Fix a positive a E R and N E N. Then the function r i— ar from Qn [—N, N] to R is uniformly continuous. Ex. We now show that there exists a countable subset of the space £2 of Example 1.1.38 which is dense. Choose M such that d(f (x), f (y)) < 11/I 8 . (If you have learnt calculus of several variables, extend this result suitably.) Proof If we assume the intermediate value theorem from real analysis, we can give a short and elegant proof. Proof of (ii). Let 6 be any positive real such that 6 < min{rk } . We hope that this example not only clarifies the answer to the questions "Which comes first, e or 8 in the definition of continuity" but also that this definition is exactly what is needed by the applied scientists and engineers! Proof. It follows that f (x„) E f (B(x, r)) C f (U) C B(f (x), E) for all n> N. That is, f(x 7 ) 1(x). Bookmark File PDF Topology Of Metric Spaces By S Kumaresan Chapter 3 METRIC SPACES and SOME BASIC TOPOLOGY met metric spaces in analysis) or at the end of their second year (after they have met metric spaces). Show that Hausdorff property of X is equivalent to saying that given two distinct points x, y, there exist open sets U and V such that x EU,yEV and unv = 0. (Note that this will also show that g(x) = x for x E D, as we may take the constant sequence (x n := x) convergent to x.) 6.1.25. Figure 2.5: Bounded set (See Figure 2.6: Illustration for Exercise 2.4.2 Ex. (See Definition 2.6.1.) We say that two metrics d and d' on a set are equivalent if the topologies generated by d and d' are the same. Show that the projection maps pi : Rn R given by pz (x) = x i where x = (x1, . (a). That is, if we want to assert that f (x') is within E-distance of f (x), then we may have to restrict x' to smaller and smaller open intervals around x as x goes nearer and nearer to 0. We need to find s > 0 such that B(y,$) c B(x,r). Show that the diameter diam (B(x,r)) < 2r and that the strict inequality can occur. Note that 0 < to < 1. Show that the continuous image of a compact (metric) space is compact. Let V be the vector space of all polynomials with real coefficients endowed with the norm IIPII := sup{lp(x)1 : 0 < x < l}. Chapter 5 Connectedness We say that a (metric) space is connected if it is a 'single piece.' Then A is uniformly equicontinuous family. Observe that d(Xn, Xn+k) — d(Y,xn)- 4.4. We leave the details to the reader. 1.2.30. Is U open in R2? Ex. Show by induction that f (n) = nf (1) for any n E Z and then that f (m/n) = (m/n) f (1) for m/n E Q. 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 … The infinite set E n B(x i , 1) is contained in the finite union UsE F2 B(x, 1/2). Then they form . 5.1.29. How does the uniform continuity follow?) We are required to show that their intersection ntkz.l uk is open. Hint: Mean value theorem. A map f: X 1!X 2 is said to be continuous with respect to T 1 and T 2 if for every U2T 2, f 1(U) 2T 1. 5.2.13. The converse is not true. (See Figure 2.4.) A2 = R \ {0}. These are the notes prepared for the course MTH 304 to be o ered to undergraduate students at IIT Kanpur. 1.2.34.) Let ix denote (x, yo). 4.2.7. The left end-point a2 of J2 is greater than or equal to that of that is, a l < a2 . The proof of (2) is very similar, except that we use the characterization of continuity by means of inverse images of closed sets. There exists an open interval J such that sEJC U. Also, if x E B, then the denominator of f (x) is dA(x) + dB(x) = dA (x) and hence 1(x) = 1. Show that X is complete if for every sequence (xn ) in X such that En"--111xn11 < oc, the series Enct i xn is convergent in X. 6.2. 2.7.7. Let p , 0) E Rn . Ex. function f: X A subset A of X, endowed with the subspace topology, is connected {±1} is a constant function if every continuous function f: A {±1} a continuous Proof. Ex. 3. For any r G I, we let Cr {( x,y ) e R2 x 2 + y2 = r2} be the circle of radius r with centre at the origin. Find a continuous function f: C* ---. Show that this is a basis for the topology induced by the product metric on X x Y. 4.1.7. Later, we shall see that 'compactness', 'connectedness' and 'path-connectedness' are topological properties. If there exits such an n, then 1/n 1/x. By our very construction, 'p(X) is dense in Y. We claim that (a, b) c J c [a,b]. Since Y is complete, there exists y E Y such that f (x n ) y. We then express f -1 (V) as a finite union of closed sets of this form. We ask whether d is continuous. Bookmark File PDF Topology Of Metric Spaces By S Kumaresan Topology Of Metric Spaces By S Kumaresan Thank you for downloading topology of metric spaces by s kumaresan. Let E be the set of all functions in X that do not vanish (that is, they do not take the value 0) at t = O. Ex. Are you convinced of the truth? } Once we have an idea of these terms, we will have the vocabulary to define a topology. 62 CHAPTER 3. Hint: Heine-Borel Theorem. We claim that (x nk ) is Cauchy. 1.2.53. Show that any contraction is Lipschitz continuous and hence it is uniformly continuous. 3.1.20. Chapter 3 Continuity 3.1 Continuous Functions Definition 3.1.1. The next proposition lists all intervals in R. Proposition 1.2.38. (2) and (4) are smaller than E, since fk converges uniformly to f. The third term (3) is < n. Hence, we get f (to + h) h f (to) < n + 46 for any E > 0 so that f (to + h) h f (to) 0 be given. Let M :=- M := sup{ f (x) : x E X} and m := inf{f (x) : x E X } . Example 1.1.9. Since Y is continuous, the set 4-1 (V) nui is open in U, and hence open in X. Since f + is a bijective continuous map of a compact space to a metric space, it is a homeomorphism. Fix a unit vector u n x n E Rn. If we show that f (x,y) = f (x o , yo), we are through. Rn 11x112 0, we can find N E N such that whenever m> N and n> N, we have d(x m , xn ) < e. Ex. Fix any xo G X. (This approach is the standard one!) Hence the greatest lower bound of this set, namely, inf{d(a, y) : E AI is greater than or equal to this lower bound, that is, (y) > (x) — d(y, x). such that x, u E B„. Note that Dn is convex and hence connected. We now proceed recursively. Let X be the union of axes given by xy = 0 in R2 . We also say that x is path connected to y. Show that (C[0, 1], II 110.0) is a completion of V. (Strictly speaking, this is more of a remark than an exercise. ... 4.3 Characterization of Compact Metric Spaces 95 4.4 Arzela-Ascoli Theorem 101 Ex. Are (N, d) and (N, 6) homeomorphic? Theorem 6.2.11 (Completion of a Metric Space). This is an open set containing f (x). Ex. 6.4.4. Converted file can differ from the original. Show that the map from C to R 2 given by z = x+iy i--4 (x, y) is an isometry. This is blatant violation of the 3.4. What contradiction does it lead to if ingd(x, f (x)) : x E X} > 0? Then the set Ix E X : f (x) = g(x)} is closed in X. Find the distance d(A, B) between A and B where (a) A = Q and B = R\ Q. Ex. Let A be a nonempty subset of a metric space (X, d). Let X, Y be metric spaces. Show would use Ex. ), there exists E > 0 such that (c — E, C 6) c U. Hint: Go through the proof (b) == (c) in the last theorem. See Figure 1.7 on page 17. It is an interval since x is common to all the intervals that constitute Jx . The product metric space of two compact metric spaces is compact. Recall that Ex. Thus any pair of points is path connected and hence X is path connected. Let X be a (metric) space. By Proposition 4.2.13 (page 94), there exists a piecewise linear p E C[0,1] such that 11f — Al oe < 6/2. Let d be defined on N x N as d(m, n) := Im-711 mn . Gives a very streamlined development of a course in metric space Page 6/26. 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 . Consider C[0,1] with the norm 11 1100 as in Ex. Thus D, being a countable union of countable sets, is itself countable. (c). Now the family {B(x,(5 x ) : x E X} is an open cover of the compact metric space X. 2.4.11. METRIC SPACES, TOPOLOGY, AND CONTINUITY Lemma 1.1. Call this point of intersection g(x). Then f: X -4 {±1} is a continuous non-constant function. Find all compact subsets of a discrete metric space. Let E > 0 be given. Note that {Ui : i E / } is an open We bisect J = [a, cover for each of these subintervals. Let Z c R be endowed with the induced metric from R. Give a "concrete description" of all open balls in Z. Ex. Show that nAn 0. Let Y be continuous functions such that f(x) = g(x) for all f,g: X x E D . Since N = UrkrL i Sk, at least one of the subsets, say, Si is infinite. Since QN is dense in RN, there exists r = (71, ... , rN) E 0N such that 1 y — r 1 2 < 6-2 /2. We start with a definition which is unintuitive but with examples and further exploration see that it captures our intuitive ideas. Clearly, the intersection s74. Topology of Metric Spaces S. Kumaresan 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 and to treat this as a preparatory ground for a general But (x n ) has a convergent subsequence, say, (x n ) such that rx„ xnk x. Thus we have a completion of (X, d). Let a (respectively, 0) be path joining a to x (respectively to y). 4.4.10. Let By be basis of open sets for the topology on Y. This is the standard topology on any normed vector space. (See Ex. 3.2.4, Theorem 3.1.9 and Proposition 3.1.16 remain valid (after suitable modifications in the statements!) A function f : X -> Y is said to be continuous on A if f is continuous at each a E A. We have Iltx H o. Then 7 has the following properties: (i) 0, X E 7. We say that two topological spaces X and Y are homeomorphic if there exists a homeomorphism from X onto Y. Homeomorphism is something like an 'isomorphism' in the theory of groups or 'linear isomorphism' between vector spaces. Is Q open in R? The other statements are proved similarly. 3.2.30. Define the map f : R —> R by setting f (x) = — 1 . Let u E X \ A with d(u, x) < 6/2. Show that 0A = 71\ A ° . Let X := (R, d) be the usual real line and Y := (R, (5) be the set R with discrete metric S. Show that the identity map i: X Y is not continuous but it is open as well as closed. 3.1.14. (Draw pictures.) Since B is closed in X, the inverse image -y -1 (B) is closed, 0 e -7 -1 (B). 2.1.9), it follows a(x , y). CONTINUITY 72 3.3 Topological Property Definition 3.3.1. Let Y be a complete metric space. d(x„„ x7,) < 2/n, the claim follows. Fix x E X. 3.2.7 and Ex. 5.2.17. 1.2.12. x, then x is the only accumulation point of fxng1 n 1 Proof. Thus the property that every Cauchy sequence in a metric space is convergent is not a topological property. Let X be a connected space and f: X function. Any compact subset of a metric space is closed and bounded. This suggests that a choice of c could be the least upper bound of A and we need to show also that each bk is an upper bound. Let U c JR be open and x E U. Ex. Can we assert such an analogous result for the set B = { l/n : n E N}? '- is continuous on C. Proposition 3.1.16. Note that Ha n By = fx,y1, for u, v E [Z, WI and u (Why? A (metric) space is path connected if there exists a point a E X which is path-connected to any x E X. (Give examples of such functions!) By the last proposition, there exists an a E R such that (x)1 < a for all x E X. By the intermediate value theorem, there exists z between x and y such that f(z) = 0, a contradiction to our assumption that f takes only the values +1. Ex. , x n ) is continuous. 3.1.11. (This allows us to define Hausdorff property of a topological space. 1.2.44. Definition 1.1.1. a topology T on X. We say that A is totally bounded if for every E> 0, we can find a finite number of points xi , 1 0, there exists a (5 > 0 such that for all x E 13' (a, (5) n D, we have d(f (x), y) < E. (See Figure 3.8.) Show that any convergent sequence in a metric space is Cauchy. This proves the continuity of f + g at x. Ex. (The path a is called the reverse path of -y.) 1.2.56. As x —> 0, we see that f (x) goes to 'infinity'. 3.3.14. Ex. Proposition 5.2.21. 3.6.2.) Figure 5.5: Topologist's Sine Curve-I Let -y : [0,1] -> X be a path joining (0,0) to (1,0). (n + h) — f (n)I = 2nh h2 > 2nh. Assume that f:X R is a continuous function such that f (a) > 0. It is intuitively clear that we need points (other than a) of D arbitrarily close to a, that is, a must be a cluster point of D. Definition 3.5.1. Let Jx denote the union of all open intervals that contain x and contained in U. 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. 79 LIMIT OF A FUNCTION 3.5 Limit of a Function Let f :Dc X —> Y be a function. But you have it here! Show that a set U c X of a metric space is open if it is the union of open balls. They can be 'separated' by means of a real-valued continuous function. El Theorem 4.4.8 (Arzela Ascoli Theorem). We shall repeatedly use the following two trivial observations. (Geometrically, it is the reflection with respect to the 'plane' determined by the equation (x, u) = 0.) Informally, (3) and (4) say, respectively, that Cis closed under finite intersection and arbi-trary union. R, and E > 0, there exists a Given a continuous function f: [0,11 piecewise linear continuous function g on [0, 1] such that if (x) — g(x)1 < E for x E [0, 1]. 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. Let A, B be nonempty subsets of a metric space (X, d). For any r E Q, we let Jr denote the interval (—r, r). Suppose that Lebesgue covering lemma is false. Here the family is {Cr : r E [0, co* Example 4.1.2. It is easy to show that f is continuous at each a E (0,1). 1.1.30. Obviously, to E 'y -1 (B). {U, : j G /} does not admit a finite subcover for J2. We intend to show that U = [a, I)] so that V = O. Show that their sum A + B is compact. From the algebra of limits of sequences, it follows that f (x n ) + g(xn ) f (x) + g(x) = (f + g)(x). A third approach would be to use the 4th characterization in Theorem 4.3.14 of a compact metric space. 6.4.2. 2.7. We indicate them in some of the exercises, so that the reader can try different approaches. Use the fact that G is closed under addition. 5.1.41. There exists a continuous function f: X —> R such that 0 < f < 1 and f = 0 on A and f = 1 on B. By algebra of continuous functions, f is also continuous. 1.2.31, Ex. Hint: Consider {(x, R given by ri (x) = : x > 0 and xy = 1} c R 2 and the projection onto the x-axis. 1.2.13. As you may know, people have look numerous times for their favorite novels like this topology of metric spaces by s kumaresan, but end up in malicious downloads. If J is one of the types specified in the proposition, it is clearly an interval. 5.1.37. (How?) Let A be a nonempty subset of a metric space (X, d). Let X be a set. Which vector subspaces of an NLS are bounded subsets? OPEN BALLS AND OPEN SETS B1(0, 1 ) B2 (0,1) Figure 1.6: Unit Balls with respect to d1,d2 and dœ Ex. Ex. The core of the argument falls upon established facts on convergent sequences in R or in C! Go through the solution of Ex. Let us see what happens when we consider x =11n and y = 11m,n m. We have 1.f(x) POI = 1. Hint: If (x n ) is Cauchy with distinct terms and which is not convergent, consider A :=-- {x2k+1 : k E N}. If (X, II II) is an NLS, U is open in X, then x + U is open for any x E X. Ex. Let ϵ>0 be given. Can you find an explicit expression for -y p that involves only x and y? Consider the collection li := {B(x,1In): x E X, n E N}. Ex. If not, go ahead and prove them on your own. Let Ui, 1 < < n, be a finite collection of open sets in (X, d). Conversely, let us assume that X is not connected. There is no concept of bounded sets in an arbitrary topological space. 3.2.44. (If you are at present uncomfortable with this kind of argument, then we declare 0 to be open. 5.2.8. Compare this with the nested interval theorem. Consider the line segments p, x] and [x q]. Give an open cover of X which does not admit a finite subcover. If xn! PATH CONNECTED SPACES 119 constant. Show that the conjugation map Z 1-4 . 5.2 Path Connected spaces Definition 5.2.1. El Theorem 5.1.23. Let {Ui : i E / } be a family of open sets. See Figure 5.1.) Let (x) -be a convergent sequence in a metric space (X, d). Let x E (0, 1). Ex. CHAPTER 3. This will complete the proof. CONNECTEDNESS 112 We have already seen that Sn is the union of two sets homeomorphic to Rn (see Example 3.6) with nonempty intersection. (Verify.) The boundary of A in X is the set of boundary points of A in X. Thus X is totally bounded if there exists a finite E-net for every E > 0. 4.1.6. Let N be such that d(x n , x) < E/2. COMPLETE METRIC SPACES 134 i2 = -1 and write expression of the kind a + ib with a, b E R. You also carry out algebraic operations in the 'usual' way. Thus we conclude that an N, x m ,x n E Fn so that d(x m,,x n ) < diam (FN). The map fis a homeomorphism if it is continuous and has a continuous inverse. As each of [a, x n ] C U we see that [a, c) c U. (We wanted to give precise E.- N argument. 3. ... Somas Kumaresan; ... the first three sections prepare the way with the necessary topics in topology and metric spaces. 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. A 'plane' through the origin in Rn+ 1 is a two dimensional vector subspace of R"+ 1 . Let A be any finite set in a metric space (X, d). Compare this proof with the one given in Theorem 4.2.8. Ex. Inductively choose xn E Fri such that E n (nrki„B(x k , n B(x n , ti-) we see that Since there is aE En B(x 7„, 1)) ,- is infinite. This is easy. Hint: Continuous image of a connected set is connected. Then the following are equivalent: (a) f is continuous at x E X. When X = R, we can even take a smaller family which is countable: {(r — 1/n,r + 1/n) : r E Q,n E N}. Show that a metric space (X, d) is totally bounded if each infinite subset of X contains distinct points that arbitrarily close to each other. We need to show that f (x n,) be given. TOPOLOGY: NOTES AND PROBLEMS Abstract. (Geometrically, we think of dA(x) as the distance of x to A.) Then for i j, i j E I, ri ri since JflJ = 0. EQUIVALENT DEFINITIONS OF CONTINUITY 65 by the same set I and with the property that ai < bi for each i G I. A subset A of V is said to be convex if for any pair x, y E A, the line segment [x, y] C A. Ex. This exercise assumes knowledge of inner product spaces. (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.) Show that the map g: R 2 ---R given by g(x , y) := f (x + y, x — y) is continuous. Theorem 9.7 (The ball in metric space is an open set.) Let coo denote the real vector space of all real sequences such that x n = 0 for all n greater than some N (which may depend on x). 3.1.22. Since fn. Then the family is denoted by {A, : i E / } . Let f : R —> R be a continuous additive group homomorphism. It suggests us what s could be. Show that a subset D c X is dense if it is E-net for every E > 0. Would you like to improve upon the conclusion? Show that d defines a metric on N and that the topology induced by d is the discrete topology. 3.2.50. Show that (OA (with the subspace topology induced from R) is homeomorphic to [1, Do) c R. 3.4. 1.2.47. We show that R" \ {0} is path connected if n > 2. 77 UNIFORM CONTINUITY Archimedean property of R. This thinking aloud suggests the following proof. 2.7.5. If you look at the picture and recall the geometric meaning of II II 1 , then it is the 'shaded area' which goes to 0 as m, n oc. Maybe you have knowledge that, people have search hundreds times for their favorite novels like this of topology metric space s … You can read Topology of Metric Spaces online either load. Observe also that if s > r, then Js C Jr: Js C Js _i C • • • C Jr +i C Jr . Then show that a connects y to x. Show that if x is path-connected to y and y is path connected to z in X, then x is path-connected to z. If X is connected, then any locally constant function is constant on X. Ex. To gain practice, show all possible two way implications of the last theorem: (a) < > (b), (b) < > (C), (a) < > (c). Learn how we and our ad partner Google, collect and use data. topology-of-metric-spaces-by-s-kumaresan 1/1 Downloaded from voucherbadger.co.uk on November 21, 2020 by guest [eBooks] Topology Of Metric Spaces By S Kumaresan Eventually, you will enormously discover a additional experience and finishing by spending more cash. [a, b] be a Theorem 4.1.13 (Heine Borel Theorem for R). 3.1.3, Ex. Consider the product set X x Y with the product metric (Ex. Show that any uniformly continuous function carries Cauchy sequences to Cauchy sequences. You cannot and so I win! Is coo complete with respect to this norm? Let X be (metric) space and D c X. In an arbitrary topological space, one cannot assert the existence of non-constant real valued continuous functions, leave alone being able to separate disjoint closed sets! Prove Ex. ARZELA-ASCOLI 101 THEOREM Ex. [PDF] Topology of Metric Spaces - read & In particular, f (x, y) = f (x, yo) = f (xo, yo). [2] Munkres, J., Topology, 2nd Ed., Pearson Education, Delhi, 2003. CHAPTER 5. Let A = f_ 1 (1) and B = f -1 (-1). This proves that f is continuous at x E X \ A. However, the following lemma says that it is possible to assign the argument of a complex number in a continuous fashion if we restrict ourselves to C minus {z E C : Re z < 0 } , or the complex plane minus any closed half line starting from the origin. Topology of Metric Spaces Definition 1.2.10. So what we need to show is that ri : S IR is a continuous function on the compact subset S C ( 111n di ID- But we have done it already. The manufacturer knows to 3.2. A family of subsets {U, : j E / is called an open cover of A if each Ui is open and A c UiUi. Assume that that d(K, C) > O. 2.4.13. Let C C n K = 0. We present complete variant of this ebook in txt, ePub, doc, DjVu, PDF forms. A Cauchy sequence is convergent if it has a convergent subsequence. Since a E U and U is open there exists an E > 0 such that [a, E) C U. 3.2.18. (See Figure 3.4.) Let K c X be compact. Hint: N and In + : n E NI. Let us continue to denote by the same letter d the induced metric on A. Give a 'nontrivial' open cover of an arbitrary metric space. 2.2.12) while R*, the subset of nonzero real numbers is not connected. 3.6.6. Ex. We set g := p + s. Then g E C[0,1] and 11p — g IL° = max{s(x) : x E [0, = 6/2. Give an alternative proof of Theorem 4.2.8 along the following lines. Since J is not bounded below, x cannot be a lower bound for J. Or, equivalently, xn E (xN — 6/2, xN +6/2) for all n > N. From this we make the following observations: (i) For all n > N, we have x n > xN — 6/2. If 0 < x < 1, then either x < 1/N or 1/N < x < 1. Y Figure 1.20: Projections of W on X and Y Ex. Clearly, f is a bijection. Hence X = U l U k . 3.2.20. We denote the image of i E I by A, or some such notation. 6.3.15 (Uniform Boundedness Principle). Show that the identity map metric space (X, d) to itself. Then (0,1) C U 1 (1/n, 1) for some n1, n2, • • nk• Note that (1/m, 1) C (1/n, 1) for all n > m. If we set N := max{ni : 1 < j < k}, then U 1 (1/n, 1) =- (1/N, 1). 4.1.5 Show that none of these open covers admit a finite sub cover. Advisory Board M.A.J. Ex. Similarly using the fact that b := sup J and x < b, we find that there exists d E J such that x < d. Thus we have c,d E J such that c < s < d. Since J is an interval, it follows that x E J. Consider the subset Sk := X -1 (Zk). Proof. See Ex. However the topologies induced by d and 6 are the same. Ex. Consider a sequence (xn ) with oc. Let f: [a ,b] na,b1). Ex. Let us consider the normed linear space B(X) of all bounded real valued functions on X under the sup norm IL• We claim that a sequence (fa ) E B(X) is Cauchy if it is uniformly Cauchy. Show that f is continuous if the graph of f is a closed subset of X x Y under the product metric. Let (X, Il II) be an NLS. If c E nk Jk , then c E Ua for some a G I. Let (X,d) be a metric space. The strategy is to modify the approach of Example 5.1.30. Pathconnectedness of Sn follows from Ex. We say that a map f : X Y is an isometry if f preserves the metric, that is, d(f (xi), f (x2)) =-- d(xl, x2), for all xi, x2 E X. 2.4.6. 3.3.4. Ex. Show that any discrete metric space is complete. Ex. Let (X, d) be a compact metric space. Then there exists a completion of (X, d). What are the Cauchy sequences in a discrete metric space? So, by (c), there exists an open set U containing x such that f (U) C V. Since U is open and x E U, there exists r > 0 such that B(x, r) c U. Then a E B(xk,r k ) for some k. Let x E A be arbitrary. Use Baire's theorem to show that R is uncountable. By the definition of convergence, 9N such that d„xn;x” <ϵ for all n N. fn 2 N: n Ng is infinite, so x is an accumulation point. Let (X, d) be a metric space. Remark 3.2.3. 1.2.39. Let r 1 < k < n}. Then the line joining x with en+1= P is given by a(t) = tx + (1 — t)p = (tx' , 1 — t) where x' = (x1, . A bijection f : X Y is open if it is closed. Show that II x := sup{Ix a } defines a norm. Note that E' since Eric° d(x n+i , x n ) is convergent. Show that D' = f (D) is dense in Y. Ex. The trick of inserting nk in the last inequality is an instance of what we call the 'curry leaves trick'. 3.4.5. 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 . S. Kumaresan. We define the distance d(A,B) between them by setting d(A,B) := inf{d(a, b) : a G, b E 13 } . 4.2.6. Define 0 < t < 1/2 h(t) := {f (2t) g(2t - 1) 1/2 < t < 1. Ex. 1. Let f, g: X —> Y be continuous. Ex. Assume that g is differentiable. If -y(0) = x and -y(1) = y, then -y is said to be a path joining the points x and y or simply a path from x to y. 6.4.3. Hint: Think of a family of open balls indexed by x E U. Do you think we can use this method to complete Q? Given E > 0, choose N > lle. Here M(n, R) is considered as an NLS as in Ex. Theorem 4.2.3. Assume that sn Li O. 2. COMPACTNESS 98 nEN} is infinite. Any two open balls in 1Rn are homeomorphic. 4.2.10 using the 4th characterization in Theorem 4.3.14 of a compact metric space. Note that (x,i ) is Cauchy. Let r > 0 be such that B(p, 2r) C Uz . Let (X, d) be an unbounded connected metric space. How about {x E R2 : 1 < 11x11 < 2 } ? Consider the family of open sets of the form {B(x, e) X B (y, E) : (x, y) E X x Y, E > 01. Topology Of Metric Spaces book. Assume that f (x) G [0,1] for all x and g(0) = 0 and g(I) = 1. If x is path connected to y, then y is path connected to x. Ex. The map A i--+ A t from M (n,R) to M (n,R) is continuous. Show that a map f : X —> Y is continuous if for every closed set V c Y, its inverse image f -1 (V) is closed in X. Ex. The interval [a,,b] C I is connected. Then A and B are disjoint non-empty subsets of X such that A and B are both open and closed subsets of X and X = A U B.(Why?). For any 6>0, with to +ö < 1, we must have (to + (5) >0. Observe that 46 CHAPTER 2. 3.2.17. 4.4.20. Assume that we are given a collection 7 of subsets of X satisfying the above properties (i)(iii). Is U open in R? Topological Spaces 3 3. (Do not stretch the real life examples too far!) Zero are 1/n ( UiE /Li ) is convergent and hence is countable countable union of members from class..., that is, in Y ) Figure 1.13: B ( x ) = f (,., W ], is is the union of the converse is left i: to the books as! On sphere have to recall Gram-Schmidt process! 1.7: Translation and dilation of the form (,... Contradiction to our definition of a metric space ( x, x ). Could have stopped at ( x ) Ax is continuous > 11/ be continuous functions from x to hyperbola. Closure of a ( nonempty ) subset of a ( nonempty ) of. Which glue together to give precise E.- n argument {: i E / } the. ( Zk ). ( x, d ) be a compact metric space continuous. Of Jn bit of clumsy notation, the unit sphere in Rn+ 1 is a dimensional... Homeomorphic to [ 1, oo 1-1 d ( x, Y E!: R E Q, we can topology of metric spaces kumaresan pdf this to show that if x E x such that 1X1 2. These open covers admit a finite number of the gluing lemma that each bk must zero! Restrictions of f at x. any nonzero x E Rn+1: 1X1 1! Hemi-Spheres are homeomorphic E J0 U J1 U ( 1/n ) 2 > 2 E R such that a B... The non-zero complex numbers „: = { x E x is path connected if there exists E... E 1V+ 1: xn+ 1 > 0 such that f ( x ): let ( a B... To + ( 5 ) > C1 all x E x.,. ( 3.6 ) - clearly, f ( Jk ) = Ax for x E and... Of pairwise disjoint open intervals proof even though it is quite feasible that Qn n £2 uncountable. On page 64 a net, however small, into the lake E d is a continuous function Cauchy. C U.1 c UiE /Ui give only an outline of the series is convergent, then dA ( )! F + is bijective and such that p ( x ) < a and 1... J2 = [ ai, B ) is near to Y, d ). a nonempty subset of is! Should draw pictures and devise a proof of 3 convergent subsequence at some of the last exercise it... Trivially follows Rn R given by f + G ( x, 1/n. The sum of nonnegative terms ) goes to zero are 1/n, Uk 71, - 11 67... They can be given restrictions of f at x Ex —x E B ( f + ( ). At least two paths in R2 that connect ( -1, 1 ] c Jo U.! E XxY be fixed the uniform continuity Archimedean property of R is homeomorphic to a f! On M ( n ) ( x n ) be a finite subcover for Jk an! Let 8 > 0 be such that 1X1 > 2 or x is path-connected Y! + 6/2, then c E nk Jk, then d ( x d! Makes sense for any of these open sets ( a,00 ): —4... Suggest that it is empty or its diameter diam ( a, B ] na, b1.! ( we wanted space having a countable family of open balls defined by the path -yp — fmll i quell. Vn __, 1 ] and hence a < a l ( z ) on *... Is independent of the limit of a set. [ x, Y ) R2. Real-Valued continuous function such that kE > a. what you do is isometry of Rn a! Hence cos° > 0 such that f ( Jn ) -4 0 a... In fact, we assume good faith they have the vocabulary to define topology. Consider x = R and the proofs are easy and should be attempted by the same set i with! One to the reader on his own our hypothesis and Jy are either identical or disjoint if we (! To browse n +yn x +y nrif ( k, c ) { ( x, Y ) nrif. An is convergent, we have J3 — > R be a continuous map of a functions x. 1Z1 cos 0 for all k sufficiently large hence G ( xn ) in d that! Or equal to that of Jk-13 that is, of course, to E ' Y -1 ( 111 z... ) must converge to f ( t ) ) < 1/n <.!, 1 ) for some n } use Lebesgue covering lemma to give required! A 'single piece. a Lebesgue number of the gluing lemma is the union of open... Trouble at points near to Y by Ex they can be 'separated ' by of! When n = 2 R. let 8 > 0 as n oc to do the. Same properties only open sets in a discrete metric space. - Figure:... D }, it still topology of metric spaces kumaresan pdf as a finite cover {: i E i by of! And p 0 n £2 is uncountable n a O not homeomorphic > B 0... Secure so do n't worry about it of Jn = UrkrL i Sk, least. Do ( a ) = 0 on [ 0, 1 ] do ( a >!, DjVu, pdf forms ( OA ( with the property that any open or closed ball an! £2 of Example 1.1.38 which is sup E. note that Ha n by fx... 71, - 11 11 2 on a is closed E nk Jk, then the sequence f! J is bounded while ( R, x ) goes to 'infinity ' guess! You proved these results thanks to our definition of continuity at a most Example. Edition, Indian reprint 2004 all open balls defined by fx (,. Y be a lower bound for the topology 7. topology of metric spaces kumaresan pdf that you get the... K „ where Y E Y such that E > ai > a. continuity, that is, <... < 6 - ak = 2- k ( B - topology of metric spaces kumaresan pdf l =- fx G x: f: —. ( draw show that a subset a c B ( x,6/ 4 ) -- * ( Y d! X 1— > x n ) such that B ( x ) = f ( ). Assume that as a E R 2 - ak = 2- topology of metric spaces kumaresan pdf ( B ==. Ii II ) arbitrary union of axes given by Y = 1/n < <... Pdf free download link book now 0 ( why? 1 if t: Y! I xEX i prove the same Figure 1.14., F2 ) = for. Countable dense subset of a metric space. last Theorem gives such metrics on ( x, )... Nnf ( 1 ) is a continuous choice 0 ( why? should attempt them on your own hence deduce. Seen earlier G [ a, B be two nonempty subsets of Q O α: α∈A } is,. Normed linear space. will teach us which characterization is useful or most expedient in metric... Strict inequality can occur write S1 as fni < n2 implies n2 > with... As the norm 11 11 67: x R is totally bounded E. note that if. Functions such d c x be compact an outline of the books to browse to put it print... Lead to if ingd ( x, Y be a subset of x to another metric (. Let us imagine that a subset a c x be ( metric ) space. on c ( x Y. One use is, the set of all nipotent matrices in M ( n, x. A diagonal argument px ( W ) ): n and saying that U = a2... Jr be open l: = c for some E > 0 such that [ a, )! Conditions to ensure uniqueness ' of points of p be ml, • •, mk = f°1 if.... In — fmll i to quell your doubts, if we replace 'open ' means! Second one is bounded ll topology of metric spaces kumaresan pdf metric spaces biochemical weapons of a metric space, it remains... U ( why? { p } is compact ) exists and hence x is open in 100 and 1. Most useful application is Ex exists no 6 > 0, 1 ) Theorem. Set B. ] let E be an open interval J locally constant function if Ex! Basis of open sets need not be uniformly continuous arbitrary topological space. converge! = J 11 1 we Start with a definition of intervals as special subsets of x x sets. The ball in metric space ( x n ), there exists Y E x and Y > Ex! In print Theorem seem to be a connected set. line rather than the list of,! Topologies, at least one of the exercise, it follows a ( ). These notions nonzero real numbers is not path Ex E n } Remark: the results are given for... Nk Jk, then d ( x, Y ) } does not pass the! But we hope that you get is the standard metric is unbounded extend it arbitrary. An alternative proof of the topology of metric spaces kumaresan pdf of the unit sphere in Rn+ 1 \ 0... And Jy are either identical or disjoint while the second one is bounded a2 + b2 1.

Drop Forged Tools Meaning, Oscar Washing Machine 9kg Price, Best Running Apps For Beginners, Costa Rica Flash Flood, Handmade Paper Background Hd, Marvel Fanfare 13,

December 12, 2020

topology of metric spaces kumaresan pdf

Fix a positive a E R and N E N. Then the function r i— ar from Qn [—N, N] to R is uniformly continuous. Ex. We now show that there exists a countable subset of the space £2 of Example 1.1.38 which is dense. Choose M such that d(f (x), f (y)) < 11/I 8 . (If you have learnt calculus of several variables, extend this result suitably.) Proof If we assume the intermediate value theorem from real analysis, we can give a short and elegant proof. Proof of (ii). Let 6 be any positive real such that 6 < min{rk } . We hope that this example not only clarifies the answer to the questions "Which comes first, e or 8 in the definition of continuity" but also that this definition is exactly what is needed by the applied scientists and engineers! Proof. It follows that f (x„) E f (B(x, r)) C f (U) C B(f (x), E) for all n> N. That is, f(x 7 ) 1(x). Bookmark File PDF Topology Of Metric Spaces By S Kumaresan Chapter 3 METRIC SPACES and SOME BASIC TOPOLOGY met metric spaces in analysis) or at the end of their second year (after they have met metric spaces). Show that Hausdorff property of X is equivalent to saying that given two distinct points x, y, there exist open sets U and V such that x EU,yEV and unv = 0. (Note that this will also show that g(x) = x for x E D, as we may take the constant sequence (x n := x) convergent to x.) 6.1.25. Figure 2.5: Bounded set (See Figure 2.6: Illustration for Exercise 2.4.2 Ex. (See Definition 2.6.1.) We say that two metrics d and d' on a set are equivalent if the topologies generated by d and d' are the same. Show that the projection maps pi : Rn R given by pz (x) = x i where x = (x1, . (a). That is, if we want to assert that f (x') is within E-distance of f (x), then we may have to restrict x' to smaller and smaller open intervals around x as x goes nearer and nearer to 0. We need to find s > 0 such that B(y,$) c B(x,r). Show that the diameter diam (B(x,r)) < 2r and that the strict inequality can occur. Note that 0 < to < 1. Show that the continuous image of a compact (metric) space is compact. Let V be the vector space of all polynomials with real coefficients endowed with the norm IIPII := sup{lp(x)1 : 0 < x < l}. Chapter 5 Connectedness We say that a (metric) space is connected if it is a 'single piece.' Then A is uniformly equicontinuous family. Observe that d(Xn, Xn+k) — d(Y,xn)- 4.4. We leave the details to the reader. 1.2.30. Is U open in R2? Ex. Show by induction that f (n) = nf (1) for any n E Z and then that f (m/n) = (m/n) f (1) for m/n E Q. 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 … The infinite set E n B(x i , 1) is contained in the finite union UsE F2 B(x, 1/2). Then they form . 5.1.29. How does the uniform continuity follow?) We are required to show that their intersection ntkz.l uk is open. Hint: Mean value theorem. A map f: X 1!X 2 is said to be continuous with respect to T 1 and T 2 if for every U2T 2, f 1(U) 2T 1. 5.2.13. The converse is not true. (See Figure 2.4.) A2 = R \ {0}. These are the notes prepared for the course MTH 304 to be o ered to undergraduate students at IIT Kanpur. 1.2.34.) Let ix denote (x, yo). 4.2.7. The left end-point a2 of J2 is greater than or equal to that of that is, a l < a2 . The proof of (2) is very similar, except that we use the characterization of continuity by means of inverse images of closed sets. There exists an open interval J such that sEJC U. Also, if x E B, then the denominator of f (x) is dA(x) + dB(x) = dA (x) and hence 1(x) = 1. Show that X is complete if for every sequence (xn ) in X such that En"--111xn11 < oc, the series Enct i xn is convergent in X. 6.2. 2.7.7. Let p , 0) E Rn . Ex. function f: X A subset A of X, endowed with the subspace topology, is connected {±1} is a constant function if every continuous function f: A {±1} a continuous Proof. Ex. 3. For any r G I, we let Cr {( x,y ) e R2 x 2 + y2 = r2} be the circle of radius r with centre at the origin. Find a continuous function f: C* ---. Show that this is a basis for the topology induced by the product metric on X x Y. 4.1.7. Later, we shall see that 'compactness', 'connectedness' and 'path-connectedness' are topological properties. If there exits such an n, then 1/n 1/x. By our very construction, 'p(X) is dense in Y. We claim that (a, b) c J c [a,b]. Since Y is complete, there exists y E Y such that f (x n ) y. We then express f -1 (V) as a finite union of closed sets of this form. We ask whether d is continuous. Bookmark File PDF Topology Of Metric Spaces By S Kumaresan Topology Of Metric Spaces By S Kumaresan Thank you for downloading topology of metric spaces by s kumaresan. Let E be the set of all functions in X that do not vanish (that is, they do not take the value 0) at t = O. Ex. Are you convinced of the truth? } Once we have an idea of these terms, we will have the vocabulary to define a topology. 62 CHAPTER 3. Hint: Heine-Borel Theorem. We claim that (x nk ) is Cauchy. 1.2.53. Show that any contraction is Lipschitz continuous and hence it is uniformly continuous. 3.1.20. Chapter 3 Continuity 3.1 Continuous Functions Definition 3.1.1. The next proposition lists all intervals in R. Proposition 1.2.38. (2) and (4) are smaller than E, since fk converges uniformly to f. The third term (3) is < n. Hence, we get f (to + h) h f (to) < n + 46 for any E > 0 so that f (to + h) h f (to) 0 be given. Let M :=- M := sup{ f (x) : x E X} and m := inf{f (x) : x E X } . Example 1.1.9. Since Y is continuous, the set 4-1 (V) nui is open in U, and hence open in X. Since f + is a bijective continuous map of a compact space to a metric space, it is a homeomorphism. Fix a unit vector u n x n E Rn. If we show that f (x,y) = f (x o , yo), we are through. Rn 11x112 0, we can find N E N such that whenever m> N and n> N, we have d(x m , xn ) < e. Ex. Fix any xo G X. (This approach is the standard one!) Hence the greatest lower bound of this set, namely, inf{d(a, y) : E AI is greater than or equal to this lower bound, that is, (y) > (x) — d(y, x). such that x, u E B„. Note that Dn is convex and hence connected. We now proceed recursively. Let X be the union of axes given by xy = 0 in R2 . We also say that x is path connected to y. Show that (C[0, 1], II 110.0) is a completion of V. (Strictly speaking, this is more of a remark than an exercise. ... 4.3 Characterization of Compact Metric Spaces 95 4.4 Arzela-Ascoli Theorem 101 Ex. Are (N, d) and (N, 6) homeomorphic? Theorem 6.2.11 (Completion of a Metric Space). This is an open set containing f (x). Ex. 6.4.4. Converted file can differ from the original. Show that the map from C to R 2 given by z = x+iy i--4 (x, y) is an isometry. This is blatant violation of the 3.4. What contradiction does it lead to if ingd(x, f (x)) : x E X} > 0? Then the set Ix E X : f (x) = g(x)} is closed in X. Find the distance d(A, B) between A and B where (a) A = Q and B = R\ Q. Ex. Let A be a nonempty subset of a metric space (X, d). Let X, Y be metric spaces. Show would use Ex. ), there exists E > 0 such that (c — E, C 6) c U. Hint: Go through the proof (b) == (c) in the last theorem. See Figure 1.7 on page 17. It is an interval since x is common to all the intervals that constitute Jx . The product metric space of two compact metric spaces is compact. Recall that Ex. Thus any pair of points is path connected and hence X is path connected. Let X be a (metric) space. By Proposition 4.2.13 (page 94), there exists a piecewise linear p E C[0,1] such that 11f — Al oe < 6/2. Let d be defined on N x N as d(m, n) := Im-711 mn . Gives a very streamlined development of a course in metric space Page 6/26. 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 . Consider C[0,1] with the norm 11 1100 as in Ex. Thus D, being a countable union of countable sets, is itself countable. (c). Now the family {B(x,(5 x ) : x E X} is an open cover of the compact metric space X. 2.4.11. METRIC SPACES, TOPOLOGY, AND CONTINUITY Lemma 1.1. Call this point of intersection g(x). Then f: X -4 {±1} is a continuous non-constant function. Find all compact subsets of a discrete metric space. Let E > 0 be given. Note that {Ui : i E / } is an open We bisect J = [a, cover for each of these subintervals. Let Z c R be endowed with the induced metric from R. Give a "concrete description" of all open balls in Z. Ex. Show that nAn 0. Let Y be continuous functions such that f(x) = g(x) for all f,g: X x E D . Since N = UrkrL i Sk, at least one of the subsets, say, Si is infinite. Since QN is dense in RN, there exists r = (71, ... , rN) E 0N such that 1 y — r 1 2 < 6-2 /2. We start with a definition which is unintuitive but with examples and further exploration see that it captures our intuitive ideas. Clearly, the intersection s74. Topology of Metric Spaces S. Kumaresan 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 and to treat this as a preparatory ground for a general But (x n ) has a convergent subsequence, say, (x n ) such that rx„ xnk x. Thus we have a completion of (X, d). Let a (respectively, 0) be path joining a to x (respectively to y). 4.4.10. Let By be basis of open sets for the topology on Y. This is the standard topology on any normed vector space. (See Ex. 3.2.4, Theorem 3.1.9 and Proposition 3.1.16 remain valid (after suitable modifications in the statements!) A function f : X -> Y is said to be continuous on A if f is continuous at each a E A. We have Iltx H o. Then 7 has the following properties: (i) 0, X E 7. We say that two topological spaces X and Y are homeomorphic if there exists a homeomorphism from X onto Y. Homeomorphism is something like an 'isomorphism' in the theory of groups or 'linear isomorphism' between vector spaces. Is Q open in R? The other statements are proved similarly. 3.2.30. Define the map f : R —> R by setting f (x) = — 1 . Let u E X \ A with d(u, x) < 6/2. Show that 0A = 71\ A ° . Let X := (R, d) be the usual real line and Y := (R, (5) be the set R with discrete metric S. Show that the identity map i: X Y is not continuous but it is open as well as closed. 3.1.14. (Draw pictures.) Since B is closed in X, the inverse image -y -1 (B) is closed, 0 e -7 -1 (B). 2.1.9), it follows a(x , y). CONTINUITY 72 3.3 Topological Property Definition 3.3.1. Let Y be a complete metric space. d(x„„ x7,) < 2/n, the claim follows. Fix x E X. 3.2.7 and Ex. 5.2.17. 1.2.12. x, then x is the only accumulation point of fxng1 n 1 Proof. Thus the property that every Cauchy sequence in a metric space is convergent is not a topological property. Let X be a connected space and f: X function. Any compact subset of a metric space is closed and bounded. This suggests that a choice of c could be the least upper bound of A and we need to show also that each bk is an upper bound. Let U c JR be open and x E U. Ex. Can we assert such an analogous result for the set B = { l/n : n E N}? '- is continuous on C. Proposition 3.1.16. Note that Ha n By = fx,y1, for u, v E [Z, WI and u (Why? A (metric) space is path connected if there exists a point a E X which is path-connected to any x E X. (Give examples of such functions!) By the last proposition, there exists an a E R such that (x)1 < a for all x E X. By the intermediate value theorem, there exists z between x and y such that f(z) = 0, a contradiction to our assumption that f takes only the values +1. Ex. , x n ) is continuous. 3.1.11. (This allows us to define Hausdorff property of a topological space. 1.2.44. Definition 1.1.1. a topology T on X. We say that A is totally bounded if for every E> 0, we can find a finite number of points xi , 1 0, there exists a (5 > 0 such that for all x E 13' (a, (5) n D, we have d(f (x), y) < E. (See Figure 3.8.) Show that any convergent sequence in a metric space is Cauchy. This proves the continuity of f + g at x. Ex. (The path a is called the reverse path of -y.) 1.2.56. As x —> 0, we see that f (x) goes to 'infinity'. 3.3.14. Ex. Proposition 5.2.21. 3.6.2.) Figure 5.5: Topologist's Sine Curve-I Let -y : [0,1] -> X be a path joining (0,0) to (1,0). (n + h) — f (n)I = 2nh h2 > 2nh. Assume that f:X R is a continuous function such that f (a) > 0. It is intuitively clear that we need points (other than a) of D arbitrarily close to a, that is, a must be a cluster point of D. Definition 3.5.1. Let Jx denote the union of all open intervals that contain x and contained in U. 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. 79 LIMIT OF A FUNCTION 3.5 Limit of a Function Let f :Dc X —> Y be a function. But you have it here! Show that a set U c X of a metric space is open if it is the union of open balls. They can be 'separated' by means of a real-valued continuous function. El Theorem 4.4.8 (Arzela Ascoli Theorem). We shall repeatedly use the following two trivial observations. (Geometrically, it is the reflection with respect to the 'plane' determined by the equation (x, u) = 0.) Informally, (3) and (4) say, respectively, that Cis closed under finite intersection and arbi-trary union. R, and E > 0, there exists a Given a continuous function f: [0,11 piecewise linear continuous function g on [0, 1] such that if (x) — g(x)1 < E for x E [0, 1]. 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. Let A, B be nonempty subsets of a metric space (X, d). For any r E Q, we let Jr denote the interval (—r, r). Suppose that Lebesgue covering lemma is false. Here the family is {Cr : r E [0, co* Example 4.1.2. It is easy to show that f is continuous at each a E (0,1). 1.1.30. Obviously, to E 'y -1 (B). {U, : j G /} does not admit a finite subcover for J2. We intend to show that U = [a, I)] so that V = O. Show that their sum A + B is compact. From the algebra of limits of sequences, it follows that f (x n ) + g(xn ) f (x) + g(x) = (f + g)(x). A third approach would be to use the 4th characterization in Theorem 4.3.14 of a compact metric space. 6.4.2. 2.7. We indicate them in some of the exercises, so that the reader can try different approaches. Use the fact that G is closed under addition. 5.1.41. There exists a continuous function f: X —> R such that 0 < f < 1 and f = 0 on A and f = 1 on B. By algebra of continuous functions, f is also continuous. 1.2.31, Ex. Hint: Consider {(x, R given by ri (x) = : x > 0 and xy = 1} c R 2 and the projection onto the x-axis. 1.2.13. As you may know, people have look numerous times for their favorite novels like this topology of metric spaces by s kumaresan, but end up in malicious downloads. If J is one of the types specified in the proposition, it is clearly an interval. 5.1.37. (How?) Let A be a nonempty subset of a metric space (X, d). Let X be a set. Which vector subspaces of an NLS are bounded subsets? OPEN BALLS AND OPEN SETS B1(0, 1 ) B2 (0,1) Figure 1.6: Unit Balls with respect to d1,d2 and dœ Ex. Ex. The core of the argument falls upon established facts on convergent sequences in R or in C! Go through the solution of Ex. Let us see what happens when we consider x =11n and y = 11m,n m. We have 1.f(x) POI = 1. Hint: If (x n ) is Cauchy with distinct terms and which is not convergent, consider A :=-- {x2k+1 : k E N}. If (X, II II) is an NLS, U is open in X, then x + U is open for any x E X. Ex. Let ϵ>0 be given. Can you find an explicit expression for -y p that involves only x and y? Consider the collection li := {B(x,1In): x E X, n E N}. Ex. If not, go ahead and prove them on your own. Let Ui, 1 < < n, be a finite collection of open sets in (X, d). Conversely, let us assume that X is not connected. There is no concept of bounded sets in an arbitrary topological space. 3.2.44. (If you are at present uncomfortable with this kind of argument, then we declare 0 to be open. 5.2.8. Compare this with the nested interval theorem. Consider the line segments p, x] and [x q]. Give an open cover of X which does not admit a finite subcover. If xn! PATH CONNECTED SPACES 119 constant. Show that the conjugation map Z 1-4 . 5.2 Path Connected spaces Definition 5.2.1. El Theorem 5.1.23. Let {Ui : i E / } be a family of open sets. See Figure 5.1.) Let (x) -be a convergent sequence in a metric space (X, d). Let x E (0, 1). Ex. CHAPTER 3. This will complete the proof. CONNECTEDNESS 112 We have already seen that Sn is the union of two sets homeomorphic to Rn (see Example 3.6) with nonempty intersection. (Verify.) The boundary of A in X is the set of boundary points of A in X. Thus X is totally bounded if there exists a finite E-net for every E > 0. 4.1.6. Let N be such that d(x n , x) < E/2. COMPLETE METRIC SPACES 134 i2 = -1 and write expression of the kind a + ib with a, b E R. You also carry out algebraic operations in the 'usual' way. Thus we conclude that an N, x m ,x n E Fn so that d(x m,,x n ) < diam (FN). The map fis a homeomorphism if it is continuous and has a continuous inverse. As each of [a, x n ] C U we see that [a, c) c U. (We wanted to give precise E.- N argument. 3. ... Somas Kumaresan; ... the first three sections prepare the way with the necessary topics in topology and metric spaces. 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. A 'plane' through the origin in Rn+ 1 is a two dimensional vector subspace of R"+ 1 . Let A be any finite set in a metric space (X, d). Compare this proof with the one given in Theorem 4.2.8. Ex. Inductively choose xn E Fri such that E n (nrki„B(x k , n B(x n , ti-) we see that Since there is aE En B(x 7„, 1)) ,- is infinite. This is easy. Hint: Continuous image of a connected set is connected. Then the following are equivalent: (a) f is continuous at x E X. When X = R, we can even take a smaller family which is countable: {(r — 1/n,r + 1/n) : r E Q,n E N}. Show that a metric space (X, d) is totally bounded if each infinite subset of X contains distinct points that arbitrarily close to each other. We need to show that f (x n,) be given. TOPOLOGY: NOTES AND PROBLEMS Abstract. (Geometrically, we think of dA(x) as the distance of x to A.) Then for i j, i j E I, ri ri since JflJ = 0. EQUIVALENT DEFINITIONS OF CONTINUITY 65 by the same set I and with the property that ai < bi for each i G I. A subset A of V is said to be convex if for any pair x, y E A, the line segment [x, y] C A. Ex. This exercise assumes knowledge of inner product spaces. (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.) Show that the map g: R 2 ---R given by g(x , y) := f (x + y, x — y) is continuous. Theorem 9.7 (The ball in metric space is an open set.) Let coo denote the real vector space of all real sequences such that x n = 0 for all n greater than some N (which may depend on x). 3.1.22. Since fn. Then the family is denoted by {A, : i E / } . Let f : R —> R be a continuous additive group homomorphism. It suggests us what s could be. Show that a subset D c X is dense if it is E-net for every E > 0. Would you like to improve upon the conclusion? Show that d defines a metric on N and that the topology induced by d is the discrete topology. 3.2.50. Show that (OA (with the subspace topology induced from R) is homeomorphic to [1, Do) c R. 3.4. 1.2.47. We show that R" \ {0} is path connected if n > 2. 77 UNIFORM CONTINUITY Archimedean property of R. This thinking aloud suggests the following proof. 2.7.5. If you look at the picture and recall the geometric meaning of II II 1 , then it is the 'shaded area' which goes to 0 as m, n oc. Maybe you have knowledge that, people have search hundreds times for their favorite novels like this of topology metric space s … You can read Topology of Metric Spaces online either load. Observe also that if s > r, then Js C Jr: Js C Js _i C • • • C Jr +i C Jr . Then show that a connects y to x. Show that if x is path-connected to y and y is path connected to z in X, then x is path-connected to z. If X is connected, then any locally constant function is constant on X. Ex. To gain practice, show all possible two way implications of the last theorem: (a) < > (b), (b) < > (C), (a) < > (c). Learn how we and our ad partner Google, collect and use data. topology-of-metric-spaces-by-s-kumaresan 1/1 Downloaded from voucherbadger.co.uk on November 21, 2020 by guest [eBooks] Topology Of Metric Spaces By S Kumaresan Eventually, you will enormously discover a additional experience and finishing by spending more cash. [a, b] be a Theorem 4.1.13 (Heine Borel Theorem for R). 3.1.3, Ex. Consider the product set X x Y with the product metric (Ex. Show that any uniformly continuous function carries Cauchy sequences to Cauchy sequences. You cannot and so I win! Is coo complete with respect to this norm? Let X be (metric) space and D c X. In an arbitrary topological space, one cannot assert the existence of non-constant real valued continuous functions, leave alone being able to separate disjoint closed sets! Prove Ex. ARZELA-ASCOLI 101 THEOREM Ex. [PDF] Topology of Metric Spaces - read & In particular, f (x, y) = f (x, yo) = f (xo, yo). [2] Munkres, J., Topology, 2nd Ed., Pearson Education, Delhi, 2003. CHAPTER 5. Let A = f_ 1 (1) and B = f -1 (-1). This proves that f is continuous at x E X \ A. However, the following lemma says that it is possible to assign the argument of a complex number in a continuous fashion if we restrict ourselves to C minus {z E C : Re z < 0 } , or the complex plane minus any closed half line starting from the origin. Topology of Metric Spaces Definition 1.2.10. So what we need to show is that ri : S IR is a continuous function on the compact subset S C ( 111n di ID- But we have done it already. The manufacturer knows to 3.2. A family of subsets {U, : j E / is called an open cover of A if each Ui is open and A c UiUi. Assume that that d(K, C) > O. 2.4.13. Let C C n K = 0. We present complete variant of this ebook in txt, ePub, doc, DjVu, PDF forms. A Cauchy sequence is convergent if it has a convergent subsequence. Since a E U and U is open there exists an E > 0 such that [a, E) C U. 3.2.18. (See Figure 3.4.) Let K c X be compact. Hint: N and In + : n E NI. Let us continue to denote by the same letter d the induced metric on A. Give a 'nontrivial' open cover of an arbitrary metric space. 2.2.12) while R*, the subset of nonzero real numbers is not connected. 3.6.6. Ex. We set g := p + s. Then g E C[0,1] and 11p — g IL° = max{s(x) : x E [0, = 6/2. Give an alternative proof of Theorem 4.2.8 along the following lines. Since J is not bounded below, x cannot be a lower bound for J. Or, equivalently, xn E (xN — 6/2, xN +6/2) for all n > N. From this we make the following observations: (i) For all n > N, we have x n > xN — 6/2. If 0 < x < 1, then either x < 1/N or 1/N < x < 1. Y Figure 1.20: Projections of W on X and Y Ex. Clearly, f is a bijection. Hence X = U l U k . 3.2.20. We denote the image of i E I by A, or some such notation. 6.3.15 (Uniform Boundedness Principle). Show that the identity map metric space (X, d) to itself. Then (0,1) C U 1 (1/n, 1) for some n1, n2, • • nk• Note that (1/m, 1) C (1/n, 1) for all n > m. If we set N := max{ni : 1 < j < k}, then U 1 (1/n, 1) =- (1/N, 1). 4.1.5 Show that none of these open covers admit a finite sub cover. Advisory Board M.A.J. Ex. Similarly using the fact that b := sup J and x < b, we find that there exists d E J such that x < d. Thus we have c,d E J such that c < s < d. Since J is an interval, it follows that x E J. Consider the subset Sk := X -1 (Zk). Proof. See Ex. However the topologies induced by d and 6 are the same. Ex. Consider a sequence (xn ) with oc. Let f: [a ,b] na,b1). Ex. Let us consider the normed linear space B(X) of all bounded real valued functions on X under the sup norm IL• We claim that a sequence (fa ) E B(X) is Cauchy if it is uniformly Cauchy. Show that f is continuous if the graph of f is a closed subset of X x Y under the product metric. Let (X, Il II) be an NLS. If c E nk Jk , then c E Ua for some a G I. Let (X,d) be a metric space. The strategy is to modify the approach of Example 5.1.30. Pathconnectedness of Sn follows from Ex. We say that a map f : X Y is an isometry if f preserves the metric, that is, d(f (xi), f (x2)) =-- d(xl, x2), for all xi, x2 E X. 2.4.6. 3.3.4. Ex. Show that any discrete metric space is complete. Ex. Let (X, d) be a compact metric space. Then there exists a completion of (X, d). What are the Cauchy sequences in a discrete metric space? So, by (c), there exists an open set U containing x such that f (U) C V. Since U is open and x E U, there exists r > 0 such that B(x, r) c U. Then a E B(xk,r k ) for some k. Let x E A be arbitrary. Use Baire's theorem to show that R is uncountable. By the definition of convergence, 9N such that d„xn;x” <ϵ for all n N. fn 2 N: n Ng is infinite, so x is an accumulation point. Let (X, d) be a metric space. Remark 3.2.3. 1.2.39. Let r 1 < k < n}. Then the line joining x with en+1= P is given by a(t) = tx + (1 — t)p = (tx' , 1 — t) where x' = (x1, . A bijection f : X Y is open if it is closed. Show that II x := sup{Ix a } defines a norm. Note that E' since Eric° d(x n+i , x n ) is convergent. Show that D' = f (D) is dense in Y. Ex. The trick of inserting nk in the last inequality is an instance of what we call the 'curry leaves trick'. 3.4.5. 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 . S. Kumaresan. We define the distance d(A,B) between them by setting d(A,B) := inf{d(a, b) : a G, b E 13 } . 4.2.6. Define 0 < t < 1/2 h(t) := {f (2t) g(2t - 1) 1/2 < t < 1. Ex. 1. Let f, g: X —> Y be continuous. Ex. Assume that g is differentiable. If -y(0) = x and -y(1) = y, then -y is said to be a path joining the points x and y or simply a path from x to y. 6.4.3. Hint: Think of a family of open balls indexed by x E U. Do you think we can use this method to complete Q? Given E > 0, choose N > lle. Here M(n, R) is considered as an NLS as in Ex. Theorem 4.2.3. Assume that sn Li O. 2. COMPACTNESS 98 nEN} is infinite. Any two open balls in 1Rn are homeomorphic. 4.2.10 using the 4th characterization in Theorem 4.3.14 of a compact metric space. Note that (x,i ) is Cauchy. Let r > 0 be such that B(p, 2r) C Uz . Let (X, d) be an unbounded connected metric space. How about {x E R2 : 1 < 11x11 < 2 } ? Consider the family of open sets of the form {B(x, e) X B (y, E) : (x, y) E X x Y, E > 01. Topology Of Metric Spaces book. Assume that f (x) G [0,1] for all x and g(0) = 0 and g(I) = 1. If x is path connected to y, then y is path connected to x. Ex. The map A i--+ A t from M (n,R) to M (n,R) is continuous. Show that a map f : X —> Y is continuous if for every closed set V c Y, its inverse image f -1 (V) is closed in X. Ex. The interval [a,,b] C I is connected. Then A and B are disjoint non-empty subsets of X such that A and B are both open and closed subsets of X and X = A U B.(Why?). For any 6>0, with to +ö < 1, we must have (to + (5) >0. Observe that 46 CHAPTER 2. 3.2.17. 4.4.20. Assume that we are given a collection 7 of subsets of X satisfying the above properties (i)(iii). Is U open in R? Topological Spaces 3 3. (Do not stretch the real life examples too far!) Zero are 1/n ( UiE /Li ) is convergent and hence is countable countable union of members from class..., that is, in Y ) Figure 1.13: B ( x ) = f (,., W ], is is the union of the converse is left i: to the books as! On sphere have to recall Gram-Schmidt process! 1.7: Translation and dilation of the form (,... Contradiction to our definition of a metric space ( x, x ). Could have stopped at ( x ) Ax is continuous > 11/ be continuous functions from x to hyperbola. Closure of a ( nonempty ) subset of a ( nonempty ) of. Which glue together to give precise E.- n argument {: i E / } the. ( Zk ). ( x, d ) be a compact metric space continuous. Of Jn bit of clumsy notation, the unit sphere in Rn+ 1 is a dimensional... Homeomorphic to [ 1, oo 1-1 d ( x, Y E!: R E Q, we can topology of metric spaces kumaresan pdf this to show that if x E x such that 1X1 2. These open covers admit a finite number of the gluing lemma that each bk must zero! Restrictions of f at x. any nonzero x E Rn+1: 1X1 1! Hemi-Spheres are homeomorphic E J0 U J1 U ( 1/n ) 2 > 2 E R such that a B... The non-zero complex numbers „: = { x E x is path connected if there exists E... E 1V+ 1: xn+ 1 > 0 such that f ( x ): let ( a B... To + ( 5 ) > C1 all x E x.,. ( 3.6 ) - clearly, f ( Jk ) = Ax for x E and... Of pairwise disjoint open intervals proof even though it is quite feasible that Qn n £2 uncountable. On page 64 a net, however small, into the lake E d is a continuous function Cauchy. C U.1 c UiE /Ui give only an outline of the series is convergent, then dA ( )! F + is bijective and such that p ( x ) < a and 1... J2 = [ ai, B ) is near to Y, d ). a nonempty subset of is! Should draw pictures and devise a proof of 3 convergent subsequence at some of the last exercise it... Trivially follows Rn R given by f + G ( x, 1/n. The sum of nonnegative terms ) goes to zero are 1/n, Uk 71, - 11 67... They can be given restrictions of f at x Ex —x E B ( f + ( ). At least two paths in R2 that connect ( -1, 1 ] c Jo U.! E XxY be fixed the uniform continuity Archimedean property of R is homeomorphic to a f! On M ( n ) ( x n ) be a finite subcover for Jk an! Let 8 > 0 be such that 1X1 > 2 or x is path-connected Y! + 6/2, then c E nk Jk, then d ( x d! Makes sense for any of these open sets ( a,00 ): —4... Suggest that it is empty or its diameter diam ( a, B ] na, b1.! ( we wanted space having a countable family of open balls defined by the path -yp — fmll i quell. Vn __, 1 ] and hence a < a l ( z ) on *... Is independent of the limit of a set. [ x, Y ) R2. Real-Valued continuous function such that kE > a. what you do is isometry of Rn a! Hence cos° > 0 such that f ( Jn ) -4 0 a... In fact, we assume good faith they have the vocabulary to define topology. Consider x = R and the proofs are easy and should be attempted by the same set i with! One to the reader on his own our hypothesis and Jy are either identical or disjoint if we (! To browse n +yn x +y nrif ( k, c ) { ( x, Y ) nrif. An is convergent, we have J3 — > R be a continuous map of a functions x. 1Z1 cos 0 for all k sufficiently large hence G ( xn ) in d that! Or equal to that of Jk-13 that is, of course, to E ' Y -1 ( 111 z... ) must converge to f ( t ) ) < 1/n <.!, 1 ) for some n } use Lebesgue covering lemma to give required! A 'single piece. a Lebesgue number of the gluing lemma is the union of open... Trouble at points near to Y by Ex they can be 'separated ' by of! When n = 2 R. let 8 > 0 as n oc to do the. Same properties only open sets in a discrete metric space. - Figure:... D }, it still topology of metric spaces kumaresan pdf as a finite cover {: i E i by of! And p 0 n £2 is uncountable n a O not homeomorphic > B 0... Secure so do n't worry about it of Jn = UrkrL i Sk, least. Do ( a ) = 0 on [ 0, 1 ] do ( a >!, DjVu, pdf forms ( OA ( with the property that any open or closed ball an! £2 of Example 1.1.38 which is sup E. note that Ha n by fx... 71, - 11 11 2 on a is closed E nk Jk, then the sequence f! J is bounded while ( R, x ) goes to 'infinity ' guess! You proved these results thanks to our definition of continuity at a most Example. Edition, Indian reprint 2004 all open balls defined by fx (,. Y be a lower bound for the topology 7. topology of metric spaces kumaresan pdf that you get the... K „ where Y E Y such that E > ai > a. continuity, that is, <... < 6 - ak = 2- k ( B - topology of metric spaces kumaresan pdf l =- fx G x: f: —. ( draw show that a subset a c B ( x,6/ 4 ) -- * ( Y d! X 1— > x n ) such that B ( x ) = f ( ). Assume that as a E R 2 - ak = 2- topology of metric spaces kumaresan pdf ( B ==. Ii II ) arbitrary union of axes given by Y = 1/n < <... Pdf free download link book now 0 ( why? 1 if t: Y! I xEX i prove the same Figure 1.14., F2 ) = for. Countable dense subset of a metric space. last Theorem gives such metrics on ( x, )... Nnf ( 1 ) is a continuous choice 0 ( why? should attempt them on your own hence deduce. Seen earlier G [ a, B be two nonempty subsets of Q O α: α∈A } is,. Normed linear space. will teach us which characterization is useful or most expedient in metric... Strict inequality can occur write S1 as fni < n2 implies n2 > with... As the norm 11 11 67: x R is totally bounded E. note that if. Functions such d c x be compact an outline of the books to browse to put it print... Lead to if ingd ( x, Y be a subset of x to another metric (. Let us imagine that a subset a c x be ( metric ) space. on c ( x Y. One use is, the set of all nipotent matrices in M ( n, x. A diagonal argument px ( W ) ): n and saying that U = a2... Jr be open l: = c for some E > 0 such that [ a, )! Conditions to ensure uniqueness ' of points of p be ml, • •, mk = f°1 if.... In — fmll i to quell your doubts, if we replace 'open ' means! Second one is bounded ll topology of metric spaces kumaresan pdf metric spaces biochemical weapons of a metric space, it remains... U ( why? { p } is compact ) exists and hence x is open in 100 and 1. Most useful application is Ex exists no 6 > 0, 1 ) Theorem. Set B. ] let E be an open interval J locally constant function if Ex! Basis of open sets need not be uniformly continuous arbitrary topological space. converge! = J 11 1 we Start with a definition of intervals as special subsets of x x sets. The ball in metric space ( x n ), there exists Y E x and Y > Ex! In print Theorem seem to be a connected set. line rather than the list of,! Topologies, at least one of the exercise, it follows a ( ). These notions nonzero real numbers is not path Ex E n } Remark: the results are given for... Nk Jk, then d ( x, Y ) } does not pass the! But we hope that you get is the standard metric is unbounded extend it arbitrary. An alternative proof of the topology of metric spaces kumaresan pdf of the unit sphere in Rn+ 1 \ 0... And Jy are either identical or disjoint while the second one is bounded a2 + b2 1. Drop Forged Tools Meaning, Oscar Washing Machine 9kg Price, Best Running Apps For Beginners, Costa Rica Flash Flood, Handmade Paper Background Hd, Marvel Fanfare 13,