1873_solutions

Explain why the sequence of sets h u n is a

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Unformatted text preview:  2    n  m 1d x1, x2 . 1 179 m To show that x n is a Cauchy sequence, suppose that  0 and choose an integer N such that N 1d x1, x2 . 1 Then whenever m N and n N, regardless of whether m  n or m  n or m  n, we have d xm, xn  . b. Suppose that f is a contraction on a complete metric space X. Prove that there exists a point x X such that f x  x. Choose a point t X. We define the sequence x n as in part a. Since X is complete and x n is a cauchy sequence in X we know that x n is convergent. We define x  n Ý xn. lim In order to show that f x  x c. Suppose that f is a contraction on a complete metric space X. Prove that there is one and only one point x X such that f x  x. 7. Prove that if X is a nonempty complete metric space and if every point of X is a limit point of X then X is uncountable. Solution: Suppose that X is a complete metric space and that every point of X is a limit point of X. To obtain a contradiction, suppose that x n is a sequence in X and that X is the range of the sequence x n . 1. Since x 1 is a limit point of X the set X x 1 must be nonempty. Choose a point y 1 X x 1 and, using the fact that X x 1 is a neighborhood of y 1 , choose a positive number  1 1 such that the closed ball x 1 , in other words, B y 1 ,  1 is included in X B y1, 1 X x1 Since y 1 is a limit point of X and B y 1 ,  1 is a neighborhood of y 1 , the ball B y 1 ,  1 must contain x 2 must be nonempty. Choose a point infinitely many points and so the set B y 1 ,  1 y2 B y1, 1 x2 1 and, using the fact that B y 1 ,  1 x 2 is a neighborhood of y 2 , choose a positive number  2 such 2 that B y2, 2 B y1, 1 x2 . 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 x n 1 . We observe that B yn, n X x 1 , x 2 , x 3 , , x n for each n. We shall obtain our contradiction by showing that the sequence y n is a Cauchy sequence and then we shall observe that the limit of the sequence y n cannot belong to the range of the sequence x n . 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 For each positive integer N, since the sequence y n is eventually in the closed ball B y N ,  N we have 180 x 1 , x 2 , x 3 , , x N y B yN, N and so y cannot belong to the range of the sequence x n and we have reached the desired contradiction. 8. This exercise is an improvement on Exercise 7 and is known as Baire’s theorem. 1. Suppose that H n is a sequence of closed subsets of a complete metric space X and that none of the sets H n have any interior points. prove that the set Ý  Hn n1 also fails to have any interior points. Soluti...
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