𝑥). Then 𝑧 ≻ 𝑥 ≻ 𝑦 by transitivity. Therefore 𝑧 ∈ ≻(𝑦). This shows that ≿(𝑥) ⊆ ≿(𝑦). To show that ≿(𝑥) ∕= ≿(𝑦), observe that 𝑥∈≻(𝑦) but that 𝑥∈/ ≻(𝑥) 1.58 Every finite ordered set has a least one maximal element (Exercise 1.28). 1.59 Kreps(1990,p.323),Luenberger(1995,p.170)andMas-Colelletal. (1995,p.313) adopt the weak Pareto order, whereas Varian (1992, p.323) distinguishes the two orders. Osborneand Rubinstein (1994,p.7) also distinguish the two orders, utilizing the weak order in defining the core (Chapter 13) but the strong Pareto order in the Nash bargaining solution (Chapter 15). 1.60 Assumethatagroup𝑆 isdecisiveover𝑥,𝑦 ∈𝑋. Let𝑎,𝑏∈𝑋 betwootherstates. We have to show that 𝑆 is decisive over 𝑎 and 𝑏. Without loss of generality, assume for all individuals 𝑎 ≿ 𝑥 and 𝑦 ≿ 𝑏. Then, the Pareto order implies that 𝑎 ≻ 𝑥 and 𝑖 𝑖 𝑦 ≻𝑏. Assume that for every 𝑖 ∈ 𝑆, 𝑥 ≿ 𝑦. Since 𝑆 is decisive over 𝑥 and 𝑦, the social 𝑖 order ranks 𝑥≿𝑦. By transitivity, 𝑎≿𝑏. By IIA, this holds irrespective of individual preferences on other alternatives. Hence, 𝑆 is decisive over 𝑎 and 𝑏. 1.61 Assume that 𝑆 is decisive. Let 𝑥, 𝑦 and 𝑧 be any three alternatives and assume 𝑥≿𝑦 for every 𝑖∈𝑆. Partition 𝑆 into two subgroups 𝑆 and 𝑆 so that 1 2 𝑥≿ 𝑧 for every 𝑖∈𝑆 and 𝑧 ≿ 𝑦 for every 𝑖∈𝑆 𝑖 1 𝑖 2 Since 𝑆 is decisive, 𝑥≿𝑦. By completeness, either 𝑥≿𝑧inwhichcase𝑆 isdecisiveover𝑥and𝑧. Bythefieldexpansionlemma(Exercise 1 1.60), 𝑆 is decisive. 1 9