DOCSLIB.ORG

Math 190: Fall 2014 Homework 2 Due 5:00Pm on Friday 10/17/2014 Problem 1: Let X Be a Set and Let B Be a Basis for a Topology On

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Math 190: Fall 2014 Homework 2 Due 5:00Pm on Friday 10/17/2014 Problem 1: Let X Be a Set and Let B Be a Basis for a Topology On Math 190: Fall 2014 Homework 2 Due 5:00pm on Friday 10/17/2014 Problem 1: Let X be a set and let B be a basis for a topology on X. Let T be the topology on X generated by B. Prove that T is the intersection of all the topologies on X which contain B. Solution: Let U denote the intersection of all the topologies that contain B. Since T consists of all unions of sets in B, we have that T contains B, so that U ⊂ T. On 0 the other hand, if T is a topology which contains B and fBαgα2I is a collection of 0 0 sets in B, then for all α 2 I we have that Bα 2 B ⊂ T . Since T is a topology, we S 0 have that α2I Bα 2 T . Since every set in T is a union of sets in B, this implies that T ⊂ T0. Since T0 was an arbitrary topology containing B, we must have that T ⊂ U. We conclude that T = U, as desired. Problem 2: Let K denote the set K = f1=n : n = 1; 2; 3;::: g. Let B denote the set of all open intervals (a; b) in R, together with all sets of the form (a; b)−K. Prove that B is the basis for a topology on R. (This is called the K-topology and denoted RK .) Solution: Given any x 2 R, we have that x 2 (x − 1; x + 1), so that the sets in B cover R. On the other hand, if x 2 (a; b) \ (c; d), we necessarily have that x 2 (max(a; c); min(b; d)). If x 2 ((a; b) − K) \ (c; d), then x 2 (max(a; c); min(b; d)) − K. Finally, if x 2 ((a; b) − K) \ ((c; d) − K), then x 2 (max(a; c); min(b; d)) − K. We conclude that B is the basis for a topology on R. Problem 3: (Exercise 13.6 in Munkres) Recall that the lower limit topology R` is the topology on R generated by the basis f[a; b): a; b 2 Rg. Prove that the topologies RK and R` on R are incomparable (that is, neither is finer than the other). Solution: Let K = f1; 1=2; 1=3;::: g and consider the set U = (−1; 1)−K ⊂ R. Then U is open in RK (since it is a basis element of RK ). We claim that U is not open in R`. If U were open in R`, since 0 2 U there would exist some basis element [a; b) of R` such that 0 2 [a; b) ⊂ U. But 0 2 [a; b) implies that a ≤ 0 and b > 0, so that [a; b) \ K is nonempty, and [a; b) 6⊂ U. We conclude that U is not open in R`, so that R` is not finer than RK . Let V = [0; 1) ⊂ R. Then V is open in R` (since is it a basis element of R`). We claim that V is not open in RK . To see this, suppose that V were open in RK . Then there would exist some basis element B of the basis B of RK given in Problem 2 such that 0 2 B ⊂ V . This basis element B would have to have one of the forms (a; b) for a a < 0 < b or (a; b) − K for a < 0 < b. In either case, we have that 2 2 B − V , so that B 6⊂ V , which is a contradiction. We conclude that RK is not finer than R`. 1 2 Problem 4: (Exercise 13.7 in Munkres) Consider the following five topologies on R: T1 = the standard topology, T2 = the topology RK , T3 = the finite complement topology, T4 = the upper limit topology, having all sets of the form (a; b] as a basis, T5 = the topology having all sets of the form (−∞; a) = fx 2 R : x < ag as a basis. Determine, for each of the topologies, which of the others it contains. Solution: We claim that the topologies T3 and T5 are incomparable, the topology T1 contains both T3 and T5, the topologies T2 contains T1, and the topology T4 contains T2. To see that T3 and T5 are incomparable, observe that T5 consists precisely of the sets T5 = f(−∞; a): a 2 Rg [ f;; Rg, as these are the unions of sets in the given basis. In particular, we have that R − f0g is in T3 but not T5. Moreover, we have that (−∞; 0) is in T5 but not T3. We conclude that T3 and T5 are incomparable. To see that T1 contains T3, let U 2 T3 and let x 2 U. We have that R − U is finite, so we can find some interval (a; b) in R such that x 2 (a; b) ⊂ R − U. Since intervals of the form (a; b) form a basis for T1, we get that T1 contains T3. To see that T1 contains T5, let (−∞; a) be a basis element for T5 and let x 2 (−∞; a). Then x 2 (x − 1; a) ⊂ (−∞; a). Since intervals of the form (c; d) form a basis for T1, we get that T1 contains T5. To see that T2 contains T1, we need only observe that the basis for T2 given in Problem 2 contains the standard open interval basis for T1. Since a topology consists exactly of the unions of sets in a given basis, we get that T2 contains T1. On the other hand, we have that U = (−1; 1) − f1; 1=2; 1=3;::: g is open in T2 but not in T1 (since it contains no interval (a; b) containing 0), so that the containment of T1 inside T2 is strict. To see that T4 contains T2, let B be a basis element of T2 and let x 2 B. We have that B is of one of the forms (a; b) or (a; b) − K, where K is as in Problem 2. If B is of the form (a; b), we have that x 2 (a; x] ⊂ (a; b). If B is of the form (a; b) − K and x ≤ 0, we have that x 2 (a; x] ⊆ (a; b)−K. If B is of the form (a; b)−K and x > 0, we can find n minimal such that 1=n < x. Then x 2 (1=n; x] ⊂ (a; b) − K. We conclude that T4 contains T2. To see that this containment is strict, observe that (−1; 0] is an open set in T4, but contains no basis element of T2 which contains 0. Problem 5: Endow R2 with dictionary order <. That is, we define (x; y) < (x0; y0) if and only if x < x0 or x = x0 and y < y0. Let T be the associated order topology on R2. Prove that T is strictly finer than the Euclidean topology on R2. (That is, prove that T is finer than the Euclidean topology, and also not equal to the Euclidean topology.) Solution: Let U ⊂ R2 be open in the Euclidean topology and let (x; y) 2 U. Then there exists > 0 such that B((x; y); ) ⊂ U. In particular, we have that I = f(x; y0): y − < y0 < y + g is contained in U. But I is an open interval in the dictionary order on R2, so that I 2 T. We conclude that T is finer than the Euclidean topology. To see 3 that T is strictly finer than the Euclidean topology, we need only observe that the line segment f0g × (−1; 1) is open in T but not open in the Euclidean topology. Problem 6: Let X be a finite ordered set. Prove that the order topology on X is the discrete topology. Solution: We can write X = fx1 < x2 < ··· < xng. For 2 ≤ i ≤ n − 1, we have that fxig = (xi−1; xi+1). We also have that fx1g = [x1; x2) and fxng = (xn−1; xn]. We conclude that every singleton subset of X is open in the order topology, and hence the order topology on X equals the discrete topology. Problem 7: (Exercise 16.4 in Munkres) A map f : X ! Y of topological spaces is called an open map if whenever U ⊂ X is open, we have that f(U) = ff(x): x 2 Ug ⊂ Y is open. Let X and Y be topological spaces and endow X × Y with the product topology. Prove that the projections π1 : X × Y ! X and π2 : X × Y ! Y defined by π1(x; y) = x and π2(x; y) = y are open maps. Solution: Let U ⊂ X × Y be an open set. For any (x; y) 2 U, there exist open sets Vx;y ⊂ X and Wx;y ⊂ Y such that (x; y) 2 Vx;y × Wx;y ⊂ U. It follows that S S π1(U) = π1( (x;y)2U Vx;y × Wx;y) = (x;y)2U Vx;y is open in X. Similarly, we have that S S π2(U) = π2( (x;y)2U Vx;y × Wx;y) = (x;y)2U Wx;y is open in Y . We conclude that π1 and π2 are open maps. Problem 8: Give an example of an infinite ordered set X such that the order topology on X is the discrete topology. Solution: Let X = Z be the set of integers with the usual order <. For any n 2 Z, the interval (n − 1; n + 1) = fng is open in the order topology, so that all singleton subsets of Z are open in the order topology.
Load more

Recommended publications
  • Lecture 13: Basis for a Topology
    Lecture 13: Basis for a Topology 1 Basis for a Topology Lemma 1.1. Let (X; T) be a topological space. Suppose that C is a collection of open sets of X such that for each open set U of X and each x in U, there is an element C 2 C such that x 2 C ⊂ U. Then C is the basis for the topology of X. Proof. In order to show that C is a basis, need to show that C satisfies the two properties of basis. To show the first property, let x be an element of the open set X. Now, since X is open, then, by hypothesis there exists an element C of C such that x 2 C ⊂ X. Thus C satisfies the first property of basis. To show the second property of basis, let x 2 X and C1;C2 be open sets in C such that x 2 C1 and x 2 C2. This implies that C1 \ C2 is also an open set in C and x 2 C1 \ C2. Then, by hypothesis, there exists an open set C3 2 C such that x 2 C3 ⊂ C1 \ C2. Thus, C satisfies the second property of basis too and hence, is indeed a basis for the topology on X. On many occasions it is much easier to show results about a topological space by arguing in terms of its basis. For example, to determine whether one topology is finer than the other, it is easier to compare the two topologies in terms of their bases.
    [Show full text]
  • Advance Topics in Topology - Point-Set
    ADVANCE TOPICS IN TOPOLOGY - POINT-SET NOTES COMPILED BY KATO LA 19 January 2012 Background Intervals: pa; bq “ tx P R | a ă x ă bu ÓÓ , / / calc. notation set theory notation / / \Open" intervals / pa; 8q ./ / / p´8; bq / / / -/ ra; bs; ra; 8q: Closed pa; bs; ra; bq: Half-openzHalf-closed Open Sets: Includes all open intervals and union of open intervals. i.e., p0; 1q Y p3; 4q. Definition: A set A of real numbers is open if @ x P A; D an open interval contain- ing x which is a subset of A. Question: Is Q, the set of all rational numbers, an open set of R? 1 1 1 - No. Consider . No interval of the form ´ "; ` " is a subset of . We can 2 2 2 Q 2 ˆ ˙ ask a similar question in R . 2 Is R open in R ?- No, because any disk around any point in R will have points above and below that point of R. Date: Spring 2012. 1 2 NOTES COMPILED BY KATO LA Definition: A set is called closed if its complement is open. In R, p0; 1q is open and p´8; 0s Y r1; 8q is closed. R is open, thus Ø is closed. r0; 1q is not open or closed. In R, the set t0u is closed: its complement is p´8; 0q Y p0; 8q. In 2 R , is tp0; 0qu closed? - Yes. Chapter 2 - Topological Spaces & Continuous Functions Definition:A topology on a set X is a collection T of subsets of X satisfying: (1) Ø;X P T (2) The union of any number of sets in T is again, in the collection (3) The intersection of any finite number of sets in T , is again in T Alternative Definition: ¨ ¨ ¨ is a collection T of subsets of X such that Ø;X P T and T is closed under arbitrary unions and finite intersections.
    [Show full text]
  • Math 553 - Topology Todd Riggs Assignment 1 Sept 10, 2014
    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 x 2 A there is an open set U containing x such that U ⊂ A. Show that A is open in X. Proof: Since for each x 2 A there exist Ux 2 T such that x 2 Ux and Ux ⊂ A, we have S A = Ux. That is, A is the union of open sets in X and thus by definition of the x2A topology on X; A is open. 13.3) Show that the collection Tc given in example 4 (pg 77) is a topology on the set X. Is the collection T1 = fUjX − U is infinite or empty or all of Xg a topology on X. Proof: PART (1) From example 4 Tc = fUjX − U is countable or is all of Xg. To show that Tc is a topology on X, we must show we meet all 3 conditions of the definition of topology. First, since X −X = ; is countable and X −; = X is all of X we have X and ; 2 Tc. Second, let Uα 2 Tc for α 2 I. We must look at two cases: (1) X − Uα = X for all α and (2) There exist α such that X − Uα is countable. Case (1): X − Uα = X S T This implies Uα = ; for all α. So X − ( Uα) = (X − Uα) = X and thus, α2I α2I S Uα 2 Tc.
    [Show full text]
  • Some Examples and Counterexamples of Advanced Compactness in Topology
    International Journal of Research in Engineering and Science (IJRES) ISSN (Online): 2320-9364, ISSN (Print): 2320-9356 www.ijres.org Volume 9 Issue 1 ǁ 2021 ǁ PP. 10-16 Some Examples and Counterexamples of Advanced Compactness in Topology Pankaj Goswami Department of Mathematics, University of Gour Banga, Malda, 732102, West Bengal, India ABSTRACT: The examples and counter examples are always usefull for better comprehension of underlying concept in a theorem or definition .Compactness has come to be one of the most importent and useful topic in advanced mathematics.This paper is an attempt to fill in some of the information that the standard textbook treatment of compactness leaves out, and giving some constructive significative counterexamples of Advanced Compactness in Topology . KEYWORDS: open cover, compact, connected, topological space, example, counterexamples, continuous function, locally compact, countably compact, limit point compact, sequentially compact, lindelöf , paracompact --------------------------------------------------------------------------------------------------------------------------------------- Date of Submission: 15-01-2021 Date of acceptance: 30-01-2021 --------------------------------------------------------------------------------------------------------------------------------------- I. INTRODUCTION We begin by presenting some definitions, notations,examples and some theorems that are essential for concept of compactness in advanced topology .For more detailed, see [1-2], [7-10] and others, in references .If X is a closed bounded subset of the real line ℝ, then any family of open sets in ℝ whose union contains X has a finite sub family whose union also contains X . If X is a metric space or topological space in its own right , then the above proposition can be thought as saying that any class of open sets in X whose union is equal to X has a finite subclass whose union is also equal to X.
    [Show full text]
  • Point Set Topology
    Point Set Topology A. Topological Spaces and Continuous Maps Definition 1.1 A topology on a set X is a collection T of subsets of X satisfying the following axioms: T 1. ∅;X ∈T. T2. {Oα | α ∈ I}⊆T =⇒ α∈IOα ∈T. T3. O; O0 ∈T =⇒ O∩O0 ∈TS. A topological space is a pair (X; T )whereXis a set and T is a topology on X. When the topology T on X under discussion is clear, we simply denote (X; T )by X.Let(X; T ) be a topological space. Members of T are called open sets. (T 3) implies that a finite intersection of open sets is open. Examples 1. Let X be a set. The power set 2X of X is a topology on X and is called the discrete topology on X. The collection I = {∅;X} is also a topology on X and is called the indiscrete topology on X. 2. Let (X; d) be a metric space. Define O ⊆ X to be open if for any x ∈ O, there exists an open ball B(x; r) lying inside O. Then, Td = {O ⊆ X | O is open} ∪ {∅} is a topology on X. Td is called the topology induced by the metric d. 3. Since Rn is a metric space with the usual metric: n d((x ; :::; x ); (y ; :::; y )) = (x − y )2; 1 n 1 n v i i ui=1 uX t Rn has a topology U induced by d. This topology on Rn is called the usual topology. 4. Let X be an infinite set.
    [Show full text]
  • MATH 411 HOMEWORK 3 SOLUTIONS 2.17.6. Let A, B, and A
    MATH 411 HOMEWORK 3 SOLUTIONS ADAM LEVINE 2.17.6. Let A, B, and Aα denote subsets of a space X. (a) Show that if A ⊂ B, then A¯ ⊂ B¯. By definition, A¯ is the intersection of all closed sets containing A. Since B¯ is a closed set that contains B and hence A, it must therefore contain A¯. (b) Show that A [ B = A¯ [ B¯. The set A [ B is a closed set that contains A [ B, so it contains both A and B, and therefore it contains both A¯ and B¯. Therefore, A [ B ⊃ A¯ [ B¯. Conversely, A¯ [ B¯ is a closed set (since it is the union of two closed sets) that contains A [ B, so it contains A [ B. (Note that we had to use the fact that finite intersections of closed sets are closed!) S S ¯ (c) Show that Aα ⊃ Aα; give an example where equality fails. S Just as in (b), the set Aα is a closed set that contains each Aα, so it contains each ¯ S ¯ Aα, and thus it contains Aα. (But we can't do the converse because an arbitrary union of closures isn't necessarily closed!) As a counterexample, let A be the closed set [ 1 ; 1] ⊂ . Then S A¯ = S A = i i R i2N i i2N i (0; 1], while S A = [0; 1]. i2N i 2.17.11. Show that the product of two Hausdorff spaces is Hausdorff. Let X and Y be Hausdorff spaces. For any two distinct points (x; y) and (x0; y0) in X × Y , we may assume that they differ in at least one coordinate.
    [Show full text]
  • Math 131: Introduction to Topology 1
    Math 131: Introduction to Topology 1 Professor Denis Auroux Fall, 2019 Contents 9/4/2019 - Introduction, Metric Spaces, Basic Notions3 9/9/2019 - Topological Spaces, Bases9 9/11/2019 - Subspaces, Products, Continuity 15 9/16/2019 - Continuity, Homeomorphisms, Limit Points 21 9/18/2019 - Sequences, Limits, Products 26 9/23/2019 - More Product Topologies, Connectedness 32 9/25/2019 - Connectedness, Path Connectedness 37 9/30/2019 - Compactness 42 10/2/2019 - Compactness, Uncountability, Metric Spaces 45 10/7/2019 - Compactness, Limit Points, Sequences 49 10/9/2019 - Compactifications and Local Compactness 53 10/16/2019 - Countability, Separability, and Normal Spaces 57 10/21/2019 - Urysohn's Lemma and the Metrization Theorem 61 1 Please email Beckham Myers at [email protected] with any corrections, questions, or comments. Any mistakes or errors are mine. 10/23/2019 - Category Theory, Paths, Homotopy 64 10/28/2019 - The Fundamental Group(oid) 70 10/30/2019 - Covering Spaces, Path Lifting 75 11/4/2019 - Fundamental Group of the Circle, Quotients and Gluing 80 11/6/2019 - The Brouwer Fixed Point Theorem 85 11/11/2019 - Antipodes and the Borsuk-Ulam Theorem 88 11/13/2019 - Deformation Retracts and Homotopy Equivalence 91 11/18/2019 - Computing the Fundamental Group 95 11/20/2019 - Equivalence of Covering Spaces and the Universal Cover 99 11/25/2019 - Universal Covering Spaces, Free Groups 104 12/2/2019 - Seifert-Van Kampen Theorem, Final Examples 109 2 9/4/2019 - Introduction, Metric Spaces, Basic Notions The instructor for this course is Professor Denis Auroux. His email is [email protected] and his office is SC539.
    [Show full text]
  • Basic Exam: Topology August 31, 2007 Answer five of the Seven Questions
    Department of Mathematics and Statistics University of Massachusetts Basic Exam: Topology August 31, 2007 Answer five of the seven questions. Indicate clearly which five ques- tions you want graded. Justify your answers. Passing standard: For Master’s level, 60% with two questions essentially complete. For Ph.D. level, 75% with three questions essentially complete. Let C(X, Y ) denote the set of continuous functions from topological spaces X to Y. Let R denote the real line with the standard topology. (1) (a) Give an example of a space which is Hausdorff, locally-connected, and not connected. (b) Give an example of a space which is Hausdorff, path-connected, and not locally path-connected. x (2) Consider the sequence fn(x) = sin( n ). Determine whether or not the sequence converges in each of the following topologies of C(R, R): the uniform topology, the compact-open topology, and the point-open (pointwise convergence) topology. (3) Let X = R/Z be the quotient space where the integers Z ⊂ R are identified to a single point. Prove that X is connected, Hausdorff, and non-compact. (4) Prove the Uniform Limit Theorem: “Let X be any topological space and Y a metric space. Let fn ∈ C(X, Y ) be a sequence which converges uniformly to f. Then f ∈ C(X, Y ).” (5) (a) Let f : X → Y be a quotient map, where Y is connected. Suppose that for all y ∈ Y , f −1(y) is connected. Show that X is connected. (b) Let g : A → B be continuous, where B is path-connected. Suppose that for all b ∈ B, g−1(b) is path-connected.
    [Show full text]
  • MATH 411 HOMEWORK 2 SOLUTIONS 2.13.3. Let X Be a Set
    MATH 411 HOMEWORK 2 SOLUTIONS ADAM LEVINE 2.13.3. Let X be a set, and let Tc = fU ⊂ X j X − U is countable or all of Xg T1 = fU ⊂ X j X − U is infinite or empty or all of Xg: Show that Tc is a topology. Is T1 a topology? For Tc, it is clear that ; and X are in Tc. If Uα is any collection of sets in Tc, then [ \ X − Uα = (X − Uα); α α which is an intersection of countable sets, hence countable. Likewise, if U1;:::;Uk are sets in Tc, then k k \ [ X − Uα = (X − Uα); i=1 i=1 which is a finite union of countable sets, hence countable. (Cf. Example 3 on page 77.) We claim that T1 is not a topology provided that X is an infinite set. (If X is finite, then T1 is just the indiscrete topology.) Choose some x0 2 X, and consider all of the 1-point sets fxg for x 6= x0. These sets all have infinite complement. On the other hand, the union S fxg equals X − fx g, which has complement fx g, x6=x0 0 0 so it is not open. 2.13.6. Show that the topologies of R` and RK are not comparable. The set [2; 3) is open in R`, but it is not open in RK ; any open set in RK containing 2 must contain an interval (2 − , 2 + ) for some > 0. The set (−1; 1) − K is open in RK , but it is not open in R`, since any open set in R` containing 0 must contain an interval [0; ) for some > 0, and therefore contain elements of K.
    [Show full text]
  • Mathematics 5853 Examination I September 28, 2004
    Mathematics 5853 Name (please print) Examination I September 28, 2004 Instructions: Give brief, clear answers. I. Let X = R and let T = {U ⊆ X | ∃M ∈ R, (M, ∞) ⊆ U} ∪ {∅} (where (M, ∞) means {r ∈ R | r > M}). (10) 1. Prove that T is a topology on X (you do not need to worry about special cases involving the empty set). X is open since (0, ∞) ⊂ X. The empty set is open by definition of T . Suppose {Uα}α∈A are open sets. Choose a (nonempty) Uα0 . For some M,(M, ∞) ⊆ Uα0 . Therefore (M, ∞) ⊆ ∪α∈AUα, so ∪α∈AUα is also open. Suppose {Uα1 ,...,Uαn } are open. For each i, choose Mi with (Mi, ∞) ⊆ Uαi . Put M equal to the n n maximum of the Mi, then (M, ∞) ⊆ ∩i=1Uαi , so ∩i=1Uαi is open. 2. Prove that with this topology, X is not Hausdorff. In fact, no two points can have disjoint open neighborhoods. For if U and V were open neighborhoods of two distinct points, then as in part 3 of the proof that T is a topology, there exists an interval (M, ∞) ⊆ U ∩ V , and U and V are not disjoint. II. Let S be a collection of subsets of a set X, such that X = ∪S∈S S. Define B = {S1 ∩ S2 ∩ · · · ∩ Sn | Si ∈ S}, (10) that is, the collection of all subsets of X that are intersections of finitely many elements of S. Verify that B is a basis. By hypothesis, X = ∪S∈S S. Let B1 = S1 ∩ · · · ∩ Sm and B2 = T1 ∩ · · · ∩ Tn be two elements of B, and suppose that x ∈ B1 ∩ B2.
    [Show full text]
  • Topology: Notes and Problems
    TOPOLOGY: NOTES AND PROBLEMS Abstract. These are the notes prepared for the course MTH 304 to be offered to undergraduate students at IIT Kanpur. Contents 1. Topology of Metric Spaces 1 2. Topological Spaces 3 3. Basis for a Topology 4 4. Topology Generated by a Basis 4 4.1. Infinitude of Prime Numbers 6 5. Product Topology 6 6. Subspace Topology 7 7. Closed Sets, Hausdorff Spaces, and Closure of a Set 9 8. Continuous Functions 12 8.1. A Theorem of Volterra Vito 15 9. Homeomorphisms 16 10. Product, Box, and Uniform Topologies 18 11. Compact Spaces 21 12. Quotient Topology 23 13. Connected and Path-connected Spaces 27 14. Compactness Revisited 30 15. Countability Axioms 31 16. Separation Axioms 33 17. Tychonoff's Theorem 36 References 37 1. Topology of Metric Spaces A function d : X × X ! R+ is a metric if for any x; y; z 2 X; (1) d(x; y) = 0 iff x = y. (2) d(x; y) = d(y; x). (3) d(x; y) ≤ d(x; z) + d(z; y): We refer to (X; d) as a metric space. 2 Exercise 1.1 : Give five of your favourite metrics on R : Exercise 1.2 : Show that C[0; 1] is a metric space with metric d1(f; g) := kf − gk1: 1 2 TOPOLOGY: NOTES AND PROBLEMS An open ball in a metric space (X; d) is given by Bd(x; R) := fy 2 X : d(y; x) < Rg: Exercise 1.3 : Let (X; d) be your favourite metric (X; d). How does open ball in (X; d) look like ? Exercise 1.4 : Visualize the open ball B(f; R) in (C[0; 1]; d1); where f is the identity function.
    [Show full text]
  • Topology Sol 2014-15.Pdf
    Topology B.Math (Hons.) II Year Semester II, 2014{15 http://www.isibang.ac.in/~statmath/oldqp/QP/Topology%202014-15.pdf Final Exam Solution Question 1. (a) Define the notions of sequential compactness and limit point compactness. (b) Prove that the Lebesgue covering lemma holds for a sequentially compact metric space. Answer. (a) Sequentially compact: Let T = (S; τ) be a topological space and X ⊆ S. Then X is called sequentially compact in T if every infinite sequence in X has a sub-sequence which converges to a point in X. Limit point compact: Let T = (S; τ) be a topological space and X ⊆ S. Then X is called limit point compact or weakly countably compact if every infinite subset of X has a limit point in X. (b) Lebesgue covering lemma: If the metric space (X; d) is compact and an open cover of X is given, then there exists a number δ > 0 such that every subset of X having diameter less than δ; is contained in some member of the cover. Now, we prove that Lebesgue covering lemma holds for a sequentially compact metric space. Proof: Let U be an open cover of X and suppose, to the contrary, that there is no such δ. Then for every n 2 N, 1=n is not a Lebesgue number, that is, there is a subset Sn ⊂ X such that diam(Sn) < 1=n but Sn is not entirely contained in any of the elements of U. (In particular, Sn is not empty.) For each n 2 N, we choose 0 one point xn 2 Sn.
    [Show full text]