1873_solutions

We need to observe that the function f belongs to s in

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Unformatted text preview: on: We want to show that the set Ý  Hn n1 cannot include any nonempty open set. Suppose that U is a nonemty open subset of X and, using the fact that H 1 has no interior points we choose a point y 1 U H 1 and using the fact that the set U H 1 is a neighborhood of y 1 we choose a positive number  1 1 such that B y1, 1 U H1. Using the fact that the open set B y 1 ,  1 cannot be included in the set H 2 (because H 2 has no interior points) we choose a point y2 B y1, 1 H2  B y1, 1 X H1 1 and, using the fact that the latter set is a neighborhood of y 2 we choose a positive number  2 such 2 that B y2, 2 B y1, 1 H2. Continuing this process we obtain a sequence y n in the space X and a sequence  n of positive numbers such that, for each n, 1 n n and B y n 1 ,  n 1 B yn, n H n 1 . We observe that n B yn, n U  Hj j1 In order to show that the set Ý U  Hn n1 is nonempty we shall show that the sequence y n is a Cauchy sequence and then we shall observe that the limit of the sequence y n must belong to the set Ý U  Hn. n1 To show that y n is a Cauchy sequence, suppose that  0. Choose an integer N  2/ . Whenever m and n are integers and m N and n N we know that both y m and y n belong to the ball B y N ,  N and so d ym, yN  d yN, yn  1  1  . d ym, yn N N Thus y n is a Cauchy sequence and, since the space X is complete, the sequence y n is convergent. We define y  n Ý yn. lim 181 For each positive integer N, since the sequence y n is eventually in the closed ball B y N ,  N we have N y B yN, N U  Hj. j1 Thus Ý y  Hn U n1 and the proof is complete. Exercises on Compact Metric Spaces 1. Prove the following converse of the Cantor intersection theorem: If X is a metric space and if every contracting sequence of nonempty closed subsets of X has a nonempty intersection then X is compact. Solution: To show that every sequence in X has a partial limit, suppose that x n is a sequence in X. For each positive integer n we define Using the fact that the sequence H n point x such that Hn  xm m n . is a contracting sequence of nonempty closed subsets of X, choose a Ý  Hn. x n1 It is easy to see that x is a partial limit of the sequence x n . Write out the details. 2. Given a metric space X, prove that the following conditions are equivalent: a. The space X is compact. b. For every family covers X. of open balls that covers the space X there exists a finite subfamily of that still It is obvious that condition a implies condition b because every open ball is an open set. Now suppose that condition b holds. Suppose that is a family of open sets that covers X. We define to be the family of open balls B that are included in at least one member of the family . Given any point x X, since covers X we know that x belongs to at least one member U of the family and, since such a set U is open, there exists an open ball B such that x B U. Therefore the family covers the space X and, using condition b, we choose finitely many members, that we shall call...
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