Introduction To Real Analysis Fall 2014 Lecture Notes

Introduction to Real AnalysisFall 2014 Lecture NotesVern I. PaulsenNovember 6, 2014

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Contents1 Metric Spaces1.1 Definition and Examples . . . . . . . . . . .1.1.1 Vector Spaces, Norms and Metrics .1.2 Open Sets . . . . . . . . . . . . . . . . . . .1.2.1 Uniformly Equivalent Metrics . . . .1.3 Closed Sets . . . . . . . . . . . . . . . . . .1.4 Convergent Sequences . . . . . . . . . . . .1.4.1 Further results on convergence in Rk1.4.2 Subsequences . . . . . . . . . . . . .1.5 Interiors, Closures, Boundaries of Sets . . .1.6 Completeness . . . . . . . . . . . . . . . . .1.7 Compact Sets . . . . . . . . . . . . . . . . .2 Finite and Infinite Sets, Countability3 Continuous Functions3.1 Functions into Euclidean space . . . . . . . . . . . .3.2 Continuity of Some Basic Functions . . . . . . . . .3.3 Continuity and Limits . . . . . . . . . . . . . . . . .3.4 Continuous Functions and Compact Sets . . . . . . .3.5 Connected Sets and the Intermediate Value Theorem55910121417212324263037.4145474951524 The Contraction Mapping Principle554.1 Application: Newton’s Method . . . . . . . . . . . . . . . . . 574.2 Application: Solution of ODE’s . . . . . . . . . . . . . . . . . 585 Riemann and Riemann-Stieltjes Integration635.1 The Riemann-Stieltjes Integral . . . . . . . . . . . . . . . . . 685.2 Properties of the Riemann-Stieltjes Integral . . . . . . . . . . 745.3 The Fundamental Theorem of Calculus . . . . . . . . . . . . . 793

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Chapter 1Metric SpacesThese notes accompany the Fall 2011 Introduction to Real Analysis course1.1Definition and ExamplesDefinition 1.1. Given a set X a metric on X is a function d : X X Rsatisfying:1. for every x, y X, d(x, y) 0,2. d(x, y) 0 if and only if x y,3. d(x, y) d(y, x),4. (triangle inequality) for every x, y, z X, d(x, y) d(x, z) d(z, y).The pair (X, d) is called a metric space.Example 1.2. Let X R and set d(x, y) x y , then d is a metric onR. We call this the usual metric on R.To 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). Since d(x, y) 0 ifand only if x y 0 if and only if x 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 1.3 (The taxi cab metric). Let X R2 . Given x (x1 , x2 ), y (y1 , y2 ), set d(x, y) x1 y1 x2 y2 , then d is a metric on R2 .5

6CHAPTER 1. METRIC SPACESWe verify (1)–(4). (1) and (3) are obvious. For (2): d(x, y) 0 iff x1 y1 x2 y2 0. But since both terms in the sum are non-negativefor the sum to be 0, each one must be 0. So d(x, y) 0 iff x1 y1 0AND x2 y2 0 iff x1 y1 and x2 y2 iff x (x1 , x2 ) (y1 , y2 ) y.Finally 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 by d1 (x, y).Example 1.4. A different metric on R2 . For x (x1 , x2 ), y (y1 , y2 ) setd (x, y) max{ x1 y1 , x2 y2 }. So the distance between two points isthe larger of these two numbers.We only check the triangle inequality. Let z (z1 , z2 ) be another point.We have two cases to check. Either x1 y1 d(x, y) OR d(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 true.Example 1.5. In this case we let X be the set of all continuous real-valuedfunctions on [0, 1]. We use three facts from Math 3333:1. if f and g are continuous on [0,1], then f g is continuous on [0,1],2. if f is continuous on [0,1], then f is continuous on [0,1],3. if h is continuous on [0,1], then there is a point 0 t0 1, so thath(t) h(t0 ) for every 0 t 1. That is h(t0 ) max{h(t) : 0 t 1}.

1.1. DEFINITION AND EXAMPLES7Now given f, g X, we set d(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 maximum.We now show that d is a metric on X. Clearly, (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 every t. But then this means that f (t) g(t) for everyt, and so f g. So d(f, g) 0 implies f g. Also f g implies d(f, g) 0so (2) holds. Clearly (3) holds. Finally to see the triangle inequality, we letf, g, h be three continuous functions on [0,1]. that is, f, g, h X. We mustshow that d(f, g) d(f, h) d(h, g).We know that there is a point t0 , 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 1.6 (Euclidean space, Euclidean metric). Let X Rn the set ofreal n-tuples. For x (a1 , . . . , an ) and y (b1 , . . . , bn ) we setpd(x, y) (a1 b1 )2 · · · (an bn )2 .This defines a metric on Rn , which we will prove shortly. This metricis called the Euclidean metric and (Rn , d) is called Euclidean space.Sometimes, we will write d2 for the Euclidean metric.It is easy to see that the Euclidean metric satisfies (1)–(3) of a metric.It is harder to prove the triangle inequality for the Euclidean metric thansome of the others that we have looked at. This requires some results first.Lemma 1.7. Let p(t) at2 bt c with a 0. If p(t) 0 for every t R, then b2 4ac.Proposition 1.8 (Cauchy-Schwarz Inequality). Let a1 , ., an , b1 , ., bn bereal numbers. Thenqq22 a1 b1 · · · an bn a1 · · · an b21 · · · b2n .

8CHAPTER 1. METRIC SPACESProof. Look at p(t) (ta1 b1 )2 · · · (tan bn )2 0 for all t.pCorollary1.9(Minkowski’sInequality).(a1 b1 )2 · · · (an bn )2 pp2222a1 · · · an b1 · · · bn .Proof. Let 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(a1 b1 · · · an bn ) b21 · · · b2nqq a21 · · · a2n 2 a21 · · · a2n b21 · · · b2n b21 · · · b2nqq ( a21 · · · a2n b21 · · · b2n )2 (RHS)2Now prove the triangle inequality.Example 1.10 (The discrete metric). Let X be any non-empty set anddefine(1 x 6 yd(x, y) .0 x yThen this is a metric on X called the discrete metric and we call (X, d)a discrete metric space.Example 1.11. When (X, d) is a metric space and Y X is a subset,then restricting the metric on X to Y gives a metric on Y, we call (Y, d) asubspace of (X,d).Problem 1.12. Let X R, define d(x, y) x y 1. Show that this isNOT a metric.Problem 1.13. Let X R, define d(x, y) x2 y 2 . Show that this isNOT a metric.Problem 1.14. Let X R, define d(x, y) x y x2 y 2 . Prove thatthis is a metric on R.Problem 1.15. Let X Rn for x (a1 , ., an ), y (b1 , ., bn ) defined1 (x, y) a1 b1 · · · an bn .Prove that this is a metric.Problem 1.16. Let X Rn for x and y as before, defined (x, y) max{ a1 b1 , ., an bn }.Prove that this is a metric.

1.1. DEFINITION AND EXAMPLES1.1.19Vector Spaces, Norms and MetricsDefinition 1.17. Let X be a real vector space. A function k · k : X R iscalled a norm provided that1. kxk 0 for all x,2. kxk 0 if and only if x 0,3. krxk r kxk for every r R and x X,4. (triangle inequality) kx yk kxk kyk.The next result summarizes the relation between this concept and norms.Proposition 1.18. Let X be a real vector space and let k · k be a norm onX. Then setting d(x, y) kx yk defines a metric on X.Proof. We have that d(x, y) kx yk 0 so property (1) of a metric holds.We have that d(x, y) 0 iff kx yk 0 iff x y 0 iff x y, soproperty (2) of a metric holds.Also, d(x, y) kx yk k( 1)(y x)k 1 · ky xk ky xk d(y, x), so property (3) of a metric holds.Finally, d(x, y) kx yk kx z z yk kx zk kz yk d(x, z) d(z, y) so property (4) of a metric holds.The metric that one gets from a norm by using this result is called themetric induced by the norm.Example 1.19. For x (a1 , ., an ) Rn if we set kxk2 then this defines a norm called the Euclidean norm.pa21 · · · a2nTo see that this is a norm, it is easy to check the first 3 propertiesneeded for a norm. To see the fourth property, use Corollary 1.9. Thus, theEuclidean metric is the metric induced by the Euclidean norm via Proposition 1.18.Example 1.20. For x (a1 , a2 ) R2 if we set kxk1 a1 a2 andkxk max{ a1 , a2 } then these both define norms on R2 and the metricsthat they induce are respectively, d1 –the taxi cab metric and d .

101.2CHAPTER 1. METRIC SPACESOpen SetsDefinition 1.21. Let (X,d) be a metric space, fix x X and r 0. Theopen ball of radius r centered at x is the setB(x; r) {y X : d(x, y) r}.Example 1.22. In R with the usual metric, B(x; r) {y : x y r} {y : x r y x r} (x r, x r).Example 1.23. In R2 with the Euclidean metric, x (x1 , x2 ), then B(x; r) {(y1 , y2 ) : (x1 y1 )2 (x2 y2 )2 r2 }, which is a disk of radius r centeredat x.Example 1.24. In R3 with the Euclidean metric, B(x; r) really is an openball of radius r. This example is where the name comes from.Example 1.25. In (R2 , d ) we haveB(x; r) {(y1 , y2 ) : x1 y1 r and x2 y2 r} {(y1 , y2 ) : x1 r y1 x1 r and x2 r y2 y2 r}.So now an open “ball” is actually an open square, centered at x with sidesof length 2r.Example 1.26. In (R2 , d1 ) we have B((0, 0); 1) {(x, y) : x 0 y 0 1} which can be seen to be the “diamond” with corners at (1,0),(0,1),(-1,0),(0,-1).Example 1.27. When X {f : [0, 1] R f is continuous} and d(f, g) max{ f (t) g(t) : 0 t 1}, then B(f ; r) {g : g is continuous and f (t) r g(t) f (t) r, t}. This can be pictured as all continuous functions gwhose graphs lie in a band of width r about the graph of f.Example 1.28. If we let R have the usual metric and let Y [0, 1] Rbe the subspace, then when we look at the metric space Y we have thatB(0; 1/2) [0, 1/2) ( 1/2, 1/2) Y.Example 1.29. When we let X be a set with the discrete metric and x X,then B(x; r) {x} when r 1. When r 1, then B(x; r) X.Definition 1.30. Given a metric space (X,d) a subset O X is called openprovided that whenever x O, then there is an r 0 such that B(x; r) O.

1.2. OPEN SETS11Showing that sets are open really requires proof, so we do a few examples.Example 1.31. In R with the usual metric, an interval of the form (a, b) {x : a x b} is an open set.Proof. Given x (a, b), let r min{x a, b x}. If y B(x; r) then x r y x r. Since y x r x (b x) b and y x r x (x a) awe have a y b. So B(x; r) (a, b). Thus, (a, b) is open.So “open intervals” really are “open sets”.Example 1.32. In R2 with the Euclidean metric, a rectangle of the formR {(y1 , y2 ) : a y1 b, c y2 d} is an open set.Proof. Given x (x1 , x2 ) R, let r min{x1 a, b x1 , x2 c, d x2 }.We claim that B(x; r) R. To see this let y (y1 , y2 ) B(x; r), then(y1 x1 )2 (y2 x2 )2 r2 .Looking at just one term in the sum, we see that (y1 x1 )2 r2 and so y1 x1 r. This implies that x1 r y1 x1 r. As before we see thatx1 r b and x1 r a, so a y1 b.Similarly, we get that c y2 d, so y R.Example 1.33. In R with the usual metric an interval of the form [a, b) isnot open.Proof. Consider the point a [a, b). Any ball about this point is of the formB(a; r) (a r, a r) and this contains points that are less than a and sonot in the set [a, b).CAREFUL! If we let Y [0, 1) R be the subspace. Then in Y the setO [0, 1/2) is open!(Explain why.)The next result justifies us calling B(x; r) an open ball.Proposition 1.34. Let (X,d) be a metric space, fix x X and r 0. ThenB(x; r) is an open set.Theorem 1.35. Let (X,d) be a metric space. Then1. the empty set is open,2. X is open,3. the union of any collection of open sets is open,4. the intersection of finitely many open sets is open.