So then n N implies s n s and s n s as required Another proof that Cauchy

So then n n implies s n s and s n s as required

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So then n > N implies | s n - s | < and s n -→ s as required. Another proof that Cauchy sequences converge. Suppose ( s n ) is a Cauchy sequence. Then ( s n ) is bounded and so lim inf s n and lim sup s n are real numbers. We’ll show that lim inf s n = lim sup s n . Let > 0. There exists an N so that n, m > N = ⇒ | s n - s m | < 2 . Fix m > N . Then n > N = s m - 2 < s n < s m + 2 and so s m - 2 inf { s n : n > N } ≤ lim inf s n lim sup s n sup { s n : n > N } ≤ s m + 2 . Thus 0 lim sup s n - lim inf s n . Since was an arbitrary positive number this implies that lim inf s n = lim sup s n . The previous theorem tells us that ( s n ) converges to lim inf s n = lim sup s n . 2 Subsequential limits Definition. Suppose ( s n ) is a sequence of real numbers. A subsequential limit is an element of R ∪ {-∞ , + ∞} , which is a limit of a subsequence of ( s n ). Theorem. Let ( s n ) be a sequence. 1. Suppose t R . Then t is a subsequential limit of ( s n ) if and only if the set { n N : | s n - t | < } is infinite for all > 0 . 2. + is a subsequential limit if and only if the sequence is unbounded above. 3. -∞ is a subsequential limit if and only if the sequence is unbounded below. 2
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Proof. 1. Suppose ( s n k ) is a subsequence converging to t . Then given an > 0, we can find an N so that k > N implies | s n k - t | < . So { n k : k > N } is an infinite subset of { n N : | s n - t | < } . Suppose conversely, that the set { n N : | s n - t | < } is infinite for all > 0 and suppose we have n 1 < n 2 < . . . < n k - 1 with | s n r - t | < 1 r for each r < k . Since the set n N : | s n - t | < 1 k is infinite, we can find n k > n k - 1 with | s n k - t | < 1 k . Continuing in this way we obtain a subsequence ( s n k ) with lim k -→∞ s n k = t . 2. Suppose ( s n ) is unbounded above. Suppose we have n 1 < n 2 < . . . < n k - 1 with s n r > r for each r < k . Since { s n : n N } is not bounded above we can find n k N so that s n k > max[ { s n : n n k - 1 } ∪ { k } ]. We necessarily have n k > n k - 1 , and s n k > k . Continuing this way we obtain a subsequence ( s n k ) with lim k -→∞ s n k = + . Suppose + is a sequential limit. Then there is a subsequence ( s n k ) diverging to . Given M > 0 we can find k so that s n k > M and so ( s n ) is unbounded above.
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  • Fall '08
  • hitrik
  • lim, Metric space, Limit of a sequence, subsequence, lim sup sn

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