Math 401 - Introduction to Real Analysis

1 math 401 - Introduction to real AnalysisTopics for Midterm I - Review1 - BijectionsA mapf:A7 Bis aninjectionif it is one-to-one, distinct elementsa1, a2 Ahavedistinct imagesf(a1)6=f(a2). The mapfis asurjectionif it is onto, every elementb Bisthe image of some element say that a mapf:A7 Bis abijectionif it isone-to-oneandonto. If a bijectionexists, we regard the two setsAandBas having the same number of elements. This allows us tocompare also sets with infinitely many - Mathematical inductionGiven a sequence of statementsP1, P2, P3, .., mathematical induction is a technique for prov-ing that all of the statements are true. Namely, one has to show that(i) The first statementP1is true.(ii) IfPkis true, then also the following statementPk+1is - Upper bound, supremumA setS IRisbounded aboveif there exists a numberusuch thatu xfor allx this caseuis called anupper bound. The smallest upper bound is calledsupremumandwritten (completeness of the real numbers).

2 If a setSis bounded above, then it has prove thatu= supS, one needs to show:(i)u xfor everyx S,(ii) For every >0, there exists a pointx Ssuch thatu < that (ii) is certainly true ifu - IntervalsAn open interval is a set of the form (a, b) ={x IR;a < x < b}.A closed interval is a set of the form [a, b] ={x IR;a x b}. Here one may havea= orb= + . In this case the interval is say that a sequence of intervalsIn= [an, bn] arenestedifI1 I2 I3 . Thishappens if and only ifa1 a2 a3 andb1 b2 b3 Theorem (intersection of nested intervals).Given a sequence of nested intervals [an, bn]which are closed and bounded, their intersection is non-empty. In addition, if their lengthsbn anshrink to zero, then the intersection contains exactly one - SequencesA sequence is a map fromNintoIR. It is usually denoted as (x1, x2, x3, ..), or (xn)n 1. Thesequence isboundedif all pointsxnare contained in a bounded interval [a, b].

3 It ismonotoneincreasingifx1 x2 x3 A sequence can be defined by directly assigning its values:xn=f(n).Alternatively, one can define the sequence by induction: (i)fix the initial valuex1, and then(ii) give a rule for computingxk+1from the previous - LimitsWe say that the sequence (xn)n 1converges tox, and writelimn xn=xif, for every >0 one can find a numberK( ) sufficiently large so thatx < xn< x+ (1)for alln K( ).Intuitively this means that, asngrows large, the numbersxnbecome closer and closer prove that limn xn=x, one has to study the inequality (1), and show that it is satisfiedfor all integersnsufficiently - Limit theoremsTheorem (sandwich).If (xn)n 1, (yn)n 1, and (zn)n 1are three sequences such thatxn yn znfor everyn 1, and iflimn xn=x= limn zn,then we also havelimn yn=x .2 Theorem (sums, products, quotients).If (xn)n 1and (yn)n 1are two sequences such thatlimn xn=x ,limn yn=y ,andc IRis a real number, thenlimn (xn+yn) =x+y ,limn c xn=c x ,limn (xn yn) =x y ,limn xnyn=xy(ify6= 0).

4 Theorem (monotone sequences). If the sequence (xn)n 1is bounded and monotone increasing,than it has a limit:limn xn= sup{xn;n 1}.Basic Problems Construct a bijection between two infinite sets. Using unique factorization, prove that certain numbers such as 5 are irrational. Work out proofs using mathematical induction. Decide whether a setS IRis bounded or not. Find its supremum. Given a setSand anumberu, prove thatu= supS. Given a sequence (xn)n 1, check if it converges or not. Prove that limn xn=x, using the definition of limit, or the basic theorems about limits. Study a sequence (xn)n 1defined inductively:xk+1=f(xk). Check if it converges and findits 401 - Introduction to real AnalysisTopics for Midterm II - Review8 - Convergence criteriaThe following theorems guarantee that a sequence (xn)n 1converges, even if we do not knowprecisely what the limit is. A sequence (xn)n 1isboundedif there exists a numberMlarge enough so thatxn [ M, M] for (monotone convergence).

5 Assume that the sequence is increasing, so thatx1 x2 x3 Then the sequence converges to some limitxif and only if it is bounded. In this caselimn xn= sup{xn;n 1}. A sequence (xn)n 1is aCauchy sequenceif, for every >0 one can find an integerH( )large enough so that|xn xm|< for alln, m > H( ). Intuitively this means that, whenm, n , the numbersxm, xnget closer and closer to each (Cauchy criterion).A sequence (xn)n 1converges to some limitxif and only if it isa Cauchy (Bolzano - Weierstrass).If the sequence (xn)n 1is bounded, then one can selectinteger numbersn1< n2< n3< , such that the subsequencexn1, xn2, xn3, ..converges to - Divergent sequencesWe say that the sequence (xn)n 1tends to + , and write limn xn= + , if for every(arbitrarily large) IRthere exists a numberK( ) such thatxn> for every integern K( ).Theorem (unbounded monotone sequences).If the sequence (xn)n 1is monotone increasingand unbounded, then limn xn= +.

6 Theorem (comparison).Ifxn ynfor everyn, and if limn xn= + , then we also havelimn yn= + .410 - SeriesGiven a sequence (xn)n 1, we consider the infiniteseries n=1xn=x1+x2+x3+ The correspondingsequence of partial sumsis defined ass1=x1,s2=x1+x2, sk=x1+x2+ +xk, If the sequence of partial sumsskhas a limit, we say that the series is convergent. We then define n=1xn= limk following theorems guarantee that a series (comparison).Assume 0 xn ynfor the series n=1ynconverges, then the series n=1xnconverges as the series n=1xndiverges, then the series n=1yndiverges as use the above theorem, it is useful to keep in mind that:the series n=11np{converges ifp >1,diverges ifp series n=0an{converges if|a|<1,diverges if|a| should also remember the formula for the partial sums1 +a+a2+ +ak=1 ak+11 ahencelimk (1 +a+a2+ +ak) = limk 1 ak+11 a=11 aif|a|< (ratio test).Assume thatlimn |xn+1||xn|=L < the series n= , the series xnconverges if the termsxnbecome smaller and smaller ( approachzero) quickly - Limits of functionsDefinition of limit:Consider a functionf:A7 IR.}}

7 We say thatlimx cf(x) =Lif, for every >0 one can find >0 such that f(x) L < for allx Asuch that|x c|< , x6=c .Theorem (sequential criterion).One has limx cf(x) =Lif and only if, for every sequencexnconverging toc, the sequencef(xn) converges (properties of limits).Assume thatlimx cf(x) =L ,limx cg(x) =M .Thenlimx c(f(x) +g(x))=L+M ,limx c(f(x) g(x))=L M ,limx ca f(x) =aL .IfM6= 0,then also limx cf(x)g(x)= (comparison).Assume thatf, g, hare three functions defined on the same domainA, withf(x) g(x) h(x) for allx A. If limx cf(x) =L= limx ch(x), then we also havelimx cg(x) = - Continuous functionsDefinition of continuous function:A functionf:A7 IRis continuous at a pointc Aiflimx cf(x) =f(c). This means that, for every >0 there exists >0 such that f(x) f(c) < for allx Asuch that|x c|< .We say that a functionf:A7 IRis continuous iffis continuous at every point of of continuous functions are:f(x) =a(constant function),f(x) =x,f(x) = sinx,f(x) = cosx,f(x) = (more continuous functions).

8 Letf, gbe continuous functions, defined on the samedomainA. Then the functionsf+g,f g,a fare also continuous. Moreover, the quotientfunctionh(x) =f(x)/g(x) is continuous at every pointxwhereg(x)6= (composition of continuous functions).Iff:A7 IRandg:B7 IRarecontinuous functions, withf(A) B, then the composed maph(x) =g(f(x))is also - Continuous functions on an [a, b] be a closed interval, and letf: [a, b]7 IRbe a continuous function. Thenthere exists pointsx andx wherefattains its minimum and its maximum values. Namelyf(x ) =m= infx [a,b]f(x),f(x ) =M= infx [a,b]f(x).Moreover, the imagef([a, b])is precisely the closed interval [m, M]. In other words,fattains allthe intermediate values between the minimum and the : [a, b]7 IRbe a continuous function. Thenfisuniformly continuous, in thesense that, given >0, one can find >0 such that f(x) f(y) < for allx, y [a, b] such that|x y|<.

9 Say that a functionf:A7 IRisLipschitz continuousif there exists a constantKsuch that f(x) f(y) K|x y|for allx, y A .In this case, the functionfis uniformly - The derivativeLetfbe a function defined in a neighborhood of a pointc. Thederivativeoffatcisf (c) = limx cf(x) f(c)x cprovided that the above limit exists. In this case we say differentiable atc, thenfis continuous atc. However, a function may becontinuous but not rules:Iff, gare differentiable at the pointc, and is any number, then( f) (c) = f (c),(f+g) (c) =f (c) +g (c)product rule: (f g)(c) =f (c)g(c) +f(c)g (c),quotient rule:(fg) (c) =f (c)g(c) f(c)g (c)g2(c)(ifg(c)6= 0),chain rule:(g f) (c) =g (f(c))f (c).Here (g f)(x) =g(f(x)) is the composed thatf, gare inverse of each other, so thaty=f(x) impliesx=g(y). Differentiatingthe equalityg(f(x)) =xusing the chain rule we obtaing (y)f (x) = 1,henceg (y) =1f (x)where the pointsx, yare related byy=f(x),x=g(y).

10 14 - The mean value theoremIf a differentiable functionf: [a, b]7 IRattains a local maximum (or a local minimum) atsome interior pointc, witha < c < b, then its derivative satisfiesf (c) = (Rolle).If a differentiable functionf: [a, b]7 IRsatisfiesf(a) =f(b), then thereexists a pointa < c < bsuch thatf (c) = value : [a, b]7 IRis a differentiable function, then there exists a pointc [a, b] such thatf(b) f(a)b a=f (c)[slope of secant line] = [slope of tangent line at the pointc]Consequences of the mean value theorem: Iff (x) = 0 for allx [a, b], thenfis constant. Iff (x) =g (x) for allx [a, b], thenf gis constant, hence there exists a numberCsuchthatf(x) =g(x) +Cfor allx. Iff (x) 0 for allx [a, b], then the functionfis increasing. That means: ifx < ythenf(x) f(y). Iff (x)>0 for allx [a, b], then the functionfis strictly increasing. That means: ifx < ythenf(x)< f(y). Iff: [a, b]7 IRis differentiable andf (x) 0 forx < candf (x)<0, thenfattains itsmaximum at the Problems Check if a sequence is convergent, or properly divergent, using the definition or a comparisonmethod.