Foundation of Mathematical Economics Solutions

5/20/2018 Foundation of Mathematical Economics Solutions

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Solutions ManualFoundations of Mathematical Economics

Michael Carter

November 15, 2002


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Solutions for Foundations of Mathematical Economicsc 2001 Michael Carter

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Chapter 1: Sets and Spaces

1.1

1, 3, 5, 7 . . . or : is odd 1.2 Every also belongs to . Every also belongs to . Hence, have

precisely the same elements.

1.3 Examples of nite 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 innite sets are

the real numbers the natural numbers 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 = 1, 2, 3, 4, 5, 6 ,= 2, 4, 6 .1.5 The player set is = Jenny, Chris . Their action spaces are

= Rock, Scissors, Paper = Jenny, Chris1.6 The set of players is =1, 2, . . . , . The strategy space of each player is the set

of feasible outputs

= + : where is the output of dam .

1.7 The player set is = 1, 2, 3. There are 23 = 8 coalitions, namely() = , 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 ( ). That is / . This implies / and / ,or

and

. Consequently,

. Conversely, assume

.

This implies that and . Consequently / and / and therefore / . This implies that ( ). The other identity is proved similarly.

1.9

=

=

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0-1 11

-1

12

Figure 1.1: The relation

(, ) : 2 + 2 = 1

1.10 The sample space of a single coin toss is , . The set of possible outcomes in

three tosses is the product

, , , =(,,), (,,), ( , , ),( , , ), ( , , ), ( , , ), ( , , ), ( , , )

A typical outcome is the sequence (,,) of two heads followed by a tail.

1.11

+=

0

where 0= (0, 0, . . . , 0) is the production plan using no inputs and producing no outputs.To see this, rst note that 0 is a feasible production plan. Therefore,0 . Also,0 + and therefore 0 +.To show that there is no other feasible production plan in+, we assume the contrary.That is, we assume there is some feasible production plan y + 0. This impliesthe existence of a plan producing a positive output with no inputs. This technologicalinfeasible, so that /.

1.12 1. Letx(). This implies that (, x). Let x x. Then (, x)(, x) and free disposability implies that (, x) . Therefore x ().

2. Again assume x

(). This implies that (,

x)

. By free disposal,

(, x) for every , which implies thatx (). () ().1.13 The domain of


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1.16 The following table lists their respective properties.

< =reexive 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= . That is, the relation is reexive,symmetric and transitive. We rst show that every belongs to some equivalenceclass. Let be any element in and let() be the class of elements equivalent to, that is

() : Since is reexive, and so (). Everybelongs to some equivalenceclass and therefore

=

()

Next, we show that the equivalence classes are either disjoint or identical, that is() =() if and only if f() () =.First, assume() () = . Then () but / (). Therefore() =().Conversely, assume() ()= and let () (). Then and bysymmetry . Also and so by transitivity . Let be any elementin() so that . Again by transitivity and therefore (). Hence() (). Similar reasoning implies that() (). Therefore() =().We conclude that the equivalence classes partition .

1.18 The set of proper coalitions is not a partition of the set of players, since any playercan belong to more than one coalition. For example, player 1 belongs to the coalitions1,1, 2and so on.

1.19

= and = and

Transitivity of implies . We need to show that . Assume otherwise, thatis assume This implies and by transitivity . But this implies that which contradicts the assumption that . Therefore we conclude that and therefore

. The other result is proved in similar fashion.

1.20 asymmetric Assume . =

while

= Therefore

=

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transitive Assume and . = and = and

Since

is transitive, we conclude that

.It remains to show that . Assume otherwise, that is assume . Weknow that and transitivity implies that , contrary to the assumptionthat . We conclude that and

and = This shows that is transitive.

1.21 reexive Since is reexive, which implies .transitive Assume and . Now

and

and Transitivity of implies

and =

and = Combining

and =

symmetric

and and

1.22 reexive Every integer is a multiple of itself, that is = 1.

transitive Assume = and = where , . Then = so that is amultiple of.

not symmetric If = , , then = 1 and / . For example, 4 is amultiple of 2 but 2 is not a multiple of 4.

1.23

[, ] = ,,, (, ) =

1.24

() = ,, () = , () = ,, () = ,

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1.25 Let be ordered by . is a minimal element there is no element whichstrictly precedes it, that is there is no element such that . is therst element if it precedes every other element, that is for all .

1.26 The maximal elements of are and . The minimal element of is . Theseare also best and worst elements respectively.

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

1.28 Assume that the elements are denoted1, 2, . . . , . We can identify the maximalelement by constructing another list using the following recursive algorithm

1 = 1

=

if11 otherwise

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

1.29

is maximal there does not exist

that is

() =: =

is best for every for every

That is, every belongs to (

) or (

) =.1.30 Let be a nonempty set of a set ordered by . is a lower bound for

if it precedes every element in , that is for all . It is a greatest lowerbound if it dominates every lower bound, that is for every lower bound of.

1.31 Any multiple of 60 is an upper bound for . Thus, the set of upper bounds ofis60, 120, 240, . . .. The least upper bound of is 60. The only lower bound is 1,hence it is the greatest lower bound.

1.32 The least upper bounds of interval [, ] are and . The least upper bound of(, ) is .

1.33

is an upper bound of for every for every ()

Similarly

is a lower bound of for every for every ()

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1.34 For every 2,

if1> 1 or1 = 1 and2> 2Since all elements 2 are comparable,is complete; it is a total order.

1.35 Assume is complete for every . Then for every , and for all =1, 2, . . . , , either or or both. Either

for all Then dene . for some Let be the rst individual with a strict preference, that is =

min(). (Completeness of ensures that is dened). Then dene

if otherwise

1.36 Let, andbe subsets of a nite set . Set inclusionisreexive since.transitive since and implies .anti-symmetric since and implies = Thereforeis a partial order.

1.37 Assume and are both least upper bounds of. That is for all and for all . Further, if is a least upper bound, . If is a leastupper bound, . By anti-symmetry, = .

1.38

= and

which implies that = by antisymmetry. Each equivalence class

() =:

comprises just a single element .

1.39 max() = and min () =.1.40 The subset2, 4, 8forms a chain. More generally, the set of integer powers of a

given number , 2, 3, . . . forms a chain.1.41 Assume and are maximal elements of the chain . Then for all

and in particular . Similarly, for all and in particular . Since is anti-symmetric, = .

1.42 1. By assumption, for every

,

() is a nonempty nite chain. Hence,

it has a unique maximal element, ().

2. Let be any node. Either is an initial node or has a unique predecessor ().Either () is an initial node, or it has a unique predecessor (()). Continuingin this way, we trace out a unique path from back to an initial node. We canbe sure of eventually reaching an initial node since is nite.

1.43

(1, 2) (3, 1) = (3, 2) and (1, 2) (3, 2) = (1, 2)

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1.44 1. is an upper bound for , , that isxy andxy . Similarly, is a lower bound for , .

2. Assume . Then is an upper bound for , , that is . If isany upper bound for , , then . Therefore, is the least upper boundfor

,

. Similarly, is a lower bound for

,

, and is greater than any

other lower bound. Conversely, assume = . Then is an upper bound for , , that is .

3. Using the preceding equivalence

= ( ) = = ( ) =

1.45 A chain is a complete partially ordered set. For every, with = ,either or. Therefore, dene the meet and join by

= if if

=

if if

is a lattice with these operations.

1.46 Assume1 and2 are lattices, and let =1 2. Consider any two elementsx = (1, 2) and y = (1, 2) in . Since 1 and 2 are lattices, 1 = 1 1 1and 2 = 2 2 2, so that b = (1, 2) = (1 1, 2 2) . Furthermoreb x and b y in the natural product order, so that b is an upper bound for thex, y. Every upper bound b = (1, 2) ofx, y must have and ,so that b b. Therefore, b is the least upper bound ofx, y, that is b = x y.Similarly, x y= (1 1, 2 2).

1.47 Letbe a subset ofand let

= : for every

be the set of upper bounds of. Then =. By assumption, has a greatestlower bound . Since every is a lower bound of , for every .Therefore is an upper bound of . Furthermore, is the least upper bound of ,since for every . This establishes that every subset ofalso has a leastupper bound. In particular, every pair of elements has a least upper and a greatestlower bound. Consequently is a complete lattice.

1.48 Without loss of generality, we will prove the closed interval case. Let [, ] be aninterval in a lattice . Recall that = inf[, ] and = sup[, ]. Choose any , in

[, ] . Since is a lattice, and = sup ,

Therefore [, ]. Similarly, [, ]. [, ] is a lattice. Similarly, for anysubset[, ] , sup ifis complete. Also, sup = sup[, ]. Thereforesup [, ]. Similarly inf[, ] so that [, ] is complete.

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1.49 1. The strong set order is

antisymmetric Let1, 2 with 1 2 and 2 1. Choose 11and 22. Since 1 2, 1 21 and 1 22. On the otherhand, since2 1, 1 = (1 (1 2) 2 and2 = 2 (1 2) 1(Exercise 1.44. Therefore

1=

2and

is antisymmetric.

transitive Let 1, 2, 3 with 1 2 and 2 3. Choose 1 1,2 2 and 3 3. Since 1 2 and 2 3, 1 2 and 2 3are in2. Therefore2= 1 (2 3) 2 which implies

1 3 = 1 (

(2 3) 3

=(

1 (2 3) 3

=2 33since 2 3. Similarly 2= (1 2) 32 and

1 3 =(

1 (1 2) 3

=1 ((1 2) 3=1 2 1

Therefore,1 3.

2. if and only if, for every 1, 2, 1 2and 1 2, whichis the case if and only ifis a sublattice.

3. Let () denote the set of all sublattices of . We have shown that isreexive, transitive and antisymmetric on (). Hence, it is a partial order on().

1.50 Assume 1 2. For any 11 and 22, 1 21 and 1 22.Therefore

sup 1 1 2 2 for every 2 2which implies that sup 1 sup 2. Similarly

inf2 1 2 1 for every 1 1which implies that inf2 inf1. Note that completeness ensures the existence ofsup and inf respectively.

1.51 An argument analogous to the preceding exercise establishes = . (Complete-ness is not required, since for any interval = inf[, ] and = sup[, ]).

To establish the converse, assume that 1 = [1, 1] and 2 = [2, 2]. Consider any1

1 and 2

2. There are two cases.

Case 1. 1 2 Sinceis a chain, 1 2 = 1 1. 1 2 = 22.Case 2. 12 Since is a chain, 1 2 = 2. Now 1 1 2 2 2.

Therefore, 2 = 1 2 1. Similarly 2 1 1 2 2. Therefore1 2 = 1 2.

We have shown that 1 2 in both cases.

1.52 Assume that is a complete relation on . This means that for every , ,either or . In particular, letting = , for . is reexive.

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1.53 Anti-symmetry implies that each indifference class contains a single element. If theconsumers preference relation was anti-symmetric, there would be no baskets of goodsbetween which the consumer was indifferent. Each indifference curve which consist asingle point.

1.54 We previously showed (Exercise 1.27) that every best element is maximal. Toprove the converse, assume that is maximal in the weakly ordered set . We have toshow that for all . Assume otherwise, that is assume there is somefor which . Since is complete, this implies that which contradicts theassumption that is maximal. Hence we conclude that for and is abest element.

1.55 False. A chain has at most one maximal element (Exercise 1.41). Here, uniquenessis ensured by anti-symmetry. A weakly ordered set in which the order is not anti-symmetric may have multiple maximal and best elements. For example, and areboth best elements in the weakly ordered set .

1.56 1. For every , either = () or = () since is complete. Consequently, () () = If() (), then and so that and .

2. For every, either = () or = () since iscomplete. Consequently, () () = and () () =.

3. For every ,() and partition () and therefore(), and()partition.

1.57 Assume and (). Then by transitivity. Therefore ().This shows that ()().Similarly, assume and (). Then by transitivity. Therefore (). This shows that() (). To show that()= (), observe that () but that / ()

1.58 Every nite ordered set has a least one maximal element (Exercise 1.28).

1.59 Kreps (1990, p.323), Luenberger (1995, p.170) and Mas-Colell et al. (1995, p.313)adopt the weak Pareto order, whereas Varian (1992, p.323) distinguishes the two or-ders. Osborne and Rubinstein (1994, p.7) also distinguish the two orders, utilizing theweak order in dening the core (Chapter 13) but the strong Pareto order in the Nashbargaining solution (Chapter 15).

1.60 Assume that a group is decisive over , . Let, be two other states.We have to show that is decisive over and . Without loss of generality, assumefor all individuals and . Then, the Pareto order implies that and .Assume that for every , . Since is decisive over and , the socialorder ranks . By transitivity, . By IIA, this holds irrespective of individual

preferences on other alternatives. Hence, is decisive over and .

1.61 Assume that is decisive. Let, and be any three alternatives and assume for every . Partitioninto two subgroups 1 and 2 so that

for every 1 and for every 2Since is decisive, . By completeness, either

in which case1is decisive over and. By the eld expansion lemma (Exercise1.60), 1 is decisive.

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which implies . In this case, 2 is decisive over and , and therefore(Exercise 1.60) decisive.

1.62 Assume is a social order which is Pareto and satises Independence of IrrelevantAlternatives. By the Pareto principle, the whole group is decisive over any pair ofalternatives. By the previous exercise, some proper subgroup is decisive. Continuingin this way, we eventually arrive at a decisive subgroup of one individual. By theField Expansion Lemma (Exercise 1.60), that individual is decisive over every pair ofalternatives. That is, the individual is a dictator.

1.63 Assume is decisive over and and is decisive over and . That is, assume

= =

Also assume

for every

for every

This implies that and (Pareto principle). Combining these preferences,transitivity implies that

which contradicts the assumption that . Therefore, the implied social ordering isintransitive.

1.64 Assume core. In particular this implies that there does not exist any ()such that. Therefore Pareto.

1.65 No state will accept a cost share which exceeds what it can achieve on its own, sothat if core then

1870

5330 860

Similarly, the combined share of the two states AP and TN should not exceed 6990,which they could achieve by proceeding without KM, that is

+ 6990Similarly

+ 1960+ 5020

Finally, the sum of the shares should equal the total cost

+ + = 6530

The core is the set of all allocations of the total cost which satisfy the precedinginequalities.

For example, the allocation ( = 1500, = 5000, = 30) does not belongto the core, since TN and KM will object to their combined share of 5030; since theycan meet their needs jointly at a total cost of 5020. One the other hand, no groupcan object to the allocation ( = 1510, = 5000, = 20), which thereforebelongs to the core.

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1.66 The usual way to model a cost allocation problem as a TP-coalitional game isto regard the potential cost savings from cooperation as the sum to be allocated. Inthis example, the total joint cost of 6530 represents a potential saving of 1530 overthe aggregate cost of 8060 if each region goes its own way. This potential saving of1530 measures (). Similarly, undertaking a joint development, AP and TN could

satisfy their combined requirements at a total cost of 6890. This compares with thestandalone costs of 7100 (= 1870 (AP) + 5330 (TN)). Hence, the potential cost savingsfrom their collaboration are 210 (= 7100 - 6890), which measures (,). Bysimilar calculations, we can compute the worth of each coalition, namely

() = 0 (,) = 210

( ) = 0 (,) = 770 () = 1530

() = 0 ( , ) = 1170

An outcome in this game is an allocation of the total cost savings () = 1530 amongstthe three players. This can be translated into nal cost shares by subtracting eachplayers share of the cost savings from their standalone cost. For example, a specic

outcome in this game is ( = 370, = 330, = 830), which corresponds tonal cost shares of 1500 for AP, 5000 for TN and 30 for KM.

1.67 Let

=x :

() for every

1. core Assume thatx . Supposex / core. This implies there exists somecoalition and outcome y () such that yx for every .

y () implies () while y xfor every implies > for every . Summing, this

implies

>

()

This contradiction establishes that x core.2. core Assume thatx core. Supposex /. This implies there exists some

coalitionsuch that

< (). Let = ()

and consider theallocation y obtained by reallocating from to, that is

=

+ / /( ) /

where = is the number of players in and = is the number in .Then > for everyso that y x for every . Further, y()since

=

+ = () andy since

=

(+ /) +/

( /( )) =

= ()

This contradicts our assumption that x / core, establishing thatx .

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1.68 The 7 unanimity games for the player set =1, 2, 3are

1() =

1 S =1,1,2,1,3, N0 otherwise

2() =

1 S =2,1,2,2,3, N0 otherwise

3() =

1 S =3,1,3,2,3, N0 otherwise

1,2() =

1 S =1,2, N0 otherwise

1,3() =

1 S =1,3, N0 otherwise

2,3() = 1 S =2,3, N

0 otherwise

() =

1 S = N

0 otherwise

1.69 Firstly, consider a simple game which is a unanimity game with essential coalitionand let be an outcome in which

0 for every = 0 for every /

and

= 1

We claim that core.Winning coalitions Ifis winning coalition, then () = 1. Furthermore, if it is a

winning coalition, it must contain, that is and

= 1 =()

Losing coalitions Ifis a losing coalition, () = 0 and

0 = ()

Therefore core and so core =.Conversely, consider a simple game which is not a unanimity game. Suppose thereexists an outcome core. Then

() = 1 (1.15)

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Since there are no veto players ( =),( ) = 1 for every player and=

( ) = 1

which implies that = 0 for every

contradicting (1.15). Thus we conclude that

core =.1.70 The excesses of the proper coalitions at x1 and x2 are

x1 x2

AP -180 -200KM -955 -950TN -395 -380AP, KM -365 -380AP, TN -365 -370KM, TN -180 -160

Therefore

(x1) = (

180,

180,

365,

365,

395,

955)

and

(x2) = (160, 200, 370, 380, 380, 950)d(x1) d(x2) which implies x1 x2.

1.71 It is a weak order on , that is is reexive, transitive and complete. Reexivityand transitivity ow from the corresponding properties of on2 . Similarly, forany x, y , either d(x) d(y) or d(y) d(x) since is complete on2 .Consequently eitherx y ory x (or both).

is not a partial order since it is not antisymmetric

d(x) d(y) and d(y) d(x) does not imply x = y

1.72

(, x) =()

so that

(, x)0

()

1.73 Assume to the contrary that x Nu but that x / core. Then, there exists acoalition with a positive decit(, x)> 0. Since core =, there exists someysuch that (, y) 0 for every Nu. Consequently, d(y) d(x) and y x, sothatx /Nu. This contradiction establishes that Nu core.

1.74 For player 1, 1 =

,

and

(, ) 1 (, )

(, ) 1 (, )

Similarly for player 2

(, ) 2 (, )

(, ) 2 (, )

Therefore, (, ) satises the requirements of the denition of a Nash equilibrium(Example 1.51).

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1.75 Ifa is the best element in (,) for every player , then

( , a) (, a) for every and afor every . Therefore, a is a Nash equilibrium.

To show that it is unique, assume that a is another Nash equilibrium. Then for everyplayer

(, a) (, a) for every which implies that a is a maximal element of. To see this, assume not. That is,assume that there exists some such that which implies

(, a) (, a) for every aIn particular

(, a)( , a)

which contradicts the assumption that a

is a Nash equilibrium. Therefore, a is an-other Nash equilibrium, then is maximal in and hence also a best element of

(Exercise 1.54), which contradicts the assumption that is the unique best element.Consequently, we conclude that a is the unique Nash equilibrium of the game.

1.76 We show that (, ) = satises the requirements of a metric, namely1. 0.2. = 0 if and only if = .3. = .

To establish the triangle inequality, we can consider various cases. For example, if

+ = = If

+ = + = = and so on.

1.77 We show that , = max=1 satises the requirements of a metric,

namely

1. max=1 02. max=1 = 0 if and only if= for all .

3. max

=1 = max

=1 4. For every, + from previous exercise. Therefore

max max( + ) max + max

1.78 For any, any neighborhood of 1/contains points of(namely 1/) and pointsnot in(1/ + ). Hence every point in is a boundary point. Also, 0 is a boundarypoint. Therefore b() = 0. Note that b(). Therefore,has no interiorpoints.

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1.79 1. Let int. Thus is a neighborhood of . Therefore, is aneighborhood of, so that is an interior point of.

2. Clearly, if , then . Therefore, assume which impliesthat is a boundary point of. Every neighborhood of contains other pointsof

. Hence

.

1.80 Assume that is open. Every has a neighborhood which is disjoint from. Hence no is a closure point of. contains all its closure points and istherefore closed.

Conversely, assume that is closed. Let be a point its complement . Since is closed and / , is not a boundary point of . This implies that has aneighborhood which is disjoint from, that is . Hence, is an interior pointof. This implies that contains only interior points, and hence is open.

1.81 Clearly is a neighborhood of every point , since () for every >0. Hence, every point is an interior point of. Similarly, every point is an interior point (there are none). Since andare open, there complementsand are closed.

Alternatively, has no boundary points, and is therefore is open. Trivialy, on the otherhand,contains all its boundary points, and is therefore closed.

1.82 Letbe a metric space. Assume is the union of two disjoint closed sets and, that is

= =

Then = is open as is = . Thereforeis not connected.

Conversely, assume thatis not connected. Then there exist disjoint open sets and such that = . But = is also closed as is = . Therefore is theunion of two disjoint closed sets.

1.83 Assume is both open and closed, . We show that we can representas the union of two disjoint open sets, and . For any, = and =. is open by assumption. It complement is open since is closed.Therefore,is not connected.

Conversely, assume that is not connected. That is, there exists two disjoint opensets and such that = . Now = , which implies that is closed since is open. Therefore is both open and closed.

1.84 Assume thatis both open and closed. Then so is andis the disjoint unionof two closed sets

=

so that

b() = = =

Conversely, assume that b() = =. This implies that Consider any .Since =, / . A fortiori, x / which implies that and therefore . is closed. Similarly we can show that so that is closed andtherefore is open. is both open and closed.

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1.85 1. Letbe a (possibly innite) collection of open sets. Let =. Letbe a point in . Then there exists some particular which contains. Since is open, is a neighborhood of. Since, is an interior point of.Since is an arbitrary point in , we have shown that every is an interiorpoint. Hence, is open.

What happens if every is empty? In this case, = and is open (Exercise1.81). The other possibility is that the collection is empty. Again =which is open.

Suppose 1, 2, . . . , is a nite collection of open sets. Let =. If=, then it is trivially open. Otherwise, let be a point in . Then for all = 1, 2, . . . , . Since the sets are open, for every, there exists an openball (, ) about. Let be the smallest radius of these open balls, thatis = min 1, 2, . . . , . Then ()(, ), so that () for all i.Hence(). is an interior point of and is open.To complete the proof, we need to deal with the trivial case in which the collectionis empty. In that case, = = and hence is open.

2. The corresponding properties of closed sets are established analogously.

1.86 1. Let0be an interior point of. This implies there exists an open ball about 0. Every is an interior point of. Hence int. 0 is aninterior point of int which is therefore open.

Let be any open subset of and be a point in . is neighborhood of,which implies that is also neighborhood of. Therefore is an interiorpoint of. Therefore int contains every open subset , and hence is thelargest open set in .

2. Let denote the closure of the set . Clearly, . To show the converse, let be a closure point of and let be a neighborhood of. Then containssome other point

= which is a closure point of. is a neighborhood of

which intersects. Hence is a closure point of.

Consequently= which implies that is closed.

Assumeis a closed subset of containing . Then

=since is closed. Hence, is a subset of every closed set containing .

1.87 Every is either an interior point or a boundary point. Consequently, theinterior of is the set of all which are not boundary points

int= b()

1.88 Assume that is closed, that is

= b() = This implies that b() . contains its boundary.Assume thatcontains its boundary, that is b(). Then

= b() = is closed.

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1.89 Assume is bounded, and let = (). Choose any . For all ,(, ) < + 1. Therefore, (, + 1). is contained in the open ball(, + 1).

Conversely, assumeis contained in the open ball (). Then for any ,

(, ) (, ) + (, )< 2by the triangle inequality. Therefore ()< 2 and the set is bounded.

1.90 Let (0). For every ,(, )< and therefore

(, 0) (, ) + (, 0)< + = 2

so that 2(0).1.91 Lety0 . For any >0, let y =y be the production plan which is units

less in every commodity. Then, for any y(y)

(y, y)< for every

and thereforey < y0. Thus (y) and so y int =.1.92 For any 1

= (, 2)> 0

Similarly, for every2 =(, 1)> 0

Let

1 = 1

/2()

2 =

2

/2()

Then 1 and 2 are open sets containing 1 and 2 respectively.

To show that 1 and 2 are disjoint, suppose to the contrary that 1 2. Then,there exist points 1 and2 such that

(, )< /2, (, )< /2

Without loss of generality, suppose that and therefore

(, )

(, ) + (, )< /2 + /2

which contradicts the denition of and shows that 1 2 =.1.93 By Exercise 1.92, there exist disjoint open sets 1 and 2 such that 11 and

2 2. Since 2 2, 22 =. 2 is a closed set which contains 1, andtherefore2 1=. =1 is the desired set.

1.94 See Figure 1.2.

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1 2

1

1/2((2, 0))

Figure 1.2: Open ball about (2, 0) relative to

1.95 Assume is connected. Suppose is not an interval. This implies that thereexists numbers ,, such that < < and, while /. Then

= ( (, )) ( (, ))representsas the union of two disjoint open sets (relative to ), contradicting theassumption that is connected.

Conversely, assume that is an interval. Suppose that is not connected. That is,= where and are nonempty disjoint closed sets. Choose and .Since and are disjoint, = . Without loss of generality, we may assume < .Sinceis an interval, [, ] = . Let

= sup

[, ]

Clearly so that. Nowbelongs to eitheror. Sinceis closed in,[, ] is closed and = sup [, ] . This implies the < . Consequently,+ for every >0 such that + . Since is closed, . This impliesthat belongs to both and contradicting the assumption that =. Weconclude that must be connected.

1.96 Assume and also . We have to show that = . Suppose not,that is suppose = (see Figure 1.3). Then (, ) = >0. Let = /3> 0. Since , there exists some such that () for all . Since ,there exists some such that

() for all . But these statements arecontradictory since () (, ) =. We conclude that the successive terms of aconvergent sequence cannot get arbitrarily close to two distinct points, so that the limit

a convergent sequence is unique.1.97 Let () be a sequence which converges to . There exists some such that

( )< 1for all . Let

= max (1 ), (2 ), . . . , (1 ), 1 Then for all,() . That is every element in the sequence () belongs to(, + 1), the open ball about of radius + 1. Therefore the sequence is bounded.

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(, )

(, )

Figure 1.3: A convergent sequence cannot have two distinct limits

1.98 The share of the th guest is

=1

2

lim = 0

However, > 0 for all . There is no limit to the number of guests who will get ashare of the cake, although the shares will get vanishingly small for large parties.

1.99 Suppose

. That is, there exists some such that (

, ) < /2 for all. Then, for all , (, ) (, ) + (, )

< /2 + /2 =

1.100 Let () be a Cauchy sequence. There exists some such that

( )< 1for all . Let

= max (1 ), (2 ), . . . , (1 ), 1 Every belongs to (, + 1), the ball of radius + 1 centered on .

1.101 Let (

) be a bounded increasing sequence inand let=

be the set ofelements of (). Let be the least upper bound of. We show that .First observe that for every (since is an upper bound). Sinceis the leastupper bound, for every >0 there exists some element such that > . Since() is increasing, we must have

< for every That is, for every >0 there exists an such that

(, )< for every .

1.102 If >1, the sequence , 2, 3, . . . is unbounded.

Otherwise, if1, 1 and the sequence is decreasing and bounded by 1.Therefore the sequence converges (Exercise 1.101). Let = lim. Then

+1 =

and therefore

= lim

+1 = lim

=

which can be satised if and only if

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1.104 The following sequence approximates the square root of any positive number

1 =

+1 =1

2( +

1.105 Let . If, then is the limit of the sequence ( , , , . . . ). If / ,

then is a boundary point of . For every , the ball (, 1/) contains a point . From the sequence of open balls (, 1/) for = 1, 2, 3, . . . , we can generateof a sequence of points which converges to.

Conversely, assume that is the limit of a sequence () of points in . Either and therefore . Or / . Since , every neighborhood of contains points of the sequence. Hence, is a boundary point ofand .

1.106 is closed if and only if= . The result follows from Exercise 1.105.

1.107 Let be a closed subset of a complete metric space . Let () be a Cauchysequence in . Since is complete, . Since is closed, (Exercise

1.106).1.108 Since () 0, cannot contain more than one point. Therefore, it suffices

to show that is nonempty. Choose some from each . Since()0, () isa Cauchy sequence. Since is complete, there exists some such that .Choose some. Since the sets are nested, the subsequence : . Since is closed, (Exercise 1.106). Since for every

=1

1.109 If player 1 picks closed balls whose radius decreases by at least half after eachpair of moves, then 1, 3, 5, . . . is a nested sequence of closed sets which has anonempty intersection (Exercise 1.108).

1.110 Let () be a sequence in with closed and compact. Since iscompact, there exists a convergent subsequence . Since is closed,we must have (Exercise 1.106). Therefore () contains a subsequence whichconverges in, so that is compact.

1.111 Let () be a Cauchy sequence in a metric space. For every > 0, there exists such that

(, )< /2 for all ,

Trivially, if () converges, it has a convergent subsequence (the whole sequence).

Conversely, assume that () has a subsequence () which converges to . That is,

there exists some such that

(, )< /2 for all

Therefore, by the triangle inequality

(, ) (, ) + (, )< /2 + /2 =

for all max ,

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1.112 We proceed sequentially as follows. Choose any 1 in . If the open ball(1, )contains, we are done. Otherwise, choose some2 / (1, ) and consider the set2

=1 (, ). If this set contains , we are done. Otherwise, choose some 3 /

2=1 (, ) and consider

3=1 (, )

The process must terminate with a nite number of open balls. Otherwise, if the processcould be continued indenitely, we could construct an innite sequence (1, 2, 3, . . . )which had no convergent subsequence. The would contradict the compactness of.

1.113 Assumeis compact. The previous exercise showed that is totally bounded.Further, since every sequence has a convergent subsequence, every Cauchy sequenceconverges (Exercise 1.111). Therefore is complete.

Conversely, assume that is complete and totally bounded and let 1 =11, 21, 31, . . . be an innite sequence of points in . Since is totally bounded, it is covered by a

nitecollection of open balls of radius 1/2. 1 has a subsequence2 =12, 22, 32, . . . all of whose points lie in one of the open balls. Similarly, 2 has a subsequence3 = 13, 23, 33, . . . all of whose points lie in an open ball of radius 1/3. Contin-uing in this fashion, we construct a sequence of subsequences, each of which lies in a

ball of smaller and smaller radius. Consequently, successive terms of the diagonalsubsequence 11, 22, 33, . . . get closer and closer together. That is, is a Cauchysequence. Sinceis complete, converges inand1 has a convergent subsequence. Hence, is compact.

1.114 1. Every big set has a least two distinct points. Hence () >0 for every .

2. Otherwise, there exists such that () 1/ for every and therefore= inf() 1/ > 0.

3. Choose a point in each . Since is compact, the sequence () has a

convergent subsequence () which converges to some point 0.4. The point 0 belongs to at least one 0 in the open cover

. Since 0 is open,

there exists some open ball (0) 0.5. Consider the concentric ball /2(0). Since (

) is a convergent subsequence,there exists some such that /2() for every .

6. Choose some 0 min , 2/ . Then 1/0 < /2 and (0) < 1/0 < /2.0 0 /2() and therefore (Exercise 1.90)0() 0.

This contradicts the assumption that is a big set. Therefore, we conclude that >0.

1.115 1. is totally bounded (Exercise 1.112). Therefore, for every > 0, thereexists a nite number of open balls() such that

=

=1

()

2. (()) = 2 < . By denition of the Lebesgue number, every () iscontained in some .

3. The collection of open balls() covers . Therefore, fore every ,there exists such that

()

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Therefore, the nite collection1, 2, . . . , covers.

1.116 For any family of subsets

=

=

Suppose to the contrary thatis a collection of closed sets with the nite intersectionproperty, but that

=. Then : is a open cover ofwhich does

not have a nite subcover. Consequently cannot be compact.

Conversely, assume every collection of closed sets with the nite intersection propertyhas a nonempty intersection. Letbe an open cover of. Let

= : That is

=which implies

=

Consequently, does not have the nite intersection property. There exists a nitesubcollection 1, 2, . . . , such that

=1

=

which implies that

=1

=

1, 2, . . . , is a nite subcover of. Thus, is compact.1.117 Every nite collection of nested (nonempty) sets has the nite intersection prop-

erty. By Exercise 1.116, the sequence has a non-empty intersection. (Note: every set is a subset of the compact set 1.)

1.118 (1) = (2) Exercises 1.114 and 1.115.(2) = (3) Exercise 1.116(3) = (1) Letbe a metric space in which every collection of closed subsets with

the nite intersection property has a nite intersection. Let () be a sequencein . For any , let be the tail of the sequence minus the rst terms, thatis

= : = + 1, + 2, . . .

The collection () has the nite intersection property since, for any nite set ofintegers1, 2, . . . ,

=1

=

where= max 1, 2, . . . , . Therefore

=1

=

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Choose any =1 . That is, for each = 1, 2, . . . . Thus, for every >0 and = 1, 2, . . . , there exists some () We construct a subsequence as follows. For = 1, 2, . . . , let be the rstterm in which belongs to1/(). Then, (

) is a subsequence of () whichconverges to. We conclude that every sequence has a convergent subsequence.

1.119 Assume () is a bounded sequence in. Without loss of generality, we canassume that [0, 1]. Divide 0 = [0, 1] into two sub-intervals [0, 1/2] and[1/2, 1]. At least one of the sub-intervals must contain an innite number of terms ofthe sequence. Call this interval 1. Continuing this process of subdivision, we obtaina nested sequence of intervals

0 1 2 . . .

each of which contains an innite number of terms of the sequence. Consequently,we can construct a subsequence () with . Furthermore, the intervals getsmaller and smaller with () 0, so that () is a Cauchy sequence. Since iscomplete, the subsequence () converges to .Note how we implicitly called on the Axiom of Choice (Remark 1.5) in choosing asubsequence from the nested sequence of intervals.

1.120 Let () be a Cauchy sequence in. That is, for every >0, there exists suchthat < for all, . () is bounded (Exercise 1.100) and hence by theBolzano-Weierstrass theorem, it has a convergent subsequence () with .Choose from the convergent subsequence such that and < /2. Bythe triangle inequality

+ < /2 + /2 =

Hence the sequence () converges to .1.121 Since 1 and 2 are linear spaces, x1 + y1

1 and x2 + y2

2, so that

(x1 + y1, x2 + y2) 1 2. Similarly (x1, x2) 1 2 for every (x1, x2)1 2. Hence, =1 2 is closed under addition and scalar multiplication.With addition and scalar multiplication dened component-wise, inherits the arith-metic properties (like associativity) of its constituent spaces. Verifying this wouldproceed identically as for. It is straightforward though tedious. The zero elementin is 0 = (01, 02) where 01 is the zero element in 1 and 02 is the zero element in2. Similarly, the inverse ofx = (x1, x2) isx= (x1, x2).

1.122 1.

x + y= x + z

x + (x + y) =x + (x + z)(x + x) + y= (x + x) + z

0 + y= 0 + z

y= z

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2.

x= y

1

(x) =

1

(y)

1

)

x=

1

)

y

x= y

3. x= ximplies

( )x= x x= 0Providedx = 0, we must have

( )x= 0xThat is

= 0 which implies = .

4.

( )x= ( + ())x=x + ()x=x x

5.

(x y) = (x + (1)y)=x + (1)y=x y

6.

0= (x+ (x))=x + (x)=x x=0

1.123 The linear hull of the vectors(1, 0), (0, 2)is

lin(1, 0), (0, 2) =

1

10

)+ 2

02

)

= 1

2)

=2

The linear hull of the vectors (1, 0), (0, 2) is the whole plane 2. Figure 1.4 illustrateshow any vector in2 can be obtained as a linear combination of(1, 0), (0, 2).

1.124 1. From the denition of,

= ()

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(2, 3)

2

3

1

1-1-2 0

Figure 1.4: Illustrating the span of (1, 0), (0, 2) .

for every . Rearranging() = +

==

+

=

2.

() =

() +

()

=

1 +

0

=

1

=()

1.125 1. Choose anyx . By homogeneity 0x= .2. For everyx ,x= (1)x by homogeneity.

1.126 Examples of subspaces in include:1. The set containing just the null vector0 is subspace.2. Letx be any element in and let be the set of all scalar multiples ofx

=

x:

is a line through the origin in and is a subspace.3. Letbe the set of all -tuples with zero rst coordinate, that is

= (1, 2, . . . , ) : 1 = 0, , = 1 For anyx, y

x + y= (0, 2, 3, . . . , ) + (0, 2, 3, . . . , )

= (0, 2+ 2, 3+ 3, . . . , + )

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Similarly

x= (0, 2, 3, . . . , )

= (0, 2, 3, . . . , )

Therefore is a subspace of. Generalizing, any set of vectors with one ormore coordinates identically zero is a subspace of.

4. We will meet some more complicated subspaces in Chapter 2.

1.127 No,x / +ifx + unlessx = 0.+ is an example of a cone (Section 1.4.5).1.128 lin is a subspace Letx, ybe two elements in lin . xis a linear combination

of elements of, that is

x= 11+22+ . . .

Similarly

y= 11+ 22+ . . .

and

x + y= (1+ 1)1+ (2+ 2)2+ + (+ ) lin

and

x= 11+ 22+ + lin

This shows that lin is closed under addition and scalar multiplication and henceis a subspace.

lin is the smallest subspace containing Letbe any subspace containing.Then contains all linear combinations of elements in , so that lin

.

Hence linis the smallest subspace containing S.

1.129 The previous exercise showed that lin is a subspace. Therefore, if= lin ,is a subspace.

Conversely, assume that is a subspace. Then is the smallest subspace containing, and therefore = lin (again by the previous exercise).

1.130 Let x, y = 1 2. Hence x, y 1 and for any , , x+y 1.Similarly x + y2 and therefore x + y. is a subspace.

1.131 Let= 1+ 2. First note that 0 = 0 + 0 . Supposex, ybelong to . Thenthere exists1, t11 and s2, t22 such that x = s1+s2 andy = t1+ t2. For any, ,

x + y= (s1+ s2) + (t1+ t2)= (s1+ t1) + (s2+ t2)

since s1+ t1 1 ands2+ t22.1.132 Let

1 =(1, 0) : 2 =(0, 1) :

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1 and2 are respectively the horizontal and vertical axes in2. Their union is not asubspace, since for example

11

)=

10

)+

01

)/1 2

However, any vector in 2 can be written as the sum of an element of1and an elementof2. Therefore, their sum is the whole space2, that is

1+ 2 =2

1.133 Assume that is linearly dependent, that is there exists x1, . . . , x and2, . . . , such that

x1 = 2x2+3x3+ . . . , x

Rearranging, this implies

1x1 2x2 3x3 . . . x= 0

Conversely, assume there exist x1, x2, . . . , xx and 1, 2, . . . , such that1x1+ 2x2 . . .+ x= 0

Assume without loss of generality that1= 0. Then

x1 = 21

x2 31

x3 . . . 1

x

which shows that

x1 lin x1

1.134 Assume(1, 1, 1), (0, 1, 1), (0, 0, 1) are linearly dependent. Then there exists1, 2, 3 such that

1

11

1

+ 2

01

1

+ 3

00

1

=

00

0

or equivalently

1= 0

1+ 2= 0

1+ 2+ 3= 0

which imply that1 = 2 = 3= 0

Therefore(1, 1, 1), (0, 1, 1), (0, 0, 1)are linearly independent.1.135 Suppose on the contrary that is linearly dependent. That is, there exists a

set of games 1, 2 , . . . , and nonzero coefficients (1, 2, . . . , ) such that(Exercise 1.133)

11+ 22+ . . . + =0 (1.16)

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Assume that the coalitions are ordered so that 1 has the smallest number of playersof any of the coalitions 1, 2, . . . , . This implies that no coalition 2, 3, . . . , isa subset of1 and

(1) = 0 for every = 2, 3, . . . , (1.17)

Using (1.39), 1 can be expressed as a linear combination of the other games,

1 = 1/1

=2

(1.18)

Substituting (1.40) this implies that

1(1) = 0

whereas

() = 1 for every

by denition. This contradiction establishes that the set is linearly independent.

1.136 Ifis a subspace, then 0 and

x1 = 0

with = 0 and x1 = 0 (Exercise 1.122). Thereforeis linearly dependent (Exercise1.133).

1.137 Suppose x has two representations, that is

x= 1x1+ 2x2+ . . . + x

x= 1x1+ 2x2+ . . . + x

Subtracting

0= (1 1)x1+ (2 2)x2+ . . . + ( )x (1.19)

Sincex1, x2, . . . , , x is linearly independent, (1.19) implies that = = 0 for all (Exercise 1.133)

1.138 Let be the set of all linearly independent subsets of a linear space . ispartially ordered by inclusion. Every chain= has an upper bound, namely

. By Zorns lemma,has a maximal element . We show that is a basisfor.

is linearly independent since . Suppose that does not span so thatlin

. Then there exists some x

lin . The set

x

is a linearly

independent and contains , which contradicts the assumption that is the maximalelement of. Consequently, we conclude that spans and hence is a basis.

1.139 Exercise 1.134 established that the set = (1, 1, 1), (0, 1, 1), (0, 0, 1) is linearlyindependent. Since dim 3 = 3, any other vectors must be linearly dependent on .That is lin =3. is a basis.By a similar argument to exercise 1.134, it is readily seen that (1, 0, 0), (0, 1, 0), (0, 0, 1)is linearly independent and hence constitutes a basis.

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1.140 Let =a1, a2, . . . , a and =b1, b2, . . . , b be two bases for a linearspace. Let

1=1 =b1, a1, a2, . . . , a is linearly dependent (since

1 lin ) and spans . Therefore, there exists

1, 2, . . . , and1 such that

1b1+ 1a1+ 2a2+ . . . + a= 0

At least one = 0. Deleting the corresponding element a, we obtain another set 1of elements

1 =b1, a1, a2, . . . , a1, a+1, . . . , awhich is also spans. Adding the second element from , we obtain the +1 elementset

2= b1, b2, a1, a2, . . . , a1, a+1, . . . , awhich again is linearly dependent and spans .

Continuing in this way, we can replace vectors in with the vectors from whilemaintaining a spanning set. This process cannot eliminate all the vectors in , becausethis would imply thatwas linearly dependent. (Otherwise, the remaining bwould belinear combinations of preceding elements of .) We conclude that necessarily .Reversing the process and replacing elements of with elements of establishes that . Together these inequalities imply that = and and have the samenumber of elements.

1.141 Suppose that the coalitions are ordered in some way, so that

() =0, 1, 2, . . . , 21

with 0= . There are 2 coalitions. Each game corresponds to a unique listof length 2 of coalitional worths

v= (0, 1, 2, . . . , 21)

with 0 = 0. That is, each game denes a vector = (0, 1, . . . , 21) 2 andconversely each vector 2 (with 0 = 0) denes a game. Therefore, the space ofall games is formally identical to the subspace of2 in which the rst componentis identically zero, which in turn is equivalent to the space 21. Thus, is a2 1-dimensional linear space.

1.142 For illustrative purposes, we present two proofs, depending upon whether thelinear space is assumed to be nite dimensional or not. In the nite dimensional case,a constructive proof is possible, which forms the basis for practical algorithms forconstructing a basis.

Let be a linearly independent set in a linear space .

is nite dimensional Let = dim . Assume has elements and denote it.

If lin= , then is a basis and we are done. Otherwise, there exists somex+1 lin . Adding x+1 to gives a new set of + 1 elements

+1 = x+1

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which is also linearly independent ( since x+1 /lin ).If lin +1 = , then +1 is a basis and we are done. Otherwise, there existssome x+2 lin +1. Adding x+2 to +1 gives a new set of + 2elements

+2 = +1 x+2 which is also linearly independent ( since x+2 /lin +2).Repeating this process, we can construct a sequence of linearly independent sets, +1, +2 . . . such that lin lin +1 lin +2 . Eventu-ally, we will reach a set which spans and hence is a basis.

is possibly innite dimensional For the general case, we can adapt the proofof the existence of a basis (Exercise 1.138), restricting to be the class of alllinearly independent subsets of containing.

1.143 Otherwise (if a set of + 1 elements was linearly independent), it could beextended to basis at least + 1 elements (exercise 1.142). This would contradict the

fundamental result that all bases have the same number of elements (Exercise 1.140).1.144 Every basis is linearly independent. Conversely, let =x1, x2, . . . , x be a

set of linearly independent elements in an -dimensional linear space. We have toshow that lin = .

Take anyx . The set x=x1, x2, . . . , x, x

must be linearly dependent (Exercise 1.143). That is there exists numbers 1, 2, . . . , , ,not all zero, such that

1x1+ 2x2+ + x+ x= 0 (1.20)

Furthermore, it must be the case that = 0 since otherwise1x1+ 2x2+ + x= 0

which contradicts the linear independence of. Solving (1.20) for x, we obtain

x= 1

1x1+ 2x2+ + x

Sincex was an arbitrary element of, we conclude that spansand hence is abasis.

1.145 A basis spans . To establish the converse, assume that =x1, x2, . . . , x is a set of elements which span . If is linearly dependent, then one element

is linearly dependent on the other elements. Without loss of generality, assume thatx1 lin x1. Deleting x1 the set

x1 =x2, x3, . . . , x

also spans. Continuing in this fashion by eliminating dependent elements, we nishwith a linearly independent set of < elements which spans . That is, we cannd a basis of < elements, which contradicts the assumption that the dimensionof is (Exercise 1.140). Thus any set of vectors which spans must be linearlyindependent and hence a basis.

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1.146 We have previously shown

that the set is linearly independent (Exercise 1.135). the space has dimension 21 (Exercise 1.141).

There are 21 distinct T-unanimity games

in . Hence spans the 21 space. Alternatively, note that any game can be written as a linear combinationof T-unanimity games (Exercise 1.75).

1.147 Let =x1, x2, . . . , x be a basis for . Since is linearly independent, (Exercise 1.143). There are two possibilities.Case 1: = . is a set of linearly independent elements in an -dimensional

space. Hence is a basis for and= lin = .

Case 2: < . Since is linearly independent but cannot be a basis for the -dimensional space, we must have= lin.

Therefore, we conclude that if is a proper subspace, it has a lower dimensionthan.

1.148 Let 1, 2, 3be the coordinates of (1, 1, 1) for the basis (1, 1, 1), (0, 1, 1), (0, 0, 1).That is

111

=1

11

1

+ 2

01

1

+ 3

00

1

which implies that1 = 1,2 = 3 = 0. Therefore (1, 0, 0) are the required coordinatesof the (1, 1, 1) with respect to the basis(1, 1, 1), (0, 1, 1), (0, 0, 1).(1, 1, 1) are the coordinates of the vector (1, 1, 1) with respect to the standard basis.

1.149 A subset of a linear space is a subspace of if

x+ y

for every x, y

and for every ,

Letting= 1 , this implies thatx + (1 )y for every x, y and

is an affine set.

Conversely, suppose thatis an affine set containing 0, that is

x + (1 )y for every x, y and Lettingy = 0, this implies that

x for every x and

so thatis homogeneous. Now letting = 12 , for everyx and y in ,

1

2x +

1

2y

and homogeneity implies

x + y= 2

1

2x +

1

2y

)

is also additive. Hence is subspace.

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1.150 For anyx , let

= x= v: v+ x

is an affine set For anyv1, v2 , there exist correspondings1, s2 such thatv

1

=s1

xand v2

=s2

xand thereforev1 + (1 )v2 =(s1 x) + (1 )(s1 x)

=s1 + (1 )s2 x+ (1 )x=s x

wheres = 1 + (1 )2 . There is an affine set. is a subspace Since x, 0 = x x . Therefore is a subspace (Exercise

1.149).

is unique Suppose that there are two subspaces1 and2 such that= 1 + x1

and= 2 + x2. Then

1+ x1 = 2+ x2

1 = 2+ (x2 x1)=2+ x

wherex = x2 x1 . Therefore 1 is parallel to 2.Since1is a subspace,0 1which implies that x 2. Since2is a subspace,this implies thatx 2 and 2+ x 2. Therefore1 = 2+ x 2. Similarly,21 and hence 1 = 2. Therefore the subspace is unique.

1.151 Letdenote the relation is parallel to , that is

= + xfor some x

The relationisreexive since = + 0transitive Assume= + x and =+ y. Then = + (x + y)

symmetric = + x = =+ (x)Therefore is an equivalence relation.

1.152 See exercises 1.130 and 1.162.

1.153 1. Exercise 1.150

2. Assume x0 . For every x

x= x0+ v= wwhich implies that . Conversely, assume=. Then x0 = 0 since is a subspace.

3. By denition, . Therefore = x .4. Letx1 / . Suppose to the contrary

linx1, =

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Then

=x0+

is an affine set (Exercise 1.150) which strictly contains . This contradicts thedenition ofas a maximal proper affine set.

5. Let x1 / . By the previous part,xlinx1, . That is, there exists such that

x= x1+ v for some v To see that is unique, suppose that there exists such that

x= x1+ v for somev

Subtracting

0= ( )x1+ (v v)which implies that = since x1 / .

1.154 Assume x, y

. That is, x, y

and

=

= ()

For any , x+ (1 )y and

+ (1 ) =

+ (1 )

=() + (1 )()=()

Henceis an affine subset of.1.155 See Exercise 1.129.

1.156 No. A straight line through any two points in+ extends outside+. Putdifferently, the affine hull of+ is the whole space.

1.157 Let

= aff x1= aff0, x2 x1, x3 x1, . . . , x x1

is a subspace (0) andaff= + x1

and

dim aff= dim

Note that the choice ofx1 is arbitrary.

is affinely dependent if and only if there exists some x such that x aff ( x). Since the choice ofx1 is arbitrary, we assume that x=x1.

xaff ( x) x ( + x1) x x x1 x x1 x x1 linx2 x1, x3 x1, . . . , x1 x1,

. . . , x+1 x1, . . . , x x1

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Therefore, is affinely dependent if and only ifx2x1, x3x1, . . . , xx1 islinearly independent.

1.158 By the previous exercise, the set =x1, x2, . . . , x is affinely dependent ifand only if the setx2 x1, x3 x1, . . . , x x1 is linearly dependent, so that thereexist numbers

2,

3, . . . ,

, not all zero, such that

2(x2 x1) + 3(x3 x1) + + (x x1) = 0

or

2x2+ 3x3+ + x

=2

x1 = 0

Let 1 =

=2 . Then

1x1+ 2x2+ . . . + x= 0

and

1+ 2+ . . .+ = 0

as required.

1.159 Let

= aff x1 = aff 0, x2 x1, x3 x1, . . . , x x1

Then

aff= x1+

If is affinely independent, every x affhas a unique representation as

x= x1+ v, v

with

v= 2(x2 x1) + 3(x3 x1) + + (x x1)

so that

x= x1+ 2(x2 x1) + 3(x3 x1) + + (x x1)

Dene1 = 1

=2 . Then

x= 1x1+ 2x2+ + xwith

1+ 2+ + = 1x is a unique affine combination of the elements of.

1.160 Assume that , (, ) . This means that < < and < < . Forevery 0 1

+ (1 ) > + (1 )

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and

+ (1 ) < + (1 )Therefore < +(1) < and+(1)(, ). (, ) is convex. Substituting for < demonstrates that [, ] is convex.Letbe an arbitrary convex set in. Assume that is not an interval. This impliesthat there exist numbers , , such that < < and, while / . Dene

=

so that

1 =

Note that 0 1 and that

+ (1 ) =

+

= /which contradicts the assumption thatis convex. We conclude that every convex setinis an interval. Note that may be a hybrid interval such (, ] or [, ) as well asan open (, ) or closed [, ] interval.

1.161 Let (, ) be a TP-coalitional game. If core(, ) = then it is trivially convex.Otherwise, assume core(, ) is nonempty and let x1 and x2 belong to core(, ).That is

1 () for every

1 =()

and therefore for any 0 1

1 () for every

1 =()

Similarly

(1 )2 (1 )() for every

(1

)2 = (1

)()

Summing these two systems

1 + (1 )2 () + (1 )() =() for every

1 + (1 )2 =() + (1 )() =()

That is, 1 + (1 )2 belongs to core(, ).

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1.162 Let be a collection of convex sets and let x, y belong to

. for every , x, y and therefore x + (1 )yfor all 0 1 (since is convex).Thereforex+ (1 )y .

1.163 Fix some output . Assume thatx1, x2 (). This implies that both (, x1)and (,

x

2) belong to the production possibility set . If is convex

(, x1) + (1 )(, x2) = (+ (1 ), x1+ (1 )x2)= (, x1+ (1 )x2)

for every [0, 1]. This implies thatx1+ (1 )x2 (). Since the choice ofwas arbitrary, this implies that () is convex for every .

1.164 Assume 1 and 2 are convex sets. Let = 1+ 2. Suppose x, y belong to .Then there exists1, t1 1 and s2, t2 2 such that x = s1+ s2 andy = t1+ t2. Forany [0, 1]

x + (1 )y= s1+ s2+ (1 )t1+ t2=s1+ (1

)t1+ s2+ (1

)t2

sinces1+ (1 )t1 1 ands2+ (1 )t2 2. The argument readily extends toany nite number of sets.

1.165 Without loss of generality, assume that = 2. Let= 1 2= 1 2.Suppose x = (1, 2) and y = (1, 2) belong to . Then

x + (1 )y= (1, 2) + (1 )(1, 2)= (1, 2) + ((1 )1, (1 )2)= (1+ (1 )1, 2+ (1 )2)

1.166 Letx, ybe points inso thatx, y . Sinceis convex,x+(1)y for every 0

1. Multiplying by

(x + (1 )y) =(x) + (1 )(y)

Therefore, is convex.

1.167 Combine Exercises 1.164 and 1.166.

1.168 The inclusion + (1 ) is true for any set (whether convex or not),since for every x

x= x + (1 )x + (1 )

The reverse inclusion+(1) follows directly from the denition of convexity.

1.169 Given any two convex sets and in a linear space, the largest convex setcontained in both is ; the smallest convex set containing both is conv .Therefore, the set of all convex sets is a lattice with

= = conv

The lattice is complete since every collection has a least upper bound conv and a greatest lower bound.

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1.170 If a set contains all convex combinations of its elements, it contains all convexcombinations of any two points, and hence is convex.

Conversely, assume that is convex. Letx be a convex combination of elements in ,that is let

x= 1x1+ 2x2+ . . . + x

where x1, x2, . . . , x and 1, 2, . . . , + with 1+ 2+ . . .+ = 1. Weneed to show that x .We proceed by induction of the number of points . Clearly, x if = 1 or = 2.To show that it is true for = 3, let

x= 1x1+ 2x2+ 3x3

where x1, x2, x3and 1, 2, 3 + with 1+ 2+3 = 1. Assume that > 0for all (otherwise = 1 or = 2) so that 1 < 1. Rewriting

x= 1x1+ 2x2+ 3x3

=1x1+ (1 1)

21 1 x2+

21 1 x3

)=1x1+ (1 1)y

where

y=

2

1 1 x2+ 21 1 x3

)

y is a convex combination of two elements x2 andx3 since

2

1 1+

2

1 1=

2+ 3

1 1= 1

and 2 + 3 = 11. Hence y . Thereforex is a convex combination of twoelements x1 and and is also in . Proceeding in this fashion, we can show that everyconvex combination belongs to , that is conv .

1.171 This is precisely analogous to Exercise 1.128. We observe that

1. convis a convex set.

2. if is any convex set containing , then conv.Therefore, convis the smallest convex set containing S.

1.172 Note rst that conv for any set . The converse for convex sets followsfrom Exercise 1.170.

1.173 Assume xconv (1+ 2). Then, there exist numbers1, 2, . . . , and vec-tors x1, x2, . . . , x in 1+ 2 such that

x= 1x1+ 2x2+ + xFor everyx, there exists x

1 1 and x2 2 such that

x = x1 + x

2

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and therefore

x=

=1

x1 +

=1

x2

=x1 + x2

where x1 =

=1 x1 1 and x2 =

=1 x

2 2. Therefore x conv1 +

conv 2.

Conversely, assume that x conv 1+ conv2. Then x = x1 + x2, where

x1 =

=1

1 , x

1 1

x2 =

=1

2 , x

2 2

and

x= x1 + x2 =

=1

1 +

=1

2 conv (1+ 2)

since x1 , x2 1+ 2 for every and .

1.174 The dimension of the input requirement set () is . Its affine hull is.1.175 1. Let

x= 1x1+ 2x2+ . . . + x (1.21)

If >dim +1, the elementsx1, x2, . . . , x are affinely dependent (Exercise1.157 and therefore there exist numbers 1, 2, . . . , , not all zero, such that

(Exercise 1.158)1x1+ 2x2+ . . . + x = 0 (1.22)

and

1+ 2+ . . .+ = 0

2. Combining (1.21) and (1.22)

x= x 0

=

=1

x

=1

x

=

=1

( )x (1.23)

for any .3. Let = min

:> 0

=

We note that

> 0 since > 0 for every .

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If > 0, then / / and therefore 0 If0 then > 0 for every > 0. Therefore 0 for every and

= 0 for = .

Therefore, (1.23) represents x as a convex combination of only 1 points.4. This process can be repeated until x is represented as a convex combination of

at most dim + 1 elements.

1.176 Assume xis not an extreme point of. Then there exists distinct x1 andx2 inS such that

x= x1+ (1 )x2Without loss of generality, assume 1/2 and let y = x2 x. Thenx + y= x2 .Furthermore

x

y= x

x2+ x

= 2x x2= 2(x1+ (1 )x2) x2= 2x1+ (1 2)x2

since 1/2.1.177 1. For anyx = (1, 2) 2, there exists some 1 [0, 1] such that

1 = 1+ (1 1)() = (21 1)In fact, 1 is dened by

1 =1+

2

Therefore (see Figure 1.5) 1

)=1

)+ (1 1)

)

1

)=1

)+ (1 1)

)

Similarly 2 = 2+ (1 2)() where

2 =2+

2

Therefore, for any x 2,

x=

12

)=2

1

)+ (1 2)

1

)

=12

)+ (1 1)2

)

+ 1(1 2)

)+ (1 1)(1 2)

)

=1

)+ 2

)+ 3

)+ 4

)

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(1, c)

(1, -c)

1

x

Figure 1.5: A cube in2

where 0 1 and

1+ 2+ 3+ 4= 12+ (1 1)2+ 1(1 2) + (1 1)(1 2)=12+ 2 12+ 1 12+ 1 1 2+ 12= 1

That is

conv

),

),

),

)

2. (a) For any point (1, 2, . . . , 1, ) which lies on face of the cube, (1, 2, . . . , 1)1 and therefore

(1, 2, . . . , 1) conv, , . . . , ) 1

so that

x conv (, , . . . , , )

Similarly, any point (1, 2, . . . , 1, ) on the opposite face lies in theconvex hull of the points (, , . . . , , ) .

(b) For any other pointx = (1, 2, . . . , ) , let

= +

2

so that

= + (1 )()

Then

x=

12. . .

1

=

12. . .

1

+ (1 )

12. . .

1

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Hence

x conv (, , . . . , )

In other words

conv (, , . . . , )

3. Let denote the set of points of the form (, , . . . , ) . Clearly,every point in is an extreme point of . Conversely, we have shown thatconv. Therefore, no point x can be an extreme point of. is the set of extreme points of.

4. Since is convex, and , conv. Consequently, = conv.1.178 Letx, y belong to is convex. For any [0, 1]

x + (1 )y since convex x + (1 )y / sinceis a face

Thus x+ (1 )y which is convex.1.179 1. Trivial.

2. Let be a collection of faces of and let =

. Choose any x, y .If the line segment between x and y intersects, then it intersects some face which implies that x, y . Therefore, x, y =

.

3. Let be a collection of faces ofand let =

. Choose anyx, y. ifthe line segment between x and y intersects , then it intersects every face which implies that x, y for every . Therefore, x, y =

.

4. Let be the collection of all faces of. This is partially ordered by inclusion.By the previous result, every nonempty subcollection has a least upper bound( ). Hence is a complete lattice (Exercise 1.47).

1.180 Letbe a polytope. Then = conv x1, x2, . . . , x . Note that every extremepoint belongs to x1, x2, . . . , x . Now choose the smallest subset whose convex hullis still , that is delete elements which can be written as convex combinations of otherelements. Suppose the minimal subset is x1, x2, . . . , x . We claim that each ofthese elements is an extreme point of, that is x1, x2, . . . , x = .Assume not, that is assume that xis not an extreme point so that there exists x, y with

x= x + (1 )y with 0< < 1 (1.24)Sincex, y convx1, x2, . . . , x

x=

=1

x y=

=1

x

Substituting in (1.24), we can write x as a convex combination ofx1, x2, . . . , x.

x=

=1

(+ (1 )

x =

=1

x

where

= + (1 )

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Note that 01, so that either < 1 or= 1. We show that both cases leadto a contradiction.

< 1. Then

x= 1

1 1=1

(+ (1 )x

which contradicts the minimality of the setx1, x2, . . . , x. = 1. Then = 0 for every =. That is

+ (1 ) = 0 for every =which implies that = for every = and thereforex = y.

Therefore, ifx1, x2, . . . , x is a minimal spanning set, every point must be an extremepoint.

1.181 Assume to the contrary that one of the vertices is not an extreme point of thesimplex. Without loss of generality, assume this is x

1. Then, there exist distinct

y, zand 0< < 1 such thatx1 = y+ (1 )z (1.25)

Now, sincey, there exist1, 2, . . . , such that

y=

=1

x,

=1

= 1

Similarly, there exist1, 2, . . . , such that

z=

=1

x,

=1

= 1

Substituting in (1.25)

x1 =

=1

x+ (1 )

=1

x

=

=1

(+ (1 )

x

Since

=1

(+ (1 )

=

=1 + (1 )

=1 = 1

x1 =

=1

(+ (1 )x

Subtracting, this implies

0=

=2

(+ (1 )

(x x1)

This establishes that the setx2 x1, x3 x1, . . . , x x1 is linearly dependent andtherefore= x1, x2, . . . , x is affinely dependent (Exercise 1.157). This contradictsthe assumption that is a simplex.

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1.182 Let be the dimension of a convex setin a linear space. Then = dim affand there exists a set x1, x2, . . . , x+1 of affinely independent points in . Dene

= conv x1, x2, . . . , x+1 Then is an -dimensional simplex contained in .

1.183 Letw = ((1), (2), . . . , ()) denote the vector of individual worths andlet denote the surplus to be distributed, that is

= ()

()

> 0 if the game is essential. For each player = 1, 2, . . . , , let

y =w + e

be the outcome in which player receives the entire surplus. (e is theth unit vector.)Note that

=

() + = () =

Eachy is an imputation since () and

=

() + = ()

Therefore y1, y2, . . . , y . Since is convex (why ?),= convy1, y2, . . . , y . Further, for every , the vectors

y y =(e e)are linearly independent. Thereforeis an

1-dimensional simplex in

.

For any x dene

= ()

so that

=() + Sincex is an imputation

0

=

(

()

/= 1

We claim that x =

y

since for each = 1, 2, . . . ,

=

() +

=() + =

Thereforexconvy1, y2, . . . , y = , that is . Since we previously showedthat, we have established that =, which is an 1 dimensional simplex in.

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1

2

1

2

1

2

3. A convex cone1. A non-convex cone 2. A convex set

Figure 1.6: A cone which is not convex, a convex set and a convex cone

1.184 See Figure 1.6.1.185 Letx = (1, 2, . . . , ) belong to+, which means that0 for every . For

every >0

x= (x1, x2, . . . , x)

and 0 for every . Therefore x +.+ is a cone in.1.186 Assume

x + y for every x, yand , + (1.26)

Letting= 0, this implies that

x for every x and +so that is a cone. To show that is convex, let x and y be any two elements in .For any [0, 1], (1.26) implies that

x+ (1 )y

Therefore is convex.

Conversely, assume that is a convex cone. For any , + and x, y

+ x +

+ y

and therefore

x+ y

1.187 Assume satises

1. for every 02. +

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By (1),is a cone. To show that it is convex, let x and y belong to. By (1), xand(1 )y belong to , and therefore x + (1 )y belongs to by (2). is convex.Conversely, assume that is a convex cone. Then

for every

0

Let x and y be any two elements in . Since is convex, = 12 x+ (1 ) 12 yand since it is a cone, 2 = x + y . Therefore

+

1.188 We have to show that is convex cone. By assumption, is convex. To showthatis a cone, let y be any production plan in . By convexity

y= y+ (1 )0 for every 0 1

Repeated use of additivity ensures that

y for every = 1, 2, . . .Combining these two conclusions implies that

y for every 0

1.189 Let denote the set of all superadditive games. Let1, 2 be twosuperadditive games. Then, for all distinct coalitions , with =

1( ) 1() + 1()2( ) 2() + 2()

Adding

(1 + 2)( ) = 1( ) + 2( )1() + 2() + 1() + 2()= (1 + 2)() + (1 + 2)()

so that1 + 2 is superadditive. Similarly, we can show that 1 is superadditive forall +. Henceis a convex cone in.

1.190 Let x belong to

=1 . Then x for every . Since each is a cone,x for every 0 and therefore x

=1 .

Let = 1+ 2 + + and assume x belongs to . Then there existx ,= 1, 2, . . . , such that

x= x1+ x2+ + xFor any 0

x= (x1+ x2+ + x)=x1+ x2+ + x

since x for every .

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1.191 1. Suppose thaty. Then, there exist,2, . . . , 80 such that

y=8

=1

y

and for the rst commodity

y1 =

8=1

1

If y= 0, at least one of the > 0 and hence y1 < 0 since 1 < 0 for =1, 2, . . . , 8.

2. Free disposal requires thaty, y y = y . Consider the productionplan y = (2, 2, 2, 2). Note that y y3 and y y6. Suppose thaty . Then there exist1, 2, . . . , 8 0 such that

y=

8=1

y

For the third commodity (component), we have

41+ 32+ 33+ 34+ 125 26+ 58 =2 (1.27)

and for the fourth commodity

22 13+ 14+ 56+ 107 28 =2 (1.28)

Adding to (1.28) to (1.27) gives

41+ 52+ 23+ 44+ 125+ 36+ 107+ 38 = 4

which is impossible given that 0. Therefore, we conclude that y / .3.

y2 = (7, 9, 3, 2) (8, 13, 3, 1) = y43y1 = (9, 18, 12, 0) (11, 19, 12, 0) =y5

y7 = (8, 5, 0, 10) (8, 6, 4, 10) = 2y62y3= (2, 4, 6, 2) (2, 4, 5, 2) = y8

4.

2y3+ y7= (

2,

4, 6,

2)+ (

8,

5, 0, 10)

= (10, 9, 6, 8)(14, 18, 6, 4) = 2y2

20y3+ 2y7 = 20(1, 2, 3, 1) + 2(8, 5, 0, 10)= (20, 40, 60, 20) + (16, 10, 0, 20)= (36, 50, 60, 0)(45, 90, 60, 0) = 15y1

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5. Eff() = cone y3, y7 .1.192 This is precisely analogous to Exercise 1.128. We observe that

1. coneis a convex cone.

2. if is any convex cone containing , then conv

.

Therefore, cone is the smallest convex cone containing S.

1.193 For any set , cone. If is a convex cone, Exercise 1.186 implies thatcone .

1.194 1. If > = dim cone = dim lin , the elements x1, x2, . . . , x arelinearly dependent and therefore there exist numbers 1, 2, . . . , , not all zero,such that (Exercise 1.134)

1x1+ 2x2+ . . . + x = 0 (1.29)

2. Combining (1.14) and (1.29)

x= x 0

=

=1

x

=1

x

=

=1

( )x (1.30)

for any .3. Let = min

:> 0

=

We note that

> 0 since > 0 for every .

If > 0, then / / and therefore 0. If0 then > 0 for every > 0. Therefore 0 for every and = 0 for = .

Therefore, (1.30) representsx as a nonnegative combination of only 1 points.4. This process can be repeated until x is represented as a convex combination of

at most points.

1.195 1. The affine hull of is parallel to the affine hull of. Therefore

dim = dimaff= dimaff

Since

00

)/aff ,

dimcone = dimaff + 1 = dim + 1

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2. For everyx conv,

x1

) conv and there exist (Exercise 1.194) + 1

points

x

1

) such that

x

1

) conv x11

),

x21

), . . .

x+1

1

) This implies that

x conv x1, x2, . . . , x+1

with x1, x2, . . . , x+1 .1.196 A subsimplex with precisely one distinguished face is completely labeled. Suppose

a subsimplex has more than one distinguished face. This means that it has verticeslabeled 1, 2, . . . , . Since it has + 1 vertices, one of these labels must be repeated(twice). The distinguished faces lie opposite the repeated vertices. There are preciselytwo distinguished faces.

1.197 1. (x, y) = x y 0.2. (x, y) =x y= 0 if and only ifx y= 0, that is x = y.3. Property (3) ensures thatx= x and thereforex y=y x so that

(x, y) =x y=y x= (y, x)

4. For anyz

(x, y) =x y=x z + z y

x

z

+

z

y

=(x, z) + (z, y)

Therefore(x, y) =x y satises the properties required of a metric.1.198 Clearlyx 0 andx = 0 if and only ifx = 0. Thirdly

x= max=1

= max=1

= x

To prove the triangle inequality, we note that for any ,

max(+ ) max + max Therefore

x= max=1

(+ ) max=1

+ max=1

=x + y

1.199 Suppose that producing one unit of good 1 requires two units of good 2 and threeunits of good 3. The production plan is (1, 2, 3) and the average net output,2,is negative. A norm is required to be nonnegative. Moreover, the quantities of inputsand outputs may balance out yielding a zero average. That is, (

=1 )/ = 0 does

not imply that= 0 for all .

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1.200

x y=x y+ y y x y + y y=

x

y

1.201 Using the previous exercisex x x x 0

1.202 First note that each term x+ y by linearity. Similarly,x +y. Fixsome > 0. There exists somex such thatx x< for all x. Similarly,there exists someysuch that y y< for all y. For allmax x, y ,

(x+ y) (x + y)= (x x) + (y y) x x + y y<

Similarly, for every xx x= x x

/2 0 as 0

1.203 Letx be a sequence in + converging to x. For every, there existsy andzsuch thatx = y + z. Sinceis compact, there exists a subsequence zconverging to z. Let y= lim . Then y since is closed. By the previousexercise,y+ z y + z. By assumption, y+ z x so that x = y + z + .+ is closed.

1.204 Yes. Apply Exercise 1.202.

1.205 The th partial sum of the series is

s= x + x + 2x + + 1xMultiplying this equation bygives

s= x + 2x + 3x + + x

Subtracting this equation from the previous one and canceling common terms gives

(1 )s= x x= (1 )xProvided that= 1

s=x x

1 = x

1 x

1 (1.31)

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1.207 The present value of the payments is the th partial sum of the geometricseries + + 2 + 3 + . . . which (using (1.31)) is given by

Present value ==

1 1.208 By Exercise 1.93, there exists an open set 1such that2 =. For every

x 1, there exists an open ball(x) such that(x)and therefore(x)2 = .The collection(x) of open balls is an open cover for 1. Since1 is compact thereexists a nite subcover, that is there exists pointsx1, x2, . . . , x in 1 such that

1

=1

(x)

Furthermore, for every x, there exists such that

(x) = x+

where is the unit ball. Let = min . = is the requir

Foundation of Mathematical Economics Solutions

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