Introduction to Real Analysis Fall 2014 Lecture Notes

1 Introduction to real AnalysisFall 2014 Lecture NotesVern I. PaulsenNovember 6, 20142 Contents1 Metric Definition and Examples .. Spaces, Norms and Metrics .. Open Sets .. Equivalent Metrics .. Closed Sets .. Convergent Sequences .. results on convergence inRk.. Interiors, Closures, Boundaries of Sets .. Completeness .. Compact Sets .. 302 Finite and Infinite Sets, Countability373 Continuous Functions into Euclidean space .. Continuity of Some Basic Functions.

2 Continuity and Limits .. Continuous Functions and Compact Sets .. Connected Sets and the Intermediate Value Theorem .. 524 The Contraction Mapping Application: Newton s Method .. Application: Solution of ODE s .. 585 Riemann and Riemann-Stieltjes The Riemann-Stieltjes Integral .. Properties of the Riemann-Stieltjes Integral .. The Fundamental Theorem of Calculus .. 7934 CONTENTSC hapter 1 Metric SpacesThese Notes accompany the fall 2011 Introduction to real Definition and ExamplesDefinition a set X ametricon X is a functiond:X X Rsatisfying:1.

3 For everyx,y X, d(x,y) 0, (x,y) = 0if and only ifx=y, (x,y) =d(y,x),4. (triangle inequality) for everyx,y,z X, d(x,y) d(x,z) +d(z,y).The pair(X,d)is called ametric setd(x,y) =|x y|,thendis a metric call this theusual prove it is a metric we verify (1) (4). For (1):d(x,y) =|x y| 0,by the definition of the absolute value functions so (1). Sinced(x,y) = 0 ifand only if|x y|= 0 if and only ifx=y,(2) follows. (3) follows sinced(x,y) =|x y|=|y x|=d(y,x).Finally, for (4),d(x,y) =|x y|=|x z+z y| |x z|+|z y|=d(x,z) +d(z,y).Example (The taxi cab metric).

4 LetX= (x1,x2),y=(y1,y2),setd(x,y) =|x1 y1|+|x2 y2|,thendis a metric 1. METRIC SPACESWe verify (1) (4). (1) and (3) are obvious. For (2):d(x,y) = 0 iff|x1 y1|+|x2 y2|= since both terms in the sum are non-negativefor the sum to be 0, each one must be 0. Sod(x,y) = 0 iff|x1 y1|= 0 AND|x2 y2|= 0 iffx1=y1andx2=y2iffx= (x1,x2) = (y1,y2) = to see (4):d(x,y) =|x1 y1|+|x2 y2|=|x1 z1+z1 y1|+|x2 z2+z2 y2| |x1 z1|+|z1 y1|+|x2 z2|+|z2 y2|=d(x,z) +d(z,y).We often denote the taxi cab metric byd1(x,y).Example different metric (x1,x2),y= (y1,y2)setd (x,y) = max{|x1 y1|,|x2 y2|}.

5 So the distance between two points isthe larger of these two only check the triangle inequality. Letz= (z1,z2) be another have two cases to check. Either|x1 y1|=d(x,y) ORd(x,y) =|x2 y2|.Case 1:d(x,y) =|x1 y1|.Now notice that|x1 z1| max{|x1 z1|,|x2 z2|}=d(x,z).Similarly,|z1 y1| max{|z1 y1|,|z2 y2|}.Hence,d(x,y) =|x1 y1|=|x1 z1+z1 y1| |x1 z1|+|z1 y1| d(x,z) +d(z,y).Case 2:d(x,y) =|x2 y2|.Now use|x2 z2| max{|x1 z1|,|x2 z2|}=d(x,z).Similarly,|z2 y2| max{|z1 y1|,|z2 y2|}.Hence,d(x,y) =|x2 y2|=|x2 z2+z2 y2| |x2 z2|+|z2 y2| d(x,z) +d(z,y).So in each case the triangle inequality is true, so it is this case we letXbe the set of all continuous real -valuedfunctions on[0,1].

6 We use three facts from Math 3333:1. iffandgare continuous on [0,1], thenf gis continuous on [0,1],2. iffis continuous on [0,1], then|f|is continuous on [0,1],3. ifhis continuous on [0,1], then there is a point0 t0 1,so thath(t) h(t0)for every0 t ish(t0) = max{h(t) : 0 t 1}. DEFINITION AND EXAMPLES7 Now givenf,g X,we setd(f,g) = max{|f(t) g(t)|: 0 t 1}.Note that by (1) and (2)|f g|is continuous and so by (3) there is apoint where it acheives its now show thatdis a metric , (1) holds. Next, ifd(f,g) = 0,then the maximum of|f(t) g(t)|is 0, so we must have that|f(t) g(t)|= 0 for then this means thatf(t) =g(t) for everyt,and sof= (f,g) = 0 impliesf= (f,g) = 0so (2) holds.

7 Clearly (3) holds. Finally to see the triangle inequality, we letf,g,hbe three continuous functions on [0,1]. that is,f,g,h mustshow thatd(f,g) d(f,h) +d(h,g).We know that there is a pointt0,0 t0 1,so thatd(f,g) = max{|f(t) g(t)|: 0 t 1}=|f(t0) g(t0)|.Hence,d(f,g) =|f(t0) g(t0)|=|f(t0) h(t0) +h(t0) g(t0)| |f(t0) h(t0)|+|h(t0) g(t0)| max{|f(t) h(t)|: 0 t 1}+ max{|h(t) g(t)|: 0 t 1}=d(f,h) +d(h,g)Example (Euclidean space, Euclidean metric).LetX=Rnthe set ofreal n-tuples. Forx= (a1,..,an)andy= (b1,..,bn)we setd(x,y) = (a1 b1)2+ + (an bn) defines a metric onRn,which we will prove shortly.

8 This metricis called theEuclidean metricand(Rn,d)is calledEuclidean , we will writed2for the Euclidean is easy to see that the Euclidean metric satisfies (1) (3) of a is harder to prove the triangle inequality for the Euclidean metric thansome of the others that we have looked at. This requires some results (t) =at2+bt+cwitha (t) 0for everyt R ,thenb2 (Cauchy-Schwarz Inequality).Leta1,..,an,b1,..,bnbereal numbers. Then|a1b1+ anbn| a21+ +a2n b21+ + 1. METRIC atp(t) = (ta1+b1)2+ + (tan+bn)2 0 for (Minkowski s Inequality).

9 (a1+b1)2+ + (an+bn)2 a21+ +a2n+ b21+ + LHS denote the left hand side, RHS the right hand side of theinequality. then(LHS)2= (a1+b1)2+ +(an+bn)2=a21+ +a2n+2(a1b1+ +anbn)+b21+ +b2n a21+ +a2n+ 2 a21 +a2n b21+ +b2n+b21+ +b2n= ( a21+ +a2n+ b21+ +b2n)2= (RHS)2 Now prove the triangle (The discrete metric).LetXbe any non-empty set anddefined(x,y) ={1x6=y0x= this is a metric onXcalled thediscrete metricand we call(X,d)adiscrete metric (X,d)is a metric space andY Xis a subset,then restricting the metric on X to Y gives a metric on Y, we call(Y,d)asubspace of (X,d).}

10 Problem ,defined(x,y) =|x y|+ that this isNOT a ,defined(x,y) =|x2 y2|.Show that this isNOT a ,defined(x,y) =|x y|+|x2 y2|.Prove thatthis is a metric (a1,..,an),y= (b1,..,bn)defined1(x,y) =|a1 b1|+ +|an bn|.Prove that this is a before, defined (x,y) = max{|a1 b1|,..,|an bn|}.Prove that this is a DEFINITION AND Vector Spaces, Norms and MetricsDefinition a real vector space. A function :X Riscalled anormprovided that1. x 0for allx,2. x = 0if and only ifx= 0,3. rx =|r| x for everyr Randx X,4. (triangle inequality) x+y x + y.