45 2.3. If we remove a single point from a circle, it still remains as a single piece. E is nonempty. Using well-ordering principle of N, we can write S1 as fni < n2 < • • 1. (See Figure 1.16.) C=J Example 3.3.6. 2.3.11. Z, y Z } . The length = b1 - a l = (b - a)/2. Show that Ill is a completion of (Q, I I). 1.2.19. Choose k so large that Xnk E B(p,r) and link < r. Now if a E A„, is any element, then, d(a,p) < d(a,xnk ) + d(xn k ,P) < 2r. Show that any constant map from a metric space to another is continuous. Let X be a complete metric space and T a family of continuous real valued functions on X. It follows that the map If x E S, then 114 = 1 and hence x satisfies: x ri(x) > 0 for x E S. What we want to claim is that it is bounded below by a positive constant C1. Ex. We shall show that f -1 (C) is closed in X. The restrictions of f to either of these sets are continuous and hence constants. (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.) First we observe that fn (x) = 1 for 1/n < x < 1. The strategy is clear. 1.2.12. Topology of Metric Spaces 1 2. Let X be a compact space. Any compact subset of a metric space is closed and bounded. (Here 40 denotes the restriction of the metric on B(X) to Y.) Then a set U C X is open if it is the union of members of B. Proof By considering the function x >—> 1 + g(x), we may assume that g: X —> [1, 2]. Show that X \A is open. Let D c X and assume that a E X is a cluster point of D. Assume that f: D \ fal —> Y be given. Assume that as a subset of C it is bounded. Browse other questions tagged metric-spaces normed-spaces or ask your own question. Can you generalize this exercise? 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 topology course. a vector space over R under the obvious operations. We let u(x) := T be the Euclidean unit vector in the direction of x. If the given Cauchy sequence (x n ) is convergent, we can take it as the convergent subsequence. You can write a book review and share your experiences. a topology T on X. 1. But we have If (x) -f = 2+ (1/n) 2 > 2! Ex. Let A E M(n, R). (Why?) Then for i j, i j E I, ri ri since JflJ = 0. the inclusion x Theorem 3.1.9 (Space of Continuous Functions). We say that a collection A of (nonempty) subsets of X has finite intersection property (f.i.p., in short) if every finite family A 1 , ... , An of elements in A has a nonempty intersection. Hence, from a standard result from real analysis, the sequence (ak) converges to c := sup{ ak } . However, this definition of open in metric spaces is the same as that as if we regard our metric space as a topological space. If we draw the graph of the function, we may observe that if lxj is very large, small increments in x produce large differences in the values of f taken at these points. 2.4.3. 4.3.11. 3.1.21. 5.1.43. Ex. 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 Hint: Think of a family of open balls indexed by x E U. Note that E' since Eric° d(x n+i , x n ) is convergent. Learn how we and our ad partner Google, collect and use data. Ex. Then there exists a path joining x to y by Ex. Contents 1. 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) When they arrived at these numbers, they made a lot of simplifying assumptions. COMPLETION OF A METRIC SPACE 131 Thus f cannot be continuous. in an arbitrary metric space? Suppose you challenge me. What contradiction does it lead to if ingd(x, f (x)) : x E X} > 0? Show that h E B(X) and that the map yo: x f x is an isometry of X into (B(X),11 Hoo). Balls are intrinsically open because 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.) The next proposition lists all intervals in R. Proposition 1.2.38. Prove that X is complete. 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. Find the initial value problem of an ordinary differential equation whose solution is the fixed point of T. Bibliography The first two books are recommended for futher study of point-set topology. The file will be sent to your Kindle account. Theorem 5.1.31. Define x n = f (x n _1). The moral of the last example is that `boundedness' is metric specific. 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. A subset U C X of a metric space is said to be d-open if for each x E U, there exists r > 0 such that B(x, r) c U. 4.1.6. Compare this result with Theorem 6.1.3 - b - a, the length of the If J = [a, is an interval, we let f(J) interval. Ex. Namely, we will discuss metric spaces, open sets, and closed sets. We shall give only an outline of the proof. Can we find this point explicitly? Then yn := f (x n ) -- y = f (x) by the continuity of f at x. Suppose we have chosen nk. We claim that (x nk ) is Cauchy. Now the onus is on the manufacturer. 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. 3.3.13. Hint: Convergence in (R N , (5 ) is the same as coordinatewise convergence as in the finite dimensional R n . Topology of Metric Spaces S. Kumaresan Alpha Science International Ltd. Harrow, U.K. S. Kumaresan ... contact me on email and receive a pdf version in the near future. For, all ak E [a, b]. Hint: If (x n ) is Cauchy with distinct terms and which is not convergent, consider A :=-- {x2k+1 : k E N}. Also, recall the triangle inequality in 11/n for any n. ! (a) (R2 ,11 112), (the standard Euclidean norm). (c) f : R R such that f(x) exists and is bounded. See Theorem 3.2.33. 4.2.10 using the 4th characterization in Theorem 4.3.14 of a compact metric space. Let D := {(x,y) Ele : X 2 + y2 < 1 } . Hint: Ex. 2.7. Hint: Vector addition is continuous. (The series Encc-i xn is convergent in X if the sequence of partial sums sn := Enk=1 Xk is convergent in X.) (See Figure 5.2 on page 113.) .- - Figure 1.12: 1.12: Open set in a metric space Ex. 0 , Ex. (Verify.) We shall assume these results in the proof of the theorem below. Ex. 1.2.31, Ex. 2.4.10. 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. 1.2.6. 6.2.6. 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. Call this point of intersection g(x). 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. (Give examples of such functions!) A topological space having the property mentioned in (b) is known as a normal space: any two disjoint closed sets can be 'separated' by means of open sets. Two metrics d1, d2 on a set X are equivalent if the identity map of X from (X, d 1 ) to (X, d2) is a homeomorphism. (How?) (Note that Ui need not be contained in A and they are required to be open in X.) Since a' defined on various [— N , NJ coincide on their common domain, it follows that we have a function x 1--- a' for all x E R. Ex. Then I ask you to give me a point x in it and I promise that I shall give you an r > 0 as required. Ex. Let (X , d) and (Y, d) be metric spaces. 1.2.62. Consider the product set X x Y with the product metric (Ex. 3.1.15. Since Ui is open, there exists r > 0 such that B(x, C U.1 C UiE /Ui. 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. What is the completion of the space of irrationals with respect to the absolute value metric? Given A, B two compact subsets of a metric space such that A n B = 0. Ex. Hint: Consider the function x 1--4 d(x , f (x)). 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. Ex. 5.1.39. Let X be a connected space and f: X function. 5.1.27. A function f : X -> Y is said to be continuous on A if f is continuous at each a E A. 1.1.34. Read Book Topology Of Metric Spaces By S Kumaresan Topology Of Metric Spaces By S Kumaresan Thank you very much for reading topology of metric spaces by s kumaresan. Start by pressing the button below! Gives a very streamlined development of a course in metric space Page 6/26. Remark 3.4.11. Ex. 1.1.34.) What are the Cauchy sequences in a discrete metric space? 2.3.9. Ex. In fact, show that there exist a E A and b E B such that d(A,B) = d(a,b). It is time that the reader went ahead on his own. 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. Let (Y, d) be another metric space. Show that Ru is an isometry. to any metric space? Fix yo E Y. Lemma 1.2.24. We say that f is continuous at x if given an open set V 3 f (x), we can find an open set U 3 x such that f (U) C V. Ex. (See Ex. 5.2. Similarly, we show that h(u) < E h(X) . Use Ex. Let p E U. 3.2.35. A metric space is an ordered pair (,) where is a set and is a metric on , i.e., a function: × → such that for any ,, ∈, the following holds: If Tx 0 = xo , then we got what we wanted. Hint: Let x E E. Choose open sets V and W in Y such that f (x) E V and g(x) E W and V n w = O. Suppose x′ is another accumulation point. We can also get to this topology from a metric, where we define d(x 1;x 2) = ˆ 0 if x 1 = x 2 1 if x 1 6=x 2 For completeness sake, we give the proof. Let X, Y be (metric) spaces. Thus, Ac B(xk,2rk) C U2 (ik ). Ex. (b) Assume that A and B are closed disjoint subsets. Prove that there exists a sequence (xn ) in X and a positive ri such that d(x m , x n ) > i7 whenever m n. Ex. 3.3.9. Next consider h. We have J3 —> 0 as n oc. Show that the set of all nipotent matrices in M (n,R) is closed. In particular, f (a) = f o -y(0) =-- f o -y(1) = f (x). Proof. Hint: You learnt to prove the uniqueness of the limit of a convergent sequence. Can you generalize this? 3.2.8. 5.2.2.) Figure 5.1: Illustration for Example 5.1.30 Example 5.1.30. 3.2.4. CONTINUITY 78 An instructive application of the theorem is the extension of the meaning of a' for r E Q to a' for any x E R for any fixed a > O. The left end-point a2 of J2 is greater than or equal to that of that is, a l < a2 . What additional condition on (X or on Y) will be needed to ensure uniqueness'? The space Rn is path connected. We are required to show that their intersection ntkz.l uk is open. 3.1.20. COMPACTNESS 86 Ex. 1.1.32.) Topology Of Metric Spaces By S Kumaresan 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 topology course. 5.2.4. (c) {(x,y) E R2 : xy = 1 } . Hint: We needed this when we wish to prove that j j satisfies II f j 1 = 0 if f = 0 in Example 1.1.10. Show that f is continuous if the graph of f is a closed subset of X x Y under the product metric. , x n ) is continuous. Let ix denote (x, yo). Ex. Proof We shall only sketch the proof, as filling in the details will be a very instructive exercise to the reader. Also d(x,x, i ) < 2/n for all n. 4 Thus B(x,3In) includes B(x„, 771„) which includes infinitely many elements of E. Thus x is a cluster point of E. Hence (3) is proved. Show that the set 0(n) of all orthogonal matrices (that is, the set, of matrices satisfying AA' = I = At A) is a bounded subset of M(n,R). of-topology-metric-space-s-kumershan 1/1 Downloaded from happyhounds.pridesource.com on December 12, 2020 by guest [eBooks] Of Topology Metric Space S Kumershan Thank you for reading of topology metric space s kumershan. Throughout the book, we need to deal with intervals in IR. Let K {x E Rn : xlI co = 1} be the boundary of Q. Consider R with the standard metric d and the metric 6(x, y) := min{d(x, y), 1}. Assume that g is differentiable. EQUIVALENT DEFINITIONS OF CONTINUITY 59 Ex. S l := lz E C: izi = 11 such that f (z) = z for z E S l . Since Y is complete, there exists y E Y such that f (x n ) y. 58 CONTINUITY Ex. Observe that you can choose the 61 's independent of f E A thanks to the fact that A is equicontinuous at X G X. Chapter 5 Connectedness We say that a (metric) space is connected if it is a 'single piece.' Now consider the 'path' [x, p] or the path [x, q] U [q, I)] connecting x and p, not passing through the origin. Ex. 1.2.43. Show that (C[0, 1], II 110.0) is a completion of V. (Strictly speaking, this is more of a remark than an exercise. Let B be an open ball in RV. Since (x n ) is Cauchy we see But then d(x, x nk ) < that x n also converges to limk x nk • Thus X is complete. Observe that d(Xn, Xn+k) — d(Y,xn)- 4.4. Let us look at a most standard example: f: [1, oo) given by f (x) = 1/x. 3.6.2.) We shall apply these two observations below for 6 = 1 and 8 = e. Let S := Ix E R: there exists infinitely many n such that x n > x}. Find an explicit expression/formula for f. Draw its graph. A function f: (X, d) —> (Y, d) is said to be Lipschitz if there exists a constant L > 0 (called a Lipschitz constant of f) such that for all xi, x2 E X, we have d(f (x i ), f (x 2 )) < Ld(x l , x2). Assume that for all x E X, there exists Cx > 0 such that If(x)1 < for all f E T. Then there exists a non-empty open set U C X and a constant C such that I f (x)f to. 5.2.4. 3.2.6. S. Kumaresan (c) Given an open set V containing f (x) in Y, we can find an open set U containing x such that f (U) C V. Proof. Consider x = n and y = n + 71,-. Ex. 6.4.14. 2.4.14. 2.4.11. What we would like to analyze is the behaviour of 6 required to prove the continuity of f at x as x varies in (0, 1). Let E > 0 be given. (e) {(x,Y) Theorem 5.1.10 shows that connectedness is a topological property. Fix then Take . Work backwards. Let X be an infinite dimensional complete normed linear space. We continue to denote the Euclidean norm by ll ll. If we assume that Y is Hausdorff then the results are true. For this value of e, the convergence of 1/n —, x implies that there exists no such that for all n > no, we have lx — 1/n1 < e. If we choose n > max{2N, no}, we find that 1 n x— — > > , a contradiction. 6.1.24. Show that the latter admits a finite subcover of [0,1]. Proof of (iii). 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topology of metric spaces kumaresan pdf
45 2.3. If we remove a single point from a circle, it still remains as a single piece. E is nonempty. Using well-ordering principle of N, we can write S1 as fni < n2 < • • 1. (See Figure 1.16.) C=J Example 3.3.6. 2.3.11. Z, y Z } . The length = b1 - a l = (b - a)/2. Show that Ill is a completion of (Q, I I). 1.2.19. Choose k so large that Xnk E B(p,r) and link < r. Now if a E A„, is any element, then, d(a,p) < d(a,xnk ) + d(xn k ,P) < 2r. Show that any constant map from a metric space to another is continuous. Let X be a complete metric space and T a family of continuous real valued functions on X. It follows that the map If x E S, then 114 = 1 and hence x satisfies: x ri(x) > 0 for x E S. What we want to claim is that it is bounded below by a positive constant C1. Ex. We shall show that f -1 (C) is closed in X. The restrictions of f to either of these sets are continuous and hence constants. (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.) First we observe that fn (x) = 1 for 1/n < x < 1. The strategy is clear. 1.2.12. Topology of Metric Spaces 1 2. Let X be a compact space. Any compact subset of a metric space is closed and bounded. (Here 40 denotes the restriction of the metric on B(X) to Y.) Then a set U C X is open if it is the union of members of B. Proof By considering the function x >—> 1 + g(x), we may assume that g: X —> [1, 2]. Show that X \A is open. Let D c X and assume that a E X is a cluster point of D. Assume that f: D \ fal —> Y be given. Assume that as a subset of C it is bounded. Browse other questions tagged metric-spaces normed-spaces or ask your own question. Can you generalize this exercise? 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 topology course. a vector space over R under the obvious operations. We let u(x) := T be the Euclidean unit vector in the direction of x. If the given Cauchy sequence (x n ) is convergent, we can take it as the convergent subsequence. You can write a book review and share your experiences. a topology T on X. 1. But we have If (x) -f = 2+ (1/n) 2 > 2! Ex. Let A E M(n, R). (Why?) Then for i j, i j E I, ri ri since JflJ = 0. the inclusion x Theorem 3.1.9 (Space of Continuous Functions). We say that a collection A of (nonempty) subsets of X has finite intersection property (f.i.p., in short) if every finite family A 1 , ... , An of elements in A has a nonempty intersection. Hence, from a standard result from real analysis, the sequence (ak) converges to c := sup{ ak } . However, this definition of open in metric spaces is the same as that as if we regard our metric space as a topological space. If we draw the graph of the function, we may observe that if lxj is very large, small increments in x produce large differences in the values of f taken at these points. 2.4.3. 4.3.11. 3.1.21. 5.1.43. Ex. 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 Hint: Think of a family of open balls indexed by x E U. Note that E' since Eric° d(x n+i , x n ) is convergent. Learn how we and our ad partner Google, collect and use data. Ex. Then there exists a path joining x to y by Ex. Contents 1. 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) When they arrived at these numbers, they made a lot of simplifying assumptions. COMPLETION OF A METRIC SPACE 131 Thus f cannot be continuous. in an arbitrary metric space? Suppose you challenge me. What contradiction does it lead to if ingd(x, f (x)) : x E X} > 0? Show that h E B(X) and that the map yo: x f x is an isometry of X into (B(X),11 Hoo). Balls are intrinsically open because 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.) The next proposition lists all intervals in R. Proposition 1.2.38. Prove that X is complete. 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. Find the initial value problem of an ordinary differential equation whose solution is the fixed point of T. Bibliography The first two books are recommended for futher study of point-set topology. The file will be sent to your Kindle account. Theorem 5.1.31. Define x n = f (x n _1). The moral of the last example is that `boundedness' is metric specific. 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. A subset U C X of a metric space is said to be d-open if for each x E U, there exists r > 0 such that B(x, r) c U. 4.1.6. Compare this result with Theorem 6.1.3 - b - a, the length of the If J = [a, is an interval, we let f(J) interval. Ex. Namely, we will discuss metric spaces, open sets, and closed sets. We shall give only an outline of the proof. Can we find this point explicitly? Then yn := f (x n ) -- y = f (x) by the continuity of f at x. Suppose we have chosen nk. We claim that (x nk ) is Cauchy. Now the onus is on the manufacturer. 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. 3.3.13. Hint: Convergence in (R N , (5 ) is the same as coordinatewise convergence as in the finite dimensional R n . Topology of Metric Spaces S. Kumaresan Alpha Science International Ltd. Harrow, U.K. S. Kumaresan ... contact me on email and receive a pdf version in the near future. For, all ak E [a, b]. Hint: If (x n ) is Cauchy with distinct terms and which is not convergent, consider A :=-- {x2k+1 : k E N}. Also, recall the triangle inequality in 11/n for any n. ! (a) (R2 ,11 112), (the standard Euclidean norm). (c) f : R R such that f(x) exists and is bounded. See Theorem 3.2.33. 4.2.10 using the 4th characterization in Theorem 4.3.14 of a compact metric space. Let D := {(x,y) Ele : X 2 + y2 < 1 } . Hint: Ex. 2.7. Hint: Vector addition is continuous. (The series Encc-i xn is convergent in X if the sequence of partial sums sn := Enk=1 Xk is convergent in X.) (See Figure 5.2 on page 113.) .- - Figure 1.12: 1.12: Open set in a metric space Ex. 0 , Ex. (Verify.) We shall assume these results in the proof of the theorem below. Ex. 1.2.31, Ex. 2.4.10. 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. 1.2.6. 6.2.6. 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. Call this point of intersection g(x). 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. (Give examples of such functions!) A topological space having the property mentioned in (b) is known as a normal space: any two disjoint closed sets can be 'separated' by means of open sets. Two metrics d1, d2 on a set X are equivalent if the identity map of X from (X, d 1 ) to (X, d2) is a homeomorphism. (How?) (Note that Ui need not be contained in A and they are required to be open in X.) Since a' defined on various [— N , NJ coincide on their common domain, it follows that we have a function x 1--- a' for all x E R. Ex. Then I ask you to give me a point x in it and I promise that I shall give you an r > 0 as required. Ex. Let (X , d) and (Y, d) be metric spaces. 1.2.62. Consider the product set X x Y with the product metric (Ex. 3.1.15. Since Ui is open, there exists r > 0 such that B(x, C U.1 C UiE /Ui. 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. What is the completion of the space of irrationals with respect to the absolute value metric? Given A, B two compact subsets of a metric space such that A n B = 0. Ex. Hint: Consider the function x 1--4 d(x , f (x)). 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. Ex. 5.1.39. Let X be a connected space and f: X function. 5.1.27. A function f : X -> Y is said to be continuous on A if f is continuous at each a E A. 1.1.34. Read Book Topology Of Metric Spaces By S Kumaresan Topology Of Metric Spaces By S Kumaresan Thank you very much for reading topology of metric spaces by s kumaresan. Start by pressing the button below! Gives a very streamlined development of a course in metric space Page 6/26. Remark 3.4.11. Ex. 1.1.34.) What are the Cauchy sequences in a discrete metric space? 2.3.9. Ex. In fact, show that there exist a E A and b E B such that d(A,B) = d(a,b). It is time that the reader went ahead on his own. 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. Let (Y, d) be another metric space. Show that Ru is an isometry. to any metric space? Fix yo E Y. Lemma 1.2.24. We say that f is continuous at x if given an open set V 3 f (x), we can find an open set U 3 x such that f (U) C V. Ex. (See Ex. 5.2. Similarly, we show that h(u) < E h(X) . Use Ex. Let p E U. 3.2.35. A metric space is an ordered pair (,) where is a set and is a metric on , i.e., a function: × → such that for any ,, ∈, the following holds: If Tx 0 = xo , then we got what we wanted. Hint: Let x E E. Choose open sets V and W in Y such that f (x) E V and g(x) E W and V n w = O. Suppose x′ is another accumulation point. We can also get to this topology from a metric, where we define d(x 1;x 2) = ˆ 0 if x 1 = x 2 1 if x 1 6=x 2 For completeness sake, we give the proof. Let X, Y be (metric) spaces. Thus, Ac B(xk,2rk) C U2 (ik ). Ex. (b) Assume that A and B are closed disjoint subsets. Prove that there exists a sequence (xn ) in X and a positive ri such that d(x m , x n ) > i7 whenever m n. Ex. 3.3.9. Next consider h. We have J3 —> 0 as n oc. Show that the set of all nipotent matrices in M (n,R) is closed. In particular, f (a) = f o -y(0) =-- f o -y(1) = f (x). Proof. Hint: You learnt to prove the uniqueness of the limit of a convergent sequence. Can you generalize this? 3.2.8. 5.2.2.) Figure 5.1: Illustration for Example 5.1.30 Example 5.1.30. 3.2.4. CONTINUITY 78 An instructive application of the theorem is the extension of the meaning of a' for r E Q to a' for any x E R for any fixed a > O. The left end-point a2 of J2 is greater than or equal to that of that is, a l < a2 . What additional condition on (X or on Y) will be needed to ensure uniqueness'? The space Rn is path connected. We are required to show that their intersection ntkz.l uk is open. 3.1.20. COMPACTNESS 86 Ex. 1.1.32.) Topology Of Metric Spaces By S Kumaresan 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 topology course. 5.2.4. (c) {(x,y) E R2 : xy = 1 } . Hint: We needed this when we wish to prove that j j satisfies II f j 1 = 0 if f = 0 in Example 1.1.10. Show that f is continuous if the graph of f is a closed subset of X x Y under the product metric. , x n ) is continuous. Let ix denote (x, yo). Ex. Proof We shall only sketch the proof, as filling in the details will be a very instructive exercise to the reader. Also d(x,x, i ) < 2/n for all n. 4 Thus B(x,3In) includes B(x„, 771„) which includes infinitely many elements of E. Thus x is a cluster point of E. Hence (3) is proved. Show that the set 0(n) of all orthogonal matrices (that is, the set, of matrices satisfying AA' = I = At A) is a bounded subset of M(n,R). of-topology-metric-space-s-kumershan 1/1 Downloaded from happyhounds.pridesource.com on December 12, 2020 by guest [eBooks] Of Topology Metric Space S Kumershan Thank you for reading of topology metric space s kumershan. Throughout the book, we need to deal with intervals in IR. Let K {x E Rn : xlI co = 1} be the boundary of Q. Consider R with the standard metric d and the metric 6(x, y) := min{d(x, y), 1}. Assume that g is differentiable. EQUIVALENT DEFINITIONS OF CONTINUITY 59 Ex. S l := lz E C: izi = 11 such that f (z) = z for z E S l . Since Y is complete, there exists y E Y such that f (x n ) y. 58 CONTINUITY Ex. Observe that you can choose the 61 's independent of f E A thanks to the fact that A is equicontinuous at X G X. Chapter 5 Connectedness We say that a (metric) space is connected if it is a 'single piece.' Now consider the 'path' [x, p] or the path [x, q] U [q, I)] connecting x and p, not passing through the origin. Ex. 1.2.43. Show that (C[0, 1], II 110.0) is a completion of V. (Strictly speaking, this is more of a remark than an exercise. Let B be an open ball in RV. Since (x n ) is Cauchy we see But then d(x, x nk ) < that x n also converges to limk x nk • Thus X is complete. Observe that d(Xn, Xn+k) — d(Y,xn)- 4.4. Let us look at a most standard example: f: [1, oo) given by f (x) = 1/x. 3.6.2.) We shall apply these two observations below for 6 = 1 and 8 = e. Let S := Ix E R: there exists infinitely many n such that x n > x}. Find an explicit expression/formula for f. Draw its graph. A function f: (X, d) —> (Y, d) is said to be Lipschitz if there exists a constant L > 0 (called a Lipschitz constant of f) such that for all xi, x2 E X, we have d(f (x i ), f (x 2 )) < Ld(x l , x2). Assume that for all x E X, there exists Cx > 0 such that If(x)1 < for all f E T. Then there exists a non-empty open set U C X and a constant C such that I f (x)f to. 5.2.4. 3.2.6. S. Kumaresan (c) Given an open set V containing f (x) in Y, we can find an open set U containing x such that f (U) C V. Proof. Consider x = n and y = n + 71,-. Ex. 6.4.14. 2.4.14. 2.4.11. What we would like to analyze is the behaviour of 6 required to prove the continuity of f at x as x varies in (0, 1). Let E > 0 be given. (e) {(x,Y) Theorem 5.1.10 shows that connectedness is a topological property. Fix then Take . Work backwards. Let X be an infinite dimensional complete normed linear space. We continue to denote the Euclidean norm by ll ll. If we assume that Y is Hausdorff then the results are true. For this value of e, the convergence of 1/n —, x implies that there exists no such that for all n > no, we have lx — 1/n1 < e. If we choose n > max{2N, no}, we find that 1 n x— — > > , a contradiction. 6.1.24. Show that the latter admits a finite subcover of [0,1]. Proof of (iii). 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