1873_solutions

# Suppose that x n is a bounded sequence of real

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Unformatted text preview: any set then the metric space Ý S is complete. We saw the definition of this space earlier. Solution: Suppose that S is a set and that f n is a Cauchy sequence in the metric space every member x of the set S we see from the fact that f gÝ |f m x f n x | Ý S . For for all m and n that the sequence f n x is a Cauchy sequence in R. Since R is complete the sequence f n x converges for every x S. For every x S we define f x  n Ý fn x lim and, in this way, we have defined a function f : S Suppose that  0. R. Now we need to show that f n f Ý 0 as n Ý. 1. Using the fact that f n is a Cauchy sequence in Ý S , choose an integer N such that the inequality fn fm Ý  holds whenever m and n are integers and m N and n N. Given any n whenever m N and since S we have |f n x f m x | lim |f n x f x |  m Ý|f n x f m x | we have |f n x fx | . Therefore if n N we have f n f  sup |f n x f x | and we have shown that f n f Ý 0 as n Ý. 178 x S N and any member x of the set Finally we need to observe that the function f belongs to Ý S , in other words, that f is bounded. Using the fact that f n f Ý 0 as n Ý, we choose an integer n such that f n f Ý  1. We see that f f Ý fn  fn f Ý fn Ý  fn Ý  1  fn Ý  Ý. 3. Prove that if x n is a Cauchy sequence in a metric space X and if x X then the sequence of numbers d x n , x converges in the metric space R. Given any positive integers m and n we have |d x m , x d x n , x | d x m , x n and it is therefore clear that the sequence d x n , x is a Cauchy sequence in R. 4. Give an example of a complete metric space X and a bounded sequence x n in X such that the sequence x n has no partial limit. Hint: Look for an example in a space of the form Ý S. We take S to be the set Z of positive integers. For each n the equation  fn Ý R by 1 if x  n fn x  From the fact that f m ÝS. S we define the function f n : S 0 if x  1 whenever m S n n we see that f n has no partial limit in the space 5. Give an example of a complete metric space X and a contracting sequence H n of nonempty closed bounded sets such that Ý  Hn  . n1 We return to the functions f n that were defined in the solution to Exercise 4. For each n we define Hn  fm m n . 6. If X is a metric space then a function f : X X is said to be a contraction on X if there exists a number   1 such that whenever x and t belong to X we have d f t ,f x d t, x . a. Suppose that f is a contraction on a metric space X and that t X. Suppose that we have x 1  t and that for every positive integer n we have x n 1  f x n . Prove that the sequence x n is a Cauchy sequence. We observe first that d x 1 , x 2 d x2, x3  d f x1 , f x2 d x3, x4  d f x2, f x3 d x 2 , x 3 2d x1, x2 and, in general, n 1d x1, x2 . d x n , x n 1 Now suppose that m and n are positive integers and that n  m. We see that d x m , x m 1  d x m 1 , x m 2    d x n 1 , x n d xm, xn   m 1 d x 1 , x 2   m d x 1 , x 2   m 1 d x 1 , x 2     n 2 d x 1 , x 2  m 1d x1, x2 1  ...
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## This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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