Since vn nin is in xn nin lim vn xn v 0 n by the

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Unformatted text preview: nd {yn }n∈IN intersect, there is a Cauchy sequence {un }n∈IN that is in both {xn }n∈IN and {yn }n∈IN . Consequently {un }n∈IN is equivalent to both {xn }n∈IN and {yn }n∈IN so that lim n→∞ un − xn V = 0 and lim n→∞ un − yn V =0 On the other hand, as {xn }n∈IN and {yn }n∈IN are different, there is a Cauchy sequence {vn }n∈IN that is in one of them (say {xn }n∈IN ) and not in the other (say {yn }n∈IN ). Since {vn }n∈IN is in {xn }n∈IN , lim vn − xn V = 0 n→∞ By the triangle inequality vn − yn But this forces lim n→∞ V ≤ vn − xn vn − yn V = 0, which contradicts the assumption that {vn }n∈IN is not in V + xn − un V + un − yn V {yn }n∈IN . Lemma 4 If {xn }n∈IN , {yn }n∈IN ∈ V then lim xn , yn n→∞ Proof: exists. V We first observe that xn , yn V − xm , ym V ≤ xn − xm , yn ≤ xn − xm V + V yn V xm , yn − ym + xm V V yn − ym (1) V Since {xn }n∈IN and {yn }n∈IN are both Cauchy, both { xm V }m∈IN and { yn V }n∈IN are bounded and b...
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This note was uploaded on 02/13/2014 for the course MATH 511 taught by Professor Joelfeldman during the Spring '13 term at The University of British Columbia.

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