{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hmw5s

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

This preview shows pages 1–2. Sign up to view the full content.

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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}