upper bounds respectively for the set A n a n a n 1 a n 2 for each n N Since y

Upper bounds respectively for the set a n a n a n 1 a

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upper bounds (respectively) for the set A n = { a n , a n +1 , a n +2 , . . . } for each n N . Since y n = sup A n , it now follows that α - < y n < α + for all n N . Since ( y n ) y , the OLT now implies that α - y α + . Since is arbitrary, we have that y = α by Theorem 1.2.6. The argument showing that lim z n = α is very similar. Exercise 2.5.1 (a) Impossible. If the subsequence is bounded, then it must have a convergent subsequence by B-W Theorem. (b) Define the sequences x n = 1 / ( n + 1) and y n = 1 + x n . Then, ( x n ) 0 and ( y n ) 1 . Moreover, x n , y n / ∈ { 0 , 1 } for all n N . Thus, the sequence ( x 1 , y 1 , x 2 , y 2 , . . . ) satisfies the request. (c) The following sequence satisfies the request: 1 , 1 , 1 2 , 1 , 1 2 , 1 3 , 1 , 1 2 , 1 3 , 1 4 , 1 , 1 2 , 1 3 , 1 4 , 1 5 , 1 , 1 2 , 1 3 , 1 4 , 1 5 , 1 6 , 1 , . . . . (d) This request is impossible because there must also be a subsequence that converges to zero. To see why this is so, let ( a n ) be a sequence that has subsequences that converge to every number in the set A = { 1 /n : n N } . We will construct a subsequence ( a n k ) of ( a n ) that converges to zero in the following way: Since ( a n ) has a subsequence converging to 1, n 1 N such that | a n 1 - 1 | < 1 . Since ( a n ) has a subsequence converging to 1 / 2 , n 2 N such that | a n 2 - 1 / 2 | < 1 / 2 . Since ( a n ) has a subsequence converging to 1 / 3 , n 3 N such that | a n 3 - 1 / 3 | < 1 / 3 . . . . Since ( a n ) has a subsequence converging to 1 /k , n k N such that | a n k - 1 /k | < 1 /k . Continuing in this fashion gives us the subsequence ( a n k ) such that | a n k - 1 /k | < 1 /k for each k N . To show that ( a n k ) 0 , let > 0 be arbitrary. Choose N N such that N > 2 / . Then, for any k N , | a n k | = a n k - 1 k + 1 k a n k - 1 k + 1 k < 2 + 2 = . Therefore, ( a n k ) 0 .
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Solutions Math 3320: Homework #4 Page 4 of 4 Exercise 2.5.5 Assume ( a n ) is a bounded sequence with the property that every convergent subsequence of ( a n ) converges to the same limit a R . For a contradiction, assume that ( a n ) does not converge to a . Then, there exists an 0 > 0 with the following property: For every N N , there exists n N such that | a - a n | ≥ 0 .
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  • Spring '15
  • AmarjitNijjar
  • Math, lim, Supremum, Limit of a sequence, Limit superior and limit inferior, Xn, subsequence

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