Take the set of all continuous functions at [0,1]
with
d(f(x),g(x)) = \int_{0}^{1} \mid f(x) - g(x) \mid dx
Let
f_n(x) = x^n , I want to show f_n is Cauchy
Given \epsilon > 0
if x= 1 ,
f_n (x) = 1 which is Cauchy
if x<1
x^n \rightarrow 0
There exist n_0 such that
\mid x^n \mid < \frac{\epsilon}{2} for all n > n_0
for m,n > n_0
\mid x^n - x^m \mid < \mid x^n\mid + \mid x^m \mid < \frac{\epsilon}{2} + \frac{\epsilon}{2}= \epsilon
d(f_n(x) , f_m(x) ) = \int_{0}^{1} \mid f_n(x) - f_m(x) \mid dx < \int_{0}^{1} \epsilon\;\; dx = \epsilon
Using the fact for f,g continuous functions with f<= g
\int_{a}^{b} f(x) \leq \int_{a}^{b} g(x)
so it is Cauchy but
f_n(x)
converges to
f(x) = \left\{ \begin{array}{11} 1 & : x=1 \\ 0 & : x\in [0,1) \end{array} \right.
f not continuous so it is not in the Metric
so f_n(x) is not converge in our metric
but it is Cauchy
Another Example
The metric on the Rational number set Q
with \mid x - y \mid
Let
P_n = \sum_{k=0}^{n} \frac{1}{k!}
this sequence is Cauchy but it is converge to e [\tex] which is not rational number <br />
Another sequence in the same metric is <br />
P_n = \left( 1 + \frac{1}{n} \right)^n which converge to e