Solutions Manual Foundations of Mathematical Economics By Michael Carter

Solutions Manualnd



Foundations of Mathematical
nd nd nd



Economics
nd




Michael ndCarter nd

, ⃝ c nd nd nd2001 n d Michael
Solutions n d for n d Foundations n d of n d Mathematical n d CarterAll ndrights
n d Economics ndreserved




Chapter 1: nd n d Sets and Spaces
nd nd




1.1
{nd1, nd3, nd5, nd7 nd. . . nd} ndor n d {nd�nd ∈ nd� n d : n d �n d is n d odd nd}
1.2 Every n d � ∈ � n d also n d belongs n d to n d �. n d ∈Every n d � � n d also n d belongs n d to
n d �. n d Hence n d �, nd� n d havendprecisely n d the n d same n d elements.


1.3 Examples n d of n d finite n d sets n d are
∙ the n d letters n d of n d the n d alphabet n d {ndA, n d B, n d C, n d . . . nd , n d Z nd}
∙ the n d set n d of n d consumers n d in n d an n d economy
∙ the n d set n d of n d goods n d in n d an n d economy
∙ the nd set nd of nd players ndin nd a
nd game.ndExamples n d of n d infinite
nd sets n d are
∙ the n d real nd numbers n d ℜ
∙ the n d natural n d numbers n d �
∙ the nd set nd of nd all nd possible nd colors
∙ the n d set n d of n d possible n d prices n d of n d copper n d on n d the n d world n d market
∙ the n d set n d of n d possible n d temperatures n d of n d liquid n d water.
1.4 nd �n d = nd {nd1, nd2, nd3, nd4, nd5, nd6 nd}, nd �n d = nd {nd2, nd4, nd6 nd}.
1.5 The n d player n d set n d is n d � n d = n d {ndJenny, ndChris nd} . ndTheir n d action n d spaces n d are
�� n d = nd{ndRock, ndScissors, ndPaper nd} � n d = nd Jenny, ndChris
1.6 The n d set n d of n d players n d is{n d � n d = n d 1,
} nd2 , . . ., nd� n d . ndThe n d strategy n d space n d of
n d each n d player n d is n d the n d set ndof n d feasible n d outputs


�� n d = nd {nd�� n d ∈ ndℜ + n d : nd �� n d ≤ nd��nd}
where n d �� ndndis ndndthe n d output n d of n d dam n d �.
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1.7 The n d player n d set n d is n d � n d = n d {1, nd2, nd3}. ndThere n d are n d 2 nd = n d 8 n d coalitions, n d namely
� (�nd) n d = n d {∅ , nd{1}, nd{2}, nd{3}, nd{1, nd2}, nd{1, nd3}, nd{2, nd3}, nd{1, nd2, nd3}}
10
There n d are n d 2 nd coalitions n d in n d a n d ten n d player n d game.
1.8 nd nd Assume nd ndthat nd nd� nd nd∈ nd(� n d ∪ nd�nd)�. nd nd ndThat nd ndis nd nd� nd nd∈/ nd nd� n d ∪ nd�nd. nd nd ndThis nd
� �
ndimplies nd nd� nd nd∈/ nd nd� nd ndand nd nd� nd nd∈/ nd nd�nd, ndor nd�nd∈ nd� nd and n d �nd∈ nd�nd . n d Consequently,
� � � �
n d �nd∈ nd� nd∩ nd�nd . n d Conversely, n d assume n d �nd∈ nd� nd∩ nd�nd . ndThis nd ndimplies nd ndthat nd nd�
� �
n d ∈ nd� nd ndand nd nd� n d ∈ nd�nd . nd ndndConsequently nd nd�nd∈/ nd nd�nd ndand nd nd�nd∈/ nd nd�nd nd and nd ndtherefore

�∈/ n d �nd∪ nd�nd. ndThis n d implies ndndthat n d �nd ∈ nd(�nd∪ nd�nd)�. ndThe n d other n d identity n d is n d proved n d similarly.
1.9
∪
�nd = nd�
�∈�
∩
�n d = nd∅
�∈�


1

, ⃝ c nd nd nd2001 n d Michael
Solutions n d for n d Foundations n d of n d Mathematical n d CarterAll ndrights
n d Economics ndreserved



�2
1




�1
-1 0 1




-1
2 2
Figure n d 1.1: nd The n d relation n d {nd(�, nd�) nd : n d � nd + nd � nd = n d 1 nd}


{ n d toss
1.10 n d The n d sample n d space n d of n d a n d single n d coin } n d is nd�, nd� n d . nd The n d set n d of
n d possible n d outcomes n d inndthree n d tosses n d is n d the n d product

{
{�, nd�nd} ×nd{�, nd�nd}×nd{�, nd�nd} nd= n d (�, nd�, nd�), nd(�, nd�, nd�nd), nd(�, nd�nd, nd�),
}
(�, nd�nd, nd�nd), nd(�, nd�, nd�), nd(�, nd�, nd�nd), nd(�, nd�, nd�), nd(�, nd�, nd�nd)


A n d typical n d outcome n d is n d the n d sequence n d (�, nd�, nd�nd) n d of n d two n d heads n d followed n d by n d a n d tail.
1.11
�nd
� n d ∩ndℜ + = n d {0}

where nd0 nd = nd(0, nd0 , . . . nd, nd0) ndis ndthe ndproduction ndplan ndusing ndno ndinputs ndand ndproducing
ndno ndoutputs. ndTo n d see n d this, n d first n d note n d that n d 0 n d is n d a n d feasible n d production

n d plan. n d Therefore, n d 0 n d ∈ nd�nd. n d Also,

0 n d ∈ ndℜ+� n d and n d therefore n d 0 n d ∈ nd�+ n d ∩ ndℜ
�nd .


To ndshow ndthat ndthere ndis ndno ndother ndfeasible ndproductionℜ nd+plan ndin nd nd nd nd nd� nd, ndwe
ndassume ndthe ndcontrary. ndThat ndis, ndwe ndassume ndthere ndis ndsome ∈ ndℜ nd d ∖ nd{
+nfeasible ndproduction
�
ndplan ndy nd nd nd nd nd nd nd nd nd nd nd nd nd nd0 nd nd. nd ndThis ndimplies ndthe } existence ndof nda ndplan
n d nd

ndproducing nda ndpositive ndoutput ndwith ndno ndinputs. ndThis ndtechnological ndinfeasible, n d so

n d that n d �nd∈/ n d �nd.


1.12 1. nd ndLet ndndx n d ∈ nd�nd(�). nd ndThis ndndimplies ndndthat ndnd(�, nd− x) n d ∈ nd�nd. nd ndLet ndndx′ nd ≥ ndx. nd nd Then ndnd(�, nd− x′ ) n d ≤
(�, nd− x) n d and n d free n d disposability n d implies ndndthat n d (�, nd− x′ ) nd ∈ nd�nd. ndTherefore n d x′ nd∈ nd�nd(�).
2. nd nd Again nd ndassume nd ndx nd n d ∈ nd �nd(�). nd nd nd ndThis nd nd implies nd nd that nd nd (�, nd− x) nd n d ∈
nd �nd. nd nd nd ndBy nd nd free nd nd disposal, nd(�′ , nd− x) nd ∈ nd�nd n d for n d every n d �′ nd≤ nd�, n d which

n d implies ndndthat n d x n d ∈ nd�nd(�′ ). nd nd�nd(�′ ) nd ⊇ nd�nd(�).


1.13 The n d domain n d of n d “<” n d is n d {1, nd2}nd= nd � n d and n d the n d range n d is n d {2, nd3}nd⫋ nd �nd.
1.14 Figure nd1.1.
1.15 The n d relation n d “is n d strictly n d higher n d than” n d is n d transitive, n d antisymmetric
n d and n d asymmetric.ndIt n d is n d not n d complete, n d reflexive n d or n d symmetric.




2

, ⃝ c nd nd nd2001 n d Michael
Solutions n d for n d Foundations n d of n d Mathematical n d CarterAll ndrights
n d Economics ndreserved


1.16 The n d following n d table n d lists n d their n d respective n d properties.
< ≤√ nd n d=√
reflexive ×nd n d
transitive √ √ nd n d √
symmetric √ nd n d √
×nd n d
√
asymmetric
anti-symmetric √ nd n d × √
nd n d ×
√
√ n d √ n d
complete ×
Note n d that n d the n d properties n d of n d symmetry n d and n d anti-symmetry n d are n d not n d mutually n d exclusive.
1.17 Let nd∼be ndan ndequivalence ndrelation ndof nda nd∕set �nd= nd. n d That ndis, nd∼the ndrelation
n dnd∅

ndis ndreflexive, ndsymmetric ndand ndtransitive. ndWe ndfirst ∈ ndshow ndthat ndevery nd� nd�
ndbelongs ndto ndsome ndequivalence ndclass. n∼d Let n d � n d be n d any n d element n d in n d � n d and

n d let n d (�) n d be n d the n d class n d of n d elements n d equivalent n d to

�, ndthat nd is
∼(�) n d ≡ nd{nd�n d ∈ nd� n d : n d �n d ∼ nd�nd}
Since ∼ is n d reflexive, n d∼��ndand ndso nd∈�nd∼ (�). n d Every n ∈d � � n d belongs n d to n d some
n d equivalencendclass n d and n d therefore
∪
�n d = ∼(�)
�∈�

Next, nd we n d show n d that n d the equivalence n d classes
n d n d are n d either n d disjoint n d or
n d identical, nd nd that n d is

∼(�) nd ∕= n d ∼(�) n d if n d and n d only n d if n d f∼(�) nd∩nd∼ (�) nd= n d ∅ .
First, n d assume n d ∼(�) nd∩nd∼ (�) nd= nd ∅ . ndThen n d �nd∈ nd∼ (�) n d but ndnd�∈
�
/ ∼( ). ndTherefore n d ∼(�) nd ∕= nd ∼(�).
Conversely, nd ndassume nd nd∼(�) n d ∩ nd∼ (�) nd nd∕= nd nd∅ ndand nd ndlet nd nd�nd nd∈ nd∼(�) n d ∩ nd∼ (�). nd nd ndThen nd
nd� nd nd∼ nd� nd ndand nd ndb yndsymmetry n d � n d ∼ nd�. nd nd ndAlso n d � n d ∼ nd� ndand ndso n d by
n d transitivity nd� n d ∼ nd�. nd nd ndLet nd� n d be n d any ndelement ndin nd nd∼(�) nd ndso nd ndthat nd nd� nd nd∼

nd�. nd nd ndAgain nd ndby nd ndtransitivity nd nd�nd nd∼ nd�nd ndand nd ndtherefore nd nd�nd nd∈ nd∼(�). nd nd ndHence

∼(�) nd ⊆ nd∼ (�). ndSimilar ndndreasoning n d implies ndndthat n d ∼(�) nd ⊆ nd∼ (�). ndTherefore n d ∼(�) nd= n d ∼(�).
We n d conclude nd that nd the n d equivalence nd classes nd partition nd �.
1.18 The ndset ndof ndproper ndcoalitions ndis nd not nda ndpartition ndof ndthe nd set ndof ndplayers,
ndsince nd any nd playerndcan nd belong nd to n d more nd than nd one n d coalition. ndFor nd example,

nd player nd 1 n d belongs nd to nd the nd coalitions

{1}, n d {1, nd2} ndand n d so n d on.
1.19

�n d ≻ nd� n d =⇒ nd �n d ≿ nd � n d and n d � n d ∕≿ nd �
� n d ∼ nd� n d =⇒ nd � n d ≿ nd � n d and n d � n d ≿ nd �
Transitivity nd of nd ≿ ndimplies n d �nd≿ nd�. ndWe n d need nd to n d show nd that nd �nd∕≿ nd�. ndAssume
n d otherwise, nd thatndis n d assume n d � n d ≿ nd � n d This n d implies n d � n d ∼ nd� n d and n d by
n d transitivity n d � n d ∼ nd�. n d But n d this n d implies n d that

� n d ≿ nd� n d which n d contradicts n d the n d assumption n d that n d �n d ≻ nd�. n d Therefore n d we n d conclude n d that n d � n d ∕≿ nd �
and n d therefore n d �nd ≻nd�. ndThe n d other n d result n d is n d proved n d in n d similar n d fashion.
1.20 asymmetric n d Assume nd �n d ≻ nd�.

Therefore
while

3