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

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

View Full Document Right Arrow Icon
± 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 ±> 0there ex ists 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
Background image of page 1

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

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/16/2011 for the course STAT 5446 taught by Professor Frade during the Fall '09 term at FSU.

Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online