CHAPTER1 PRELIMINARIES
Wesuggestthatthischapterbetreatedasreviewandcoveredquickly,without detailedclassroomdiscussion.Foronereason,manyoftheseideaswillbealready familiartothestudents—atleastinformally.Further,webelievethat,inpractice, thosenotionsofimportancearebestlearnedinthearenaofrealanalysis,where theiruseandsignificancearemoreapparent.Dwellingontheformalaspectof setsandfunctionsdoesnotcontributeverygreatlytothestudents’understanding ofrealanalysis.
Ifthestudentshavealreadystudiedabstractalgebra,numbertheoryorcombinatorics,theyshouldbefamiliarwiththeuseofmathematicalinduction.Ifnot, thensometimeshouldbespentonmathematicalinduction.
Thethirdsectiondealswithfinite,infiniteandcountablesets.Thesenotions areimportantandshouldbebrieflyintroduced.However,webelievethatitis notnecessarytogointotheproofsoftheseresultsatthistime.
Section1.1
Studentsareusuallyfamiliarwiththenotationsandoperationsofsetalgebra, sothatabriefreviewisquiteadequate.Oneitemthatshouldbementionedis thattwosets A and B areoftenprovedtobeequalbyshowingthat:(i)if x ∈ A, then x ∈ B ,and(ii)if x ∈ B ,then x ∈ A.Thistypeofelement-wiseargumentis verycommoninrealanalysis,sincemanipulationswithsetidentitiesisoftennot suitablewhenthesetsarecomplicated.
Studentsareoftennotfamiliarwiththenotionsoffunctionsthatareinjective (=one-one)orsurjective(=onto).
2.Thesetsareequalto(a)
3.If A ⊆ B ,then x ∈ A implies x ∈ B ,whence x ∈ A ∩ B ,sothat A ⊆ A ∩ B ⊆ A
Thus,if A ⊆ B ,then A = A ∩ B . Conversely,if A = A ∩ B ,then x ∈ A implies x ∈ A ∩ B ,whence x ∈ B
Thusif A = A ∩ B ,then A ⊆ B
4.If x isin A \ (B ∩ C ),then x isin A but x/ ∈ B ∩ C ,sothat x ∈ A and x is eithernotin B ornotinC.Thereforeeither x ∈ A \ B or x ∈ A \ C ,which impliesthat x ∈ (A \ B ) ∪ (A \ C ).Thus A \ (B ∩ C ) ⊆ (A \ B ) ∪ (A \ C ).
Conversely,if x isin(A \ B ) ∪ (A \ C ),then x ∈ A \ B or x ∈ A \ C .Thus
x ∈ A andeither x/ ∈ B or x/ ∈ C ,whichimpliesthat x ∈ A but x/ ∈ B ∩ C , sothat x ∈ A \ (B ∩ C ).Thus(A \ B ) ∪ (A \ C ) ⊆ A \ (B ∩ C )
Sincethesets A \ (B ∩ C )and(A \ B ) ∪ (A \ C )containthesameelements, theyareequal.
5.(a)If x ∈ A ∩ (B ∪ C ), then x ∈ A and x ∈ B ∪ C .Henceweeitherhave (i) x ∈ A and x ∈ B ,orwehave(ii) x ∈ A and x ∈ C .Therefore,either
x ∈ A ∩ B or x ∈ A ∩ C ,sothat x ∈ (A ∩ B ) ∪ (A ∩ C ).Thisshowsthat
A ∩ (B ∪ C )isasubsetof(A ∩ B ) ∪ (A ∩ C ).
Conversely,let y beanelementof(A ∩ B ) ∪ (A ∩ C ).Theneither(j) y ∈
A ∩ B ,or(jj) y ∈ A ∩ C .Itfollowsthat y ∈ A andeither y ∈ B or y ∈ C
Therefore, y ∈ A and y ∈ B ∪ C ,sothat y ∈ A ∩ (B ∪ C ).Hence(A ∩ B ) ∪ (A ∩ C )isasubsetof A ∩ (B ∪ C ).
InviewofDefinition1.1.1,weconcludethatthesets A ∩ (B ∪ C )and (A ∩ B ) ∪ (A ∩ C )areequal.
(b)Similarto(a).
6.Theset D istheunionof {x : x ∈ A and x/ ∈ B } and {x : x/ ∈ A and x ∈ B }.
7.Here An = {n +1, 2(n +1),...
8.(a)Thegraphconsistsoffourhorizontallinesegments. (b)Thegraphconsistsofthreeverticallinesegments.
9.No.Forexample,both(0,1)and(0, 1)belongto C
, 3],so h(E )= g (f (E ))= g ([2, 3])= {y 2 :2 ≤ y ≤ 3} =[4, 9]. (b) g 1 (G)= {y :0 ≤ y 2 ≤ 4} =[ 2, 2],so h 1 (G)= f 1 (g 1 (G))= f 1 ([ 2, 2])= {x : 2 ≤ x +2 ≤ 2} =[ 4, 0].
12.If0isremovedfrom E and F ,thentheirintersectionisempty,butthe intersectionoftheimagesunder f is {y :0 <y ≤ 1}
13. E \ F = {x : 1 ≤ x< 0}, f (E ) \ f (F )isempty,and f (E \ F )= {y :0 <y ≤ 1}
14.If y ∈ f (E ∩ F ),thenthereexists x ∈ E ∩ F suchthat y = f (x).Since x ∈ E implies y ∈ f (E ),and x ∈ F implies y ∈ f (F ),wehave y ∈ f (E ) ∩ f (F ).This proves f (E ∩ F ) ⊆ f (E ) ∩ f (F ).
15.If x ∈ f 1 (G) ∩ f 1 (H ),then x ∈ f 1 (G)and x ∈ f 1 (H ),sothat f (x) ∈ G and f (x) ∈ H .Then f (x) ∈ G ∩ H, andhence x ∈ f 1 (G ∩ H ).Thisshows
that f 1 (G) ∩ f 1 (H ) ⊆ f 1 (G ∩ H ).Theoppositeinclusionisshownin Example1.1.8(b).Theproofforunionsissimilar.
16.If f (a)= f (b),then a/√a2 +1= b/√b2 +1,fromwhichitfollowsthat a2 = b2 . Since a and b musthavethesamesign,weget a = b,andhence f isinjective. If 1 <y< 1,then x := y/ 1 y 2 satisfies f (x)= y (why?),sothat f takes R ontotheset {y : 1 <y< 1}.If x> 0,then x = √x2 < √x2 +1,soitfollows that f (x) ∈{y :0 <y< 1}
17.Onebijectionisthefamiliarlinearfunctionthatmaps a to0and b to1, namely, f (x):=(x a)/(b a) Showthatthisfunctionworks.
18.(a)Let f (x)=2x, g (x)=3x. (b)Let f (x)= x2 , g (x)= x, h(x)=1.(Manyexamplesarepossible.)
19.(a)If x ∈ f 1 (f (E )),then f (x) ∈ f (E ),sothatthereexists x1 ∈ E such that f (x1 )= f (x).If f isinjective,then x1 = x,whence x ∈ E .Therefore, f 1 (f (E )) ⊆ E .Since E ⊆ f 1 (f (E ))holdsforany f ,wehavesetequality when f isinjective.SeeExample1.1.8(a)foranexample.
(b)If y ∈ H and f issurjective,thenthereexists x ∈ A suchthat f (x)= y . Then x ∈ f 1 (H )sothat y ∈ f (f 1 (H )).Therefore H ⊆ f (f 1 (H )).Since f (f 1 (H )) ⊆ H forany f ,wehavesetequalitywhen f issurjective.See Example1.1.8(a)foranexample.
20.(a)Since y = f (x)ifandonlyif x = f 1 (y ), itfollowsthat f 1 (f (x))= x and f (f 1 (y ))= y
(b)Since f isinjective,then f 1 isinjectiveon R(f ).Andsince f issurjective,then f 1 isdefinedon R(f )= B .
21.If g (f (x1 ))= g (f (x2 )), then f (x1 )= f (x2 ),sothat x1 = x2 ,whichimpliesthat g ◦ f isinjective.If w ∈ C ,thereexists y ∈ B suchthat g (y )= w ,andthere exists x ∈ A suchthat f (x)= y .Then g (f (x))= w ,sothat g ◦ f issurjective. Thus g ◦ f isabijection.
22.(a)If f (x1 )= f (x2 ),then g (f (x1 ))= g (f (x2 )),whichimplies x1 = x2 ,since g ◦ f isinjective.Thus f isinjective.
(b)Given w ∈ C ,since g ◦ f issurjective,thereexists x ∈ A suchthat g (f (x))= w .If y := f (x),then y ∈ B and g (y )= w .Thus g issurjective.
23.Wehave x ∈ f 1 (g 1 (H )) ⇐⇒ f (x) ∈ g 1 (H ) ⇐⇒ g (f (x)) ∈ H ⇐⇒ x ∈ (g ◦ f ) 1 (H ).
24.If g (f (x))= x forall x ∈ D (f ),then g ◦ f isinjective,andExercise22(a) impliesthat f isinjectiveon D (f ).If f (g (y ))= y forall y ∈ D (g ),then Exercise22(b)impliesthat f maps D (f )onto D (g ) Thus f isabijectionof D (f )onto D (g ),and g = f 1 .
Section1.2
ThemethodofproofknownasMathematicalInductionisusedfrequentlyinreal analysis,butinmanysituationsthedetailsfollowaroutinepatternsandare
lefttothereaderbymeansofaphrasesuchas:“TheproofisbyMathematical Induction”.Sincemaystudentshaveonlyahazyideaofwhatisinvolved,itmay beagoodideatospendsometimeexplainingandillustratingwhatconstitutesa proofbyinduction.
Painsshouldbetakentoemphasizethattheinductionhypothesisdoes not entail“assumingwhatistobeproved”.Theinductivestepconcernsthevalidity ofgoingfromtheassertionfor k ∈ N tothatfor k +1.Thetruthoffalsityofthe individualassertionisnotanissuehere.
SampleAssignment:Exercises1,2,6,11,13,14,20.
PartialSolutions:
1.Theassertionistruefor n =1because1/(1 2)=1/(1+1) Ifitistrue for n = k ,thenitfollowsfor k +1because k/(k +1)+1/[(k +1)(k +2)]= (k +1)/(k +2).
2.Thestatementistruefor n =1because[
4.Itistruefor n =1since1=(4 1)/3.Ifitistruefor n = k ,thenadd (2
k +2 [(k +1)(k +2)]/2.
6.If n =1,then13 +5 1=6isdivisibleby6.If k 3 +5k isdivisibleby6, then(k +1)3 +5(k +1)=(k 3 +5k )+3k (k +1)+6isalso,because k (k +1) isalwayseven(why?)sothat3k (k +1)isdivisibleby6,andhencethesum isdivisibleby6.
7.If52k 1isdivisibleby8,thenitfollowsthat52(k +1) 1=(52k 1)+24 52k isalsodivisibleby8.
8.5k +1 4(k +1) 1=5 · 5k 4k 5=(5k 4k 1)+4(5k 1).Nowshowthat 5k 1isalwaysdivisibleby4.
9.If k 3 +(k +1)3 +(k +2)3 isdivisibleby9,then(k +1)3 +(k +2)3 +(k +3)3 = k 3 +(k +1)3 +(k +2)3 +9(k 2 +3k +3)isalsodivisibleby9.
10.Thesumisequalto n/(2n +1)
ever2 <k +1,including k =2, 3, eventhoughthestatementisfalseforthese values.]
15.For
16.Itistruefor n =1and n ≥ 5,butfalsefor n =2, 3, 4.Theinequality 2k +1 < 2k ,wichholdsfor k ≥ 3, isneededintheinductionargument.[The inductivestepisvalidfor n =3, 4eventhoughtheinequality n2 < 2n isfalse forthesevalues.]
17. m =6triviallydivides n3 n for n =1,anditisthelargestintegertodivide 23 2=6.If k 3 k isdivisibleby6,thensince k 2 + k iseven(why?),it followsthat(k +1)3 (k +1)=(k 3 k )+3(k 2 + k )isalsodivisibleby6.
18. √k +1/√k +1=(√k √k +1+1)/√k +1 > (k
19.Firstnotethatsince2 ∈ S ,thenthenumber1=2 1belongsto S .If m/ ∈ S, then m< 2m ∈ S ,so2m 1 ∈ S ,etc.
20.If1 ≤ xk 1 ≤ 2and1 ≤ xk ≤ 2,then2 ≤
1 + xk ≤ 4,sothat1 ≤ xk +1 = (xk 1 + xk )/2 ≤ 2.
Section1.3
Everystudentofadvancedmathematicsneedstoknowthemeaningofthewords “finite”,“infinite”,“countable”and“uncountable”.Formoststudentsatthis levelitisquiteenoughtolearnthedefinitionsandreadthestatementsofthe theoremsinthissection,buttoskiptheproofs.Probablyeveryinstructorwill wanttoshowthat Q iscountableand R isuncountable(seeSection2.5).
Somestudentswillnotbeabletocomprehendthatproofsarenecessaryfor “obvious”statementsaboutfinitesets.Otherswillfindthematerialabsolutely fascinatingandwanttoprolongthediscussionforever.Theteachermustavoid gettingboggeddowninaprotracteddiscussionofcardinalnumbers.
SampleAssignment:Exercises1,5,7,9,11.
PartialSolutions:
1.If T1 = ∅ isfinite,thenthedefinitionofafinitesetappliesto T2 = Nn for some n.If f isabijectionof T1 onto T2 ,andif g isabijectionof T2 onto Nn , then(byExercise1.1.21)thecomposite g ◦ f isabijectionof T1 onto Nn ,so that T1 isfinite.
2.Part(b)Let f beabijectionof Nm onto A andlet C = {f (k )} forsome k ∈ Nm .Define g on Nm 1 by g (i):= f (i)for i =1,...,k 1,and g (i):= f (i +1)for i = k,...,m 1.Then g isabijectionof Nm 1 onto A\C .(Why?)
Part(c)Firstnotethattheunionoftwofinitesetsisafiniteset.Nownote thatif C/B werefinite,then C = B ∪ (C \ B )wouldalsobefinite.
3.(a)Theelement1canbemappedintoanyofthethreeelementsof T ,and 2canthenbemappedintoanyofthetworemainingelementsof T ,after whichtheelement3canbemappedintoonlyoneelementof T. Hencethere are6=3 · 2 · 1differentinjectionsof S into T
(b)Suppose a mapsinto1.If b alsomapsinto1,then c mustmapinto2;if b mapsinto2,then c canmapintoeither1or2.Thusthereare3surjections thatmap a into1,andthereare3othersurjectionsthatmap a into2.
4. f (n):=2n +13,n ∈ N
5. f (1):=0,f (2n):= n,f (2n +1):= n for n ∈ N.
6.ThebijectionofExample1.3.7(a)isoneexample.Anotheristheshiftdefined by f (n):= n +1thatmaps N onto N \{1}
7.If T1 isdenumerable,take T2 = N.If f isabijectionof T1 onto T2 ,andif g isabijectionof T2 onto N,then(byExercise1.1.21) g ◦ f isabijectionof T1 onto N,sothat T1 isdenumerable.
8.Let An := {n} for n ∈ N,so An = N
9.If S ∩ T = ∅ and f : N → S,g : N → T arebijectionsonto S and T ,respectively, let h(n):= f ((n +1)/2)if n isoddand h(n):= g (n/2)if n iseven.Itisreadily seenthat h isabijectionof N onto S ∪ T ;hence S ∪ T isdenumerable.What if S ∩ T = ∅?
m + n 1=9and m =6imply n =4.Then h(6, 4)= 1 2 · 8 · 9+6=42
h(m, 3)= 1 2 (m +1)(m +2)+ m =19,sothat m2 +5m 36=0.Thus m =4.
11.(a) P ({1, 2})= {∅, {1}, {2}, {1, 2}} has22 =4elements.
(b) P ({1, 2, 3})has23 =8elements.
(c) P ({1, 2, 3, 4})has24 =16elements.
12.Let Sn+1 := {x1 ,...,xn ,xn+1 } = Sn ∪{xn+1 } have n +1elements.Thena subsetof Sn+1 either(i)contains xn+1 ,or(ii)doesnotcontain xn+1 .The inductionhypothesisimpliesthatthereare2n subsetsoftype(i),sinceeach suchsubsetistheunionof {xn+1 } andasubsetof Sn .Therearealso2n subsetsoftype(ii).Thusthereisatotalof2n +2n =2 2n =2n +1 subsets of Sn+1
13.Foreach m ∈ N,thecollectionofallsubsetsof Nm isfinite.(SeeExercise12.)
Everyfinitesubsetof N isasubsetof Nm forasufficientlylarge m.Therefore Theorem1.3.12impliesthat F (N)= ∞ m=1 P (Nm )iscountable.
CHAPTER2
THEREALNUMBERS
Studentswillbefamiliarwithmuchofthefactualcontentofthefirstfewsections, buttheprocessofdeducingthesefactsfromabasiclistofaxiomswillbenew tomostofthem.Theabilitytoconstructproofsusuallyimprovesgradually duringthecourse,andtherearemuchmoresignificanttopicsforthcoming.Afew selectedtheoremsshouldbeprovedindetail,sincesomeexperienceinwriting formalproofsisimportanttostudentsatthisstage.However,oneshouldnot spendtoomuchtimeonthismaterial.
Sections2.3and2.4ontheCompletenessPropertyformtheheartofthis chapter. Thesesectionsshouldbecoveredthoroughly. AlsotheNestedIntervals PropertyinSection2.5shouldbetreatedcarefully.
Section2.1
OnegoalofSection2.1istoacquaintstudentswiththeideaofdeducingconsequencesfromalistofbasicaxioms.Studentswhohavenotencounteredthistype offormalreasoningmaybesomewhatuncomfortableatfirst,sincetheyoften regardtheseresultsas“obvious”.Sincethereismuchmoretocome,asampling ofresultswillsufficeatthisstage,makingitclearthatitisonlyasampling. Theclassicproofoftheirrationalityof √2shouldcertainlybeincludedinthe discussion,andstudentsshouldbeaskedtomodifythisargumentfor √3, etc.
SampleAssignment:Exercises1(a,b),2(a,b),3(a,b),6,13,16(a,b),20,23.
PartialSolutions:
1.(a)Applyappropriatealgebraicpropertiestoget b =0+ b =( a + a)+ b =
a +(a + b)= a +0= a.
(b)Apply(a)to( a)+ a =0with b = a toconcludethat a = ( a)
(c)Apply(a)totheequation a +( 1)a = a(1+( 1))= a · 0=0toconclude that( 1)a = a.
(d)Apply(c)with a = 1toget( 1)( 1)= ( 1).Thenapply(b)with a =1toget( 1)( 1)=1.
2.(a) (a + b)=( 1)(a + b)=( 1)a +( 1)b =( a)+( b).
(b)( a) ( b)=(( 1)a) (( 1)b)=( 1)( 1)(ab)= ab.
(c)Notethat( a)( (1/a))= a(1/a)=1.
(d) (a/b)=( 1)(a(1/b))=(( 1)a)(1/b)=( a)/b.
3.(a)Add 5tobothsidesof2x +5=8anduse (A2),(A4),(A3)toget2x =3.
Thenmultiplybothsidesby1/2toget x =3/2.
(b)Write x2 2x = x(x 2)=0andapplyTheorem2.1.3(b).Alternatively, notethat x =0satisfiestheequation,andif x =0,thenmultiplicationby 1/x gives x =2.
(c)Add 3tobothsidesandfactortoget x2 4=(x 2)(x +2)=0.Now apply2.1.3(b)toget x =2or x = 2.
(d)Apply2.1.3(b)toshowthat(x 1)(x +2)=0ifandonlyif x =1or x = 2.
4.Clearly a =0satisfies a a = a. If a =0and a a = a,then(a a)(1/a)= a(1/a), sothat a = a(a(1/a))= a(1/a)=1.
5.If(1/a)(1/b)ismultipliedby ab, theresultis1.Therefore,Theorem2.1.3(a) impliesthat1/(ab)=(1/a)(1/b).
6.Notethatif q ∈ Z andif3q 2 iseven,then q 2 iseven,sothat q iseven.Hence, if(p/q )2 =6,thenitfollowsthat p iseven,say p =2m,whence2m2 =3q 2 ,so that q isalsoeven.
7.If p ∈ N,therearethreepossibilities:forsome m ∈ N ∪{0},(i) p =3m, (ii) p =3m +1,or(iii) p =3m +2.Ineithercase(ii)or(iii),wehave p2 = 3h +1forsome h ∈ N ∪{0}
8.(a)Let x = m/n,y = p/q ,where m,n =0,p,q =0areintegers.Then x + y = (mq + np)/nq and xy = mp/nq arerational.
(b)If s := x + y ∈ Q,then y = s x ∈ Q,acontradiction.If t := xy ∈ Q and x =0,then y = t/x ∈ Q,acontradiction.
(b)If x = s + t√2 =0isin K ,then s t√2 =0(why?)and
10(a)If c = d,then2.1.7(b)implies a + c<b + d.If c<d,then a + c< b + c<b + d
(b)If c = d =0,then ac = bd =0.If c> 0,then0 <ac bytheTrichotomy Propertyand ac<bc followsfrom2.1.7(c).Ifalso c ≤ d,then ac ≤ ad<bd. Thus0 ≤ ac ≤ bd holdsinallcases.
11.(a)If a> 0,then a =0bytheTrichotomyProperty,sothat1/a exists.If
1/a =0,then1= a (1/a)= a 0=0,whichcontradicts(M3).If1/a< 0,then 2.1.7(c)impliesthat1= a(1/a) < 0,whichcontradicts2.1.8(b).Thus1/a> 0, and2.1.3(a)impliesthat1/(1/a)= a
(b)If a<b,then2a = a + a<a + b,andalso a + b<b + b =2b.Therefore, 2a<a + b< 2b,which,since 1 2 > 0(by2.1.8(c)andpart(a)),impliesthat a< 1 2 (a + b) <b.
12.Let a =1and b =2.If c = 3and d = 1,then ac<bd.Ontheotherhand, if c = 3and d = 2, then bd<ac.(Manyotherexamplesarepossible.)
13.If a =0,then2.1.8(a)impliesthat a2 > 0;since b2 ≥ 0,itfollowsthat a2 + b2 > 0.
14.If0 ≤ a<b,then2.1.7(c)implies ab<b2 .If a =0,then0= a2 = ab<b2 If a> 0,then a2 <ab by2.1.7(c).Thus a2 ≤ ab<b2 .If a =0,b =1,then 0= a2 = ab<b =1.
15.(a)If0 <a<b,then2.1.7(c)impliesthat0 <a2 <ab<b2 .ThenbyExample
2.1.13(a),weinferthat a = √a2 < √ab< √b2 = b
(b)If0 <a<b,then ab> 0sothat1/ab> 0,andthus1/a 1/b = (1/ab)(b a) > 0.
16.(a)Tosolve(x 4)(x +1) > 0,lookattwocases.Case1: x 4 > 0and x +1 > 0,whichgives x> 4.Case2: x 4 < 0and x +1 < 0,whichgives
x< 1.Thuswehave {x : x> 4or x< 1}
(b)1 <x2 < 4hasthesolutionset {x :1 <x< 2or 2 <x< 1}
(c)Theinequalityis1/x x =(1 x)(1+ x)/x< 0.If x> 0,thisisequivalentto(1 x)(1+ x) < 0,whichissatisfiedif x> 1.If x< 0,thenwesolve
(1 x)(1+ x) > 0,andget 1 <x< 0.Thusweget {x : 1 <x< 0or x> 1}
(d)thesolutionsetis {x : x< 0or x> 1}.
17.If a> 0,wecantake ε0 := a> 0andobtain0 <ε0 ≤ a,acontradiction.
18.If b<a andif ε0 :=(a b)/2,then ε0 > 0and a = b +2ε0 >b + ε0 .
19.Theinequalityisequivalentto0 ≤ a2 2ab + b2 =(a b)2
20.(a)If0 <c< 1,then2.1.7(c)impliesthat0 <c2 <c,whence0 <c2 <c< 1.
(b)Since c> 0,then2.1.7(c)impliesthat c<c2 ,whence1 <c<c2 .
21.(a)Let S := {n ∈ N :0 <n< 1}.If S isnotempty,theWell-OrderingProperty of N impliesthereisaleastelement m in S .However,0 <m< 1impliesthat 0 <m2 <m,andsince m2 isalsoin S ,thisisacontradictiontothefactthat m istheleastelementof S
(b)If n =2p =2q 1forsome p,q in N,then2(q p)=1,sothat0 <q p< 1. Thiscontradicts(a).
22.(a)Let x := c 1 > 0andapplyBernoulli’sInequality2.1.13(c)toget cn =
(1+ x)n ≥ 1+ nx ≥ 1+ x = c forall n ∈ N,and cn > 1+ x = c for n> 1.
(b)Let b :=1/c andusepart(a).
23.If0 <a<b and ak <bk ,then2.1.7(c)impliesthat ak +1 <abk <bk +1 so Inductionapplies.If am <bm forsome m ∈ N,thehypothesisthat0 <b ≤ a leadstoacontradiction.
24.(a)If m>n,then k := m n ∈ N,soExercise22(a)impliesthat ck ≥ c> 1. Butsince ck = cm n ,thisimpliesthat cm >cn .Conversely,thehypothesis that cm >cn and m ≤ n leadtoacontradiction.
(b)Let b :=1/c andusepart(a).
BartleandSherbert
25.Let b := c1/mn .Weclaimthat b> 1;forif b ≤ 1,thenExercise22(b)implies that1 <c = bmn ≤ b ≤ 1,acontradiction.ThereforeExercise24(a)implies that c1/n = bm >bn = c1/m ifandonlyif m>n
26.Fix m ∈ N anduseMathematicalInductiontoprovethat am + n = am an and (am )n = amn forall n ∈ N.Then,foragiven n ∈ N,provethattheequalities arevalidforall m ∈ N.
Section2.2
Thenotionofabsolutevalueofarealnumberisdefinedintermsofthebasicorder propertiesof R.Wehaveputitinaseparatesectiontogiveitemphasis.Many studentsneedextraworktobecomecomfortablewithmanipulationsinvolving absolutevalues,especiallywheninequalitiesareinvolved.
Wehavealsousedthissectiontogivestudentsanearlyintroductiontothe notionofthe ε-neighborhoodofapoint.Asapreviewoftheroleof ε-neighborhoods,wehaverecastTheorem2.1.9intermsof ε-neighborhhoodsin Theorem2.2.8.
SampleAssignment:Exercises1,4,5,6(a,b),8(a,b),9,12(a,b),15.
PartialSolutions:
/|b|.If b< 0,then1/b< 0,sothat |1/b| = (1/b)=1/( b)=1/|b|
2.Firstshowthat ab ≥ 0ifanonlyif |ab| = ab.Thenshowthat(|a| + |b|)2 = (a + b)2 ifandonlyif |ab| = ab.
3.If x ≤ y ≤ z ,then |x y | + |y z | =(y x)+(z y )= z x = |z x|.To establishtheconverse,showthat y<x and y>z areimpossible.Forexample, if y<x ≤ z ,itfollowsfromwhatwehaveshownandthegivenrelationship that |x y | =0,sothat y = x,acontradiction.
4. |x a| <ε ⇐⇒−ε<x a<ε ⇐⇒ a ε<x<a + ε
5.If a<x<b and b< y< a,itfollowsthat a b<x y<b a.Since a b = (b a), theargumentin2.2.2(c)givestheconclusion |x y | <b a. Thedistancebetween x and y islessthanorequalto b a 6.(a) |
7.Case1: x ≥ 2 ⇒ (x +1)+(x 2)=2x 1=7,so x =4.
Case2: 1 <x< 2 ⇒ (x +1)+(2 x)=3 =7, sonosolution.
Case3: x ≤−1 ⇒ ( x 1)+(2 x)= 2x +1=7,so x = 3. Combiningthesecases,weget x =4or x = 3.
8.(a)If x> 1/2,then x +1=2x 1,sothat x =2.If x ≤ 1/2,then x +1= 2x +1,sothat x =0.Therearetwosolutions {0, 2}
(b)If x ≥ 5,theequationimplies x = 4,sonosolutions.If x< 5,then x =2.
9.(a)If x ≥ 2,theinequalitybecomes 2 ≤ 1.If x ≤ 2,theinequalityis x ≥ 1/2, sothiscasecontributes1/2 ≤ x ≤ 2.Combiningthecasesgivesusall x ≥ 1/2.
(b) x ≥ 0yields x ≤ 1/2,sothatweget0 ≤ x ≤ 1/2. x ≤ 0yields x ≥−1,so that 1 ≤ x ≤ 0.Combiningcases,weget 1 ≤ x ≤ 1/2.
10.(a)Eitherconsiderthethreecases: x< 1, 1 ≤ x ≤ 1and1 <x;or,square bothsidestoget 2x> 2x.Eitherapproachgives x< 0.
(b)Considerthethreecases x ≥ 0, 1 ≤ x< 0and x< 1toget 3/2 < x< 1/2.
11. y = f (x)where f (x):= 1for x< 0,f (x):=2x 1for0 ≤ x ≤ 1,and f (x):=1 for x> 1.
12.Case1: x ≥ 1 ⇒ 4 < (x +2)+(x 1) < 5, so3/2 <x< 2. Case2: 2 <x< 1 ⇒ 4 < (x +2)+(1 x) < 5,sothereisnosolution.
Case3: x< 2 ⇒ 4 < ( x 2)+(1 x) < 5,so 3 <x< 5/2. Thusthesolutionsetis {x : 3 <x< 5/2or3/2 <x< 2}.
13. |2x 3| < 5 ⇐⇒−1 <x< 4,and |x +1| > 2 ⇐⇒ x< 3or x> 1.Thetwo inequalitiesaresatisfiedsimultaneouslybypointsintheintersection {x : 1 <x< 4}
14.(a) |x| = |y |⇐⇒ x2 =
y = x or y = x.Thus {(x,y ): y = x or y = x}
)(x + y )=0
(b)Considerfourcases.If x ≥ 0,y ≥ 0,wegetthelinesegmentjoiningthe points(0,1)and(1,0).If x ≤ 0,y ≥ 0,wegetthelinesegmentjoining( 1, 0) and(0, 1),andsoon.
(c)Thehyperbolas y =2/x and y = 2/x
(d)Considerfourcasescorrespondingtothefourquadrants.Thegraph consistsofaportionofalinesegmentineachquadrant.Forexample,if x ≥ 0,y ≥ 0,weobtaintheportionoftheline y = x 2inthisquadrant.
15.(a)If y ≥ 0,then y ≤ x ≤ y andwegettheregionintheupperhalf-planeon orbetweenthelines y = x and y = x.If y ≤ 0,thenwegettheregioninthe lowerhalf-planeonorbetweenthelines y = x and y = x.
(b)Thisistheregiononandinsidethediamondwithvertices(1,0),(0,1), ( 1, 0)and(0, 1).
16.Fortheintersection,let γ bethesmallerof ε and δ .Fortheunion,let γ be thelargerof ε and δ
17.Chooseany ε> 0suchthat ε< |a b|.
18.(a)If a ≤ b,thenmax{a,b} = b = 1 2 [a + b +(b a)]andmin{a,b} = a = 1 2 [a + b (b a)].
(b)If a =min {a,b,c},thenmin{min{a,b},c} = a =min{a,b,c}.Similarly,if b or c ismin{a,b,c}.
19.If a ≤ b ≤ c,thenmid{a,b,c} = b =min{b,c,c} =min{max{a,b},max{b,c}, max{c,a}}.Theothercasesaresimilar.
Section2.3
Thissectioncompletesthedescriptionoftherealnumbersystembyintroducing thefundamentalcompletenesspropertyintheformoftheSupremumProperty. Thispropertyisvitaltorealanalysisandstudentsshouldattainaworkingunderstandingofit.Effortexpendedinthissectionandtheonefollowingwillberichly rewardedlater.
SampleAssignment:Exercises1,2,5,6,9,10,12,14.
PartialSolutions:
1.Anynegativenumberor0isalowerbound.Forany x ≥ 0,thelargernumber x +1isin S1 ,sothat x isnotanupperboundof S1 .Since0 ≤ x forall x ∈ S1 , then u =0isalowerboundof S1 .If v> 0,then v isnotalowerboundof S1 because v/2 ∈ S1 and v/2 <v .Thereforeinf S1 =0.
2. S2 haslowerbounds,sothatinf S2 exists.Theargumentusedfor S1 also showsthatinf S2 =0,butthatinf S2 doesnotbelongto S2 . S2 doesnot haveupperbounds,sothatsup S2 doesnotexists.
3.Since1/n ≤ 1forall n ∈ N,then1isanupperboundfor S3 .But1isa memberof S3 ,sothat1=sup S3 .(SeeExercise7below.)
4.sup S4 =2andinf S4 =1/2.(Notethatbotharemembersof S4 .)
5.Itisinterestingtocomparealgebraicandgeometricapproachestothese problems.
(a)inf A = 5/2,sup A doesnotexist,
(b)sup B =2,inf B = 1,
(c)sup C =1,inf B doesnotexist,
(d)sup D =1+ √6,inf D =1 √6.
6.If S isboundedbelow,then S := {−s : s ∈ S } isboundedabove,sothat u :=sup S exists.If v ≤ s forall s ∈ S ,then v ≥−s forall s ∈ S ,sothat v ≥ u,andhence v ≤−u.Thusinf S = u
7.Let u ∈ S beanupperboundof S. If v isanotherupperboundof S ,then u ≤ v .Hence u =sup S .
8.If t>u and t ∈ S ,then u isnotanupperboundof S
9.Let u :=sup S .Since u isanupperboundof S, sois u +1/n forall n ∈ N
Since u isthesupremumof S and u 1/n<u,thenthereexists s0 ∈ S with u 1/n<s0 ,whence u 1/n isnotanupperboundof S.
10.Let u :=sup A, v :=sup B and w :=sup{u,v }.Then w isanupperboundof A ∪ B ,becauseif x ∈ A,then x ≤ u ≤ w ,andif x ∈ B ,then x ≤ v ≤ w .If z is
anyupperboundof A ∪ B ,then z isanupperboundof A andof B ,sothat u ≤ z and v ≤ z .Hence w ≤ z .Therefore, w =sup(A ∪ B ).
11.Sincesup S isanupperboundof S ,itisanupperboundof S0 ,andhence sup S0 ≤ sup S .
12.Considertwocases.If u ≥ s∗ ,then u =sup(S ∪{u}). If u<s∗ ,thenthere exists s ∈ S suchthat u<s ≤ s∗ ,sothat s∗ =sup(S ∪{u}).
13.If S1 := {x1 }, showthat x1 =sup S1 .If Sk := {x1 ,...,xk } issuchthatsup Sk ∈ Sk ,thenprecedingexerciseimpliesthatsup{x1 ,...,xk ,xk +1 } isthe largerofsup Sk and xk +1 andsoisin Sk +1 .
14.If w =inf S and ε> 0,then w + ε isnotalowerboundsothatthereexists t in S suchthat t<w + ε.If w isalowerboundof S thatsatisfiesthestated condition,andif z>w ,let ε = z w> 0.Thenthereis t in S suchthat t<w + ε = z ,sothat z isnotalowerboundof S. Thus, w =inf S
Section2.4
Thissectionexhibitshowthesupremumisusedinpractice,andcontainssome importantpropertiesof R thatwilloftenbeusedlater.TheArchimedeanProperties2.4.3–2.4.6andtheDensityProperties2.4.8and2.4.9arethemostsignificant. Theexercisesalsocontainsomeresultsthatwillbeusedlater.
SampleAssignment:Exercises1,2,4(b),5,7,10,12,13,14.
PartialSolutions:
1.Since1 1/n< 1forall n ∈ N,thenumber1isanupperbound.Toshow that1isthesupremum,itmustbeshownthatforeach ε> 0thereexists n ∈ N suchthat1 1/n> 1 ε,whichisequivalentto1/n<ε.Applythe ArchimedeanProperty2.4.3or2.4.5.
2.inf S = 1andsup S =1.Toseethelatternotethat1/n 1/m ≤ 1forall m,n ∈ N.Ontheotherhandif ε> 0thereexists m ∈ N suchthat1/m<ε, sothat1/1 1/m> 1 ε.
3.Supposethat u ∈ R isnotthesupremumof S .Theneither(i) u isnotan upperboundof S (sothatthereexists s1 ∈ S with u<s1 ,whencewetake n ∈ N with1/n<s1 u toshowthat u +1/n isnotanupperboundof S ),or (ii)thereexistsanupperbound u1 of S with u1 <u (inwhichcasewetake 1/n<u u1 toshowthat u 1/n isnotanupperboundof S ).
4.(a)Let u :=sup S and a> 0.Then x ≤ u forall x ∈ S ,whence ax ≤ au forall x ∈ S ,whenceitfollowsthat au isanupperboundof aS .If v isanotherupper boundof aS ,then ax ≤ v forall x ∈ S ,whence x ≤ v/a forall x ∈ S ,showing that v/a isanupperboundfor S sothat u ≤ v/a,fromwhichweconclude that au ≤ v .Therefore au =sup(aS ).Thestatementabouttheinfimumis provedsimilarly.
BartleandSherbert
(b)Let u :=sup S and b< 0.If x ∈ S ,then(since b< 0) bu ≤ bx sothat bu isalowerboundof bS .If v ≤ bx forall x ∈ S ,then x ≤ v/b (since b< 0), sothat v/b isanupperboundfor S .Hence u ≤ v/b whence v ≤ bu.Therefore bu =inf(bS ).
5.If x ∈ S ,then0 ≤ x ≤ u,sothat x2 ≤ u2 whichimpliessup T ≤ u2 .If t isany upperboundof T ,then x ∈ S implies x2 ≤ t sothat x ≤ √t.Itfollowsthat u ≤ √t,sothat u2 ≤ t.Thus u2 ≤ sup T.
6.Let u :=sup f (X ).Then f (x) ≤ u forall x ∈ X ,sothat a + f (x) ≤ a + u for all x ∈ X ,whencesup{a + f (x): x ∈ X }≤ a + u.If w<a + u,then w a<u, sothatthereexists xw ∈ X with w a<f (xw ),whence w<a + f (xw ),and thus w isnotanupperboundfor {a + f (x): x ∈ X }.
7.Let u :=sup S , v :=sup B , w :=sup(A + B ).If x ∈ A and y ∈ B ,then x + y ≤ u + v ,sothat w ≤ u + v .Now,fix y ∈ B andnotethat x ≤ w y forall x ∈ A;thus w y isanupperboundfor A sothat u ≤ w y .Then y ≤ w u forall y ∈ B ,so v ≤ w u andhence u + v ≤ w .Combiningthese inequalities,wehave w = u + v
8.If u :=sup f (X )and v :=sup g (X ),then f (x) ≤ u and g (x) ≤ v forall x ∈ X , whence f (x)+ g (x) ≤ u + v forall x ∈ X .Thus u + v isanupperbound fortheset {f (x)+ g (x): x ∈ X }.Thereforesup{f (x)+ g (x): x ∈ X }≤ u + v .
9.(a) f (x)=2x +1, inf{f (x): x ∈ X } =1. (b) g (y )= y ,sup{g (y ): y ∈ Y } =1.
10.(a) f (x)=1for x ∈ X .(b) g (y )=0for y ∈ Y
11.If x ∈ X , y ∈ Y ,then g (y ) ≤ h(x,y ) ≤ f (x).Ifwefix y ∈ Y andtakethe infimumover x ∈ X ,thenweget g (y ) ≤ inf{f (x): x ∈ X } foreach y ∈ Y . Nowtakethesupremumover y ∈ Y .
12.Let S := {h(x,y ): x ∈ X,y ∈ Y }.Wehave h(x,y ) ≤ F (x)forall x ∈ X,y ∈ Y
sothatsup S ≤ sup{F (x): x ∈ X }.If w< sup{F (x): x ∈ X },thenthere exists x0 ∈ X with w<F (x0 )=sup {h(x0 ,y ): y ∈ Y },whencethereexists y0 ∈ Y with w<h(x0 ,y0 ) Thus w isnotanupperboundof S ,andso w< sup S. Sincethisistrueforany w suchthat w< sup{F (x): x ∈ X }, weconcludethatsup{F (x): x ∈ X }≤ sup S.
13.If x ∈ Z,take n := x +1.If x/ ∈ Z,wehavetwocases:(i) x> 0(whichis coveredbyCor.2.4.6),and(ii) x< 0.Incase(ii),let z := x anduse2.4.6. If n1 <n2 areintegers,then n1 ≤ n2 1sothesets {y : n1 1 ≤ y<n1 } and {y : n2 1 ≤ y<n2 } aredisjoint;thustheinteger n suchthat n 1 ≤ x<n isunique.
14.Notethat n< 2n (whence1/2n < 1/n)forany n ∈ N.
15.Let S3 := {s ∈ R :0 ≤ s,s2 < 3}.Showthat S3 isnonemptyandbounded by3andlet y :=sup S3 .If y 2 < 3and1/n< (3 y 2 )/(2y +1)showthat
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y +1/n ∈ S3 .If y 2 > 3and1/m< (y 2 3)/2y showthat y 1/m ∈ S3 . Therefore y 2 =3.
16.Case1:If a> 1,let Sa := {s ∈ R :0 ≤ s,s2 <a}.Showthat Sa isnonempty andboundedaboveby a andlet z :=sup Sa .Nowshowthat z 2 = a. Case2:If0 <a< 1,thereexists k ∈ N suchthat k 2 a> 1(why?).If z 2 = k 2 a, then(z/k )2 = a
17.Consider T := {t ∈ R :0 ≤ t,t3 < 2}.If t> 2,then t3 > 2sothat t/ ∈ T .Hence y :=sup T exists.If y 3 < 2,choose1/n< (2 y 3 )/(3y 2 +3y +1)andshow that(y +1/n)3 < 2,acontradiction,andsoon.
18.If x< 0 <y ,thenwecantake r =0.If x<y< 0,weapply2.4.8toobtaina rationalnumberbetween y and x.
19.Thereexists r ∈ Q suchthat x/u<r<y/u.
Section2.5
AnotherimportantconsequenceoftheSupremumPropertyof R istheNested IntervalsProperty2.5.2.Itisaninterestingfactthatifweassumethevalidityof both theArchimedeanProperty2.4.3andtheNestedIntervalsProperty,thenwe canprovetheSupremumProperty.Hencethesetwopropertiescouldbetaken asthecompletenessaxiomfor R.However,establishingthislogicalequivalence wouldconsumevaluabletimeandnotsignificantlyadvancethestudyofrealanalysis,sowewillnotdoso.(Thereareotherpropertiesthatcouldbetakenasthe completenessaxiom.)
Thediscussionofbinaryanddecimalrepresentationsisincludedtogivethe studentaconcreteillustrationoftheratherabstractideasdevelopedtothispoint. However,thismaterialisnotvitalforwhatfollowsandcanbeomittedortreated lightly.Wehavekeptthisdiscussioninformaltoavoidgettingburiedintechnical detailsthatarenotcentraltothecourse.
SampleAssignment:Exercises3,4,5,6,7,8,10,11.
PartialSolutions:
1.Notethat[a,b] ⊆ [a ,b ]ifandonlyif a ≤ a ≤ b ≤ b
2. S hasanupperbound b andalowerbound a ifandonlyif S iscontainedin theinterval[a,b].
3.Sinceinf S isalowerboundfor S andsup S isanupperboundfor S ,then S ⊆ IS .Moreover,if S ⊆ [a,b], then a isalowerboundfor S and b isan upperboundfor S ,sothat[a,b] ⊇ IS
4.Because z isneitheralowerboundoranupperboundof S
5.If z ∈ R,then z isnotalowerboundof S sothereexists xz ∈ S suchthat xz ≤ z .Also z isnotanupperboundof S sothereexists yz ∈ S suchthat z ≤ yz .Since z belongsto[xz ,yz ], itfollowsfromtheproperty(1)that z ∈ S