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hmw5s - 18.100B Fall 2002 Homework 5 Due by Noon Tuesday...

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18.100B, Fall 2002, Homework 5 Due by Noon, Tuesday October 8. Rudin: (1) Chapter 3, Problem 1 Solution: The sequence is supposed to be in R n . We use the triangle inequality in the form | b | = | b a + a | ≤ | a b | + | a | which implies that | b | − | a | ≤ | a b | . Reversing the roles of a and b we also see that | a | − | b | ≤ | a b | and so || a | − | b || | a b | . If { s n } converges to s then given > 0 there exists N such that n > N implies | s n s | < . By the triangle inequality || s n |−| s || | s n s | so {| s n |} converges to | s | . (2) Chapter 3, Problem 20 Solution: Let { p n } be a Cauchy sequence in a metric space X. By as- sumption, some subsequence { p n ( k ) } converges to p X. Thus, given > 0 there exits K such that k > K implies that d ( p, p n ( k ) < / 2 for all k > K. By the Cauchy condition, given > 0 there exists M such that n, m > M implies d ( x n , x m ) < / 2 . Now, consider N = n ( l ) for some l K such that n ( l ) > M, which exists since n ( k ) → ∞ with k. For this choice, n > N = d ( p n , p ) d ( p n , p n ( l ) ) + d ( p n ( l ) , p ) < shows that { p n } converges to p.
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