# Cauchy sequence

• May 22nd 2009, 01:02 PM
pkr
Cauchy sequence
HI, got an exam soon and need help understanding this question;

http://i97.photobucket.com/albums/l2...bbbbbbbb-1.jpg

So i know what a Cauchy sequence is i.e d(x_n,x_m)<=epsilon, but have no idea how to apply that to this problem.
• May 23rd 2009, 03:30 AM
Opalg
Quote:

Originally Posted by pkr
HI, got an exam soon and need help understanding this question;

http://i97.photobucket.com/albums/l2...bbbbbbbb-1.jpg

So i know what a Cauchy sequence is i.e d(x_n,x_m)<=epsilon, but have no idea how to apply that to this problem.

This question is wrongly worded, because it claims that the functions $f_n(x) = \begin{cases}0&\text{if }x\in[1/n,1] \\ nx&\text{if }x\in[0,1/n)\end{cases}$ are in the space $\{f:[0,1]\to\mathbb{R}\,|\,f\text{ continuous}\}$, whereas in fact these functions are obviously discontinuous at x = 1/n.

Suppose we correct the question by redefining the functions to be continuous, say $f_n(x) = \begin{cases}1&\text{if }x\in[1/n,1], \\ nx&\text{if }x\in[0,1/n).\end{cases}$ To show that they form a Cauchy sequence for the given metric, the easiest method is to show that they form a convergent sequence, with the limit function being the constant function g(x)=1.

The function $g(x) - f_n(x)$ is then zero except on the interval [0,1/n], where it is equal to 1– nx. Therefore $d(g,f_n) = \int_0^{1/n}(1-nx)\,dx = \frac1{2n}\to0$ as $n\to\infty$. Therefore the sequence is convergent and hence Cauchy.