Analysis III. Exam 1

Size: px
Start display at page:

Download "Analysis III. Exam 1"

Transcription

1 Analysis III Math 414 Spring 27 Professor Ben Richert Exam 1 Solutions Problem 1 Let X be the set of all continuous real valued functions on [, 1], and let ρ : X X R be the function ρ(f, g) = sup f g (1) Prove that (X, ρ) is a metric space Solution We are required to show that: (a) for all f, g X, ρ(f, g) = ρ(g, f), (b) ρ(f, g) = if and only if f = g, and (c) for all f, g, h X, ρ(f, g) ρ(f, h) + ρ(g, h) The first fact is obvious since ρ(f, g) = sup f g = sup g f = ρ(g, f) The second fact is similarly immediate since ρ(f, g) = t [, 1], that is, if f = g, and obviously f = g implies that ρ(f, g) = sup observe that ρ(f, g) = sup f g = sup f h + h g sup where the last inequality holds since sup f h + h g f h + h g sup f g = can only occur if f = g for all f g = sup = Finally we ( f h + h g ) for each t [, 1] by the usual triangle inequality We finish by observing that ( ) f h + h g sup f h + sup h g = ρ(f, h) + ρ(h, g) = ρ(f, h) + ρ(g, h) as required (2) Prove that the collection P of polynomials with domains taken to be [, 1] is dense in X Solution We need to show that the closure of P in X is all of X Since the closure of X is the union of P with all its accumulation points, it is enough to show that any f X is an accumulation point of P That is, it is enough to show that for f X and ɛ >, there is p P such that ρ(p, f) < ɛ By the Weierstrass approximation theorem, there does exist a polynomial p such that f(x) p(x) < ɛ for all x [, 1] It follows immediately that ρ(p, f) = sup f p < ɛ, and we are finished (3) Suppose that f j = j ( 1) k x 2k+1 (2k + 1)! for j = 1, 2, Show that f j sin x as elements of X k= Solution Let ɛ > be given We need to demonstrate that there is N > such that n > N implies ρ(sin x, f n ) = sin x f n (x) < ɛ sup

2 We know that sin x is infinitely differentiable on ( 2, 2), and hence by Taylor s approximation theorem, (with f(x) = sin x for the sake of notation) j sin x = f (k) () xk k! + f (j+1) (x t)j dt j! Of course, The pattern is that dk dx k sin x x= = and thus for each j N, = sin x = j ( 1) k x 2k+1 (2k + 1)! + k= This implies that sin x f j (x) = k= d dx sin x x= = cos() = 1 d 2 dx 2 sin x x= = sin() = d 3 dx 3 sin x x= = cos() = 1 d 4 dx 4 sin x x= = sin() = d 5 dx 5 sin x x= = cos() = 1 2j+1 k= 1 if k = 4l if k = 4l + 3 if k is even f (k) () xk k! + dt = f j (x) + dt f (2j+2) Since f (2j+2) = ± sin t or ± cos t, it follows that f (2j+2) 1 and hence f (2j+2) (x t) 2j+1 dt (x t) 2j+1 dt But (x t) 2j+1 is positive on [, x], so (x t) 2j+1 dt = (x t) 2j+1 (x t)2j+2 dt = (2j + 2)! since x [, 1] Now let N be such that (2N + 2)! > 1 It follows that for all n > N, ɛ x dt (x t) 2j+1 dt dt = ()2j+2 (2j + 2)! + (x)2j+2 (2j + 2)! 1 (2j + 2)! sin x f n (x) 1 (2n + 2)! 1 (2N + 2)! < ɛ, and hence ρ(sin x, f n ) < ɛ for all n > N This completes the proof (4) Suppose that F is a closed, infinite, equicontinuous, equibounded family of functions in X Prove that F is compact in X

3 Solution We must demonstrate that any infinite {f n } F has a subsequence which converges in F Since F is closed, it contains its limit points, so it is simply enough to identify a convergent subsequence Of course, {f n } is an equicontinuous, equibounded family of functions, and hence, by the Arzela-Ascoli theorem, has an accumulation point, call it f Choose n 1 N such that ρ(f n1, f) < 1, an integer n 2 > n 1 such that ρ(f n2, f) < 1/2, and in general, n j > n j 1 such that ρ(f nj, f) < 1/j (in each case we can do this because f is an accumulation point, and hence there are infinitely many sequence elements of index larger than any given N in B(f, ɛ)) It follows that the subsequence {f nj } converges to f (given ɛ > let N = 1 ɛ, whence ρ(f n, f) < 1 n < 1 ɛ for all n > N as required) We N conclude that any sequence of F has a subsequence which converges in F, which completes the proof (5) Suppose that F is a closed family of functions of X with f(s) f < s t for all f F, s, t [, 1] Suppose further that there is a function g : [, 1] R such that F is compact 1 g(x) dx exists and g(x) f(x) for all f F Prove that Solution Since g is Riemann integrable on [, 1] it is bounded there, and since f(x) g(x) for all x [, 1], f F, it follows that F is equibounded Note also that F is equicontinuous To wit, for ɛ > given, let δ = ɛ 2 Then for all s, t [, 1] such that s t < δ, we have f(s) f < s t < δ = ɛ 2 = ɛ for all f F It follows that F is a closed, equicontinuous, equibounded family, so by the previous problem, it is compact Problem 2 In class we proved that countable unions of measurable sets are measurable Repeat this construction, only this time from the point of view of intersections Note: I expect you to redevelop the theory from the definitions, focusing on intersections where in class we considered unions Solution We want to show that if E j is measurable for all j N, then j=1e j is measurable Defintion 1 Given E R let m (E) = inf I j where the infimum is taken over coverings of E by collections {I j } E ji j of open intervals and I j is the usual length Lemma 1 If A B R, the m (A) m (B) Proof This is clear since A B I j for each open covering of B Lemma 2 For sets A, B R, m (A B) m (A) + m (B) Proof This is clear If A j I j and B i J i, then A B j I j i J i j Defintion 2 A set E R is said to be measurable if for all A R, m (A) = m (A E) + m (A E c ) Lemma 3 A set E R is measurable if and only if E c is measurable Proof This follows because the equation m (A) = m (A E) + m (A E c ) is symmetric in E and E c Lemma 4 To show that E R is measurable, it is enough to show that m (A) m (A E) + m (A E c ) for an arbitrary A R Proof Since A = (A E) (A E c ), it follows by lemma (2) that m (A) m (A E) + m (A E c ) holds automatically Lemma 5 Suppose that E 1 and E 2 are measurable sets Then E 1 E 2 is measurable Proof Let A R, and note that A (E 1 E 2 ) c = ((A E 1 ) E 2 ) (A E c 1) Then m (A (E 1 E 2 )) + m (A (E 1 E 2 ) c ) m ((A E 1 ) E 2 ) + m ((A E 1 ) E c 2) + m (A E c 1) = m (A E 1 ) + m (A E1) c = m (A) The first inequality holds by lemma (2), the second holds because E 2 is measurable, and the last holds because E 1 is measurable This is enough by lemma (4) Lemma 6 Suppose that E 1 and E 2 are measurable sets Then E 1 E 2 is measurable Proof This follows immediately from lemma (5) and lemma (3) Lemma 7 Suppose that E 1,, E n is a collection of measurable sets Then n j=1e j is measurable

4 Proof We do induction In fact, if n = 1 this is a tautology If n = 2, then the result follows from lemma (5) Suppose that n > 2 By induction, we can conclude that E := E 1 E n 1 is measurable, and again by lemma (5) E E n is measurable Since E E n = E 1 E n, we are finished Lemma 8 Suppose that {E i } is a collection of measurable sets Then there are measurable sets {F i } such that F i F j = R for i j E j = F j Proof Let F 1 = E 1 F 2 = E 2 E1 c F 3 = E 3 (E 2 E 1 ) c F j = E j (E j 1 E 1 ) c By lemmas (3,6,7), F i is measurable for all i If j > i, then ( F j F i = E j (E j 1 E 1 ) c) (E i (E i 1 E 1 ) c) = E j E c j 1 E c i E c 1 E i E c i 1 E c 1 = R since both E i and Ei c appear in the union Furthermore, note that since E i F i, it follows that E i F i If x F i, then by definition x E i (E i 1 E 1 ) c Of course, it is now enough to show that x E i for all i, that is, it is enough to show that the set A = {t x E t } is empty Note that 1 A by definition If A, then there is a smallest element, 1 < a A Thus, x E a 1 E 1, but x E a But x F a = E a (E a 1 E 1 ) c, which must then imply that x (E a 1 E 1 ) c, and this is a contradiction Lemma 9 Suppose that F 1,, F n are measurable sets and F i F j = R for i j Then F c n (F 1 F n 1 ) c = Proof This is clear since (F c n (F 1 F n 1 ) c ) c = (F n (F 1 F n 1 )) = (F n F 1 ) (F n F 1 ) = R R = R Lemma 1 Suppose that E 1 and E 2 are measurable sets and E 1 E 2 = The m (E 1 ) + m (E 2 ) = m (E 1 E 2 ) Proof Since E 1 is measurable, we have But E 1 E 2 =, and E 2 E c 1 = E 2, so the result is proved m (E 2 E 1 ) + m (E 2 E c 1) = m (E 2 ) Lemma 11 Suppose that F 1,, F n are measurable sets F i F j = R for i j Then m (A (F 1 F n ) c ) = Proof We do induction on n If n = 1 this is a tautology If n > 1, then write m (A (F 1 F n ) c ) = m (A ((F 1 F n 1 ) c F c n)) = m ((A F c n) (A ((F 1 F n 1 ) c )) = m (A Fn) c + m (A (F 1 F n 1 ) c ) n 1 n = m (A Fn) c + m (A Fi c ) = m (A Fi c ) n m (A Fi c ) where the third equality holds because Fn c and (F 1 F n 1 ) c are measurable and disjoint (see lemmas (9,1)), and the fourth inequality holds by the induction hypothesis Lemma 12 Suppose that {E j } is a countable collection of measurable sets Then m (A Ej c ) m (A ( E j ) c )

5 Proof Let ɛ > be given and suppose that m (A Ej c ) = e j Then (by the definition of outer measure) there is {I (j) i } such that i I (j) i (A Ej c ) and I (j) i < e j + ɛ/2 j Of course i so m (A ( j E j ) c ) j completes the proof i I (j) i = j j i I (j) i j (A E c j ) = A ( j E j ) c, e j + ɛ/2 j = ɛ + j e j = ɛ + j m (A E c j ) Since this is true for each ɛ >, this Lemma 13 Suppose that {E j } is a countable collection of measurable sets Then j E j is measurable Proof First, we find {F j } such that F i F j = R for i j and E j = F j (as guaranteed to exist by lemma (8)) Let G = F i = E i and G n = F 1 F n Then G n is measurable for all n N by lemma (7), and hence m (A) = m (A G n ) + m (A G c n) for all A R Since G n = F 1 F n F i = G, it follows that m (A G n ) m (A G) (by lemma (1)), and thus Now by lemma (11), we can write and this holds for all n We conclude that But by lemma (12), m (A) = m (A G n ) + m (A G c n) m (A G) + m (A G c n) m (A) m (A G) + m (A G c n) = m (A G) + m (A) m (A G) + n m (A Fi c ) m (A Fi c ) m (A Fi c ) m (A ( Fi c )) = m (A ( F i ) c ), and this certainly completes the proof, since now we have that m (A) m (A G) + m (A Fi c ) m (A G) + m (A ( F i ) c ) = m (A G) + m (A G c ), which is enough by lemma (4) Problem 3 Let {f j } be a sequence of continuous, real valued functions on R with the property that {f j (x)} is unbounded for all x Q Use the Baire Category theorem to prove that it cannot then be true that whenever t is irrational then the sequence {f j } is bounded Note: this is a somewhat challenging problem You need to start by deciding what complete metric space the Baire Category theorem should give you information about Solution We will show that if the sequence {f j } is bounded at each irrational t, then we can write R as a countable union of nowhere dense sets, which contradicts the Baire Category theorem (since R is a complete metric space) For j N, let { } A j = t R Q { f j } is bounded (not strictly) by j Let {q j } be an enumeration of Q, and let Q j = {q j } By hypothesis, every t R Q is in some A j, and of course j=1q j = Q, so clearly ( j=1q j ) ( i j=nftya j ) = R Also, this is a countable union, so to obtain the contradiction it is enough to show that Q j and A j for all j N are nowhere dense That Q j is nowhere dense is immediate, since Q j = {q j } consists of a single point For A j, we first show that it is closed Suppose that r is an accumulation point of A j We need to demonstrate that f N (r) < j for all N N, that is, we need to show that r A j So fix N N Then given ɛ >, there is δ > (since f N is

6 continuous) such that x r < δ implies that f N (x) f N (r) < ɛ But r is an accumulation point of A j, so there is t A j such that t r < δ and hence f N (r) < f N + ɛ j + ɛ Since ɛ was arbitrary, we conclude that f N (r) j But N was also arbitrary Hence the conclusion that {f j (r)} is bounded above by j and thus r A j as required Next we show that A j contains no open balls (this is enough since A j is its own closure) In fact, this is easy because (by definition) Q A j =, but there are rationals arbitrarily close to any irrational To be precise, suppose that t A j and ɛ > is given Then the ball B(t, ɛ) = {x R x t < ɛ} Aj because (by the density of the rationals in the reals) there is q Q such that q t < ɛ, but q A j We conclude that A j is nowhere dense By the Baire Category theorem, this completes the proof

THEOREMS, ETC., FOR MATH 515

THEOREMS, ETC., FOR MATH 515 THEOREMS, ETC., FOR MATH 515 Proposition 1 (=comment on page 17). If A is an algebra, then any finite union or finite intersection of sets in A is also in A. Proposition 2 (=Proposition 1.1). For every

More information

Math 209B Homework 2

Math 209B Homework 2 Math 29B Homework 2 Edward Burkard Note: All vector spaces are over the field F = R or C 4.6. Two Compactness Theorems. 4. Point Set Topology Exercise 6 The product of countably many sequentally compact

More information

Real Analysis Notes. Thomas Goller

Real Analysis Notes. Thomas Goller Real Analysis Notes Thomas Goller September 4, 2011 Contents 1 Abstract Measure Spaces 2 1.1 Basic Definitions........................... 2 1.2 Measurable Functions........................ 2 1.3 Integration..............................

More information

2.2 Some Consequences of the Completeness Axiom

2.2 Some Consequences of the Completeness Axiom 60 CHAPTER 2. IMPORTANT PROPERTIES OF R 2.2 Some Consequences of the Completeness Axiom In this section, we use the fact that R is complete to establish some important results. First, we will prove that

More information

Problem Set 2: Solutions Math 201A: Fall 2016

Problem Set 2: Solutions Math 201A: Fall 2016 Problem Set 2: s Math 201A: Fall 2016 Problem 1. (a) Prove that a closed subset of a complete metric space is complete. (b) Prove that a closed subset of a compact metric space is compact. (c) Prove that

More information

CLASS NOTES FOR APRIL 14, 2000

CLASS NOTES FOR APRIL 14, 2000 CLASS NOTES FOR APRIL 14, 2000 Announcement: Section 1.2, Questions 3,5 have been deferred from Assignment 1 to Assignment 2. Section 1.4, Question 5 has been dropped entirely. 1. Review of Wednesday class

More information

CHAPTER I THE RIESZ REPRESENTATION THEOREM

CHAPTER I THE RIESZ REPRESENTATION THEOREM CHAPTER I THE RIESZ REPRESENTATION THEOREM We begin our study by identifying certain special kinds of linear functionals on certain special vector spaces of functions. We describe these linear functionals

More information

ANALYSIS WORKSHEET II: METRIC SPACES

ANALYSIS WORKSHEET II: METRIC SPACES ANALYSIS WORKSHEET II: METRIC SPACES Definition 1. A metric space (X, d) is a space X of objects (called points), together with a distance function or metric d : X X [0, ), which associates to each pair

More information

Contents Ordered Fields... 2 Ordered sets and fields... 2 Construction of the Reals 1: Dedekind Cuts... 2 Metric Spaces... 3

Contents Ordered Fields... 2 Ordered sets and fields... 2 Construction of the Reals 1: Dedekind Cuts... 2 Metric Spaces... 3 Analysis Math Notes Study Guide Real Analysis Contents Ordered Fields 2 Ordered sets and fields 2 Construction of the Reals 1: Dedekind Cuts 2 Metric Spaces 3 Metric Spaces 3 Definitions 4 Separability

More information

Maths 212: Homework Solutions

Maths 212: Homework Solutions Maths 212: Homework Solutions 1. The definition of A ensures that x π for all x A, so π is an upper bound of A. To show it is the least upper bound, suppose x < π and consider two cases. If x < 1, then

More information

Math 118B Solutions. Charles Martin. March 6, d i (x i, y i ) + d i (y i, z i ) = d(x, y) + d(y, z). i=1

Math 118B Solutions. Charles Martin. March 6, d i (x i, y i ) + d i (y i, z i ) = d(x, y) + d(y, z). i=1 Math 8B Solutions Charles Martin March 6, Homework Problems. Let (X i, d i ), i n, be finitely many metric spaces. Construct a metric on the product space X = X X n. Proof. Denote points in X as x = (x,

More information

Spaces of continuous functions

Spaces of continuous functions Chapter 2 Spaces of continuous functions 2.8 Baire s Category Theorem Recall that a subset A of a metric space (X, d) is dense if for all x X there is a sequence from A converging to x. An equivalent definition

More information

Math 320-2: Midterm 2 Practice Solutions Northwestern University, Winter 2015

Math 320-2: Midterm 2 Practice Solutions Northwestern University, Winter 2015 Math 30-: Midterm Practice Solutions Northwestern University, Winter 015 1. Give an example of each of the following. No justification is needed. (a) A metric on R with respect to which R is bounded. (b)

More information

Math 4317 : Real Analysis I Mid-Term Exam 1 25 September 2012

Math 4317 : Real Analysis I Mid-Term Exam 1 25 September 2012 Instructions: Answer all of the problems. Math 4317 : Real Analysis I Mid-Term Exam 1 25 September 2012 Definitions (2 points each) 1. State the definition of a metric space. A metric space (X, d) is set

More information

Functional Analysis Exercise Class

Functional Analysis Exercise Class Functional Analysis Exercise Class Week 2 November 6 November Deadline to hand in the homeworks: your exercise class on week 9 November 13 November Exercises (1) Let X be the following space of piecewise

More information

Metric Spaces and Topology

Metric Spaces and Topology Chapter 2 Metric Spaces and Topology From an engineering perspective, the most important way to construct a topology on a set is to define the topology in terms of a metric on the set. This approach underlies

More information

converges as well if x < 1. 1 x n x n 1 1 = 2 a nx n

converges as well if x < 1. 1 x n x n 1 1 = 2 a nx n Solve the following 6 problems. 1. Prove that if series n=1 a nx n converges for all x such that x < 1, then the series n=1 a n xn 1 x converges as well if x < 1. n For x < 1, x n 0 as n, so there exists

More information

Part III. 10 Topological Space Basics. Topological Spaces

Part III. 10 Topological Space Basics. Topological Spaces Part III 10 Topological Space Basics Topological Spaces Using the metric space results above as motivation we will axiomatize the notion of being an open set to more general settings. Definition 10.1.

More information

Real Analysis Math 131AH Rudin, Chapter #1. Dominique Abdi

Real Analysis Math 131AH Rudin, Chapter #1. Dominique Abdi Real Analysis Math 3AH Rudin, Chapter # Dominique Abdi.. If r is rational (r 0) and x is irrational, prove that r + x and rx are irrational. Solution. Assume the contrary, that r+x and rx are rational.

More information

MATH 202B - Problem Set 5

MATH 202B - Problem Set 5 MATH 202B - Problem Set 5 Walid Krichene (23265217) March 6, 2013 (5.1) Show that there exists a continuous function F : [0, 1] R which is monotonic on no interval of positive length. proof We know there

More information

REAL VARIABLES: PROBLEM SET 1. = x limsup E k

REAL VARIABLES: PROBLEM SET 1. = x limsup E k REAL VARIABLES: PROBLEM SET 1 BEN ELDER 1. Problem 1.1a First let s prove that limsup E k consists of those points which belong to infinitely many E k. From equation 1.1: limsup E k = E k For limsup E

More information

The Heine-Borel and Arzela-Ascoli Theorems

The Heine-Borel and Arzela-Ascoli Theorems The Heine-Borel and Arzela-Ascoli Theorems David Jekel October 29, 2016 This paper explains two important results about compactness, the Heine- Borel theorem and the Arzela-Ascoli theorem. We prove them

More information

From now on, we will represent a metric space with (X, d). Here are some examples: i=1 (x i y i ) p ) 1 p, p 1.

From now on, we will represent a metric space with (X, d). Here are some examples: i=1 (x i y i ) p ) 1 p, p 1. Chapter 1 Metric spaces 1.1 Metric and convergence We will begin with some basic concepts. Definition 1.1. (Metric space) Metric space is a set X, with a metric satisfying: 1. d(x, y) 0, d(x, y) = 0 x

More information

REAL AND COMPLEX ANALYSIS

REAL AND COMPLEX ANALYSIS REAL AND COMPLE ANALYSIS Third Edition Walter Rudin Professor of Mathematics University of Wisconsin, Madison Version 1.1 No rights reserved. Any part of this work can be reproduced or transmitted in any

More information

Metric Space Topology (Spring 2016) Selected Homework Solutions. HW1 Q1.2. Suppose that d is a metric on a set X. Prove that the inequality d(x, y)

Metric Space Topology (Spring 2016) Selected Homework Solutions. HW1 Q1.2. Suppose that d is a metric on a set X. Prove that the inequality d(x, y) Metric Space Topology (Spring 2016) Selected Homework Solutions HW1 Q1.2. Suppose that d is a metric on a set X. Prove that the inequality d(x, y) d(z, w) d(x, z) + d(y, w) holds for all w, x, y, z X.

More information

MAT 570 REAL ANALYSIS LECTURE NOTES. Contents. 1. Sets Functions Countability Axiom of choice Equivalence relations 9

MAT 570 REAL ANALYSIS LECTURE NOTES. Contents. 1. Sets Functions Countability Axiom of choice Equivalence relations 9 MAT 570 REAL ANALYSIS LECTURE NOTES PROFESSOR: JOHN QUIGG SEMESTER: FALL 204 Contents. Sets 2 2. Functions 5 3. Countability 7 4. Axiom of choice 8 5. Equivalence relations 9 6. Real numbers 9 7. Extended

More information

Analysis Finite and Infinite Sets The Real Numbers The Cantor Set

Analysis Finite and Infinite Sets The Real Numbers The Cantor Set Analysis Finite and Infinite Sets Definition. An initial segment is {n N n n 0 }. Definition. A finite set can be put into one-to-one correspondence with an initial segment. The empty set is also considered

More information

P-adic Functions - Part 1

P-adic Functions - Part 1 P-adic Functions - Part 1 Nicolae Ciocan 22.11.2011 1 Locally constant functions Motivation: Another big difference between p-adic analysis and real analysis is the existence of nontrivial locally constant

More information

NAME: Mathematics 205A, Fall 2008, Final Examination. Answer Key

NAME: Mathematics 205A, Fall 2008, Final Examination. Answer Key NAME: Mathematics 205A, Fall 2008, Final Examination Answer Key 1 1. [25 points] Let X be a set with 2 or more elements. Show that there are topologies U and V on X such that the identity map J : (X, U)

More information

Consequences of the Completeness Property

Consequences of the Completeness Property Consequences of the Completeness Property Philippe B. Laval KSU Today Philippe B. Laval (KSU) Consequences of the Completeness Property Today 1 / 10 Introduction In this section, we use the fact that R

More information

HW 4 SOLUTIONS. , x + x x 1 ) 2

HW 4 SOLUTIONS. , x + x x 1 ) 2 HW 4 SOLUTIONS The Way of Analysis p. 98: 1.) Suppose that A is open. Show that A minus a finite set is still open. This follows by induction as long as A minus one point x is still open. To see that A

More information

Solutions Final Exam May. 14, 2014

Solutions Final Exam May. 14, 2014 Solutions Final Exam May. 14, 2014 1. Determine whether the following statements are true or false. Justify your answer (i.e., prove the claim, derive a contradiction or give a counter-example). (a) (10

More information

REVIEW OF ESSENTIAL MATH 346 TOPICS

REVIEW OF ESSENTIAL MATH 346 TOPICS REVIEW OF ESSENTIAL MATH 346 TOPICS 1. AXIOMATIC STRUCTURE OF R Doğan Çömez The real number system is a complete ordered field, i.e., it is a set R which is endowed with addition and multiplication operations

More information

10 Compactness in function spaces: Ascoli-Arzelá theorem

10 Compactness in function spaces: Ascoli-Arzelá theorem 10 Compactness in function spaces: Ascoli-Arzelá theorem Recall that if (X, d) isacompactmetricspacethenthespacec(x) isavector space consisting of all continuous function f : X R. The space C(X) is equipped

More information

Continuous Functions on Metric Spaces

Continuous Functions on Metric Spaces Continuous Functions on Metric Spaces Math 201A, Fall 2016 1 Continuous functions Definition 1. Let (X, d X ) and (Y, d Y ) be metric spaces. A function f : X Y is continuous at a X if for every ɛ > 0

More information

Walker Ray Econ 204 Problem Set 3 Suggested Solutions August 6, 2015

Walker Ray Econ 204 Problem Set 3 Suggested Solutions August 6, 2015 Problem 1. Take any mapping f from a metric space X into a metric space Y. Prove that f is continuous if and only if f(a) f(a). (Hint: use the closed set characterization of continuity). I make use of

More information

MATH 722, COMPLEX ANALYSIS, SPRING 2009 PART 5

MATH 722, COMPLEX ANALYSIS, SPRING 2009 PART 5 MATH 722, COMPLEX ANALYSIS, SPRING 2009 PART 5.. The Arzela-Ascoli Theorem.. The Riemann mapping theorem Let X be a metric space, and let F be a family of continuous complex-valued functions on X. We have

More information

Continuity. Matt Rosenzweig

Continuity. Matt Rosenzweig Continuity Matt Rosenzweig Contents 1 Continuity 1 1.1 Rudin Chapter 4 Exercises........................................ 1 1.1.1 Exercise 1............................................. 1 1.1.2 Exercise

More information

A List of Problems in Real Analysis

A List of Problems in Real Analysis A List of Problems in Real Analysis W.Yessen & T.Ma December 3, 218 This document was first created by Will Yessen, who was a graduate student at UCI. Timmy Ma, who was also a graduate student at UCI,

More information

2 Topology of a Metric Space

2 Topology of a Metric Space 2 Topology of a Metric Space The real number system has two types of properties. The first type are algebraic properties, dealing with addition, multiplication and so on. The other type, called topological

More information

van Rooij, Schikhof: A Second Course on Real Functions

van Rooij, Schikhof: A Second Course on Real Functions vanrooijschikhof.tex April 25, 2018 van Rooij, Schikhof: A Second Course on Real Functions Notes from [vrs]. Introduction A monotone function is Riemann integrable. A continuous function is Riemann integrable.

More information

7 Complete metric spaces and function spaces

7 Complete metric spaces and function spaces 7 Complete metric spaces and function spaces 7.1 Completeness Let (X, d) be a metric space. Definition 7.1. A sequence (x n ) n N in X is a Cauchy sequence if for any ɛ > 0, there is N N such that n, m

More information

Solutions to Tutorial 8 (Week 9)

Solutions to Tutorial 8 (Week 9) The University of Sydney School of Mathematics and Statistics Solutions to Tutorial 8 (Week 9) MATH3961: Metric Spaces (Advanced) Semester 1, 2018 Web Page: http://www.maths.usyd.edu.au/u/ug/sm/math3961/

More information

In N we can do addition, but in order to do subtraction we need to extend N to the integers

In N we can do addition, but in order to do subtraction we need to extend N to the integers Chapter The Real Numbers.. Some Preliminaries Discussion: The Irrationality of 2. We begin with the natural numbers N = {, 2, 3, }. In N we can do addition, but in order to do subtraction we need to extend

More information

Analysis Qualifying Exam

Analysis Qualifying Exam Analysis Qualifying Exam Spring 2017 Problem 1: Let f be differentiable on R. Suppose that there exists M > 0 such that f(k) M for each integer k, and f (x) M for all x R. Show that f is bounded, i.e.,

More information

Lecture Notes in Advanced Calculus 1 (80315) Raz Kupferman Institute of Mathematics The Hebrew University

Lecture Notes in Advanced Calculus 1 (80315) Raz Kupferman Institute of Mathematics The Hebrew University Lecture Notes in Advanced Calculus 1 (80315) Raz Kupferman Institute of Mathematics The Hebrew University February 7, 2007 2 Contents 1 Metric Spaces 1 1.1 Basic definitions...........................

More information

1 Topology Definition of a topology Basis (Base) of a topology The subspace topology & the product topology on X Y 3

1 Topology Definition of a topology Basis (Base) of a topology The subspace topology & the product topology on X Y 3 Index Page 1 Topology 2 1.1 Definition of a topology 2 1.2 Basis (Base) of a topology 2 1.3 The subspace topology & the product topology on X Y 3 1.4 Basic topology concepts: limit points, closed sets,

More information

Measure and Category. Marianna Csörnyei. ucahmcs

Measure and Category. Marianna Csörnyei.   ucahmcs Measure and Category Marianna Csörnyei mari@math.ucl.ac.uk http:/www.ucl.ac.uk/ ucahmcs 1 / 96 A (very short) Introduction to Cardinals The cardinality of a set A is equal to the cardinality of a set B,

More information

Lebesgue Measure. Dung Le 1

Lebesgue Measure. Dung Le 1 Lebesgue Measure Dung Le 1 1 Introduction How do we measure the size of a set in IR? Let s start with the simplest ones: intervals. Obviously, the natural candidate for a measure of an interval is its

More information

Introduction to Proofs in Analysis. updated December 5, By Edoh Y. Amiran Following the outline of notes by Donald Chalice INTRODUCTION

Introduction to Proofs in Analysis. updated December 5, By Edoh Y. Amiran Following the outline of notes by Donald Chalice INTRODUCTION Introduction to Proofs in Analysis updated December 5, 2016 By Edoh Y. Amiran Following the outline of notes by Donald Chalice INTRODUCTION Purpose. These notes intend to introduce four main notions from

More information

MATH 566 LECTURE NOTES 6: NORMAL FAMILIES AND THE THEOREMS OF PICARD

MATH 566 LECTURE NOTES 6: NORMAL FAMILIES AND THE THEOREMS OF PICARD MATH 566 LECTURE NOTES 6: NORMAL FAMILIES AND THE THEOREMS OF PICARD TSOGTGEREL GANTUMUR 1. Introduction Suppose that we want to solve the equation f(z) = β where f is a nonconstant entire function and

More information

Austin Mohr Math 730 Homework. f(x) = y for some x λ Λ

Austin Mohr Math 730 Homework. f(x) = y for some x λ Λ Austin Mohr Math 730 Homework In the following problems, let Λ be an indexing set and let A and B λ for λ Λ be arbitrary sets. Problem 1B1 ( ) Show A B λ = (A B λ ). λ Λ λ Λ Proof. ( ) x A B λ λ Λ x A

More information

MATH 54 - TOPOLOGY SUMMER 2015 FINAL EXAMINATION. Problem 1

MATH 54 - TOPOLOGY SUMMER 2015 FINAL EXAMINATION. Problem 1 MATH 54 - TOPOLOGY SUMMER 2015 FINAL EXAMINATION ELEMENTS OF SOLUTION Problem 1 1. Let X be a Hausdorff space and K 1, K 2 disjoint compact subsets of X. Prove that there exist disjoint open sets U 1 and

More information

Exercises from other sources REAL NUMBERS 2,...,

Exercises from other sources REAL NUMBERS 2,..., Exercises from other sources REAL NUMBERS 1. Find the supremum and infimum of the following sets: a) {1, b) c) 12, 13, 14, }, { 1 3, 4 9, 13 27, 40 } 81,, { 2, 2 + 2, 2 + 2 + } 2,..., d) {n N : n 2 < 10},

More information

Course 212: Academic Year Section 1: Metric Spaces

Course 212: Academic Year Section 1: Metric Spaces Course 212: Academic Year 1991-2 Section 1: Metric Spaces D. R. Wilkins Contents 1 Metric Spaces 3 1.1 Distance Functions and Metric Spaces............. 3 1.2 Convergence and Continuity in Metric Spaces.........

More information

1/12/05: sec 3.1 and my article: How good is the Lebesgue measure?, Math. Intelligencer 11(2) (1989),

1/12/05: sec 3.1 and my article: How good is the Lebesgue measure?, Math. Intelligencer 11(2) (1989), Real Analysis 2, Math 651, Spring 2005 April 26, 2005 1 Real Analysis 2, Math 651, Spring 2005 Krzysztof Chris Ciesielski 1/12/05: sec 3.1 and my article: How good is the Lebesgue measure?, Math. Intelligencer

More information

g 2 (x) (1/3)M 1 = (1/3)(2/3)M.

g 2 (x) (1/3)M 1 = (1/3)(2/3)M. COMPACTNESS If C R n is closed and bounded, then by B-W it is sequentially compact: any sequence of points in C has a subsequence converging to a point in C Conversely, any sequentially compact C R n is

More information

Definition 2.1. A metric (or distance function) defined on a non-empty set X is a function d: X X R that satisfies: For all x, y, and z in X :

Definition 2.1. A metric (or distance function) defined on a non-empty set X is a function d: X X R that satisfies: For all x, y, and z in X : MATH 337 Metric Spaces Dr. Neal, WKU Let X be a non-empty set. The elements of X shall be called points. We shall define the general means of determining the distance between two points. Throughout we

More information

MASTERS EXAMINATION IN MATHEMATICS SOLUTIONS

MASTERS EXAMINATION IN MATHEMATICS SOLUTIONS MASTERS EXAMINATION IN MATHEMATICS PURE MATHEMATICS OPTION SPRING 010 SOLUTIONS Algebra A1. Let F be a finite field. Prove that F [x] contains infinitely many prime ideals. Solution: The ring F [x] of

More information

Math 117: Continuity of Functions

Math 117: Continuity of Functions Math 117: Continuity of Functions John Douglas Moore November 21, 2008 We finally get to the topic of ɛ δ proofs, which in some sense is the goal of the course. It may appear somewhat laborious to use

More information

Topology. Xiaolong Han. Department of Mathematics, California State University, Northridge, CA 91330, USA address:

Topology. Xiaolong Han. Department of Mathematics, California State University, Northridge, CA 91330, USA  address: Topology Xiaolong Han Department of Mathematics, California State University, Northridge, CA 91330, USA E-mail address: Xiaolong.Han@csun.edu Remark. You are entitled to a reward of 1 point toward a homework

More information

Problem set 1, Real Analysis I, Spring, 2015.

Problem set 1, Real Analysis I, Spring, 2015. Problem set 1, Real Analysis I, Spring, 015. (1) Let f n : D R be a sequence of functions with domain D R n. Recall that f n f uniformly if and only if for all ɛ > 0, there is an N = N(ɛ) so that if n

More information

Set, functions and Euclidean space. Seungjin Han

Set, functions and Euclidean space. Seungjin Han Set, functions and Euclidean space Seungjin Han September, 2018 1 Some Basics LOGIC A is necessary for B : If B holds, then A holds. B A A B is the contraposition of B A. A is sufficient for B: If A holds,

More information

Advanced Calculus Math 127B, Winter 2005 Solutions: Final. nx2 1 + n 2 x, g n(x) = n2 x

Advanced Calculus Math 127B, Winter 2005 Solutions: Final. nx2 1 + n 2 x, g n(x) = n2 x . Define f n, g n : [, ] R by f n (x) = Advanced Calculus Math 27B, Winter 25 Solutions: Final nx2 + n 2 x, g n(x) = n2 x 2 + n 2 x. 2 Show that the sequences (f n ), (g n ) converge pointwise on [, ],

More information

A LITTLE REAL ANALYSIS AND TOPOLOGY

A LITTLE REAL ANALYSIS AND TOPOLOGY A LITTLE REAL ANALYSIS AND TOPOLOGY 1. NOTATION Before we begin some notational definitions are useful. (1) Z = {, 3, 2, 1, 0, 1, 2, 3, }is the set of integers. (2) Q = { a b : aεz, bεz {0}} is the set

More information

Economics 204 Fall 2011 Problem Set 1 Suggested Solutions

Economics 204 Fall 2011 Problem Set 1 Suggested Solutions Economics 204 Fall 2011 Problem Set 1 Suggested Solutions 1. Suppose k is a positive integer. Use induction to prove the following two statements. (a) For all n N 0, the inequality (k 2 + n)! k 2n holds.

More information

POL502: Foundations. Kosuke Imai Department of Politics, Princeton University. October 10, 2005

POL502: Foundations. Kosuke Imai Department of Politics, Princeton University. October 10, 2005 POL502: Foundations Kosuke Imai Department of Politics, Princeton University October 10, 2005 Our first task is to develop the foundations that are necessary for the materials covered in this course. 1

More information

MATH 31BH Homework 1 Solutions

MATH 31BH Homework 1 Solutions MATH 3BH Homework Solutions January 0, 04 Problem.5. (a) (x, y)-plane in R 3 is closed and not open. To see that this plane is not open, notice that any ball around the origin (0, 0, 0) will contain points

More information

In N we can do addition, but in order to do subtraction we need to extend N to the integers

In N we can do addition, but in order to do subtraction we need to extend N to the integers Chapter 1 The Real Numbers 1.1. Some Preliminaries Discussion: The Irrationality of 2. We begin with the natural numbers N = {1, 2, 3, }. In N we can do addition, but in order to do subtraction we need

More information

MATH 140B - HW 5 SOLUTIONS

MATH 140B - HW 5 SOLUTIONS MATH 140B - HW 5 SOLUTIONS Problem 1 (WR Ch 7 #8). If I (x) = { 0 (x 0), 1 (x > 0), if {x n } is a sequence of distinct points of (a,b), and if c n converges, prove that the series f (x) = c n I (x x n

More information

Lebesgue Measure on R n

Lebesgue Measure on R n CHAPTER 2 Lebesgue Measure on R n Our goal is to construct a notion of the volume, or Lebesgue measure, of rather general subsets of R n that reduces to the usual volume of elementary geometrical sets

More information

Locally convex spaces, the hyperplane separation theorem, and the Krein-Milman theorem

Locally convex spaces, the hyperplane separation theorem, and the Krein-Milman theorem 56 Chapter 7 Locally convex spaces, the hyperplane separation theorem, and the Krein-Milman theorem Recall that C(X) is not a normed linear space when X is not compact. On the other hand we could use semi

More information

Logical Connectives and Quantifiers

Logical Connectives and Quantifiers Chapter 1 Logical Connectives and Quantifiers 1.1 Logical Connectives 1.2 Quantifiers 1.3 Techniques of Proof: I 1.4 Techniques of Proof: II Theorem 1. Let f be a continuous function. If 1 f(x)dx 0, then

More information

Measures and Measure Spaces

Measures and Measure Spaces Chapter 2 Measures and Measure Spaces In summarizing the flaws of the Riemann integral we can focus on two main points: 1) Many nice functions are not Riemann integrable. 2) The Riemann integral does not

More information

1. Theorem. (Archimedean Property) Let x be any real number. There exists a positive integer n greater than x.

1. Theorem. (Archimedean Property) Let x be any real number. There exists a positive integer n greater than x. Advanced Calculus I, Dr. Block, Chapter 2 notes. Theorem. (Archimedean Property) Let x be any real number. There exists a positive integer n greater than x. 2. Definition. A sequence is a real-valued function

More information

2. Metric Spaces. 2.1 Definitions etc.

2. Metric Spaces. 2.1 Definitions etc. 2. Metric Spaces 2.1 Definitions etc. The procedure in Section for regarding R as a topological space may be generalized to many other sets in which there is some kind of distance (formally, sets with

More information

Real Analysis Problems

Real Analysis Problems Real Analysis Problems Cristian E. Gutiérrez September 14, 29 1 1 CONTINUITY 1 Continuity Problem 1.1 Let r n be the sequence of rational numbers and Prove that f(x) = 1. f is continuous on the irrationals.

More information

Bootcamp. Christoph Thiele. Summer As in the case of separability we have the following two observations: Lemma 1 Finite sets are compact.

Bootcamp. Christoph Thiele. Summer As in the case of separability we have the following two observations: Lemma 1 Finite sets are compact. Bootcamp Christoph Thiele Summer 212.1 Compactness Definition 1 A metric space is called compact, if every cover of the space has a finite subcover. As in the case of separability we have the following

More information

Folland: Real Analysis, Chapter 4 Sébastien Picard

Folland: Real Analysis, Chapter 4 Sébastien Picard Folland: Real Analysis, Chapter 4 Sébastien Picard Problem 4.19 If {X α } is a family of topological spaces, X α X α (with the product topology) is uniquely determined up to homeomorphism by the following

More information

Math General Topology Fall 2012 Homework 6 Solutions

Math General Topology Fall 2012 Homework 6 Solutions Math 535 - General Topology Fall 202 Homework 6 Solutions Problem. Let F be the field R or C of real or complex numbers. Let n and denote by F[x, x 2,..., x n ] the set of all polynomials in n variables

More information

Lecture 5 - Hausdorff and Gromov-Hausdorff Distance

Lecture 5 - Hausdorff and Gromov-Hausdorff Distance Lecture 5 - Hausdorff and Gromov-Hausdorff Distance August 1, 2011 1 Definition and Basic Properties Given a metric space X, the set of closed sets of X supports a metric, the Hausdorff metric. If A is

More information

Functional Analysis I

Functional Analysis I Functional Analysis I Course Notes by Stefan Richter Transcribed and Annotated by Gregory Zitelli Polar Decomposition Definition. An operator W B(H) is called a partial isometry if W x = X for all x (ker

More information

Bounded Derivatives Which Are Not Riemann Integrable. Elliot M. Granath. A thesis submitted in partial fulfillment of the requirements

Bounded Derivatives Which Are Not Riemann Integrable. Elliot M. Granath. A thesis submitted in partial fulfillment of the requirements Bounded Derivatives Which Are Not Riemann Integrable by Elliot M. Granath A thesis submitted in partial fulfillment of the requirements for graduation with Honors in Mathematics. Whitman College 2017 Certificate

More information

MA651 Topology. Lecture 9. Compactness 2.

MA651 Topology. Lecture 9. Compactness 2. MA651 Topology. Lecture 9. Compactness 2. This text is based on the following books: Topology by James Dugundgji Fundamental concepts of topology by Peter O Neil Elements of Mathematics: General Topology

More information

MTG 5316/4302 FALL 2018 REVIEW FINAL

MTG 5316/4302 FALL 2018 REVIEW FINAL MTG 5316/4302 FALL 2018 REVIEW FINAL JAMES KEESLING Problem 1. Define open set in a metric space X. Define what it means for a set A X to be connected in a metric space X. Problem 2. Show that if a set

More information

That is, there is an element

That is, there is an element Section 3.1: Mathematical Induction Let N denote the set of natural numbers (positive integers). N = {1, 2, 3, 4, } Axiom: If S is a nonempty subset of N, then S has a least element. That is, there is

More information

Baire category theorem and nowhere differentiable continuous functions

Baire category theorem and nowhere differentiable continuous functions Baire category theorem and nowhere differentiable continuous functions Takako Nemoto JAIST third CORE meeting 26 January 2018 Abstract In constructive mathematics, Baire Category Theorem has at least the

More information

The Arzelà-Ascoli Theorem

The Arzelà-Ascoli Theorem John Nachbar Washington University March 27, 2016 The Arzelà-Ascoli Theorem The Arzelà-Ascoli Theorem gives sufficient conditions for compactness in certain function spaces. Among other things, it helps

More information

Most Continuous Functions are Nowhere Differentiable

Most Continuous Functions are Nowhere Differentiable Most Continuous Functions are Nowhere Differentiable Spring 2004 The Space of Continuous Functions Let K = [0, 1] and let C(K) be the set of all continuous functions f : K R. Definition 1 For f C(K) we

More information

This chapter contains a very bare summary of some basic facts from topology.

This chapter contains a very bare summary of some basic facts from topology. Chapter 2 Topological Spaces This chapter contains a very bare summary of some basic facts from topology. 2.1 Definition of Topology A topology O on a set X is a collection of subsets of X satisfying the

More information

Analysis Comprehensive Exam Questions Fall F(x) = 1 x. f(t)dt. t 1 2. tf 2 (t)dt. and g(t, x) = 2 t. 2 t

Analysis Comprehensive Exam Questions Fall F(x) = 1 x. f(t)dt. t 1 2. tf 2 (t)dt. and g(t, x) = 2 t. 2 t Analysis Comprehensive Exam Questions Fall 2. Let f L 2 (, ) be given. (a) Prove that ( x 2 f(t) dt) 2 x x t f(t) 2 dt. (b) Given part (a), prove that F L 2 (, ) 2 f L 2 (, ), where F(x) = x (a) Using

More information

Problem Set 5. 2 n k. Then a nk (x) = 1+( 1)k

Problem Set 5. 2 n k. Then a nk (x) = 1+( 1)k Problem Set 5 1. (Folland 2.43) For x [, 1), let 1 a n (x)2 n (a n (x) = or 1) be the base-2 expansion of x. (If x is a dyadic rational, choose the expansion such that a n (x) = for large n.) Then the

More information

Austin Mohr Math 704 Homework 6

Austin Mohr Math 704 Homework 6 Austin Mohr Math 704 Homework 6 Problem 1 Integrability of f on R does not necessarily imply the convergence of f(x) to 0 as x. a. There exists a positive continuous function f on R so that f is integrable

More information

MATH 102 INTRODUCTION TO MATHEMATICAL ANALYSIS. 1. Some Fundamentals

MATH 102 INTRODUCTION TO MATHEMATICAL ANALYSIS. 1. Some Fundamentals MATH 02 INTRODUCTION TO MATHEMATICAL ANALYSIS Properties of Real Numbers Some Fundamentals The whole course will be based entirely on the study of sequence of numbers and functions defined on the real

More information

Chapter 4. Measure Theory. 1. Measure Spaces

Chapter 4. Measure Theory. 1. Measure Spaces Chapter 4. Measure Theory 1. Measure Spaces Let X be a nonempty set. A collection S of subsets of X is said to be an algebra on X if S has the following properties: 1. X S; 2. if A S, then A c S; 3. if

More information

Connectedness. Proposition 2.2. The following are equivalent for a topological space (X, T ).

Connectedness. Proposition 2.2. The following are equivalent for a topological space (X, T ). Connectedness 1 Motivation Connectedness is the sort of topological property that students love. Its definition is intuitive and easy to understand, and it is a powerful tool in proofs of well-known results.

More information

Math 321 Final Examination April 1995 Notation used in this exam: N. (1) S N (f,x) = f(t)e int dt e inx.

Math 321 Final Examination April 1995 Notation used in this exam: N. (1) S N (f,x) = f(t)e int dt e inx. Math 321 Final Examination April 1995 Notation used in this exam: N 1 π (1) S N (f,x) = f(t)e int dt e inx. 2π n= N π (2) C(X, R) is the space of bounded real-valued functions on the metric space X, equipped

More information

5 Set Operations, Functions, and Counting

5 Set Operations, Functions, and Counting 5 Set Operations, Functions, and Counting Let N denote the positive integers, N 0 := N {0} be the non-negative integers and Z = N 0 ( N) the positive and negative integers including 0, Q the rational numbers,

More information

SOME QUESTIONS FOR MATH 766, SPRING Question 1. Let C([0, 1]) be the set of all continuous functions on [0, 1] endowed with the norm

SOME QUESTIONS FOR MATH 766, SPRING Question 1. Let C([0, 1]) be the set of all continuous functions on [0, 1] endowed with the norm SOME QUESTIONS FOR MATH 766, SPRING 2016 SHUANGLIN SHAO Question 1. Let C([0, 1]) be the set of all continuous functions on [0, 1] endowed with the norm f C = sup f(x). 0 x 1 Prove that C([0, 1]) is a

More information

Review Notes for the Basic Qualifying Exam

Review Notes for the Basic Qualifying Exam Review Notes for the Basic Qualifying Exam Topics: Analysis & Linear Algebra Fall 2016 Spring 2017 Created by Howard Purpose: This document is a compilation of notes generated to prepare for the Basic

More information