Instant ebooks textbook Mathematical analysis: volume i 1, january 1 2024 edition teo lee peng downl
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Mathematical Analysis: Volume II 1, January 1 2024
A proposition,usuallydenotedby p,isadeclarativesentencethatiseither trueorfalse,butnotboth.
Definition1.2NegationofaProposition
If p isaproposition, ¬p isthe negation of p.Theproposition p istrueif andonlyifthenegation ¬p isfalse.
Fromtwopropositions p and q,wecanapplylogicaloperatorsandobtaina compoundproposition.
Definition1.3ConjunctionofPropositions
If p and q arepropositions, p ∧ q isthe conjunction of p and q,readas"p and q".Theproposition p ∧ q istrueifandonlyifboth p and q aretrue.
Definition1.4DisjunctionofPropositions
If p and q arepropositions, p ∨ q isthe disjunction of p and q,readas"p or q".Theproposition p ∨ q istrueifandonlyifeither p istrueor q istrue.
Definition1.5ImplicationofPropositions
If p and q arepropositions,theproposition p → q isreadas"p implies q". Itisfalseifandonlyif p istruebut q isfalse.
p → q canalsobereadas"if p then q or"p onlyif q".Inmathematics,we usuallywrite p =⇒ q insteadof p → q
Definition1.6DoubleImplication
If p and q arepropositions,theproposition p ←→ q isreadas"p ifand onlyif q".Itistheconjunctionof p → q and q → p.Hence,itistrueifand onlyifboth p and q aretrue,orboth p and q arefalse.
Thestament“p ifandonlyif q”isoftenexpressedas p ⇐⇒ q Twocompoundpropositions p and q aresaidtobelogicallyequivalent,denoted by p ≡ q,providedthat p istrueifandonlyif q istrue.
Whenthedomainsfor x and y areboththesetofrealnumbers,thefirststatement istrue,whilethesecondstatementisfalse.
Foraset A,weusethenotation x ∈ A todenote x isanelementoftheset A; andthenotation x/ ∈ A todenote x isnotanelementof A
Definition1.7EqualSets
Twosets A and B areequaliftheyhavethesameelements.Inlogical expression, A = B ifandonlyif
Definition1.8Subset
If A and B aresets,wesaythat A isa subset of B,denotedby A ⊂ B, ifeveryelementof A isanelementof B.Inlogicalexpression, A ⊂ B meansthat
x ∈ A =⇒ x ∈ B.
1. ¬ (∀xP (x)) ≡∃x ¬P (x)
2. ¬ (∃xP (x)) ≡∀x ¬P (x)
Chapter1.TheRealNumbers4
When A isasubsetof B,wewillalsosaythat A iscontainedin B,or B contains A.
Wesaythat A isa propersubset of B if A isasubsetof B and A = B.In sometextbooks,thesymbol"⊆"isusedtodenotesubset,andthesymbol"⊂" isreservedforpropersubset.Inthisbook,wewillnotmakesuchadistinction. Wheneverwewrite A ⊂ B,itmeans A isasubsetof B,notnecessaryaproper subset.
If A and B aresets,the union of A and B istheset A ∪ B whichcontains allelementsthatareeitherin A orin B.Inlogicalexpression,
Definition1.10IntersectionofSets
If A and B aresets,the intersection of A and B istheset A ∩ B which containsallelementsthatareinboth A and B.Inlogicalexpression,
Definition1.11DifferenceofSets
If A and B aresets,thedifferenceof A and B istheset A \ B which containsallelementsthatarein A andnotin B.Inlogicalexpression,
Definition1.12ComplementofaSet
If A isasetthatiscontainedinauniversalset U ,the complement of A in U istheset AC whichcontainsallelementsthatarein U butnotin A.In logicalexpression,
Chapter1.TheRealNumbers5
Sinceauniversalsetcanvaryfromcontexttocontext,wewillusuallyavoid usingthenotation AC anduse U \ A insteadforthecomplementof A in U .The advantageofusingthenotation AC isthatDeMorgan’slawtakesamoresuccint form.
Proposition1.4DeMorgan’sLawforSets
If A and B aresetsinauniversalset U ,and AC and BC aretheir complementsin U ,then
1. (A ∪ B)C = AC ∩ BC
2. (A ∩ B)C = AC ∪ BC
Definition1.13Functions
When A and B aresets,a function f from A to B,denotedby f : A → B, isacorrespondencethatassignseveryelementof A auniqueelementin B.
If a isin A,the image of a underthefunction f isdenotedby f (a),andit isanelementof B.
A iscalledthe domain of f ,and B iscalledthe codomain of f
Definition1.14ImageofaSet
If f : A → B isafunctionand C isasubsetof A,theimageof C under f istheset
f (C)= {f (c) | c ∈ C}
f (A) iscalledtherangeof f
Definition1.15PreimageofaSet
If f : A → B isafunctionand D isasubsetof B,thepreimageof D under f istheset
f 1(D)= {a ∈ A | f (a) ∈ D} .
Noticethat f 1(D) isanotation,itdoesnotmeanthatthefunction f hasan inverse.
Wesaythatafunction f : A → B isan injection,orthefunction f : A → B is injective,orthefunction f : A → B is one-to-one,ifnopairof distinctelementsof A aremappedtothesameelementof B.Namely,
Wesaythatafunction f : A → B isa surjection,orthefunction f : A → B is surjective,orthefunction f : A → B is onto,ifeveryelementof B istheimageofsomeelementin A.Namely,
Equivalently, f : A → B issurjectiveiftherangeof f is B.Namely, f (A)= B
Definition1.18Bijection
Wesaythatafunction f : A → B isa bijection,orthefunction f : A → B isbijective,ifitisbothinjectiveandsurjective.
Abijectionisalsocalleda one-to-onecorrespondence. Finally,wewouldliketomakearemarkaboutsomenotations.If f : A → B isafunctionwithdomain A,and C isasubsetof A,therestrictionof f to C is thefunction f |C : C → B definedby f |C (c)= f (c) forall c ∈ C.Whenno confusionarises,wewilloftendenotethisfunctionsimplyas f : C → B
Thesetof naturalnumbers N isthesetthatcontainsthecountingnumbers, 1,2,3 ,whicharealsocalledpositiveintegers.
N isaninductiveset.Thenumber1isthesmallestelementofthisset.If n is anaturalnumber,then n +1 isalsoanaturalnumber. Thenumber0correspondstonothing.
Foreverypositiveinteger n, n isanumberwhichproduces0whenaddsto n.Thisnumber n iscalledthenegativeof n,ortheadditiveinverseof n 1, 2, 3, ...,arecallednegativeintegers.
Definition1.20Integers
Thesetof integers Z isthesetthatcontainsallpositiveintegers,negative integersand0.
Thesetofcomplexnumbers C isthesetthatcontainsallnumbersofthe form a + ib,where a and b arerealnumbers,and i isthepurelyimaginary numbersuchthat i2 = 1.Itcontainsthesetofrealnumbers R asasubset. Additionandmultiplicationcanbeextendedtothesetofcomplexnumbers.These twooperationsoncomplexnumbersalsosatisfyallthepropertieslistedabove. Nevertheless,weshallfocusonthesetofrealnumbersinthiscourse.
Givenarealnumber x,the absolutevalue of x,denotedby |x|,isdefined tobethenonnegativenumber
Inparticular, |− x
Forexample, |
Theabsolutevalue |x| canbeinterpretedasthedistancebetweenthenumber x andthenumber 0 onthenumberline.Foranytworealnumbers x and y, |x y| isthedistancebetween x and y.Hence,theabsolutevaluecanbeusedtoexpress aninterval.
Chapter1.TheRealNumbers12
IntervalsDefinedbyAbsoluteValues
Let a bearealnumber.
1. If r isapositivenumber, |x a| <r ⇐⇒−r<x a<r ⇐⇒ x ∈ (a r,a + r).
|x 5|≤ 2 implies 3 ≤ x ≤ 7.Thismeansthat x ispositive.The inequality x ≥ 3 thenimpliesthat x2 ≥ 9,andtheinequality x ≤ 7 implies that x2 ≤ 49.Therefore, 9 ≤ x 2 ≤ 49.
Finally,wehavetheusefulCauchy’sinequality.
Chapter1.TheRealNumbers14
Proposition1.9Cauchy’sInequality
Thisisjustaconsequenceof
AnimmediateconsequenceofCauchy’sinequalityisthearithmeticmeangeometricmeaninequality.Foranynonnegativenumbers a and b,thegeometric meanof a and b is √ab,andthearithmeticmeanis
Proposition1.10
Chapter1.TheRealNumbers15
Exercises1.2
Question1
Useinductiontoshowthatforanypositiveinteger n, n! ≥ 2n 1
Let S beanonemptysubsetofrealnumbersthatisboundedabove,and let US bethesetofupperboundsof S.Then US isanonemptysetthatis boundedbelow.If US hasasmallestelement u,wesaythat u isthe least upperbound or supremum of S,anddenoteitby u =sup S.
Let S beanonemptysubsetofrealnumbers.Then S hasamaximumifand onlyif S isboundedaboveand sup S isin S
Onenaturalquestiontoaskis,if S isanonemptysubsetofrealnumbersthat isboundedabove,does S necessarilyhavealeastupperbound.Thecompleteness axiomassertsthatthisistrue.
CompletenessAxiom
If S isanonemptysubsetofrealnumbersthatisboundedabove,then S hasaleastupperbound.
tobeasetthatcontainsthesetofrationalnumbers,satisfyingallproperties ofadditionandmultiplicationoperations,aswellasthecompletenessaxiom. However,thisisatediousconstructionandwilldriftustoofar. Toshowthatthecompletenessaxiomisnotcompletelytrivial,weshowin Example 1.6 thatifweonlyconsiderthesetofrationalnumbers,wecanfinda subsetofrationalnumbers A thatisboundedabovebutdoesnothavealeastupper boundinthesetofrationalnumbers.Welookatthefollowingexamplefirst.
Example1.5
Definethesetofrealnumbers S by
= x ∈ R | x 2 < 2
Showthat S isnonemptyandisboundedabove.Concludethattheset A = x ∈ Q | x 2 < 2 isalsononemptyandisboundedabovebyarationalnumber.
Solution
Thenumber1isin S,andso S isnonempty.Forany x ∈ S, x2 < 2 < 4, andhence x< 2.Thisshowsthat S isboundedaboveby2.Since1and2 arerationalnumbers,thesamereasoningshowsthattheset A isnonempty andisboundedabovebyarationalnumber.
Example1.6
Considertheset
= x ∈ Q | x 2 < 2
ByExample 1.5, A isanonemptysubsetofrationalnumbersthatis boundedaboveby2.Let UA bethesetofupperboundsof A in Q.Namely,
Showthat UA doesnothaveasmallestelement.
Chapter1.TheRealNumbers21
Solution
Weuseproofbycontradiction.Assumethat UA hasasmallestelement c1, whichisanupperboundof A thatissmallerthanorequaltoanyupper boundof A.Thenforany x ∈ A, x 2 ≤ c1.
Since 1 isin A, c1 isapositiverationalnumber.Hence,therearepoitive integers p and q suchthat c1 = p q .
Sincetherearenorationalnumberswhosesquareis2,wemusthaveeither c2 1 < 2 or c2 1 > 2
Definethepositiverationalnumber c2 by
Noticethat
and
Case1: c2 1 < 2
Inthiscase, p2 < 2q2.Itfollowsthat c1 <c2 and c2 2 < 2.Butthen c1 and c2 arebothin A,and c2 isanelementin A thatislargerthan c1,which contradictsto c1 isanupperboundof A.Hence,wecannothave c2 1 < 2.
Chapter1.TheRealNumbers22
Case2: c2 1 > 2.
Inthiscase, p2 > 2q2.Itfollowsthat c1 >c2 and c2 2 > 2.Since c2 2 > 2,we findthatforany x ∈ A,
Thus,
Inparticular, c2 isalsoanupperboundof A.Namely, c2 isin UA.Butthen c1 and c2 arebothin UA and c1 >c2.Thiscontradictsto c1 isthesmallest elementin UA.Hence,wecannothave c2 1 > 2
SincebothCase1andCase2leadtocontradictions,weconcludethat UA doesnothaveasmallestelement.
Inthesolutionabove,theconstructionofthepositiverationalnumber c2 seems abitadhoc.Infact,wecandefine c2 by c2 = mp +2nq np + mq foranypositiveintegers m and n with m2 > 2n2.Thentheproofstillworks.
Usecompletenessaxiomtoshowthatthereisapositiverealnumber c such that
Definethesetofrealnumbers S by S
Example 1.5 assertsthat S isanonemptysubsetofrealnumbersthatis boundedabove.Completenessaxiomassertsthat S hasaleastupperbound c.
Chapter1.TheRealNumbers23
Since 1 isin S, c ≥ 1.Wearegoingtoprovethat c2 =2 usingproofby contradiction.If c2 =2,then c2 < 2 or c2 > 2.
Case1: c2 < 2
Let d =2 c2.Then 0 <d ≤ 1.Definethenumber c1 by
1 = c + d 4c .
Then c1 >c,and
.
Thisimpliesthat c1 isanelementof S thatislargerthan c,whichcontradicts to c isanupperboundof S
Case2: c2 > 2.
Let d = c2 2.Then d> 0.Definethenumber c1 by c1 = c d 2c
Then c1 <c,and
Thisimpliesthat c1 isanupperboundof S thatissmallerthan c,which contradictsto c istheleastupperboundof S. Sinceweobtainacontradictionif c2 =2,wemusthave c2 =2.
Infact,thecompletenessaxiomcanbeusedtoshowthatforanypositivereal number a,thereisapositiverealnumber c suchthat c 2 = a.
Wedenotethisnumber c as √a,calledthepositivesquarerootof a.Thenumber b = √a isanotherrealnumbersuchthat b2 = a
Moregenerally,if n isapositiveinteger, a isapositiverealnumber,thenthere isapositiverealnumber c suchthat cn = a.Wedenotethisnumber c by
c = n √a,
Chapter1.TheRealNumbers24
calledthepositive nth-rootof a Usingtheinterplaybetweenasetanditsnegative,wecandefinethegreatest lowerboundofasetthatisboundedbelow.
Definition1.27GreatestLowerBound
Let S beanonemptysubsetofrealnumbersthatisboundedbelow,and let LS bethesetoflowerboundsof S.Then LS isanonemptysetthatis boundedabove.If LS hasalargestelement ℓ,wesaythat ℓ isthe greatest lowerbound or infimum of S,anddenoteitby
=inf S.
Fromthecompletenessaxiom,wehavethefollowing.
Theorem1.13
If S isanonemptysubsetofrealnumbersthatisboundedbelow,then S hasagreatestlowerbound.
Foranonemptyset S thatisbounded,ithasaleastupperbound sup S anda greatestlowerbound inf S.Thefollowingisquiteobvious.
Proposition1.14
If S isaboundednonemptysubsetofrealnumbers,ithasaleastupper bound sup S andagreatestlowerbound inf S.Moreover, inf S ≤ sup S, and inf S =sup S ifandonlyif S containsexactlyoneelement.