TheRealNumbers
CONTENTS
Chapter2
Section2.1:TheRealandExtendedRealNumberSystem.
2.1.1. (a)If 1 0 = x,thenbymultiplyingby x,wehave1= x 0=0.(b)If 0 0 = x,thenagain,0= x 0= y 0,so 0 0 = y aswell.
2.1.2. Since x · 1 x =1 =0,and x =0,itmustbethecasethat 1 x =0.If x> 0but 1 x < 0,then1= x · 1 x < 0,incontradictiontoProposition 2.6,part(4).
2.1.3. Supposewithoutlossofgeneralitythat x<y.Then x = x + x 2 < x + y 2 < y + y 2 = y.
2.1.4. (a)Intheproofofpart(3)ofProposition2.8,replace < and > with ≤ and ≥.(b)Intheproofofpart(4)ofProposition2.8,replace < and > with ≤ and ≥
2.1.6. If x,y ≥ 0,then xy ≥ 0and |xy| = xy = |x||y|.If x,y< 0,then again xy> 0,andso |xy| = xy.Since x,y< 0, |x| = x,and |y| = y,so |x||y| =( x)( y)= xy,andagaintheresultholds. Supposeexactlyoneof x or y isnegative,andtheotherisnonnegative;withoutlossofgenerality,suppose x< 0and y ≥ 0.Then xy ≤ 0,so |xy| = xy (or0),while |x| = x and |y| = y (or0),so that |x||y| =( x)y = xy (or0).
2.1.7. If x ≥ 0,then |x| = x andtheresultistrue.If x< 0then |x| = x and x< 0 < x,andso x< |x| inthatcase.
2.1.8. Wemusthave a and b tohavethesamesign.
2.1.9. (a)Followingthehint,replacing b = c a wehave
(b)Followingthehint,replacing b = a c wehave
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IntroductiontoAnalysis:SolutionsManual (c)ThisfollowsdirectlyfromthepreviouspartsandAssertion(3) ofProposition2.8.
2.1.10. Writing0=1 1,wewouldhave ∞· 0= ∞(1 1)= ∞−∞,the lattercannotbedefined.
2.1.11. If x = ∞ ∞ ,then
So x =0.Butif x =0= ∞ ∞ ,itmustbethecasethat0 ·∞ = ∞, andtheformercannotbedefined.
2.1.12. Let x,y ∈ R ,weconstructaneighborhood U of x andaneighborhood V of y sothat U ∩ V = ∅.If x and y arebothreal, thentheproofofTheorem2.14.Supposethenthatoneof x or y isinfinite,supposewithoutlossofgeneralitythatitis x.If y is finite,thenchooseourneighborhoodof y tohaveradius1,sothat V =(y 1,y +1).If x = ∞,thenchoose U =(y +2, ∞).If x = −∞, choose U =(−∞,y 2).If x and y arebothinfinite,say x = −∞ and y = ∞,thenchoose U =(−∞, 0,and V =(0, ∞).
Section2.2:TheSupremumandInfimum.
2.2.1. (a)sup(S)=4, inf(S)= 7.(b)sup(S)= ∞, inf(S)=0.(c) sup(S)= ∞, inf(S)= −∞.(d)sup(S)= ∞, inf(S)= −∞
2.2.2. Suppose S isbounded,andthat M isanupperboundfor S,and m isalowerboundfor S.Let N =max{|m|, |M |}. Then N ≥|M |, and N ≥|m|,sothat N ≤ M ≤ N and N ≤ m ≤ N .So,for every s ∈ S,wehave N ≤ m ≤ s ≤ M ≤ N, so0 ≤|s|≤ N.
Conversely,suppose N isanupperboundfor |S|.Then,forevery |s|∈|S|,wehave |s|≤ N .Butthismeansthat N ≤ s ≤ N ,and so N and N areupperandlowerboundsfor S,respectively.
2.2.3. b isanupperboundfor S since S ⊆ [a,b]and b isanupperbound for[a,b].Thussup(S) ≤ b.Since S isnonempty,thereexistsan x ∈ S,sosince S ⊆ [a,b], a ≤ x ≤ b.So,sup(S) ≥ x ≥ a.So, sup(S) ∈ [a,b].
2.2.4. (a)Let α =sup(S)forconvenience.Noticethatforevery s ∈ S, s ≤ α,andsosince a> 0, as ≤ aα.So, aα isanupperbound for aS.Nowsuppose b<aα.Then b a <α,andso b a isnotan upperboundfor S.Thereforethereexistsan s ∈ S with b a <s,so b<as.So, b isnotanupperboundfor aS.(b)Let β =inf(S)for convenience.Noticeforevery s ∈ S,wehave s ≥ β.Since a> 0,we have as ≥ aβ.So, aβ isalowerboundfor S.Nowsuppose b>aβ.
Then b a >β,andso b a isnotalowerboundfor S.Thereforethere existsan s ∈ S sothat b a >s,so b>as.So, b isnotalowerbound for S
2.2.5. Let α =sup(S),andlet s ∈ S.Then s ≤ α andsosince a< 0we have as ≥ aα.So, aα isalowerboundfor aS.Nowlet b>aα.Then b a <α,andsothereexistsan s ∈ S with b a <s,sothat b>as,so that b isnotalowerboundfor aS.Toprovethenextassertion,let a = 1.
2.2.6. Suppose,withoutlossofgenerality,that α =sup(A)= max{sup(A), sup(B)}.Let x ∈ A ∪ B,then x ∈ A or x ∈ B.If x ∈ A,then x ≤ α.If x ∈ B,then x ≤ sup(B) ≤ α.So α isan upperboundfor A ∪ B.Nowsuppose b<α.Since α =sup(A),and b<α,thereexistsanelement a ∈ A with b<a.Therefore, b isnot anupperboundfor A ∪ B since a ∈ A ⊆ A ∪ B,and b<a.
2.2.7. Suppose,withoutlossofgenerality,that β =inf(A)= min{inf(A), inf(B)} Let x ∈ A ∪ B.Then x ∈ A or x ∈ B.If x ∈ A, then x ≥ inf(A)= β.If x ∈ B,then x ≥ inf(B) ≥ inf(A)= β.So, β isalowerboundfor A ∪ B.Nowlet b>β.Since β =inf(A), thereexistsan a ∈ A sothat b>a.So, b isnotalowerboundfor A ∪ B,since a ∈ A ⊆ A ∪ B,and b>a
2.2.8. (a)Since A and B areeachbounded,thereexistsupperbounds MA and MB for A and B,respectively.Inaddition,thereexistslower bounds mA and mB for A and B,respectively.Foreach a ∈ A and b ∈ B,wehave
mA + mB ≤ a + b ≤ MA + MB
Therefore, A + B isbounded.(b)Let α =sup(A)and β =sup(B), bothrealnumbers.Let a ∈ A,and b ∈ B.Then a + b ≤ α + β, so α + β isanupperboundfor A + B.Since α + β isanupper boundfor A + B, α + β iseitherthesup(A + B),or,sup(A + B) < α + β.Ifthelatterweretrue,therewouldexist a + b ∈ A + B with α + β<a + b.Butforthistobetrue,oneof a or b mustbegreater than α or β,respectively.Eitherpossibilityisacontradiction,so α + β =sup(A + B).
2.2.9. Let A = {1, 1},and B = {−2, 1}.Thensup(A)=1, sup(B)=1, but AB = {−2, 1, 1, 2},sosup(AB)=2 =1 1=1
2.2.10. (a)Let > 0.Noticethatsup(S) isnotanupperboundfor S sinceitisstrictlylessthansup(S),thesmallestupperbound.So, thereexistsan s ∈ S withsup(S) <s.That s ≤ sup(S)follows fromthefactthatsup(S)isanupperboundfor S.(b)Let > 0. Noticethatinf(S)+ isnotalowerboundfor S sinceitisstrictly greaterthaninf(S),thelargestupperbound.So,thereexistsan
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IntroductiontoAnalysis:SolutionsManual s ∈ S with s< inf(S)+ .Thatinf(S) ≤ s followsfromthefact thatinf(S)isalowerboundfor S
2.2.11. Wedeterminewhetherornotinf(U )isgreaterorlessthansup(S). Supposeinf(U ) < sup(S).Theninf(U )isnotanupperboundfor S, andsothereexistsan s ∈ S withinf(U ) <s.Considerthenumber 1 2 (inf(U )+ s):thisislargerthaninf(U ),andsothereexistsa u ∈ U with u< 1 2 (inf(U )+ s) <s.Thiscontradictsthefactthatevery elementin U isanupperboundfor s.Theargumentthatinf(U )is notgreaterthansup(S)issimilar.
2.2.12. Let x ∈ A.Then f (x)+ g(x) ≤ sup(f (A))+sup(g(A)).Therefore, sup(f (A))+sup(g(A))isanupperboundfor(f + g)(A),andthe resultfollows,sincesup((f + g)(A))isthesmallestupperboundof (f + g)(A).Anexamplethatillustratesinequalityis A =[ 1, 1], f (x)= x,and g(x)= x.Wehave(f + g)(x)=0forall x,but sup(f (A))=sup(g(A))=1.
2.2.13. WeuseMathematicalInductiononthecardinalityof S.Suppose S = {x1} containsoneelement,thensup(S)= x1 ∈ S.Nowsuppose S = {x1,...,xn,xn+1},andanyfinitesetcontaining n orfewer elementshasasupremumwithintheset.Consider S −{xn+1} Thissethas n elements,andthereforesup(S −{xn+1})= xi ∈ S forsome i =1,...,n.Now S =(S −{xn+1}) ∪{xn+1},andsoby exercise6,sup(S)iseither xi or xn+1,whicheverislarger.
Section2.3:TheCompletenessAxiom.
2.3.1. (a)Theinfis0.Thisisclearlyalowerbound,andif0 < ,the ArchimedeanPropertyassertstheexistenceof N ∈ N forwhich 1 N < ,so isnotalowerboundforthisset.(b)Thesupis1.This isclearlyanupperbound,andif b< 1,then0 < 1 b,andso 0 < √1 b,soagainbytheArchimedeanProperty,thereexistsan N ∈ N forwhich 1 N < √1 b,sothat b< 1 1 N 2 .(c)Thesupis1. Thisisclearlyanupperbound,andif b< 1thereexistsarational numberbetweenthem,since Q isdensein R.
2.3.2. Theinfexistsin R .Supposethatinf(S)= γ, andinf(S)= δ,with γ<δ.Botharegreatestlowerbounds.Since γ issuch,and δ>γ, δ mustnotbealowerbound,whichisacontradictiontoitbeing aninfimumofthisset.
2.3.3. Supposethat N wereboundedaboveby M .Thenconsiderapplying theArchimedeanPropertyto = 1 M .Thereexistsan N ∈ N for which 1 N < 1 M ,or, M<N ,contradicting M beinganupperbound for N
2.3.4. Let > 0begiven.Since S isnotboundedabove,thereexistsan s ∈ S forwhich 1 <s.Orrather, 1 s < .
TheRealNumbers 11
2.3.5. Let s ∈ S bethelargestelementin S,andconsider x = s +1and y = s +2.Everyelementin s isstrictlysmallerthan x (whichis thesmallerof x and y),andsothereisnoelementof S between x and y.Thus, S isnotdensein R.
2.3.6. Anexampleis N,whichiscountablydense,butthereisnonatural numberbetween 1 4 and 1 2 .
2.3.7. Let α begiven.Forany n ∈ N, α 1 n <α,sobythedensenessof Q in R,thereexistsan rn ∈ Q sothat α 1 n <rn <α.Since 1 n can bearbitrarilycloseto0bytheArchimedeanProperty, α 1 n can bemadearbitrarilycloseto α,andso rn canbefoundarbitrarily closeto α
2.3.8. Let a<b begiven,andsupposetherewereonlyfinitelymany rationalnumbersbetween a and b,andthat r isthelargestofthese. Then r<b andsothereexistsarationalnumber q sothat r<q<b. This q isalsobetween a and b andcontradictstheassumptionthat r wasthelargestoftherationalnumbersbetween a and b.
2.3.9. Wefirstclaimthatif q isrational,then √2q isnotrational.If √2q = r ∈ Q,then √2= r q .But r q isrationalsinceboth r and q are,but √2isnotrational.Nowlet a<b begiven.Since Q isdense in R,thereexistsa q ∈ Q forwhich a √2 <q< b √2 .So, √2q isan irrationalnumber,and a< √2q<b