a) 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.
|
|
- Charlotte Newman
- 6 years ago
- Views:
Transcription
1 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 I mean lim nk x nk a) 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. Let the metric space in which our Cauchy sequences x,y are defined be denoted F. First, note that a distance function is defined from F F to R +. So for any two points a, b F, d(a, b) R +. Now, in order to show that d(x, y) is well defined, we have to show that the sequence (d(x n, y n )) n N is Cauchy, so lim n d (x n, y n ) converges in R. Ok, so pick some fixed arbitrary ɛ > 0 R. We want to show that N such that n, m > N = d(x n, y n ) d(x m, y m ) < ɛ. Note that for any m, n N, d(x n, y n ) d(x n, x m ) + d(x m, y n ) via the triangle inequality, which is d(x n, x m ) + d(x m, y n ) via the triangle inquality again, d(x n, x m ) + d(x m, y m ) + d(y n, y m ), by the triangle inequality used twice in 1 step; first to separate out the absolute values, and then to separate out the d s. Note that we can use the triangle inequality to split the d s and the absolute values because the d s represent a distance in F and the absolute values represent the distance of the sequence (d(a, b)) R Well, x is a Cauchy sequence, so by definition for any fixed value ɛ/2, N 1 such that n, m > N 1 = x n x m < ɛ/2, and similarly, since y is a Cauchy sequence, for any fixed value ɛ/2, N 2 such that n, m > N 2 = y n y m < ɛ/2. Letting N = max {N 1, N 2 }, we see that for n, m > N d(x n, y n ) d(x n, x m ) + d(x m, y m ) + d(y n, y m ) < ɛ/2+ d(x m, y m ) +ɛ/2 = d(x m, y m ) +ɛ. But by subtracting, this implies that d(x n, y n ) d(x m, y m ) < ɛ. But since distances in R are not negative, d(x n, y n ) = d(x n, y n ) and d(x m, y m ) = d(x m, y m ), so d(x n, y n ) d(x m, y m ) = d(x n, y n ) d(x m, y m ) < ɛ = d(x n, y n ) d(x m, y m ) < ɛ. This proves the claim that d(x, y) is a Cauchy sequence. So d(x,y) converges and is always well defined. b) Prove the antisymmetry axiom for our construction of the real numbers using Cauchy sequences. 1
2 Let us take two arbitrary Cauchy sequences x and y such that x y and x y. If x y then by the order defined in Harrison s lecture notes on Cauchy sequences, N 1 N such that for n > N 1 = x n y n. If also x y, then again by our definition of order on Cauchy sequences, N 2 N such that for n > N 2 = x n y n. So take N = max {N 1, N 2 }. Then for n > N, x n y n and x n y n, which implies x n = y n by antisymmetry in Q. So then for n > N, d(x n, y n ) = 0. It follows that lim n d(x n, y n ) = 0, since the sequence d(x n, y n ) becomes a sequence of 0 s after a certain cutoff N. This shows that x y.. c) Claim: Let x n be a Cauchy sequence with a subsequence x nk that converges to x. Then x n converges to x. Assume x nk converges to x. In a metric space, this means that lim nk d(x nk, x) = 0. We want to show that the limit of x n = x, which is the same (in a metric space) as saying that lim n d(x n, x) = 0. Well, we know that d(x n, x) d(x n, x nk )+d(x nk, x) by the triangle inequality. Now, any term x nk of the sequence (x nk ) is part of the overall sequence (x n ), so let us denote any term of the sequence (x nk ) by x m. Since the sequence (x n ) is Cauchy, N 1 such that n, m > N 1 = d(x n, x m ) < ɛ/2 by the definition of Cauchy sequence. So in particular, for our choice of n k = m, n, n k > N 1 = d(x n, x nk ) < ɛ/2 Furthermore, the sequence (x nk ) converges to x, so by the definition of convergence, for any x m (x nk ), N 2 such that m > N 2 = d(x m, x) < ɛ/2. So if we pick N = max {N 1, N 2 }, n > N = d(x n, x) d(x n, x nk ) + d(x nk, x) < ɛ/2 + ɛ/2 = ɛ. So then by definition, x n converges to x. Note: We did not denote d(a,b) as a b for Cauchy sequences a and b, since for problem c, the distance function of the metric space was not specified. d) Claim: The sequence x n converges to x every subsequence of x n converges to x. 1) ( = direction) 2
3 Claim: If a sequence x n converges, then every subsequence x nk converges to the same limit. First, note that n k k, k N. The proof of this is by induction. First, n k, k N is a subset of N because it is an indexing set of a subsequence, so for our base case, letting k = 1: n 1 1, since 1 is the lower bound of N. Inductive step: Inductive Hypothesis: suppose n i i, for some fixed arbitrary choice of i N. Well, n i+1 > n i, since these n k, k N are index for the subsequence x nk, so n i+1 denotes the term of the subsequence after n i. n i i by the inductive hypothesis, so n i+1 N > i, meaning n i+1 i+1, because i+1 is the next term in N. Thus, by induction, we have shown that n k k, k N. So now, let x = lim n x n, and let ɛ > 0. Then N such that n > N = x n x < ɛ. Then, k > N = n k > N by our earlier inductive argument, = x nk x < ɛ. So then by the definition of convergence, lim n x nk = x. 2) ( = direction) of x n converges to x. Then the set of sub- Suppose every subsequence x nk sequential limits S = x. We want to show three things now. Subclaim 1: lim sup x n S Subclaim 2: lim inf x n S. Subclaim 3: If lim sup x n = lim inf x n = x, then lim x n = x. Since S consists only of x, subclaim 1 and 2 show that lim sup x n = x = lim inf x n, so by subclaim 3, lim x n = x. So we have proven the claim. Subclaim 1 and 2: We need to construct subsequences x nk and x nj such that limsupx n = lim n x nk and liminfx n = lim n x nj. First, suppose x n is unbounded above. Then = limsupx n. Then let n 1 = 1. Given n 1 <... < n k, let n k+1 > max { k + 1, x nk+1 }. I can do this because x n is unbounded above. Then this subsequence is also unbounded 3
4 above,(i.e. lim n x nk = ), so S, meaning limsupx n S. Similarly, if x n is unbounded below, then = liminfx n. Let n 1 = 1, and given n 1 >... > n k, let n k+1 < min { k, x nk+1 }. I can do this because x n is unbounded below. Then this subsequence is also unbounded below,(i.e. lim n x nk = ), so S, meaning liminfx n S. The remaining cases are that x n is either bounded above or below. Since these cases are similar, without loss of generality, let x n be bounded above, (i.e. limsupx n = x, x ). Let ɛ > 0. Then by the definition of limsup, there are infinite points x n in the open epsilon ball around x. Let p be such a point, so let d(x, x n1 ) = ɛ 1 < ɛ. Then take ɛ 2 such that 0 < ɛ 2 < ɛ 1. There must be a point x n2 in this ball, since for some N, x sup {x n : n > N} < ɛ for any epsilon. Keep doing this to get a subsequence of x n that converges to x. Then limsupx n S. So we have proven subclaim 1, and w.l.o.g subclaim 2 as well. Remark: In Ross, pg 78, we find out that limsupx n = sups and liminfx n = infs. Subclaim 3: lim sup x n = x =lim inf x n. We want to show that lim x n = x. Let ɛ > 0. lim sup x n = lim N sup {x n : n > N} by definition. So we know there N1 N such that: x sup {x n : n > N1} < ɛ. This follows directly from the definition of limit. But then, this means precisely that x n < x + ɛ, n > N1 by manipulation of the inequality. In the same way, using the definition of lim inf, we know that since lim inf x n = x n = lim N inf {x n : n > N} by definition, N2 N such that: x inf {x n : n > N2} < ɛ,meaning that x n > x ɛ, n > N2 by manipulation of the inequality. So if we take N = max {N1, N2}, we get that x ɛ < x n < x + ɛ, n > N = n > N, x n x < ɛ, which means lim x n = x by definition. e) Claim: Let x n and y n be two real sequences. Suppose lim sup (x n ) < or lim sup (y n ) >. Then lim sup (x n + y n ) lim sup (x n ) + lim sup (y n ) Case 1: Suppose lim sup (x n ) = and lim sup (y n ) =. 4
5 Then lim sup (x n ) + lim sup (y n ) = + = (by definition). On the other hand, lim sup (x n + y n ) = lim n sup {x n + y n : n N} =, since for n N, the suprema of the x n and the suprema of the y n are both individually approaching. So since, lim sup (x n + y n ) lim sup (x n ) + lim sup (y n ). Case 2: Suppose lim sup (x n ) = and lim sup (y n ) =. Then lim sup (x n ) + lim sup (y n ) = = (by definition). On the other hand, lim sup (x n + y n ) = lim n sup {x n + y n : n N} =, since for n N, the suprema of the x n and the suprema of the y n are both individually approaching. So since, lim sup (x n + y n ) lim sup (x n ) + lim sup (y n ). Case 3: Suppose lim sup (x n ) < and lim sup (y n ) >. By definition of the supremum of a set, N N, x k sup {x n : n N} for all k N, and similarly, N N, y k sup {y n : n N} for all k N. Then x k +y k sup {x n : n N}+sup {y n : n N}, for all k N, meaning sup {x n : n N}+sup {y n : n N} is an upper bound for the set {x k + y k : k N}. Then sup {x n : n N} + sup {y n : n N} sup {x k + y k : k N}. Since this is true for all N N, as long as we have that k N, I can take limits as N on both sides and still preserve the equality. So: lim N (sup {x n : n N}+sup {y n : n N}) lim N (sup {x k + y k : k N}). But this is precisely an inequality for the lim sup of the three sequences, (x n ), (y n ) and (x n + y n ). So by definition, this implies that: limsup(x k + y k ) limsup(x k ) + limsup(y k ) 5
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 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 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 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 informationd(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 informationEcon Lecture 3. Outline. 1. Metric Spaces and Normed Spaces 2. Convergence of Sequences in Metric Spaces 3. Sequences in R and R n
Econ 204 2011 Lecture 3 Outline 1. Metric Spaces and Normed Spaces 2. Convergence of Sequences in Metric Spaces 3. Sequences in R and R n 1 Metric Spaces and Metrics Generalize distance and length notions
More informationWe are now going to go back to the concept of sequences, and look at some properties of sequences in R
4 Lecture 4 4. Real Sequences We are now going to go back to the concept of sequences, and look at some properties of sequences in R Definition 3 A real sequence is increasing if + for all, and strictly
More informationp-adic Analysis Compared to Real Lecture 1
p-adic Analysis Compared to Real Lecture 1 Felix Hensel, Waltraud Lederle, Simone Montemezzani October 12, 2011 1 Normed Fields & non-archimedean Norms Definition 1.1. A metric on a non-empty set X is
More informationLecture 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 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 informationSequences. Limits of Sequences. Definition. A real-valued sequence s is any function s : N R.
Sequences Limits of Sequences. Definition. A real-valued sequence s is any function s : N R. Usually, instead of using the notation s(n), we write s n for the value of this function calculated at n. We
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 informationMATH 51H Section 4. October 16, Recall what it means for a function between metric spaces to be continuous:
MATH 51H Section 4 October 16, 2015 1 Continuity Recall what it means for a function between metric spaces to be continuous: Definition. Let (X, d X ), (Y, d Y ) be metric spaces. A function f : X Y is
More informationMath General Topology Fall 2012 Homework 1 Solutions
Math 535 - General Topology Fall 2012 Homework 1 Solutions Definition. Let V be a (real or complex) vector space. A norm on V is a function : V R satisfying: 1. Positivity: x 0 for all x V and moreover
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 information1. Supremum and Infimum Remark: In this sections, all the subsets of R are assumed to be nonempty.
1. Supremum and Infimum Remark: In this sections, all the subsets of R are assumed to be nonempty. Let E be a subset of R. We say that E is bounded above if there exists a real number U such that x U for
More informationCOMPLETION OF A METRIC SPACE
COMPLETION OF A METRIC SPACE HOW ANY INCOMPLETE METRIC SPACE CAN BE COMPLETED REBECCA AND TRACE Given any incomplete metric space (X,d), ( X, d X) a completion, with (X,d) ( X, d X) where X complete, and
More informationScalar multiplication and addition of sequences 9
8 Sequences 1.2.7. Proposition. Every subsequence of a convergent sequence (a n ) n N converges to lim n a n. Proof. If (a nk ) k N is a subsequence of (a n ) n N, then n k k for every k. Hence if ε >
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 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 410 Homework 6 Due Monday, October 26
Math 40 Homework 6 Due Monday, October 26. Let c be any constant and assume that lim s n = s and lim t n = t. Prove that: a) lim c s n = c s We talked about these in class: We want to show that for all
More informationHomework 4, 5, 6 Solutions. > 0, and so a n 0 = n + 1 n = ( n+1 n)( n+1+ n) 1 if n is odd 1/n if n is even diverges.
2..2(a) lim a n = 0. Homework 4, 5, 6 Solutions Proof. Let ɛ > 0. Then for n n = 2+ 2ɛ we have 2n 3 4+ ɛ 3 > ɛ > 0, so 0 < 2n 3 < ɛ, and thus a n 0 = 2n 3 < ɛ. 2..2(g) lim ( n + n) = 0. Proof. Let ɛ >
More informationLecture 3. Econ August 12
Lecture 3 Econ 2001 2015 August 12 Lecture 3 Outline 1 Metric and Metric Spaces 2 Norm and Normed Spaces 3 Sequences and Subsequences 4 Convergence 5 Monotone and Bounded Sequences Announcements: - Friday
More informationLecture 4: Completion of a Metric Space
15 Lecture 4: Completion of a Metric Space Closure vs. Completeness. Recall the statement of Lemma??(b): A subspace M of a metric space X is closed if and only if every convergent sequence {x n } X satisfying
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 informationCharacterisation of Accumulation Points. Convergence in Metric Spaces. Characterisation of Closed Sets. Characterisation of Closed Sets
Convergence in Metric Spaces Functional Analysis Lecture 3: Convergence and Continuity in Metric Spaces Bengt Ove Turesson September 4, 2016 Suppose that (X, d) is a metric space. A sequence (x n ) X is
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 informationSet, 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 informationMATH 117 LECTURE NOTES
MATH 117 LECTURE NOTES XIN ZHOU Abstract. This is the set of lecture notes for Math 117 during Fall quarter of 2017 at UC Santa Barbara. The lectures follow closely the textbook [1]. Contents 1. The set
More informationMath 426 Homework 4 Due 3 November 2017
Math 46 Homework 4 Due 3 November 017 1. Given a metric space X,d) and two subsets A,B, we define the distance between them, dista,b), as the infimum inf a A, b B da,b). a) Prove that if A is compact and
More informationReal Variables: Solutions to Homework 3
Real Variables: Solutions to Homework 3 September 3, 011 Exercise 0.1. Chapter 3, # : Show that the cantor set C consists of all x such that x has some triadic expansion for which every is either 0 or.
More informationMATH 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 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 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 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 informationMATH 101, FALL 2018: SUPPLEMENTARY NOTES ON THE REAL LINE
MATH 101, FALL 2018: SUPPLEMENTARY NOTES ON THE REAL LINE SEBASTIEN VASEY These notes describe the material for November 26, 2018 (while similar content is in Abbott s book, the presentation here is different).
More informationThis exam contains 5 pages (including this cover page) and 4 questions. The total number of points is 100. Grade Table
MAT25-2 Summer Session 2 207 Practice Final August 24th, 207 Time Limit: Hour 40 Minutes Name: Instructor: Nathaniel Gallup This exam contains 5 pages (including this cover page) and 4 questions. The total
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 LM (24543) Lectures 01
Math 32300 LM (24543) Lectures 01 Ethan Akin Office: NAC 6/287 Phone: 650-5136 Email: ethanakin@earthlink.net Spring, 2018 Contents Introduction, Ross Chapter 1 and Appendix The Natural Numbers N and The
More informationCopyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.
.1 Limits of Sequences. CHAPTER.1.0. a) True. If converges, then there is an M > 0 such that M. Choose by Archimedes an N N such that N > M/ε. Then n N implies /n M/n M/N < ε. b) False. = n does not converge,
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 informationMA651 Topology. Lecture 10. Metric Spaces.
MA65 Topology. Lecture 0. Metric Spaces. This text is based on the following books: Topology by James Dugundgji Fundamental concepts of topology by Peter O Neil Linear Algebra and Analysis by Marc Zamansky
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 informationAN EXPLORATION OF THE METRIZABILITY OF TOPOLOGICAL SPACES
AN EXPLORATION OF THE METRIZABILITY OF TOPOLOGICAL SPACES DUSTIN HEDMARK Abstract. A study of the conditions under which a topological space is metrizable, concluding with a proof of the Nagata Smirnov
More informationContinuity. Chapter 4
Chapter 4 Continuity Throughout this chapter D is a nonempty subset of the real numbers. We recall the definition of a function. Definition 4.1. A function from D into R, denoted f : D R, is a subset of
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 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 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 informationTHE REAL NUMBERS Chapter #4
FOUNDATIONS OF ANALYSIS FALL 2008 TRUE/FALSE QUESTIONS THE REAL NUMBERS Chapter #4 (1) Every element in a field has a multiplicative inverse. (2) In a field the additive inverse of 1 is 0. (3) In a field
More informationCLASSICAL PROBABILITY MODES OF CONVERGENCE AND INEQUALITIES
CLASSICAL PROBABILITY 2008 2. MODES OF CONVERGENCE AND INEQUALITIES JOHN MORIARTY In many interesting and important situations, the object of interest is influenced by many random factors. If we can construct
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 informationFINAL EXAM Math 25 Temple-F06
FINAL EXAM Math 25 Temple-F06 Write solutions on the paper provided. Put your name on this exam sheet, and staple it to the front of your finished exam. Do Not Write On This Exam Sheet. Problem 1. (Short
More informationCHAPTER 1. Metric Spaces. 1. Definition and examples
CHAPTER Metric Spaces. Definition and examples Metric spaces generalize and clarify the notion of distance in the real line. The definitions will provide us with a useful tool for more general applications
More informationClass Notes for MATH 255.
Class Notes for MATH 255. by S. W. Drury Copyright c 2006, by S. W. Drury. Contents 0 LimSup and LimInf Metric Spaces and Analysis in Several Variables 6. Metric Spaces........................... 6.2 Normed
More informationProblem Set 1: Solutions Math 201A: Fall Problem 1. Let (X, d) be a metric space. (a) Prove the reverse triangle inequality: for every x, y, z X
Problem Set 1: s Math 201A: Fall 2016 Problem 1. Let (X, d) be a metric space. (a) Prove the reverse triangle inequality: for every x, y, z X d(x, y) d(x, z) d(z, y). (b) Prove that if x n x and y n y
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 informationSolutions Manual for Homework Sets Math 401. Dr Vignon S. Oussa
1 Solutions Manual for Homework Sets Math 401 Dr Vignon S. Oussa Solutions Homework Set 0 Math 401 Fall 2015 1. (Direct Proof) Assume that x and y are odd integers. Then there exist integers u and v such
More informationMath 117: Topology of the Real Numbers
Math 117: Topology of the Real Numbers John Douglas Moore November 10, 2008 The goal of these notes is to highlight the most important topics presented in Chapter 3 of the text [1] and to provide a few
More informationEC 521 MATHEMATICAL METHODS FOR ECONOMICS. Lecture 1: Preliminaries
EC 521 MATHEMATICAL METHODS FOR ECONOMICS Lecture 1: Preliminaries Murat YILMAZ Boğaziçi University In this lecture we provide some basic facts from both Linear Algebra and Real Analysis, which are going
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 informationMetric 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 informationYour first day at work MATH 806 (Fall 2015)
Your first day at work MATH 806 (Fall 2015) 1. Let X be a set (with no particular algebraic structure). A function d : X X R is called a metric on X (and then X is called a metric space) when d satisfies
More informationREAL 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 informationMath 127C, Spring 2006 Final Exam Solutions. x 2 ), g(y 1, y 2 ) = ( y 1 y 2, y1 2 + y2) 2. (g f) (0) = g (f(0))f (0).
Math 27C, Spring 26 Final Exam Solutions. Define f : R 2 R 2 and g : R 2 R 2 by f(x, x 2 (sin x 2 x, e x x 2, g(y, y 2 ( y y 2, y 2 + y2 2. Use the chain rule to compute the matrix of (g f (,. By the chain
More informationA lower bound for X is an element z F such that
Math 316, Intro to Analysis Completeness. Definition 1 (Upper bounds). Let F be an ordered field. For a subset X F an upper bound for X is an element y F such that A lower bound for X is an element z F
More informationApplied Analysis (APPM 5440): Final exam 1:30pm 4:00pm, Dec. 14, Closed books.
Applied Analysis APPM 44: Final exam 1:3pm 4:pm, Dec. 14, 29. Closed books. Problem 1: 2p Set I = [, 1]. Prove that there is a continuous function u on I such that 1 ux 1 x sin ut 2 dt = cosx, x I. Define
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 informationCOUNTABLE PRODUCTS ELENA GUREVICH
COUNTABLE PRODUCTS ELENA GUREVICH Abstract. In this paper, we extend our study to countably infinite products of topological spaces.. The Cantor Set Let us constract a very curios (but usefull) set known
More informationHilbert spaces. 1. Cauchy-Schwarz-Bunyakowsky inequality
(October 29, 2016) Hilbert spaces Paul Garrett garrett@math.umn.edu http://www.math.umn.edu/ garrett/ [This document is http://www.math.umn.edu/ garrett/m/fun/notes 2016-17/03 hsp.pdf] Hilbert spaces are
More informationMeasure and Integration: Solutions of CW2
Measure and Integration: s of CW2 Fall 206 [G. Holzegel] December 9, 206 Problem of Sheet 5 a) Left (f n ) and (g n ) be sequences of integrable functions with f n (x) f (x) and g n (x) g (x) for almost
More informationContinuity. Chapter 4
Chapter 4 Continuity Throughout this chapter D is a nonempty subset of the real numbers. We recall the definition of a function. Definition 4.1. A function from D into R, denoted f : D R, is a subset of
More informationHAUSDORFF DISTANCE AND CONVEXITY JOSEPH SANTOS. A thesis submitted to the. Graduate School-Camden. Rutgers, The State University of New Jersey
HAUSDORFF DISTANCE AND CONVEXITY By JOSEPH SANTOS A thesis submitted to the Graduate School-Camden Rutgers, The State University of New Jersey In partial fulfullment of the requirements For the degree
More informationM17 MAT25-21 HOMEWORK 6
M17 MAT25-21 HOMEWORK 6 DUE 10:00AM WEDNESDAY SEPTEMBER 13TH 1. To Hand In Double Series. The exercises in this section will guide you to complete the proof of the following theorem: Theorem 1: Absolute
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 informationExercise Solutions to Functional Analysis
Exercise Solutions to Functional Analysis Note: References refer to M. Schechter, Principles of Functional Analysis Exersize that. Let φ,..., φ n be an orthonormal set in a Hilbert space H. Show n f n
More informationLecture 4 Lebesgue spaces and inequalities
Lecture 4: Lebesgue spaces and inequalities 1 of 10 Course: Theory of Probability I Term: Fall 2013 Instructor: Gordan Zitkovic Lecture 4 Lebesgue spaces and inequalities Lebesgue spaces We have seen how
More informationIn class midterm Exam - Answer key
Fall 2013 In class midterm Exam - Answer key ARE211 Problem 1 (20 points). Metrics: Let B be the set of all sequences x = (x 1,x 2,...). Define d(x,y) = sup{ x i y i : i = 1,2,...}. a) Prove that d is
More informationReal Analysis - Notes and After Notes Fall 2008
Real Analysis - Notes and After Notes Fall 2008 October 29, 2008 1 Introduction into proof August 20, 2008 First we will go through some simple proofs to learn how one writes a rigorous proof. Let start
More information5 Compact linear operators
5 Compact linear operators One of the most important results of Linear Algebra is that for every selfadjoint linear map A on a finite-dimensional space, there exists a basis consisting of eigenvectors.
More informationA lower bound for X is an element z F such that
Math 316, Intro to Analysis Completeness. Definition 1 (Upper bounds). Let F be an ordered field. For a subset X F an upper bound for X is an element y F such that A lower bound for X is an element z F
More informationFUNCTIONAL ANALYSIS-NORMED SPACE
MAT641- MSC Mathematics, MNIT Jaipur FUNCTIONAL ANALYSIS-NORMED SPACE DR. RITU AGARWAL MALAVIYA NATIONAL INSTITUTE OF TECHNOLOGY JAIPUR 1. Normed space Norm generalizes the concept of length in an arbitrary
More informationCauchy Sequences. x n = 1 ( ) 2 1 1, . As you well know, k! n 1. 1 k! = e, = k! k=0. k = k=1
Cauchy Sequences The Definition. I will introduce the main idea by contrasting three sequences of rational numbers. In each case, the universal set of numbers will be the set Q of rational numbers; all
More informationChapter 7. Metric Spaces December 5, Metric spaces
Chapter 7 Metric Spaces December 5, 2011 7.1 Metric spaces Note: 1.5 lectures As mentioned in the introduction, the main idea in analysis is to take limits. In chapter 2 we learned to take limits of sequences
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 informationEssential Background for Real Analysis I (MATH 5210)
Background Material 1 Essential Background for Real Analysis I (MATH 5210) Note. These notes contain several definitions, theorems, and examples from Analysis I (MATH 4217/5217) which you must know for
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 informationMATH 426, TOPOLOGY. p 1.
MATH 426, TOPOLOGY THE p-norms In this document we assume an extended real line, where is an element greater than all real numbers; the interval notation [1, ] will be used to mean [1, ) { }. 1. THE p
More informationCOMPLETE METRIC SPACES AND THE CONTRACTION MAPPING THEOREM
COMPLETE METRIC SPACES AND THE CONTRACTION MAPPING THEOREM A metric space (M, d) is a set M with a metric d(x, y), x, y M that has the properties d(x, y) = d(y, x), x, y M d(x, y) d(x, z) + d(z, y), x,
More informationMany of you got these steps reversed or otherwise out of order.
Problem 1. Let (X, d X ) and (Y, d Y ) be metric spaces. Suppose that there is a bijection f : X Y such that for all x 1, x 2 X. 1 10 d X(x 1, x 2 ) d Y (f(x 1 ), f(x 2 )) 10d X (x 1, x 2 ) Show that if
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 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 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 informationBolzano Weierstrass Theorems I
Bolzano Weierstrass Theorems I James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University September 8, 2017 Outline The Bolzano Weierstrass Theorem Extensions
More informationWe are going to discuss what it means for a sequence to converge in three stages: First, we define what it means for a sequence to converge to zero
Chapter Limits of Sequences Calculus Student: lim s n = 0 means the s n are getting closer and closer to zero but never gets there. Instructor: ARGHHHHH! Exercise. Think of a better response for the instructor.
More informationCourse 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 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 information4130 HOMEWORK 4. , a 2
4130 HOMEWORK 4 Due Tuesday March 2 (1) Let N N denote the set of all sequences of natural numbers. That is, N N = {(a 1, a 2, a 3,...) : a i N}. Show that N N = P(N). We use the Schröder-Bernstein Theorem.
More informationAustin 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 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 information