converges as well if x < 1. 1 x n x n 1 1 = 2 a nx n
|
|
- Barnard Porter
- 6 years ago
- Views:
Transcription
1 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 an N such that x n 1 2 for all n N. Thus, n N : a x n n 1 x n x n a n 1 1 = 2 a nx n 2 Since n=1 a nx n converges for all x such that x < 1, the radius of convergence of this power series is at least 1. Thus, the convergence is absolute for all x < 1 by a standard result. Furthermore, 2 n=1 a nx n = 2an x n also converges absolutely for all x < 1. Therefore, by the comparison test 1, for all x < 1, the series converges. n=1 a n xn 1 x n 1 Theorem 3.25a in Rudin 1
2 2. Let X be a nonempty set and d be a metric on X. Prove the standard theorem that the set of all limit points of X is closed. Solution 1: Let A be the set of all limit points of X. We show that A is closed by showing that it contains all of its limit points. If A is empty, then it is closed by definition. Otherwise, consider any limit point x of A. Since A X, this implies that x is a limit point of X, so x A by construction. Solution 2: Let A be the set of all limit points of X. By a standard result, proving A is closed is equivalent to proving that A c is open. We prove that A c is open by showing that all of its points are interior to A c. If A c is either equal to the empty set or equal to X, then it is open by definition. Otherwise, consider any x A c. By assumption, x is not a limit point of X, which implies it is an isolated point by definition. This implies there exists an r > 0 such that B r x = {x}, so B r x A c. Thus, x is an interior point of A c. Since x A c was arbitrary, this proves A c is open and that A must be closed. Solution 3 proving a more general result: Let E X, E denote the set of all limit points of E, and E denote the set of all limit points from E. We prove that E is closed by showing that E E. If E is empty, then E is closed by definition, so assume E is not empty and let x E. By definition, for any r > 0 there exists y E such that 0 < dx, y < r/2. By definition of E, there exists z E such that 0 < dy, z < dx, y < r/2 which implies z x. Thus, 0 < dx, z dx, y + dy, z < r, so x is a limit point of E, which implies x E and it follows that E E. Since X X, the result holds for X as well. 2
3 3. Let X be a nonempty set and d be a metric on X. We say that K X is sequentially compact if for every sequence {x n } K there exists a subsequence {a nk } that converges to a point x K. For a fixed ɛ > 0, we call {x α } α A X an ɛ-net of K X if the family of open balls {B ɛ x α } α A is an open cover of K. We say that K X is totally bounded if there exists a finite ɛ-net for every ɛ > 0. Use these definitions to prove the standard theorem that a nonempty sequentially compact subset of a metric space is complete and totally bounded. Suppose K X is sequentially compact. Consider any Cauchy sequence {x n } K. The sequential compactness of K implies there exists a convergent subsequence {x nk }. By a standard result, the Cauchy sequence must also converge to the same limit as the convergent subsequence 2. Assume K is not totally bounded, so that there exists an ɛ > 0 such that K has no finite ɛ-net. This means that every finite subset {x 1,..., x n } K has the property that n B ɛx i is not a cover for K. In other words, K\ n B ɛx i is not empty. Pick any x 1 K, and choose x 2 from K\B ɛ x 1. Having chosen the first n points, choose x n+1 from K\ n B ɛx i. This inductively constructs a sequence {x n } K such that dx n, x m > ɛ for any n m, which implies that any subsequence is not Cauchy. Thus, there cannot exist a convergent subsequence, a contradiction of K being sequentially compact. Therefore, K is totally bounded. 2 This is also easily proved by an ɛ/2 argument if the result is not remembered. 3
4 4. Let X be a nonempty set and d be a metric on X. Suppose f is a continuous function on A X to R n for some n N. Using only the definitions of a set being compact and a function being uniformly continuous, prove that if A is compact, then f is uniformly continuous, and provide a counterexample to the converse. Suppose A is compact and f : A R n is continuous. Let ɛ > 0. For any x A, since f is continuous, there exists δ x > 0 such that for all y B δx x, fx fy < ɛ/2. The set {B δx/2x} x A forms an open cover of A. Since A is compact, there exists a finite subcover that we denote by {B δxi /2x i } 1 i k. Let δ = min{δ x1 /2,..., δ xk /2}. Consider any y, z A such that dy, z < δ. Then, there exists some x i such that y, z B δxi x i. By the triangle inequality, fy fz fy fx i + fx i fz < ɛ. For a counterexample, suppose that f : R R n is the zero map, i.e., fx = 0 R n for all x R. Then, for any ɛ > 0 and x, y R, fx fy = 0 < ɛ, which shows that f is uniformly continuous, but R is not compact since it is not bounded. 4
5 5. Let a < b be real numbers and f : [a, b] R. For a partition P = {a = x 0 < x 1 < < x n = b} of [a, b], the upper and lower Darboux sums of f on P are defined as Uf, P = sup fx x i x i 1, x [x i 1,x i] and Lf, P = inf fy x i x i 1, y [x i 1,x i] respectively. We say that f is Riemann integrable on [a, b] if for every ɛ > 0 there exists a partition P of [a, b] such that Uf, P Lf, P < ɛ. Suppose that f : [a, b] R is bounded. Using the above definitions, prove that if f is Riemann integrable, then f 2 is Riemann integrable, and provide a counterexample to the converse. Hint: You may find it useful to exploit the fact that for any set A, and any real-valued function f defined on A that sup x A fx inf y A fy = sup x,y A fx fy. Since f is bounded on [a, b] there exists M > 0 such that fx < M for all x [a, b]. Thus, for any x, y [a, b], we have that f 2 x f 2 y = fx fy fx + fy 2M fx fy. Let ɛ > 0. Since f is Riemann integrable, there exists a partition P on [a, b] such that Uf, P Lf, P < ɛ/2m. Then, for this partition P, we have that Uf 2, P Lf 2, P = sup f 2 x inf f 2 x x i x i 1 [x i 1,x i] [x i 1,x i] = sup f 2 x f 2 y x i x i 1 x,y [x i 1,x i] 2M sup fx fy x i x i 1 x,y [x i 1,x i] = 2M sup fx inf fy x i x i 1 x [x i 1,x i] y [x i 1,x i] = 2M[Uf, P Lf, P ] < ɛ. 5
6 For a counterexample, suppose f : [0, 1] R is defined by fx = 1 whenever x [0, 1] Q, and fx = 1 otherwise. Then, f 2 x = 1 for all x [0, 1], which is clearly Riemann integrable, but, for any partition P of [0, 1], we have Uf, P = 1 1 = Lf, P, so f is not Riemann integrable. 6
7 6. Let C 1 [a, b] denote the space of real-valued continuously differentiable functions on [a, b] where a < b are real numbers. Define the metric d on C 1 [a, b] as follows where f, g C 1 [a, b] df, g = sup fx gx + sup f x g x. [a,b] [a,b] Suppose {f n } C 1 [a, b], d is a bounded sequence. Prove that if {f n} is equicontinuous, then there exists a subsequence of {f n } that converges in C 1 [a, b], d. Since {f n } is bounded in C 1 [a, b], there exists an M such that {f n } B M 0 where 0 indicates the zero function. By the definition of the metric, this implies that both {f n } and {f n} are uniformly bounded. Since {f n} is equicontinuous, the Arzelà-Ascoli theorem implies that there exists a subsequence {f n k } that converges uniformly to g C[a, b]. Since {f nk } is uniformly bounded, {f nk a} is a bounded set of real numbers, so there exists a convergent subsequence {f nkl a}. Let fa define this limit. By the fundamental theorem of calculus, for each l, f nkl x = f nkl a + x a f n kl y dy. Since {f n kl } also converges uniformly to g, then we can take the limit as l above for each x [a, b] and bring the limit through the integral by a standard result to define fx by fx = lim l f nkl x = fa + x a gy dy. By the fundamental theorem of calculus, we have that f = gx, so f C 1 [a, b]. 7
d(x n, x) d(x n, x nk ) + d(x nk, x) where we chose any fixed k > N
Problem 1. Let f : A R R have the property that for every x A, there exists ɛ > 0 such that f(t) > ɛ if t (x ɛ, x + ɛ) A. If the set A is compact, prove there exists c > 0 such that f(x) > c for all x
More informationMath 140A - Fall Final Exam
Math 140A - Fall 2014 - Final Exam Problem 1. Let {a n } n 1 be an increasing sequence of real numbers. (i) If {a n } has a bounded subsequence, show that {a n } is itself bounded. (ii) If {a n } has a
More informationMath 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 information1. Let A R be a nonempty set that is bounded from above, and let a be the least upper bound of A. Show that there exists a sequence {a n } n N
Applied Analysis prelim July 15, 216, with solutions Solve 4 of the problems 1-5 and 2 of the problems 6-8. We will only grade the first 4 problems attempted from1-5 and the first 2 attempted from problems
More informationMATH 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 informationAssignment-10. (Due 11/21) Solution: Any continuous function on a compact set is uniformly continuous.
Assignment-1 (Due 11/21) 1. Consider the sequence of functions f n (x) = x n on [, 1]. (a) Show that each function f n is uniformly continuous on [, 1]. Solution: Any continuous function on a compact set
More informationSummer Jump-Start Program for Analysis, 2012 Song-Ying Li
Summer Jump-Start Program for Analysis, 01 Song-Ying Li 1 Lecture 6: Uniformly continuity and sequence of functions 1.1 Uniform Continuity Definition 1.1 Let (X, d 1 ) and (Y, d ) are metric spaces and
More informationProblem Set 5: Solutions Math 201A: Fall 2016
Problem Set 5: s Math 21A: Fall 216 Problem 1. Define f : [1, ) [1, ) by f(x) = x + 1/x. Show that f(x) f(y) < x y for all x, y [1, ) with x y, but f has no fixed point. Why doesn t this example contradict
More informationAnalysis 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 informationThe 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 informationMath LM (24543) Lectures 02
Math 32300 LM (24543) Lectures 02 Ethan Akin Office: NAC 6/287 Phone: 650-5136 Email: ethanakin@earthlink.net Spring, 2018 Contents Continuity, Ross Chapter 3 Uniform Continuity and Compactness Connectedness
More informationProblem 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 informationSolutions 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 informationBootcamp. 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 informationMath 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 informationFrom 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 informationExam 2 extra practice problems
Exam 2 extra practice problems (1) If (X, d) is connected and f : X R is a continuous function such that f(x) = 1 for all x X, show that f must be constant. Solution: Since f(x) = 1 for every x X, either
More informationMAT 544 Problem Set 2 Solutions
MAT 544 Problem Set 2 Solutions Problems. Problem 1 A metric space is separable if it contains a dense subset which is finite or countably infinite. Prove that every totally bounded metric space X is separable.
More informationMath 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 informationProblem 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 informationContinuous 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 informationANALYSIS 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 informationSolutions Final Exam May. 14, 2014
Solutions Final Exam May. 14, 2014 1. (a) (10 points) State the formal definition of a Cauchy sequence of real numbers. A sequence, {a n } n N, of real numbers, is Cauchy if and only if for every ɛ > 0,
More informationSOLUTIONS TO SOME PROBLEMS
23 FUNCTIONAL ANALYSIS Spring 23 SOLUTIONS TO SOME PROBLEMS Warning:These solutions may contain errors!! PREPARED BY SULEYMAN ULUSOY PROBLEM 1. Prove that a necessary and sufficient condition that the
More informationLecture Notes on Metric Spaces
Lecture Notes on Metric Spaces Math 117: Summer 2007 John Douglas Moore Our goal of these notes is to explain a few facts regarding metric spaces not included in the first few chapters of the text [1],
More informationMath 328 Course Notes
Math 328 Course Notes Ian Robertson March 3, 2006 3 Properties of C[0, 1]: Sup-norm and Completeness In this chapter we are going to examine the vector space of all continuous functions defined on the
More informationMATH 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 informationEconomics 204 Fall 2011 Problem Set 2 Suggested Solutions
Economics 24 Fall 211 Problem Set 2 Suggested Solutions 1. Determine whether the following sets are open, closed, both or neither under the topology induced by the usual metric. (Hint: think about limit
More informationMetric 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 informationLecture 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 information7 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 informationReal 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 informationProblem set 5, Real Analysis I, Spring, otherwise. (a) Verify that f is integrable. Solution: Compute since f is even, 1 x (log 1/ x ) 2 dx 1
Problem set 5, Real Analysis I, Spring, 25. (5) Consider the function on R defined by f(x) { x (log / x ) 2 if x /2, otherwise. (a) Verify that f is integrable. Solution: Compute since f is even, R f /2
More informationHOMEWORK ASSIGNMENT 6
HOMEWORK ASSIGNMENT 6 DUE 15 MARCH, 2016 1) Suppose f, g : A R are uniformly continuous on A. Show that f + g is uniformly continuous on A. Solution First we note: In order to show that f + g is uniformly
More informationReal 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 informationAnalysis 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 informationAdvanced 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 informationMetric Spaces Math 413 Honors Project
Metric Spaces Math 413 Honors Project 1 Metric Spaces Definition 1.1 Let X be a set. A metric on X is a function d : X X R such that for all x, y, z X: i) d(x, y) = d(y, x); ii) d(x, y) = 0 if and only
More informationMAT 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 informationExercises 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 informationMATH 131A: REAL ANALYSIS (BIG IDEAS)
MATH 131A: REAL ANALYSIS (BIG IDEAS) Theorem 1 (The Triangle Inequality). For all x, y R we have x + y x + y. Proposition 2 (The Archimedean property). For each x R there exists an n N such that n > x.
More informationPrinciples of Real Analysis I Fall VII. Sequences of Functions
21-355 Principles of Real Analysis I Fall 2004 VII. Sequences of Functions In Section II, we studied sequences of real numbers. It is very useful to consider extensions of this concept. More generally,
More informationLogical 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 informationMcGill University Math 354: Honors Analysis 3
Practice problems McGill University Math 354: Honors Analysis 3 not for credit Problem 1. Determine whether the family of F = {f n } functions f n (x) = x n is uniformly equicontinuous. 1st Solution: The
More informationg 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 informationMid Term-1 : Practice problems
Mid Term-1 : Practice problems These problems are meant only to provide practice; they do not necessarily reflect the difficulty level of the problems in the exam. The actual exam problems are likely to
More informationa) Let x,y be Cauchy sequences in some metric space. Define d(x, y) = lim n d (x n, y n ). Show that this function is well-defined.
Problem 3) Remark: for this problem, if I write the notation lim x n, it should always be assumed that I mean lim n x n, and similarly if I write the notation lim x nk it should always be assumed that
More informationAnalysis III Theorems, Propositions & Lemmas... Oh My!
Analysis III Theorems, Propositions & Lemmas... Oh My! Rob Gibson October 25, 2010 Proposition 1. If x = (x 1, x 2,...), y = (y 1, y 2,...), then is a distance. ( d(x, y) = x k y k p Proposition 2. In
More informationReview 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 informationWeek 5 Lectures 13-15
Week 5 Lectures 13-15 Lecture 13 Definition 29 Let Y be a subset X. A subset A Y is open in Y if there exists an open set U in X such that A = U Y. It is not difficult to show that the collection of all
More informationA Glimpse into Topology
A Glimpse into Topology by Vishal Malik A project submitted to the Department of Mathematical Sciences in conformity with the requirements for Math 430 (203-4) Lakehead University Thunder Bay, Ontario,
More informationAdvanced Calculus: MATH 410 Professor David Levermore 28 November 2006
Advanced Calculus: MATH 410 Professor David Levermore 28 November 2006 1. Uniform Continuity Uniform continuity is a very useful concept. Here we introduce it in the context of real-valued functions with
More information1 Homework. Recommended Reading:
Analysis MT43C Notes/Problems/Homework Recommended Reading: R. G. Bartle, D. R. Sherbert Introduction to real analysis, principal reference M. Spivak Calculus W. Rudin Principles of mathematical analysis
More informationMidterm 1. Every element of the set of functions is continuous
Econ 200 Mathematics for Economists Midterm Question.- Consider the set of functions F C(0, ) dened by { } F = f C(0, ) f(x) = ax b, a A R and b B R That is, F is a subset of the set of continuous functions
More informationIowa State University. Instructor: Alex Roitershtein Summer Homework #5. Solutions
Math 50 Iowa State University Introduction to Real Analysis Department of Mathematics Instructor: Alex Roitershtein Summer 205 Homework #5 Solutions. Let α and c be real numbers, c > 0, and f is defined
More informationHonours Analysis III
Honours Analysis III Math 354 Prof. Dmitry Jacobson Notes Taken By: R. Gibson Fall 2010 1 Contents 1 Overview 3 1.1 p-adic Distance............................................ 4 2 Introduction 5 2.1 Normed
More informationx 0 + f(x), exist as extended real numbers. Show that f is upper semicontinuous This shows ( ɛ, ɛ) B α. Thus
Homework 3 Solutions, Real Analysis I, Fall, 2010. (9) Let f : (, ) [, ] be a function whose restriction to (, 0) (0, ) is continuous. Assume the one-sided limits p = lim x 0 f(x), q = lim x 0 + f(x) exist
More informationMath 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 informationReal Analysis. Joe Patten August 12, 2018
Real Analysis Joe Patten August 12, 2018 1 Relations and Functions 1.1 Relations A (binary) relation, R, from set A to set B is a subset of A B. Since R is a subset of A B, it is a set of ordered pairs.
More informationMA 1124 Solutions 14 th May 2012
MA 1124 Solutions 14 th May 2012 1 (a) Use True/False Tables to prove (i) P = Q Q = P The definition of P = Q is given by P Q P = Q T T T T F F F T T F F T So Q P Q = P F F T T F F F T T T T T Since the
More information4.6 Montel's Theorem. Robert Oeckl CA NOTES 7 17/11/2009 1
Robert Oeckl CA NOTES 7 17/11/2009 1 4.6 Montel's Theorem Let X be a topological space. We denote by C(X) the set of complex valued continuous functions on X. Denition 4.26. A topological space is called
More informationThus f is continuous at x 0. Matthew Straughn Math 402 Homework 6
Matthew Straughn Math 402 Homework 6 Homework 6 (p. 452) 14.3.3, 14.3.4, 14.3.5, 14.3.8 (p. 455) 14.4.3* (p. 458) 14.5.3 (p. 460) 14.6.1 (p. 472) 14.7.2* Lemma 1. If (f (n) ) converges uniformly to some
More informationTHE INVERSE FUNCTION THEOREM
THE INVERSE FUNCTION THEOREM W. PATRICK HOOPER The implicit function theorem is the following result: Theorem 1. Let f be a C 1 function from a neighborhood of a point a R n into R n. Suppose A = Df(a)
More informationThe uniform metric on product spaces
The uniform metric on product spaces Jordan Bell jordan.bell@gmail.com Department of Mathematics, University of Toronto April 3, 2014 1 Metric topology If (X, d) is a metric space, a X, and r > 0, then
More informationMaths 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 information4. (alternate topological criterion) For each closed set V Y, its preimage f 1 (V ) is closed in X.
Chapter 2 Functions 2.1 Continuous functions Definition 2.1. Let (X, d X ) and (Y,d Y ) be metric spaces. A function f : X! Y is continuous at x 2 X if for each " > 0 there exists >0 with the property
More informationFunctional Analysis. Franck Sueur Metric spaces Definitions Completeness Compactness Separability...
Functional Analysis Franck Sueur 2018-2019 Contents 1 Metric spaces 1 1.1 Definitions........................................ 1 1.2 Completeness...................................... 3 1.3 Compactness......................................
More informationMath 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 informationREVIEW 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 informationCompact operators on Banach spaces
Compact operators on Banach spaces Jordan Bell jordan.bell@gmail.com Department of Mathematics, University of Toronto November 12, 2017 1 Introduction In this note I prove several things about compact
More informationMath 5210, Definitions and Theorems on Metric Spaces
Math 5210, Definitions and Theorems on Metric Spaces Let (X, d) be a metric space. We will use the following definitions (see Rudin, chap 2, particularly 2.18) 1. Let p X and r R, r > 0, The ball of radius
More informationWalker 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 informationSummer Jump-Start Program for Analysis, 2012 Song-Ying Li. 1 Lecture 7: Equicontinuity and Series of functions
Summer Jump-Start Program for Analysis, 0 Song-Ying Li Lecture 7: Equicontinuity and Series of functions. Equicontinuity Definition. Let (X, d) be a metric space, K X and K is a compact subset of X. C(K)
More informationWhat to remember about metric spaces
Division of the Humanities and Social Sciences What to remember about metric spaces KC Border These notes are (I hope) a gentle introduction to the topological concepts used in economic theory. If the
More informationFolland: 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 information1 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 informationReal Analysis Chapter 4 Solutions Jonathan Conder
2. Let x, y X and suppose that x y. Then {x} c is open in the cofinite topology and contains y but not x. The cofinite topology on X is therefore T 1. Since X is infinite it contains two distinct points
More informationMetric Spaces Math 413 Honors Project
Metric Spaces Math 413 Honors Project 1 Metric Spaces Definition 1.1 Let X be a set. A metric on X is a function d : X X R such that for all x, y, z X: i) d(x, y) = d(y, x); ii) d(x, y) = 0 if and only
More informationBy (a), B ε (x) is a closed subset (which
Solutions to Homework #3. 1. Given a metric space (X, d), a point x X and ε > 0, define B ε (x) = {y X : d(y, x) ε}, called the closed ball of radius ε centered at x. (a) Prove that B ε (x) is always a
More informationAnalysis III. Exam 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)
More informationMost 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 informationThe 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 informationMATH NEW HOMEWORK AND SOLUTIONS TO PREVIOUS HOMEWORKS AND EXAMS
MATH. 4433. NEW HOMEWORK AND SOLUTIONS TO PREVIOUS HOMEWORKS AND EXAMS TOMASZ PRZEBINDA. Final project, due 0:00 am, /0/208 via e-mail.. State the Fundamental Theorem of Algebra. Recall that a subset K
More informationEconomics 204 Fall 2012 Problem Set 3 Suggested Solutions
Economics 204 Fall 2012 Problem Set 3 Suggested Solutions 1. Give an example of each of the following (and prove that your example indeed works): (a) A complete metric space that is bounded but not compact.
More informationQuick Tour of the Topology of R. Steven Hurder, Dave Marker, & John Wood 1
Quick Tour of the Topology of R Steven Hurder, Dave Marker, & John Wood 1 1 Department of Mathematics, University of Illinois at Chicago April 17, 2003 Preface i Chapter 1. The Topology of R 1 1. Open
More informationMASTERS 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 informationReal 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 informationDefinition 6.1. A metric space (X, d) is complete if every Cauchy sequence tends to a limit in X.
Chapter 6 Completeness Lecture 18 Recall from Definition 2.22 that a Cauchy sequence in (X, d) is a sequence whose terms get closer and closer together, without any limit being specified. In the Euclidean
More informationPart A. Metric and Topological Spaces
Part A Metric and Topological Spaces 1 Lecture A1 Introduction to the Course This course is an introduction to the basic ideas of topology and metric space theory for first-year graduate students. Topology
More informationContinuity. 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 informationMath 104: Homework 7 solutions
Math 04: Homework 7 solutions. (a) The derivative of f () = is f () = 2 which is unbounded as 0. Since f () is continuous on [0, ], it is uniformly continous on this interval by Theorem 9.2. Hence for
More information36 CHAPTER 2. COMPLEX-VALUED FUNCTIONS. In this case, we denote lim z z0 f(z) = α.
36 CHAPTER 2. COMPLEX-VALUED FUNCTIONS In this case, we denote lim z z0 f(z) = α. A complex-valued function f defined in A is called continuous at z 0 A if lim z z 0 f(z) = f(z 0 ). Theorem 2.1.1 Let A
More informationImmerse Metric Space Homework
Immerse Metric Space Homework (Exercises -2). In R n, define d(x, y) = x y +... + x n y n. Show that d is a metric that induces the usual topology. Sketch the basis elements when n = 2. Solution: Steps
More informationBonus Homework. Math 766 Spring ) For E 1,E 2 R n, define E 1 + E 2 = {x + y : x E 1,y E 2 }.
Bonus Homework Math 766 Spring ) For E,E R n, define E + E = {x + y : x E,y E }. (a) Prove that if E and E are compact, then E + E is compact. Proof: Since E + E R n, it is sufficient to prove that E +
More informationContents 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 informationOptimization, WS 2017/18. Part 1. Convergence in Metric Spaces
Lecturer Dan Rust, Optimization, WS 2017/18 Tutor Mima Stanojkovski Part 1. Convergence in Metric Spaces (about 7 Lectures) Supporting Literature: Angel de la Fuente, Mathematical Methods and Models for
More informationMATH 104 : Final Exam
MATH 104 : Final Exam 10 May, 2017 Name: You have 3 hours to answer the questions. You are allowed one page (front and back) worth of notes. The page should not be larger than a standard US letter size.
More informationMath 117: Infinite Sequences
Math 7: Infinite Sequences John Douglas Moore November, 008 The three main theorems in the theory of infinite sequences are the Monotone Convergence Theorem, the Cauchy Sequence Theorem and the Subsequence
More informationFUNCTIONAL ANALYSIS LECTURE NOTES: COMPACT SETS AND FINITE-DIMENSIONAL SPACES. 1. Compact Sets
FUNCTIONAL ANALYSIS LECTURE NOTES: COMPACT SETS AND FINITE-DIMENSIONAL SPACES CHRISTOPHER HEIL 1. Compact Sets Definition 1.1 (Compact and Totally Bounded Sets). Let X be a metric space, and let E X be
More informationIntroduction 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