For any r E Q, we let Jr denote the interval (—r, r). 2.7.5. 2.3.10. 0 CHAPTER 3. Show that any linear map T: Rn -> X is continuous. If, furthermore, we assume that lim f(Jn ) -4 0, then nn Jn contains precisely one point. In this exercise, the product sets are given the product metrics. Clearly, 0 < f < 1. . 6.1.) If A is totally bounded subset of a metric space, show that its closure A is also totally bounded. EQUIVALENT DEFINITIONS OF CONTINUITY 63 Ex. Given > 0, choose N such that d(x,„ x) < E for n > N. What can you say about d(x m , xn ) if m, n > N? 4.3.25. Hence f: X —> [1,2] is an extension of g. We plan to show that f is continuous on the closed sets A and X \ A. We shall see a proof of this statement later. Then f: X -4 {±1} is a continuous non-constant function. Can you improve upon this result? The left end-point a2 of J2 is greater than or equal to that of that is, a l < a2 . Let A be a nonempty subset of a metric space (X, d). Ex. Assume that E > 0 is given and that there exists no 6 > 0 with the required property. We claim that c G E. For each n E N, c— 1/n is not an upper bound c. Clearly for E. We can therefore find x n E E such that c— 1/n 0. Let (X,11) be an NLS, x E X and 0 0 such that B(x, r) n B(y,r) = O. There exists an open interval J such that sEJC U. The diameter diam (A) of A is defined by diam (A) := sup{d(x, y) : x, y G A}, in the extended real number system. Let us look at (i). A naive attempt would runs as follows. We need to show that given any x E £2 and 6 > 0, we can find y E D such that 11x — 02 < 6' Since x E £2, there exists N E N such that the tail Eoe N+1 1xn 1 2 < 62 /2. From (i), we see that xN — 1 E S. Hence S is nonempty. 'Recall that Er7 an is convergent if the sequence of partial sums sr, := E ak is convergent. Thus any nonzero x E Rn is path-connected to p and hence Rn {0} is path-connected by Lemma 5.2.14. Define 5 on X as in Ex. 0, then there (iii) If f : X RI is continuous at x E X and if f (x) exists r > 0 such that f(x') 0 for all x' E B(x,r) and the function g(x') := 11 f (xl) from B(x,r) to R is continuous at x. Analogous results hold if we replace R by C. Proof Let x i, x. (6.1) Let E be the set of left endpoints of Jn . (a) (R2 ,11 112), (the standard Euclidean norm). Using this bijection, we transfer the metric on X to Y. Ex. Consider the map 1+: Dn Sr+l given by f + (u) = (u1 , . The trick of inserting nk in the last inequality is an instance of what we call the 'curry leaves trick'. Thus any pair of points is path connected and hence X is path connected. Ex. Ex. Hence (iii) follows. Then there exists a path joining x to y by Ex. in an arbitrary metric space? a topology T on X. E X. Let x E X and r> 0 be arbitrary. Let E > 0 be given. Thus we conclude that an N, x m ,x n E Fn so that d(x m,,x n ) < diam (FN). Using this, we can write U as the union of a family, say, {Ji : j E I} of pairwise disjoint open intervals. Hence the claim is proved. A metric space (Y, d) is said to be a completion of (X, d) if there exists a map f : X -- Y such that (i) f is an isometry of X into Y and (ii) the image 1(X) is dense in Y. Ex. From the algebra of limits of sequences, it follows that f (x n ) + g(xn ) f (x) + g(x) = (f + g)(x). We claim that (—co, b) c J c (—oo,b1. We therefore conclude that the Cauchy sequence (1/n) is not convergent in (0, 1). We define f (x) := g(x) if x E A infIg(a)d(a,x):aeill if x A. dA(x) Note that since 0 < g(a) < 2 for a E A, we have 1 < f(x) < 2 for x E X \ A. Download Topology Of Metric Spaces By S Kumaresan book pdf free download link or read online here in PDF. Then, for any 6 = there is a subset A n with diameter less than 1/n and such that it is not a subset of U, for any i. COMPACTNESS 82 is called the union of the subfamily. Since a E U and U is open there exists an E > 0 such that [a, E) C U. So, you need something from linear algebra also! Note that Dn is convex and hence connected. Consider y(f) := f(0) as a map y: (C[0,1],11 11) -> IR. 2.7.4. Let (X, d) be an unbounded connected metric space. (c) Do (a) and (b) remain true if we replace 'open' by 'closed'? Is a nonempty finite subset of IR open? We shall give only an outline of the proof. 4.4 Arzela-Ascoli Theorem The theorem of the title gives an immensely useful criterion of compactness of subsets of C(X) where X is a compact metric space and C(X) is given the sup norm metric. CONTINUITY 66 Ex. Since J is unbounded there are three possibilities: (i) It is bounded above but not below, (ii) It is bounded below but not above and (iii) it is neither bounded above nor below. 58 CONTINUITY Ex. The satisfactory book, fiction, history, novel, Let U := {(x, y) ER Figure 1.17: (a, b) x (c, d) is open in R2 : x >0 86 y> 0 } . 1 } . Let n> 1. Proof. Assume that we are given a collection 7 of subsets of X satisfying the above properties (i)(iii). CONVERGENCE nk > k. (For, 1 < n i < n2 implies n2 > 2. (a) { (x, y) e 110 x2 + y2 = 1 1 . In Example 4.1.2, if we take A = N, then the union of the subfamily : n E NI is R. Note that it is also the union of the family {Jr : r E [OE 00}! A thorough knowledge of this will be needed to understand open covers etc. We do not give a proof of this result. Ex. We claim that x E Fn for all n. Fix n E N. Then the sequence (xk)k> n is a sequence in Fn and it converges to x. One may be tempted to believe that if we set 8 := mintSi : 1 < i < nl, it might work. In particular, if A is closed, then dA(x) = 0 if x E A. Hint: Continuous image of a connected set is connected. +f I fn fin I 0 /m 1 1 fn fm + f fn fin I + 1 1/n il/in = 1 1 + 12 + 13 + 14 . Show that any discrete metric space is complete. Let (xn, Yn) (x , y). We select E := X - 1/N > 0. That is, if p(x) = ax n + + • • • +aix + ao, then ai E R for 0 < j < n, a, 0 and n is odd, then there exists a E R such that p(a) = 0. What are all the open sets in a finite metric space? Topology of Metric Spaces gives a very streamlined development of a course in metric space topology emphasizing only the most useful concepts, concrete spaces and geometric ideas to encourage geometric thinking, to treat this as a preparatory ground for a general topology course, to use this course as a surrogate for real analysis and to help the students gain some perspective of modern … We show that (x n ) is a Cauchy sequence so that (x n ) converges to an x G X. When is a finite subset of a metric space connected? If 0 < x < 1, then either x < 1/N or 1/N < x < 1. We need to prove that f is uniformly continuous on X. Choose N1 such that d(x,,x n ) < 612 for m,n > N1 . Ex. We say that a subset J C R is an interval if for every x, y E J, and for every z such that x < z < y, it follows that z E J. The reader should draw pictures and devise a proof using the strategy. Topology Of Metric Spaces book. The details should be worked out by the reader. Then the sequence (xn ) has a convergent subsequence (x n,) such that x n, p in X. See Figure 6.3. Thus, Ih(u) — h(x)I < E for u E (X \ A) n B(x, 0. See Figure 2.5. Proof of (i). See (c) of Proposition 2.3.6. Let 74 CHAPTER 3. Hence a < c < b. What are the Cauchy sequences in a discrete metric space? (iii) The intersection of any finite number of members of 7 lies again in 7. (i)One example of a topology on any set Xis the topology T = P(X) = the power set of X(all subsets of Xare in T , all subsets declared to be open). Let f , g : X —p Ili be continuous functions on a metric space X. Given any NLS, can you think of a family of isometries? 4.4.11. Compare Ex. Lemma 3.2.46 (Gluing Lemma). Clearly, I . E R2 (X2/a2) (y2/b2))=1} for some a> b > 0. 1.2.73.) Now consider the 'path' [x, p] or the path [x, q] U [q, I)] connecting x and p, not passing through the origin. Show that 1(x) = g(x) for some x E [0,1]. TOPOLOGY: NOTES AND PROBLEMS Abstract. We now proceed recursively. Let (x n ) be a Cauchy sequence in R. Let 8 > 0 be arbitrary. 0 0 follows from that of g. At x = 0, we have 11/(x) - 1( 0 )11 = 1 .f (i)11 = X 11g(x) 11 1 x 1 _. I x 1 1 , 1 g(x I ) since 11 g(x)11 > 1. 3.3.13 and Ex. In particular, if f : X —> X is continuous, the fixed point set fx E X : f (x) =- xl is closed. Let (x n ) be Cauchy in X and assume that (x„,) is a convergent subsequence, converging to x E X. Basic Topology, Springer International Edition, Indian reprint 2004. Most often for theoretical purpose, characterization (2) of the theorem seem to be efficient. Proof. The strategy is to show that any two distinct points of S 2 C R 3 lie in a great circle, that is, the intersection of S 2 and a plane through the origin in R3 . = 0 on [-1, 0], we see that Ji = f°1 If I. But what happens in real life is that there is no guarantee the input will be exactly or precisely the x needed for the process f. The consumer also realizes this and he therefore sets a tolerance level of error. Ex. We denote it by 0A. (b) > (c): Let V be given as in (c). What contradiction does it lead to if ingd(x, f (x)) : x E X} > 0? We say that a metric d is complete if the metric space (X, d) is complete. Consider the product set X x Y with the product metric (Ex. Example 5.2.23 (Topologist's Sine Curve - I). CHAPTER 4. We write -y (t) = (-y (t) , -y2(t)). Page 6/24 2.4.11. Show that the set {x E X; dA (x) < el is open for any e > 0 (i) directly and (ii) by using the continuity of dA • Theorem 3.2.33 (Urysohn's Lemma). Prove the following: A (metric) space is compact if every family of closed sets with f.i.p. CONNECTED SPACES let B„ := [x, U [u, y]. The converse is not true. Show that no nonempty open subset of R is homeomorphic to an open subset of R2 . From (3.3), we have g (x) - 6 < g (a) < g (x) + E for xE AnB (x , (5) . A (metric) space X is said to be path connected if X for any pair of points x and y in X, there exists a path -y: [0,1] such that -y(0) = x and -y(1) = y. Ex. 3.2.50. Ex. Ex. . Can you think of a generalization? 3.2.11. From this point of view, in a topological space we still want to tell whether two … Any two open balls in 1Rn are homeomorphic. JXJJ < 4112 C211X11 1 , Hint: Ex. Thus (do is isometry of X into B(X). =0 -" (z..) Figure 6.3: Reflection of y Ex. Since c E U and U is open there exists an (relatively) open subset containing c lying in U. 3.2.14 on page 63 and Example 3.2.21 on page 64. Show that f is continuous. CHAPTER 4. Once you arrive at a complete proof, you may refer to the proof below. Let x E e l = (1, 0, , 0) E Rn and q = e2 = (0, 1, 0, be any nonzero vector. (n + h) — f (n)I = 2nh h2 > 2nh. Any polynomial with real coefficients and of odd degree has a real root. 5.1.36. Let X be endowed with discrete metric. For the given E. , there exists N G N such that L., v'kn,m-Fi Mk < E for n > m > N. Hence we have, for n > rn > N, n Il sn — sm 11 00 = sup E fk (X) xEX k=m-1-1 n 0 be given. Balls are intrinsically open because a and since a is the greatest lower bound of J, x is not a lower bound for J. Show that y is continuous. Prove that X is complete. Repeat the argument replacing J by J1 and so on. CI Ex. To gain practice, show all possible two way implications of the last theorem: (a) < > (b), (b) < > (C), (a) < > (c). We say that a function f: [a, b] R is linear if it is of the form f (t) = at + for some a, E R. Show that f is determined by its values at two (distinct) points in [a, b]. Show that the squaring map A 1—+ A 2 on M (n,R) is continuous. What is dg(x)? Once we have an idea of these terms, we will have the vocabulary to define a topology. (Compare this with the case of sequences. First of note that dA (x) + dB(x) 0 for any x E X. (Why?) Let A E M(n, R). 3.4.6. Ex. For any 6>0, with to +ö < 1, we must have (to + (5) >0. Ex. Ex. Hint: With the notation of the exercise, select a finite cover {Bi., : 1 < j < n}. There exists a continuous function f: X —> R such that 0 < f < 1 and f = 0 on A and f = 1 on B. Go through the solution of Ex. 6 Complete Metric Spaces 6.1 Examples of Complete Metric Spaces 6.2 Completion of a Metric Space 6.3 Baire Category Theorem 6.4 Banach's Contraction Principle 122 122 131 137 . Ex. If k > n then an < ak < bk. Check your pictures with those in Figure 1.6. See Theorem 4.4.8 and Ex. Since X is path-connected, there exists a path -y : [0, 1] —> X such that -y(0) = a and -y(1) = x. See Ex. (b) {(x,y) E R2 : y= x2 } . (Here IV is considered as a set of column vectors, i.e., matrices of type n x 1 so that the matrix product Ax makes sense.) Case (1): If J = 0, there is nothing to prove. CHAPTER 4. 3.1.6 and Theorem 3.1.9 to conclude that any polynomial function p(xi,... , x n ) in the variables xl, , x n will be continuous. Ex. 45 2.3. 3.1.11. We show that R" \ {0} is path connected if n > 2. 79 LIMIT OF A FUNCTION 3.5 Limit of a Function Let f :Dc X —> Y be a function. 3.2.27. Let A be a subset in a metric space (X, d). (b) A = Q and B is any nonempty subset of R. (c) A is the rectangular hyperbola xy = 1 and B is the union of axes xy = 0? Ex. Let X be a connected space and f: X function. 1.2.58 say about the metrics d1 and doo on C([0,1])? We need to impose extra conditions to ensure such a possibility. There exists a positive integer N = N(6) such that for all m > N and n > N, we have I x n — x nd < 6/2. 2) be topological spaces. Ex. Remark 3.3.11. Ex. Let (X, d) be a compact metric space. We show that (C[0,1], ll Ili) is not complete. We know how to achieve this! 2.4.13. (1,1 ...-f cll --- .---- . Let I be another nonempty set. Let D c X and assume that a E X is a cluster point of D. Assume that f: D \ fal —> Y be given. Since U is closed in [a, b] (with respect to the subspace topology) and c E [a, b], we see that c = urn x n E U. Show that a circle or a line or a parabola in homeomorphic to a hyperbola. Ex. Ex. 3.1.20. Thus, if we take 6 = 11n, then there exists an x' such that d(x, x') < In but d(f (x), f (x')) > E. Let us call this x' as x T, to emphasize its dependence on 5 = 1/n. Show that the map x Ax is continuous. C=J Example 3.3.6. -;—t 2- ,n, > 3,n E N}. It is not closed. When is it closed? 3. Hence by intermediate value theorem there exists a E [-0,0] such that p(a) = O. Ex. Thus, the error tolerance E of the output is given and then we find the error tolerance 6 of the input. (c) f : R R such that f(x) exists and is bounded. Hint: Consider {(x, R given by ri (x) = : x > 0 and xy = 1} c R 2 and the projection onto the x-axis. 3.4.7. Theorem 3.2.49 (Tietze). Then (c) Let d(x n ,4) -4 0 and d(yn ,yn' ) lim d(x n , yn ) = lim d(4, yn' ) Proof. This is the content of the next lemma. Hint: Let U be open in 100 and x E U. 1 1 2 d(x,„,x„) < d(x7,2 , a) + d(a, xn ) < — + — < — for m < n. m n m Thus (x n ) is Cauchy. Show that any convex set in an NLS is path-connected. (It is precisely the unit vector (in the norm 11 11 00 ) in the direction of x.) Let us see what happens when we consider x =11n and y = 11m,n m. We have 1.f(x) POI = 1. Now redo Ex. Theorem 3.4.10. Define x n = f (x n _1). Now if the series is convergent, then the sequence (sa) is Cauchy. Use this to show that T is continuous if there exists a constant C > 0 such that II Tx j < Cljx1 for all x E X. Ex. Let C C n K = 0. We shall see later that no two of a circle x2 + y2 = 1, a parabola y = x2 and a hyperbola x2 — y2 = 1 are homeomorphic. Theorem 4.1.16. In 1906 Maurice Fréchet introduced metric spaces in his work Sur quelques points du calcul fonctionnel. Conversely, let us assume that X is not connected. Then there exists 6> 0 such that f (x) > f(a)/2 for all x E B (a, ( 5 / 2). Since f + is a bijective continuous map of a compact space to a metric space, it is a homeomorphism. Ex. 1.2.41. If not, there is a sequence 0. Show that there exists y E X such that d(x,y) = r. Ex. 3.2.2. We say that a subset A C X is an E-net if dA(x) < E for any x E X. 4.2.4. characterization of a be closed and K be compact in Rn. Tags: S. Kumaresan, Topology of Metric Spaces (ebook) ISBN-13: 9781842652503 Additional ISBNs: 9781842652503, 1842652508 Author: S. Kumaresan Edition: Publisher: Published: 0000 Delivery: delivery within 48 hours Format: PDF/EPUB (High Quality, No missing contents and Printable) Compatible Devices: Can be read on any devices (Kindle, Android/IOS devices, Windows, MAC) 3.6.4. 3.3.8. Then by density of D in X, there exists a sequence (x n ) in D such that x n x. Let us consider the following sets: 1. Let D„ := fu E : I u Il < 11. Ex. 62 CHAPTER 3. We claim that S is nonempty, bounded above and that sup S is the limit of the given sequence. 3.5.4. For, all ak E [a, b]. 2.3.3. By Archimedean property of R, N is not bounded above in R. Hence 1/x is not an upper bound for N. It follows that there exists N E N such that N > 1/x or 1/N < z < 1. 1. Ex. We estimate If(x) POI = 1 and y> 1. 6.3.12. Any convergent sequence in (X, d) is Cauchy. Let X and Y be metric spaces. We shall only give a broad outline of the proof and leave the details for the reader to work out. If you are searched for the ebook by S. Kumaresan Topology of Metric Spaces in pdf form, in that case you come on to the loyal website. Then f is a constant function. ( 0 f (a)—E= f (;) I f (a) ) k ( = f (a) 2 f (a)-1-f = 3fa) Figure 3.4: Illustration for Exercise 3.2.4 Part (c) of Theorem 3.2.1 suggests the following definition. (See Ex. Can you think of applications'? A map f : X said to be open if the image f(U) is open for every open set U c X. Show that the set SL(n,R) of matrices in M(n,R) with determinant one is a closed subset of M (n, R). Then a E A implies d(x , a) d(x , u) + d(u, a) < + d(u,a). . 1 For positive integers n> m > N, we such that have n---1 E77 N d(Xk, X k+i) G E. d(Xk, X k+i d(x m , xn) < .m Hence (xn ) is Cauchy. Show that a (nonempty) subset of TR is an interval if it is convex. Because of this, the first third of the course presents a rapid overview of metric spaces Consider the open cover {B(x o ,n) : n G N}. Similarly the map ix : Y —> X x Y defined by ix (y) := (x, y) is continuous for every point x in X. Prove that an open interval in 1I8 is open. 3.6.5. 4.3.18—Ex. Let {Ui : j E / } be a family of open sets in a metric space (X, d). It is an interval since x is common to all the intervals that constitute Jx . This is absurd, since 1/2N E (0,1) but (1/2N) (/ (1/N, 1). If X is connected, then any locally constant function is constant on X. Ex. 2.3.9. Let the notation be as in the last exercise. Remark 3.1.10. We then conclude that the set A n (u1LElz.w1 B,i ) is uncountable. Hint: N and In + : n E NI. Is this open? '- is continuous on C. Proposition 3.1.16. Choose k1 such that if k > k1, then d(x, x n,) < E/2. A standard proof of Proposition 4.2.1 and Theorem 4.2.3 uses Ex. The map yol _. The reader is encouraged to supply the details. Show that In Exercises 1.2.27-1.2.31, you are required to find "the best possible" rp such that B(p,rp) c U, for p E U, in case you claim that U is open. Ex. Let U c JR be open and x E U. Since Uk is open and x E Uk, there exists rk > 0 such that min{ rk : B(x,rk) C Uk. Find the boundary of B minus a finite number of points. It follows that 77(x) > C 1 II x ll for 71( Ai ) > C1 all x E Rn. Let U (respectively, V) be an open set in X (respectively, Y). We now modify the argument and complete the proof. Find a continuous function f: C* ---. Thus Hfx —f = d(x , y). 44 CHAPTER 2. Let X be a compact space. 1.2.69. Let f : X ---4 X be such that d(f (x), f (y)) < d(x, y) for all x, y e X with x y. Definition 1.2.10. Hint: Start with an open cover II which does not admit a finite subcover. Let a E A. CONTINUITY 76 Note that while the continuity of f at a point is discussed, we are concerned only with the values (behaviour) of the function near the point under consideration. Define 6(p, q) := {4,0) + d(q,0), p q 0, P= for p, q E R2 . We claim that x n —> x. Digitalisiert von der TIB, Hannover, 2013. 2.7.8. Choose any x 1 E X. Inductively choose x n such that xn Unk i i1 B(x,,E/2). It is continuous if it is linear on each of the closed subintervals. As f is an upper bound for S and xN — E/2 E S (by (i)) we infer that xN — E/2 < L. Since f is the least upper bound for S and xN + E/2 is an upper bound for S (from (ii)) we see that f < x N + E/2. Show that ix is continuous. Hint: Mean value theorem. You on sphere have to recall Gram-Schmidt process!) Show that cp: (X , d) --* (Y, p) is an isometry. This El completes the proof. BOUNDARY OF A SET 53 (i) X is the union of members of B. All rights reserved. 4.2.11. 4.1.14. (a) == (b): We shall prove this by contradiction. 3.1.13. . Let X be an infinite set. Any two norms on Rn are equivalent, that is, the topologies induced by these norms are the same. 5.1.45. Which of the following sets are connected subsets of R2 ? Ex. Ex. 5.1 Connected Spaces Definition 5.1.1. (2) X is complete and totally bounded. Let p := (a, b) E R2 and A be the x-axis. Hence there exists x2 E F2 such that E n B(x i , 1) n B(x 2 , 1/2) is infinite. Thus the given family is an open cover of (0,1). We turn this idea into a proof. Let E > 0 be given. Since N is arbitrary, it follows that f(x) = 0 for x G (0,1]. We ask whether d is continuous. Hence we conclude that f (x) makes sense for any x E X. 1.2.42. Hence conclude that d is equivalent to the standard metric on N considered as a subset of R. Is (N, d) complete? Now x E Ua for some a and hence there is an r > 0 such that B(x,r) C U. Show that U := {(x, y) Figure 1.14.) 10 CHAPTER 9. 1.2.61. The joint map from M(n, R) x Rn to Rn given by (A, x) Ax is continuous. > (3): Let E be an infinite subset of X. This contradictions shows that the Cauchy D sequence (f a ) is not convergent in (C[-1, 1 ],11 110. Hence each of the terms (being a sum of nonnegative terms) goes to zero. From his experience, he may know that if the input x' is within 6 distance from x, the process will produce f (x') which may meet the acceptance level of the consumer.
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