1873_solutions

Words 2 n then whenever n n we have 156 3 3 2 n is

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Unformatted text preview: ill hold whenever n N. Then, whenever n N we have x ,x  x 5 ,x  5 . xn Now to prove that condition b implies condition a we assume that condition b holds. To prove that condition a holds, suppose that  0. Using the fact that /5 is a positive number, we choose an integer N such that the condition xn holds whenever n ,x  5 5 5 N we have x N. Thus for every n xn x 5 5 5 ,x  5 5 x ,x  . 9. Prove that n 2  3n  1 2n 2  n  4 1 2 as n Ý. We begin by observing that if n is any positive integer then 5n  5 . n 2  3n  1 5n 2 1 2 4n 4n 2 2n 2  n  4 4n 2  2n  8 Now suppose that  0. The inequality n 2  3n  1 1 2 2n 2  n  4 will hold when 5 4n which requires that n 5 4 Choose an integer N such that N  5/ 4 and observe that, whenever n N we have 5 5. n 2  3n  1 1 2 4n 4N 2n 2  n  4 10. Given that x n is a sequence in a metric space X and that x X, prove that x n x as n Ý if and only if d xn, x 0 The condition x n x as n Ý says that for every  0 there exists an integer N such that whenever n N we have d x n , x  . The condition d x n , x 0 as n Ý says that for every  0 there exists an integer N such that whenever n N we have |d x n , x 0 |  . These two conditions clearly say the same thing. as n Ý. 11. Suppose that x n is a sequence in a metric space X, that x X, and that y n is a sequence of real numbers converging to 0. Suppose that the inequality d x, x n yn holds for every n. Prove that x n x as n Ý. Suppose that  0. Using the fact that y n 0 as n Ý, choose an integer N such that the inequality y n  will hold whenever n N. We see at once that the inequality d x n , x  also holds whenever n N. 159 12. Given that x n and y n are sequences of real numbers, that x n y n for each n and that y n Ý as n Ý, prove that x n Ý as n Ý.. Suppose that w is a real number. Using the fact that y n Ý as n Ý we choose an integer N such that the inequality y n  w holds whenever n N. Then whenever n N we have x n y n  w. 13. Suppose that x n and y n are sequences in a metric space X and that x and y are points of X. a. Prove that for every n we have d x, y d x, x n  d x n , y n  d y n , y and deduce that d x, x n  d y n , y . d x, y d x n , y n Then prove that for every n we have d x, y d x, x n  d y n , y d xn, yn and deduce finally that d x, y | d x, x n  d y n , y . |d x n , y n Given any n we have d x, y d x, x n  d x n , y d x, x n  d x n , y n  d y n , y and the inequality d x, x n  d y n , y . d x, y d x n , y n follows at once. In the same way we can show that d x, y d x n , x  d y, y n . d xn, yn Therefore d x, y | d x, x n  d y n , y . |d x n , y n d x, y as n Ý. b. Prove that if x n x and y n y as n Ý then d x n , y n We assume that x n x and y n y as n Ý. Suppose that  0. Choose an integer N 1 such that the inequality d x n , x  2 holds whenever n N 1 . Choose an integer N 2 such that the inequality d y n , y  2 holds whenever n N 2 . We define N to be the larger of the integers N 1 and N 2 . Then whenever n N we...
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