Hence l a k n l so a k n l Since k n N we can say for n N a n l Assume that lim

Hence l a k n l so a k n l since k n n we can say for

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Hence, l - < a k n < l + so | a k n - l | < . Since k n N , we can say for n N , | a n - l | < . ( ) Assume that lim a n = l . Then given > 0, there exists an N N such that n N implies | a n - l | < / 2. This means for every a n N , l - / 2 < a n N < l + / 2. Now, since l - / 2 is a lower bound for a n N , it follows by the definition of greatest lower bound that l - < l - / 2 inf { a n : n N } . Similarly, sup { a n : n N } ≤ l + / 2 < l + . Thus, l - < inf { a n : n N } ≤ a n N sup { a n : n N } < l + . Thus, | inf { a n : n N } - l | < and | sup { a n : n N } - l | < . Hence lim inf a n = lim sup a n = l . 2.5.1 Give an example or argue that one does not exist. (a) A sequence that has a subsequence that is bounded but contains no subsequence that converges. Impossible. Let a n be a sequence and let ( a n k ) be a bounded subsequence of a n . Then by the the Bolzano-Weierstrass Theorem, there is some subsequence, call it ( a n k i ), of ( a n k ) that converges. But ( a n k i ) is also a subsequence of ( a n ), and it converges. (b) A sequence that does not contain 0 or 1, as a term but contains subsequences converging to each of these values. If ( a n ) = (1 / 2 , 1 - 1 / 2 , 1 / 3 , 1 - 1 / 3 , 1 / 4 , 1 - 1 / 4 · · · ), then the subsequence of odd indexed terms (1 / 2 , 1 / 3 , 1 / 4 , · · · ) converges to 0 and the subsequence of even indexed terms (1 - 1 / 2 , 1 - 1 / 3 , 1 - 1 / 4 , · · · ) converges to 1. (c) A sequence that contains subsequences converging to every point in the infinite set A = { 1 , 1 / 2 , 1 / 3 , 1 / 4 , 1 / 5 , · · · ). Let ( a n ) = (1 , 1 / 2 , 1 , 1 / 2 , 1 / 3 , 1 , 1 / 2 , 1 / 3 , 1 / 4 , 1 , 1 / 2 , 1 / 3 , 1 / 4 , 1 / 5 , · · · ). Then this sequence con- tains a subsequence ( a n k ) k N such that ( a n k ) = ( p, p, p, p, · · · ) where p is any fixed element of A . Since ( a n k ) converges to p and p was an arbitrary element of A , we see that given any element of A , we can find a subsequence that converges to it. (d) A sequence that contains subsequences converging to every point in the infinite set { 1 , 1 / 2 , 1 / 3 , 1 / 4 , 1 / 5 , · · · } , and no subsequences converging to points outside of this set.
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  • Spring '11
  • DanRomik
  • Calculus, lim, Supremum, Limit of a sequence, Limit superior and limit inferior, subsequence

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