1873_solutions

That a set s of real numbers is nonempty and bounded

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Unformatted text preview: ve that the point z  1 t x  ty belongs to the set B x,  S x . Therefore x is a limit point of S. 19. For the purpose of this exercise we shall call a subset S of a metric space X compressed if for every number  0 there exist two different points x and y in S such that d x, y  . 131 a. Prove that whenever a subset S of a metric space has a limit point, it must be compressed. This is just Exercise 17 again. b. Give an example of a compressed subset of the metric space R that has no limit point. The set S  ZÞ n  1 n Z n is compressed but, since no interval of length 1 can contain more then three members of S , the set S has no limit point. c. Give an example of a compressed subset of the metric space 0, 1 that has no limit point in 0, 1 . The set 1 n Z  has the desired properties. n d. Prove that if a metric space X is compressed and x X then the set X x is also compressed. To obtain a contradiction, suppose that the set X x fails to be compressed and choose   0 such that whenever a and b are different points in the set X x we have d a, b . Since X is compressed and since the inequality d a, b   never holds when a and b are 2 different points in the set X x there must exist a point a X x such that d a, x   . 2 Choose such a point a. Again, using the fact that X is compressed, choose a point b X x such that d b, x  d a, x . We see at once that b a. Moreover d a, b d a, x  d x, b       2 2 contradicting the choice of . e. Prove that if a metric space X is compressed and F is a finite subset of X then the set X F is compressed. The set X F can be obtained by removing the points of F from X one at a time and applying part d each time. f. Prove that a metric space X is totally bounded if and only if every infinite subset of X is compressed. The desired result follows from an earlier stated property of total boundedness. 20. Given that S is a subset of a metric space X and that every infinite subset of S has a limit point, prove that every infinite subset of the set S must have a limit point. Suppose that E is an infinite subset of S. Using the fact that E is infinite we choose a sequence x n of points of E such that whenever m n we have x m x n . For each positive integer n we choose a point y n S such that d x n , y n  1 . n In the event that the set y n n Z  is finite there must be a point y S such that y n  y for infinitely many values of n; and we choose such a point y. In the event that the set y n n Z  is infinite, it must have a limit point and we choose such a limit point and call it y. In either event we have found a point y X such that whenever   0 there are infinitely many values of n for which d y, y n  . We complete the proof by showing that this point y must be a limit point of E. Suppose that   0. There are infinitely many integers n  2 for which d y, y n   . For each of these integers n we  2 have d xn, yn  d yn, y  1     d xn, y n 2 and since the sequence x n is one-one we conclude that the...
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