131aW02-4a

131aW02-4a - creasing sequences). If 6 = S R and b = sup S,...

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Handout #4 Assignment 4 Due Wed. Feb. 13 1. p. 79: 3,5,6,7,8 2. p. 86:1,2,3,6,7,9,10 Our completeness axiom: If S is a non-empty subset of R , and S is bounded above (i.e., S b for some b ) , then S has a least upper bound b 0 = sup S. (You fomulate the corresponding result for non-empty sets that are bounded below). Here are theorems about sequences and their limts that you should be able to prove (including the relevant definitions): If x n is a convergent sequence, then it must be bounded. If x n L and x n 6 = 0 and L 6 = 0, then there is a constant c > 0 such that | x n | ≥ c for all n . If x n L and for all n , x n 0, then L 0. Note this was accidentally omitted from the original version of this list. The usual limit theorems (such as x n L and y n M implies x n + y n L + M ) If x n is an increasing sequence, and x n b, then x n b 0 = sup { x n } . (You should be able to state and prove the corresponding result for de-
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Unformatted text preview: creasing sequences). If 6 = S R and b = sup S, then there is a sequence x n S such that x n b . (You should be able to state and prove the corresponding result for the inmum). If x n is a convergent sequence, then it must be Cauchy. If x n is a Cauchy sequence, then it must be bounded. If x n L ,. then for any subsequence x n k , x n k L. Every sequence has a monotonic subsequence. If x n is a Cauchy sequence and a subsequence x n k L, then x n L. If x n is a Cauchy sequence, then it must converge. If x n is a bounded sequence, then it has a convergent subsequence. [This is often called the Balzano-Weierstrass Theorem ] 1...
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