Problems - Section 17-2, 4, 6c, 9, 10, 13, 14; Section 18-1, 3, 4, 6, 8, 10; Section 19-1, 3, 5, 7, 8, 9;
|
|
- Emily McBride
- 5 years ago
- Views:
Transcription
1 Math Topology Todd Riggs Assignment 2 Sept 17, 2014 Problems - Section 17-2, 4, 6c, 9, 10, 13, 14; Section 18-1, 3, 4, 6, 8, 10; Section 19-1, 3, 5, 7, 8, 9; 17.2) Show that if A is closed in Y and Y is closed in X, then A is closed in X. If we can show that X A is open in X, then A will be closed in X. Since Y is closed in X, we have X Y open in X. Similarly, we have Y A open in Y. By the subspace topology of Y, there exist an open subset U of of X such that U Y = Y A and hence (Y A) U. So X A = (X Y ) U and since unions of open sets are open, X A is open in X as desired. 17.4) Show that if U is open in X and A is closed in X, then U A is open in X, and A U is closed in X. To show U A is open, we must show X (U A) is closed. Notice that X (U A) = X U A which is a finite union of closed sets and thus closed. Similarly, for A U closed, we will show X (A U) is open. As above, we have X (A U) = X A U which is a union of open sets and thus open. 17.6c) Let A, B, and A α denote subsets of a space X. Prove the following: (c) A α A α ; give an example where equality fails. Let x A α. Then x A αo for some α o and, by Theorem 17.5, for every open set U, containing x, we have U A αo. So U { A α } = and as such x A α. Thus A α A α as desired. Example where equality fails: Let A n = { 1 } where n N. Then n 1 A n = { 1 n N}. So A n α = { 1 } and n A α = { 1 n N}, but A n α = { 1 n N} {0}. n Page 1 of 8
2 17.9) Let A X and B Y. Show that in the space X Y, A B = A B. Let (a, b) A B. So, for all open neighborhoods U that contain (a, b), U (A B). Now (a, b) will be in A B iff a A and b B. This will be true iff for all open W containing a and for all open V containing b we have W A and V B. Let W be an open set containing a and V an open set containing b. Then W V is open in X Y and contains (a, b). So (W V ) (A B) = (W A) (V B) and as such W A and V B. Now since W and V are arbitrary open sets containing a and b respectively, a A and b B. Thus (a, b) A B and A B A B. Conversely, let (a, b) A B. Then a A and b B. So for all open U containing a and all open V containing b, we have U A and V B. Now (a, b) will be contained in A B iff for all open W in X Y containing (a, b), W (A B). Let W be open in X Y and contain (a, b). Since W is open, there exists U V,where U is open in X and V is open in Y, such that (a, b) U V W. Then U open in X and a U implies U A and similarly V B. So, W (A B) (U V ) (A B) = (U A) (V B). Hence, (a, b) A B and thus A B = A B ) Show that every order topology is Hausdorff. Let X be a non-empty set, with more than one element, with the order topology. By definition of Hausdorff we must show for each pair x 1, x 2 of distinct points of X that there exists neighborhoods U 1 and U 2 of x 1 and x 2, respectively, that are disjoint. We will look at three cases: (1) X is finite (or bounded), (2) X is unbounded. Case (1): X is finite (or bounded). Let X = {x 1, x 2,...} where x 1 is the least, (or similarly greatest), and x i < x i+1. Then by definition of the order topology there exists disjoint neighborhoods nbr(x 1 ) = [x 1, x i ) and nbr(x i ) = (x i 1, x n ], where 1 < i n, and X is Hausdorff. Case (2): X is unbounded. Let X be an unbounded ordered set. If X is unbounded than X must have an infinite number of elements. So for all element x i, x j an ordering is possible, say... < x i < x j <.... Then neighborhoods (, x j ) of x i and (x i, ) of x j are disjoint and X is again Hausdorff. Page 2 of 8
3 17.13) Show that X is Hausdorff iff the diagonal = {x x x X} is closed in X X. Let X be a Hausdorff Space. To show the diagonal is closed we will show (X X) is open. Let (x, y) (X X). Since (x, y) we have x y. Furthermore, since X is Haudorff, there exists open neighborhoods U and V of x and y, respectively, such that U V =. Now (U V ) = since if not, there exists (x 1, x 1 ) (U V ) implying x 1 U V and contradicting U V =. So (x, y) (U V ) (X X) and U V is open in X X. Hence, (X X) is open as desired. Conversely, Suppose X X is closed so that (X X) is open. Let x, y X be such that x y. So (x, y) (X X) and since (X X) is open, there exists open neighborhoods U and V, of x and y respectively, such that (U V ) (X X). Now U V = since if x 1 U V then (x 1, x 1 ) U V and (x 1, x 1 ) would be in. Thus we have found neighborhoods U and V, of x and y respectively, such that U V =. That is, X is Hausdorff ) In the finite complement topology on R, to what point or points does the sequence x n = 1 n converge? Claim: The sequence x n = 1 converges to all x X. n In the finite complement topology, U is open in a topological space X if U = X or X U is finite (pg 77). Let U be an open neighborhood of x X. Then if X U is finite, x n U for all but finitely many n and by definition, x n converges to x. 18.1) Prove that for functions f : R R, the ɛ δ definition of continuity implies the open set definition. Let f : R R be a continuous function based on the ɛ δ definition and V be an open set in the range space. We must show that f 1 (V ) is open. Let x o f 1 (V ) so that f(x o ) V. Since V is open and f(x o ) V, there exists an ɛ > 0 such that (f(x o ) ɛ, f(x o ) + ɛ) V. From the ɛ δ definition, choose δ > 0, depending on x o and ɛ, such that if x R satisfies x x o < δ, then f(x) f(x o ) < ɛ. Now let x (x o δ, x o + δ) so that x x o < δ. Then f(x) f(x o ) < ɛ and f(x) (f(x o ) ɛ, f(x o ) + ɛ) V. So f(x) V implies x f 1 (V ) and f 1 (V ) is open as desired. Page 3 of 8
4 18.3) Let X and X denote a singleset in the two topologies T and T, respectively. Let i : X X be the idenity function. (a) Show that i is continuous iff T is finer than T. (b) Show that i is a homeomorphism iff T = T. (a) Suppose i is continuous and let U T. Since i : X X is continuous and U is open in X, we have i 1 (U) = U is open in X. That is U T and T is finer than T. Conversely, assume T is finer than T. Then for all U T we have U T. So given an open set U in X we have i 1 (U) = U is open in X. Hence i is continuous. (b) Suppose i is a homeomorphism. Then i : X X and i 1 : X X are continuous. So for U open in X we have i 1 (U) = U is open in X. That is T T. Likewise, for W open in X we have (i 1 ) 1 (W ) = i(w ) = W is open in X and T T. Thus T = T. Conversely, let T = T. Since i is the identity, it is bijective. So if U is open in X we have i 1 (U) = U is open in X. That is, i : X X is continuous. Likewise, if W is open in X we have (i 1 ) 1 (W ) = i(w ) = W is open in X and i 1 : X X is continuous. Thus i is a homeomorphism. 18.4) Given x 0 X and y 0 Y, show that the maps f : X X Y and g : Y X Y defined by f(x) = x y 0 and g(y) = x 0 y are imbeddings. We need to show that the restriction of f and g to their ranges leads to homeomorphisms. That is, f : X X {y o }, f (x) = f(x) and g : Y {x o } Y, g (y) = g(y) are homeomorphisms. Since f and g are bijective with their range, it needs to be shown that they and their inverses are continuous. For the continuity of f let W be open in X {y o }. By definition of subspace topology, W = (X {y 0 }) (U V ), where U V is open in X Y. So U is open in X and V is open in Y. Now (f ) 1 (W ) = (f ) 1 ((X U) ({y o } V ) = X U = U which is open in X. Hence f is continuous. Similarly, for the continuity of g, let W be open in{x o } Y. By definition of subspace topology, W = ({x 0 } Y ) (U V ) = ({x o } U) (Y V ), where U V is open in X Y so that U is open in X and V is open in Y. Now (g ) 1 (W ) = (g ) 1 (({x o } U) (Y V ) = Y V = V which is open in Y. Hence g is continuous. Now we need to show (f ) 1 : X {y o } X and (g ) 1 : {x o } Y Y are continuous. For (f ) 1 if U is open in X, then ((f ) 1 ) 1 (U) = f (U) = U {y o } = (X {y o }) (U Y ) which is open in X {y o }. Similarly, for (g ) 1, if V is open in Y, then ((g ) 1 ) 1 (V ) = g (V ) = {x o } V = ({x o } Y ) (X V ) which is open in {x o } Y. Since f and g are continuous bijective mappings with continuous inverses, they are homeomorphisms. Thus f and g imbeddings as desired. Page 4 of 8
5 18.6) Find a function f : R R that is continuous at precisely one point. Claim: The piecewise function f : R R defined by { x : x Q f(x) = 0 : x Q is continuous only at 0. First we will show f is continuous at 0 and than that it is not continuous at any other point. Let U be a neighborhood of f(0) = 0. (We will be utilizing Theorem 18.1(4) pg 104.) We need to show there is a neighborhood V of 0 such that f(v ) U. Since U is an open set containing 0, there exists ɛ such that 0 ( ɛ, ɛ) U. Let V = ( ɛ, ɛ). Then V is a neighborhood of 0 such that f(v ) ( ɛ, ɛ) U. So, f is continuous at 0. For the second part, let x R such that x 0. If x Q, set ɛ = x, and let U = 2 (x ɛ, x+ɛ). So U is a neighborhood of f(x) = x that does not contain 0. Now given any neighborhood V of x, it contains irrational numbers, so 0 f(v ). We cannot find a neighborhood V of x such that f(v ) U. Hence f is not continuous at x. If x R Q then U = ( x, x ) is a neighborhood of f(x) = 0. Given any neighborhood 2 2 V of x, we can choose ɛ > 0 sufficiently small so that x (x ɛ, x + ɛ) V and ɛ < x. Between x and x + ɛ we can find q Q. So f(q) = q f(v ) but q U, so 2 f(v ) is not contained in U. Hence f is not continuous at x and our claim is proved. 18.8) Let Y be an ordered set in the order topology. Let f, g : X Y be continuous. (a) Show that the set {x f(x) g(x)} is closed in X. (b) Let h : X Y be the function h(x) = min{f(x), g(x)}. Show that h is continuous. (Hint: Use pasting lemma) PART A By problem we have that Y is Hausdorff. To show {x f(x) g(x)} is closed we will show that the complement is open. That is, X {x f(x) g(x)} = {x f(x) > g(x)} = A is open. Let x A so that f(x) > g(x). Since Y is Hausdorff and f and g are continuous, we have disjoint neighborhoods U and V of f(x) and g(x) in Y, respectfully, such that f 1 (U) and g 1 (V ) are open in X. Now we have f 1 (U) g 1 (V ) is open in X and since x f 1 (U) g 1 (V ) there exists a neighborhood N such that x N (f 1 (U) g 1 (V )). Moreover, since U and V are disjoint, f(n) g(n) =. That is, for all x N we have f(x) > g(x) implying N A and we have A open as desired. PART B Using the pasting lemma, with f continuous on A = {x f(x) g(x)}, and g continuous on B = {x f(x) g(x)}, where B is closed from part (a) as well as A by a similar argument to (a), we have X = A B and A B = {x f(x) = g(x)}. Page 5 of 8
6 Now, by noticing that h(x) h (x) = { f(x) : x A g(x) : x B we have h continuous ) Let f : A B and g : C D be continuous functions. Let us define a map f g : A C B D by the equation (f g)(a c) = f(a) g(c). Show that f g is continuous. Let a A and c C. Additionally, let U = U 1 U 2 be an open neighborhood of f(a) g(c) in B D where f(a) U 1 B and g(c) U 2 D. Then since f and g are continuous, there exists basis elements B A A and B C C such that f 1 (f(a)) = a B A and g 1 (g(c)) = c B C. So B = B A B C is a basis element of A C and as such open. That is (f(a) g(c)) 1 = f 1 (f(a)) g 1 (g(c)) B is open and thus f g is continuous. 19.1) Prove Theorem 19.2 The given set, B α, is a basis element in the box topology. We must show this α J is open in the generated topology. Letting α J U α be a basis element in the product topology and x α J U α, then x α will be in U α for all alpha. Since B α is a basis for X α, there exists B α such that x α B α U α. That is, x B α U α and thus open in the generated topology. Conversely, the given set, α J B α, is a basis element in the product topology. Let x U α where U α is a basis element such that all but finitely many U α = X α. Then, for all U α X α we choose basis elements B α1,..., B α2 such that x αj B αj U αj. So, as above, x B α U α and thus open in the generated topology. 19.3) Prove Theorem 19.4 Since the box topology is finer than the product topology (pg 115), it will suffice to show the theorem holds under the box topology. Let x and y be distinct elements in X α such that each X α is Hausdorff. Since x and y are distinct, there is at least one coordinate term such that x i y i. So for each x i y i there are open neighborhoods U i and V i, where x i U i and y i V i, that are subsets of X i and U i V i =. Now define open neighborhoods U and V in X α by U = U α and V = V α. Then U V = (U α V α ) = (U 1 V 1 )... (U i V i )... = (U 1 V 1 )... ( )... =. Thus X α is Hausdorff as desired. Page 6 of 8
7 19.5) One of the implications stated in Theorem 19.6 holds for the box topology. Which one? By example 2 pg 117, if each f α is continuous then f is not necessarily continuous. So, If f : A X α is continuous than f α : A X α is continuous for all alpha. Let f : A X α be continuous and U = U α be open in X α where U α is open in X α. Then since f 1 (U) is open in A, we have fα 1 (U α ) open for all alpha and f α is continuous. 19.7) Let R be the subset R ω consisting of all sequences that are eventually zero, that is, all sequences (x 1, x 2,...) such that x i = 0 for only finitely many values of i. What is the closure of R in R ω in the box and product topologies? Justify. Claim: (a) R is closed in the box topology and (b) The closure of R in the product topology is R ω. (a) We will show R ω R is open. The box topology has open sets of the form U = U 1 U Let {x n } n N = (x 1, x 2,...) (R ω R ). Then, since there are infinitely many nonzero elements, only finitely many zero elements, we can find an open neighborhood U nj about x nj that does not contain zero. So V = U 1... U n1 U n2... will be an open neighborhood containing (x n ) n N such that V is a proper subset of R. (b) We will show that for any open neighborhood U R ω containing (x n ) n N that U R. Notice, for U = U 1 U 2... U j R R open in the product topology, that (x n ) n N = (x 1, x 2,..., x j, 0, 0,...), for some j is an element of U and an element of R. That is, U R as desired. 19.8) Given the sequences (a 1, a 2,...) and (b 1, b 2,...) of real numbers with a i > 0 for all i, define h : R ω R ω by the equation h((x 1, x 2,...)) = (a 1 x 1 + b 1, a 2 x 2 + b 2,...). Show that if R ω is given the product topology, h is a homeomorphism of R ω with itself. What happens if R ω is given the box topology? Clearly h is bijective. By theorem 19.6, h is continuous in the product topology since the composition with the projection onto the n-th factor is continuous. In the box topology let h n = h n (x) = a n x + b n be the n-th coordinate function of h. Then, for U = U 1 U 2... open in the box topology, we have h 1 (U) = h 1 1 (U 1 ) h 1 2 (U 2 )... open since h n is continuous for all n. Page 7 of 8
8 19.9) Show that the choice axiom is equivalent to the statement that for any indexed family {A α } α J of nonempty sets, with J 0, the cartesian product A α is not empty. α J The axiom of choice (pg 59) states that from any collection of nonempty sets A = {A α } α J that there exists a choice function c : A such that c(a α ) = a α A α. So, c(a α ) = (a α ) α J = A α. ADDITIONAL PROBLEM(S) A α α J 17.3) Show that if A is closed in X and B is closed in Y, then A B is closed in X Y. As in problem 17.2, we need to show that (X Y ) (A B) is open in X Y. Since A is close in X and B is closed in Y, X A and Y B are open in X and Y respectively. So by the product topology, (X A) (Y B) = (X Y ) (A B) is open in X Y as desired. Page 8 of 8
Problems: Section 13-1, 3, 4, 5, 6; Section 16-1, 5, 8, 9;
Math 553 - Topology Todd Riggs Assignment 1 Sept 10, 2014 Problems: Section 13-1, 3, 4, 5, 6; Section 16-1, 5, 8, 9; 13.1) Let X be a topological space and let A be a subset of X. Suppose that for each
More informationTopology. 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 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 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 3T03 - Topology
Math 3T03 - Topology Sang Woo Park April 5, 2018 Contents 1 Introduction to topology 2 1.1 What is topology?.......................... 2 1.2 Set theory............................... 3 2 Functions 4 3
More informationMATH 220 (all sections) Homework #12 not to be turned in posted Friday, November 24, 2017
MATH 220 (all sections) Homework #12 not to be turned in posted Friday, November 24, 2017 Definition: A set A is finite if there exists a nonnegative integer c such that there exists a bijection from A
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 information3 Hausdorff and Connected Spaces
3 Hausdorff and Connected Spaces In this chapter we address the question of when two spaces are homeomorphic. This is done by examining two properties that are shared by any pair of homeomorphic spaces.
More informationMH 7500 THEOREMS. (iii) A = A; (iv) A B = A B. Theorem 5. If {A α : α Λ} is any collection of subsets of a space X, then
MH 7500 THEOREMS Definition. A topological space is an ordered pair (X, T ), where X is a set and T is a collection of subsets of X such that (i) T and X T ; (ii) U V T whenever U, V T ; (iii) U T whenever
More informationSets and Functions. MATH 464/506, Real Analysis. J. Robert Buchanan. Summer Department of Mathematics. J. Robert Buchanan Sets and Functions
Sets and Functions MATH 464/506, Real Analysis J. Robert Buchanan Department of Mathematics Summer 2007 Notation x A means that element x is a member of set A. x / A means that x is not a member of A.
More information5 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 informationMAT3500/ Mandatory assignment 2013 Solutions
MAT3500/4500 - Mandatory assignment 2013 s Problem 1 Let X be a topological space, A and B be subsets of X. Recall the definition of the boundary Bd A of a set A. Prove that Bd (A B) (Bd A) (Bd B). Discuss
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 informationAxioms of separation
Axioms of separation These notes discuss the same topic as Sections 31, 32, 33, 34, 35, and also 7, 10 of Munkres book. Some notions (hereditarily normal, perfectly normal, collectionwise normal, monotonically
More information1. Simplify the following. Solution: = {0} Hint: glossary: there is for all : such that & and
Topology MT434P Problems/Homework Recommended Reading: Munkres, J.R. Topology Hatcher, A. Algebraic Topology, http://www.math.cornell.edu/ hatcher/at/atpage.html For those who have a lot of outstanding
More informationMath General Topology Fall 2012 Homework 8 Solutions
Math 535 - General Topology Fall 2012 Homework 8 Solutions Problem 1. (Willard Exercise 19B.1) Show that the one-point compactification of R n is homeomorphic to the n-dimensional sphere S n. Note that
More informationIn 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 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 information4 Countability axioms
4 COUNTABILITY AXIOMS 4 Countability axioms Definition 4.1. Let X be a topological space X is said to be first countable if for any x X, there is a countable basis for the neighborhoods of x. X is said
More informationB 1 = {B(x, r) x = (x 1, x 2 ) H, 0 < r < x 2 }. (a) Show that B = B 1 B 2 is a basis for a topology on X.
Math 6342/7350: Topology and Geometry Sample Preliminary Exam Questions 1. For each of the following topological spaces X i, determine whether X i and X i X i are homeomorphic. (a) X 1 = [0, 1] (b) X 2
More informationTopology Part of the Qualify Exams of Department of Mathematics, Texas A&M University Prepared by Zhang, Zecheng
Topology Part of the Qualify Exams of Department of Mathematics, Texas A&M University Prepared by Zhang, Zecheng Remark 0.1. This is a solution Manuel to the topology questions of the Topology Geometry
More informationSETS AND FUNCTIONS JOSHUA BALLEW
SETS AND FUNCTIONS JOSHUA BALLEW 1. Sets As a review, we begin by considering a naive look at set theory. For our purposes, we define a set as a collection of objects. Except for certain sets like N, Z,
More informationS. Mrówka introduced a topological space ψ whose underlying set is the. natural numbers together with an infinite maximal almost disjoint family(madf)
PAYNE, CATHERINE ANN, M.A. On ψ (κ, M) spaces with κ = ω 1. (2010) Directed by Dr. Jerry Vaughan. 30pp. S. Mrówka introduced a topological space ψ whose underlying set is the natural numbers together with
More informationMATH 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 informationMath 203A - Solution Set 1
Math 203A - Solution Set 1 Problem 1. Show that the Zariski topology on A 2 is not the product of the Zariski topologies on A 1 A 1. Answer: Clearly, the diagonal Z = {(x, y) : x y = 0} A 2 is closed in
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 informationSOLUTIONS TO EXERCISES FOR. MATHEMATICS 205A Part 1. I. Foundational material
SOLUTIONS TO EXERCISES FOR MATHEMATICS 205A Part 1 Fall 2014 I. Foundational material I.1 : Basic set theory Problems from Munkres, 9, p. 64 2. (a (c For each of the first three parts, choose a 1 1 correspondence
More informationTopology, Math 581, Fall 2017 last updated: November 24, Topology 1, Math 581, Fall 2017: Notes and homework Krzysztof Chris Ciesielski
Topology, Math 581, Fall 2017 last updated: November 24, 2017 1 Topology 1, Math 581, Fall 2017: Notes and homework Krzysztof Chris Ciesielski Class of August 17: Course and syllabus overview. Topology
More informationAustin 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 informationIn 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 informationINDECOMPOSABILITY IN INVERSE LIMITS WITH SET-VALUED FUNCTIONS
INDECOMPOSABILITY IN INVERSE LIMITS WITH SET-VALUED FUNCTIONS JAMES P. KELLY AND JONATHAN MEDDAUGH Abstract. In this paper, we develop a sufficient condition for the inverse limit of upper semi-continuous
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 information1 The Cantor Set and the Devil s Staircase
Math 4181 Name: Dr. Franz Rothe November 10, 014 14FALL\4181_fall14candev.tex For extra space, use the back pages. 1 The Cantor Set and the Devil s Staircase 10 Problem 1. For any maps f : X Y and g :
More informationChapter 1. Sets and Mappings
Chapter 1. Sets and Mappings 1. Sets A set is considered to be a collection of objects (elements). If A is a set and x is an element of the set A, we say x is a member of A or x belongs to A, and we write
More informationSolutions 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 informationNAME: 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 information3 COUNTABILITY AND CONNECTEDNESS AXIOMS
3 COUNTABILITY AND CONNECTEDNESS AXIOMS Definition 3.1 Let X be a topological space. A subset D of X is dense in X iff D = X. X is separable iff it contains a countable dense subset. X satisfies the first
More informationMath General Topology Fall 2012 Homework 13 Solutions
Math 535 - General Topology Fall 2012 Homework 13 Solutions Note: In this problem set, function spaces are endowed with the compact-open topology unless otherwise noted. Problem 1. Let X be a compact topological
More informationSOLUTIONS TO THE FINAL EXAM
SOLUTIONS TO THE FINAL EXAM Short questions 1 point each) Give a brief definition for each of the following six concepts: 1) normal for topological spaces) 2) path connected 3) homeomorphism 4) covering
More informationMAS3706 Topology. Revision Lectures, May I do not answer enquiries as to what material will be in the exam.
MAS3706 Topology Revision Lectures, May 208 Z.A.Lykova It is essential that you read and try to understand the lecture notes from the beginning to the end. Many questions from the exam paper will be similar
More informationCHODOUNSKY, DAVID, M.A. Relative Topological Properties. (2006) Directed by Dr. Jerry Vaughan. 48pp.
CHODOUNSKY, DAVID, M.A. Relative Topological Properties. (2006) Directed by Dr. Jerry Vaughan. 48pp. In this thesis we study the concepts of relative topological properties and give some basic facts and
More informationMath 190: Fall 2014 Homework 4 Solutions Due 5:00pm on Friday 11/7/2014
Math 90: Fall 04 Homework 4 Solutions Due 5:00pm on Friday /7/04 Problem : Recall that S n denotes the n-dimensional unit sphere: S n = {(x 0, x,..., x n ) R n+ : x 0 + x + + x n = }. Let N S n denote
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 informationSpring -07 TOPOLOGY III. Conventions
Spring -07 TOPOLOGY III Conventions In the following, a space means a topological space (unless specified otherwise). We usually denote a space by a symbol like X instead of writing, say, (X, τ), and we
More informationTOPOLOGY TAKE-HOME CLAY SHONKWILER
TOPOLOGY TAKE-HOME CLAY SHONKWILER 1. The Discrete Topology Let Y = {0, 1} have the discrete topology. Show that for any topological space X the following are equivalent. (a) X has the discrete topology.
More informationA Short Review of Cardinality
Christopher Heil A Short Review of Cardinality November 14, 2017 c 2017 Christopher Heil Chapter 1 Cardinality We will give a short review of the definition of cardinality and prove some facts about the
More informationMT804 Analysis Homework II
MT804 Analysis Homework II Eudoxus October 6, 2008 p. 135 4.5.1, 4.5.2 p. 136 4.5.3 part a only) p. 140 4.6.1 Exercise 4.5.1 Use the Intermediate Value Theorem to prove that every polynomial of with real
More informationCHAPTER 7. Connectedness
CHAPTER 7 Connectedness 7.1. Connected topological spaces Definition 7.1. A topological space (X, T X ) is said to be connected if there is no continuous surjection f : X {0, 1} where the two point set
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 informationThis 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 informationMAT 530: Topology&Geometry, I Fall 2005
MAT 530: Topology&Geometry, I Fall 2005 Midterm Solutions Note: These solutions are more detailed than solutions sufficient for full credit. Let X denote the set {a, b, c}. The collections Problem 1 5+5
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 informationIntroduction to Topology
Introduction to Topology Randall R. Holmes Auburn University Typeset by AMS-TEX Chapter 1. Metric Spaces 1. Definition and Examples. As the course progresses we will need to review some basic notions about
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 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 informationHomework 5. Solutions
Homework 5. Solutions 1. Let (X,T) be a topological space and let A,B be subsets of X. Show that the closure of their union is given by A B = A B. Since A B is a closed set that contains A B and A B is
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 information2. The Concept of Convergence: Ultrafilters and Nets
2. The Concept of Convergence: Ultrafilters and Nets NOTE: AS OF 2008, SOME OF THIS STUFF IS A BIT OUT- DATED AND HAS A FEW TYPOS. I WILL REVISE THIS MATE- RIAL SOMETIME. In this lecture we discuss two
More informationSelected problems from past exams
Discrete Structures CS2800 Prelim 1 s Selected problems from past exams 1. True/false. For each of the following statements, indicate whether the statement is true or false. Give a one or two sentence
More informationExercises for Unit VI (Infinite constructions in set theory)
Exercises for Unit VI (Infinite constructions in set theory) VI.1 : Indexed families and set theoretic operations (Halmos, 4, 8 9; Lipschutz, 5.3 5.4) Lipschutz : 5.3 5.6, 5.29 5.32, 9.14 1. Generalize
More informationMath 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 informationTOPOLOGY HW 2. x x ± y
TOPOLOGY HW 2 CLAY SHONKWILER 20.9 Show that the euclidean metric d on R n is a metric, as follows: If x, y R n and c R, define x + y = (x 1 + y 1,..., x n + y n ), cx = (cx 1,..., cx n ), x y = x 1 y
More informationMATH 3300 Test 1. Name: Student Id:
Name: Student Id: There are nine problems (check that you have 9 pages). Solutions are expected to be short. In the case of proofs, one or two short paragraphs should be the average length. Write your
More informationMathematics 220 Workshop Cardinality. Some harder problems on cardinality.
Some harder problems on cardinality. These are two series of problems with specific goals: the first goal is to prove that the cardinality of the set of irrational numbers is continuum, and the second
More informationTopology Homework Assignment 1 Solutions
Topology Homework Assignment 1 Solutions 1. Prove that R n with the usual topology satisfies the axioms for a topological space. Let U denote the usual topology on R n. 1(a) R n U because if x R n, then
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 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 information1 The Local-to-Global Lemma
Point-Set Topology Connectedness: Lecture 2 1 The Local-to-Global Lemma In the world of advanced mathematics, we are often interested in comparing the local properties of a space to its global properties.
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 informationNotas de Aula Grupos Profinitos. Martino Garonzi. Universidade de Brasília. Primeiro semestre 2018
Notas de Aula Grupos Profinitos Martino Garonzi Universidade de Brasília Primeiro semestre 2018 1 Le risposte uccidono le domande. 2 Contents 1 Topology 4 2 Profinite spaces 6 3 Topological groups 10 4
More informationINTRODUCTION TO TOPOLOGY, MATH 141, PRACTICE PROBLEMS
INTRODUCTION TO TOPOLOGY, MATH 141, PRACTICE PROBLEMS Problem 1. Give an example of a non-metrizable topological space. Explain. Problem 2. Introduce a topology on N by declaring that open sets are, N,
More informationSolve EACH of the exercises 1-3
Topology Ph.D. Entrance Exam, August 2011 Write a solution of each exercise on a separate page. Solve EACH of the exercises 1-3 Ex. 1. Let X and Y be Hausdorff topological spaces and let f: X Y be continuous.
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 informationChapter 2 Metric Spaces
Chapter 2 Metric Spaces The purpose of this chapter is to present a summary of some basic properties of metric and topological spaces that play an important role in the main body of the book. 2.1 Metrics
More informationREAL 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 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 421, Homework #9 Solutions
Math 41, Homework #9 Solutions (1) (a) A set E R n is said to be path connected if for any pair of points x E and y E there exists a continuous function γ : [0, 1] R n satisfying γ(0) = x, γ(1) = y, and
More informationDO FIVE OUT OF SIX ON EACH SET PROBLEM SET
DO FIVE OUT OF SIX ON EACH SET PROBLEM SET 1. THE AXIOM OF FOUNDATION Early on in the book (page 6) it is indicated that throughout the formal development set is going to mean pure set, or set whose elements,
More informationSolutions to Homework Assignment 2
Solutions to Homework Assignment Real Analysis I February, 03 Notes: (a) Be aware that there maybe some typos in the solutions. If you find any, please let me know. (b) As is usual in proofs, most problems
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 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 informationINVERSE FUNCTION THEOREM and SURFACES IN R n
INVERSE FUNCTION THEOREM and SURFACES IN R n Let f C k (U; R n ), with U R n open. Assume df(a) GL(R n ), where a U. The Inverse Function Theorem says there is an open neighborhood V U of a in R n so that
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 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 informationNotes on ordinals and cardinals
Notes on ordinals and cardinals Reed Solomon 1 Background Terminology We will use the following notation for the common number systems: N = {0, 1, 2,...} = the natural numbers Z = {..., 2, 1, 0, 1, 2,...}
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 informationA. S. Kechris: Classical Descriptive Set Theory; Corrections and Updates (October 1, 2018)
A. S. Kechris: Classical Descriptive Set Theory; Corrections and Updates (October 1, 2018) Page 3, line 9-: add after spaces,with d i < 1, Page 8, line 11: D ϕ D(ϕ) Page 22, line 6: x 0 e x e Page 22:
More informationTOPOLOGICAL GROUPS MATH 519
TOPOLOGICAL GROUPS MATH 519 The purpose of these notes is to give a mostly self-contained topological background for the study of the representations of locally compact totally disconnected groups, as
More informationSets and Functions. (As we will see, in describing a set the order in which elements are listed is irrelevant).
Sets and Functions 1. The language of sets Informally, a set is any collection of objects. The objects may be mathematical objects such as numbers, functions and even sets, or letters or symbols of any
More informationThus, X is connected by Problem 4. Case 3: X = (a, b]. This case is analogous to Case 2. Case 4: X = (a, b). Choose ε < b a
Solutions to Homework #6 1. Complete the proof of the backwards direction of Theorem 12.2 from class (which asserts the any interval in R is connected). Solution: Let X R be a closed interval. Case 1:
More informationMTG 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 informationMATH 409 Advanced Calculus I Lecture 10: Continuity. Properties of continuous functions.
MATH 409 Advanced Calculus I Lecture 10: Continuity. Properties of continuous functions. Continuity Definition. Given a set E R, a function f : E R, and a point c E, the function f is continuous at c if
More informationExtension of continuous functions in digital spaces with the Khalimsky topology
Extension of continuous functions in digital spaces with the Khalimsky topology Erik Melin Uppsala University, Department of Mathematics Box 480, SE-751 06 Uppsala, Sweden melin@math.uu.se http://www.math.uu.se/~melin
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 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 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 informationINVERSE LIMITS AND PROFINITE GROUPS
INVERSE LIMITS AND PROFINITE GROUPS BRIAN OSSERMAN We discuss the inverse limit construction, and consider the special case of inverse limits of finite groups, which should best be considered as topological
More informationSets, Models and Proofs. I. Moerdijk and J. van Oosten Department of Mathematics Utrecht University
Sets, Models and Proofs I. Moerdijk and J. van Oosten Department of Mathematics Utrecht University 2000; revised, 2006 Contents 1 Sets 1 1.1 Cardinal Numbers........................ 2 1.1.1 The Continuum
More informationMath 203A - Solution Set 1
Math 203A - Solution Set 1 Problem 1. Show that the Zariski topology on A 2 is not the product of the Zariski topologies on A 1 A 1. Answer: Clearly, the diagonal Z = {(x, y) : x y = 0} A 2 is closed in
More informationCW complexes. Soren Hansen. This note is meant to give a short introduction to CW complexes.
CW complexes Soren Hansen This note is meant to give a short introduction to CW complexes. 1. Notation and conventions In the following a space is a topological space and a map f : X Y between topological
More information