Completion Notes

1 by the construction of n the second term is smaller

Info iconThis preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: maller than 4 whenever ≥ Nn,m . (n) ˜ ◦ By hypothesis, the sequence {X }n∈IN is Cauchy. So there is a natural number N such that ε ˜ the third term is smaller than 4 whenever n, m ≥ N . 1 ◦ Finally, if ≥ m , the last term is smaller than m . ˜ Choose any natural number N ≥ max N , 4 . I claim that xn − xm V < ε whenever n, m ≥ N . ε To see this, let n, m ≥ N . Now choose to be any natural number bigger than max Nn,m , n , m . ε Then the four terms in (2) are each smaller than 4 . (n) Proof that X = lim X : Let ε > 0. By definition n→∞ X − X(n) H = lim m→∞ xm − x(n) m V ( m) x = lim m→∞ m − x(n) m V By the triangle inequality ( m) x m − x(n) m ( m) V ≤x m (n) −x n (n) V +x n − x(n) m V (3) (n) ◦ Since the sequence xn = x n n∈IN is Cauchy, there is a natural number N such that the ε first term is smaller than 2 whenever n, m ≥ N . 1 ◦ By the construction of n , the second term is smaller than n whenever m ≥ n . Choose any natural number N ≥ max N , 2 . I claim that X − X(n) H < ε whenever n ≥ N . ε To see this, let n ≥ N . Now choose m to be any natural number bigger than max N , n . Then ε the two terms in (3) are each smaller than 2 . September 7, 2011...
View Full Document

This note was uploaded on 02/13/2014 for the course MATH 511 taught by Professor Joelfeldman during the Spring '13 term at UBC.

Ask a homework question - tutors are online