Solutions Manual Foundations of Mathematical Economics By Michael Carter

Solutions Manual
Foundations of Mathematical Economics

Michael Carter

, ⃝ caaa2001a Michaela Carter
Solutionsa fora Foundationsa ofa Mathematicala Economics Allarightsareserved




Chapter 1: Sets and Spaces
a a a a




1.1
{a1,a3,a5,a7a. . . a}aora {a�a ∈ a�a :a �a isa odda}
1.2 Everya � ∈ �a alsoa belongsa toa �.a Everya � ∈
�a alsoa belongsa toa �.a Hencea �,a�a haveapreciselya thea samea elements.
1.3 Examplesa ofa finitea setsa are
∙ thea lettersa ofa thea alphabeta {aA,a B,a C,a . . . a ,a Za}
∙ thea seta ofa consumersa ina ana economy
∙ thea seta ofa goodsa ina ana economy
∙ thea seta ofa playersaina aa game
.aExamplesa ofa infinitea setsa are
∙ thea reala numbersa ℜ
∙ thea naturala numbersa �
∙ thea seta ofa alla possiblea colors
∙ thea seta ofa possiblea pricesa ofa coppera ona thea worlda market
∙ thea seta ofa possiblea temperaturesa ofa liquida water.
1.4a �a =a {a1,a2,a3,a4,a5,a6a},a �a =a {a2,a4,a6a}.
1.5 Thea playera seta isa �a =a {aJenny,aChrisa} . aTheira actiona spacesa are
��a =a{aRock,aScissors,aPapera} �a =a Jenny,aChris
1.6 Thea seta ofa playersa isa �a =a 1,
{ a2 , . .. , a�a }. a Thea strategya spacea ofa eacha playera isa thea seta
ofa feasiblea outputs
��a =a {a��a ∈ aℜ +a :a ��a ≤ a��a}
wherea ��aaisaathea outputa ofa dama �.
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1.7 Thea playera seta isa �a =a {1,a2,a3}. aTherea area 2 a =a 8a coalitions,a namely
� (�a)a =a {∅ ,a{1},a{2},a{3},a{1,a2},a{1,a3},a{2,a3},a{1,a2,a3}}
10
Therea area 2 a coalitionsa ina aa tena playera game.
1.8aa Assumeaathataa�aa∈ a(�a ∪ a�a)� .aaaThataaisaa�aa∈/aa�a ∪ a�a.aaaThisaaimpliesaa�aa∈/aa�aaandaa�aa∈/aa�a,ao
ra�a∈ a��aanda �a∈ a�a�.a Consequently,a �a∈ a��a∩ a�a�.a Conversely,a assumea �a∈ a��a∩a�a�.aThisaaim
pliesaathataa�a ∈ a� �aaandaa�a ∈ a�a� .aaaConsequentlyaa�a∈/aa�aaandaa�a∈/aa�aa andaatherefore
�∈/a �a ∪ a�a. aThisa impliesaathata �a ∈ a(�a∪ a�a)� . aThea othera identitya isa proveda similarly.
1.9
∪
�a =a�
�∈�
∩
�a =a∅
�∈�


1

, ⃝ caaa2001a Michaela Carter
Solutionsa fora Foundationsa ofa Mathematicala Economics Allarightsareserved


�2
1




�1
-1 0 1




-1
2 2
Figurea 1.1:a Thea relationa {a(�,a�)a :a � a +a � a =a 1a}


1.10a Thea samplea spacea ofa aa singlea coina tossa isa�,{a�a .a The
} a seta ofa possiblea outcomesa inathr
eea tossesa isa thea product
{
{�,a�a} ×a{�,a�a} ×a{�,a�a}a=a (�,a�,a�),a(�,a�,a�a),a(�,a�a,a�),
}
(�,a�a,a�a),a(�,a�,a�),a(�,a�,a�a),a(�,a�,a�),a(�,a�,a�a)


Aa typicala outcomea isa thea sequencea (�,a�,a�a)a ofa twoa headsa followeda bya aa tail.
1.11

�a ∩aℜ+�a =a {0}

wherea0a =a(0,a0 , . . . a,a0)aisatheaproductionaplanausinganoainputsaandaproducinganoaoutputs.a
Toa seea this,a firsta notea thata 0a isa aa feasiblea productiona plan.a Therefore,a 0a ∈ a�a.a Also,
0a ∈ aℜ �+a anda thereforea 0a ∈ a�a ∩aℜ �a . +
Toashowathatathereaisanoaotherafeasibleaproductionaplanainaaaaa�ℜa,a+weaassumeatheacontrary.aTh
atais,aweaassumeathereaisasomeafeasibleaproductionaplanayaaaaaaaa∈�aaaaaa
aℜ a +
∖0a{aaa }.aaThisaimpliesathea
existenceaofaaaplanaproducingaaapositiveaoutputawithanoainputs.aThisatechnologicalainfeas
ible,a soa thata �a∈/a �a.
1.12 1. aaLetaaxa ∈ a�a(�). aaThisaaimpliesaathataa(�,a− x)a ∈ a�a. aaLetaax′a ≥ ax.aa Thenaa(�,a− x′ )a ≤
(�,a− x)a anda freea disposabilitya impliesaathata (�,a− x′ )a ∈ a�a. aThereforea x′a∈ a�a(�).
2.aa Againaa assumeaa xaa ∈ a �a(�).aaaaThisaa impliesaa thataa (�,a− x)aa ∈ a �a.aaaaByaa freeaa disposal,a(
� ′ ,a− x)a ∈ a�aa fora everya � ′a≤ a� ,a whicha impliesaathata xa ∈ a�a(� ′ ).aa�a(� ′ )a ⊇ a�a(�).
1.13 Thea domaina ofa “<”a isa {1,a2}a=a �a anda thea rangea isa {2,a3}a⫋a �a.
1.14 Figurea1.1.
1.15 Thea relationa “isa strictlya highera than”a isa transitive,a antisymmetrica anda asymmetric
.aIta isa nota complete,a reflexivea ora symmetric.




2

, ⃝ caaa2001a Michaela Carter
Solutionsa fora Foundationsa ofa Mathematicala Economics Allarightsareserved


1.16 Thea followinga tablea listsa theira respectivea properties.
< ≤√aa √=
reflexive ×aa
transitive √ √aa √
symmetric √aa √
×aa
√
asymmetric
anti-symmetric √aa × aa ×
√ √
√a √a
complete ×
Notea thata thea propertiesa ofa symmetrya anda anti-symmetrya area nota mutuallya exclusive.
1.17 Letabe ∼ aanaequivalencearelationaofaaaseta�a=a∕.a a ∅Thatais,athearelationaisareflexive,
∼ asym

metricaandatransitive.aWeafirstashowathataeverya�a�abelongs ∈ atoasomeaequivalence aclass.

a Leta �a bea anya elementa ina � a anda leta (�)a be
∼a thea classa ofa elementsa equivalenta to
�,athata is
∼(�)a ≡a{a�a ∈ a�a :a �a ∼ a�a}
Since ∼ isa reflexive,a � ∼ �aandasoa� ∈ a∼ (�).a Everya � ∈
�a belongsa toa somea equivalenceaclassa anda therefore
∪
�a = ∼(�)
�∈�

Next,a wea showa thata thea equivalencea classesa area eithera disjointa ora identical,aathata is
∼(�)a ∕=a ∼(�)a ifa anda onlya ifa f∼(�)a∩a∼ (�) a=a ∅ .
First,a assumea ∼(�)a∩a∼ (�) a=a ∅ . aThena �a∈ a∼ (�)a butaa�∈
�/ ∼( ). aThereforea ∼(�)a ∕=a ∼(�).
Conversely,aaassumeaa∼(�)a ∩a∼ (�)aa∕=aa∅ aandaaletaa�aa∈ a∼(�)a ∩a∼ (�).aaaThenaa�aa∼ a�aaandaabyasymmet
rya �a ∼ a�.aaaAlsoa �a ∼ a�aandasoa bya transitivitya�a ∼ a�.aaaLeta�a bea anyaelementainaa∼(�)aasoaa
thataa�aa∼ a�.aaaAgainaabyaatransitivityaa�aa∼ a�aaandaathereforeaa�aa∈ a∼(�).aaaHence
∼(�)a ⊆ a∼ (�). aSimilaraareasoninga impliesaathata ∼(�)a ⊆ a∼ (�). aThereforea ∼(�) a=a ∼(�).
Wea concludea thata thea equivalencea classesa partitiona �.
1.18 Theasetaofaproperacoalitionsaisa notaaapartitionaofathea setaofaplayers,asinceaanya playe
racana belonga toa morea thana onea coalition.aFora example,a playera1a belongsa toa thea coalition
s
{1},a {1,a2}aanda soa on.
1.19

�a ≻a�a =⇒ a �a ≿a �a anda �a ∕≿a �
�a ∼ a�a =⇒ a �a ≿a �a anda �a ≿a �
Transitivitya ofa ≿aimpliesa �a≿a� . aWea needa toa showa thata �a∕≿a� . aAssumea otherwise,a thatais
a assumea �a ≿a �a Thisa impliesa �a ∼a�a anda bya transitivitya �a ∼a�.a Buta thisa impliesa that

�a ≿a�a whicha contradictsa thea assumptiona thata �a ≻a� . a Thereforea wea concludea thata �a ∕≿a �
anda thereforea �a ≻a� . aThea othera resulta isa proveda ina similara fashion.
1.20 asymmetrica Assumea �a ≻a�.

�a ≻a�a =⇒ a �a ∕≿a�
while

�a ≻a�a =⇒ a �a ≿a �
Therefore
�a ≻a�a =⇒ a �a ∕≻a�
3