Complete spaces and Cauchy sequences

[alyafey22]

I know that a metric space is complete if every Cauchy sequence converges that will surely designate compact metric spaces as complete spaces . I need to see examples of metric spaces which are not complete.

Thanks in advance !

 

[Amer]

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 [tex] e [\tex] which is not rational number
Another sequence in the same metric is
$$P_n = \left( 1 + \frac{1}{n} \right)^n$$ which converge to e

 

[alyafey22]

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 [tex] e [\tex] which is not rational number
Another sequence in the same metric is
$$P_n = \left( 1 + \frac{1}{n} \right)^n$$ which converge to e

Wow very nice , thanks for the examples :)

 

[Fantini]

There is always the simple ones: consider the interval $(0,1)$. The sequence $1/n$ is a Cauchy sequence but does not converge in $(0,1)$. :)

 

[Fernando Revilla]

Another classical example is the vector space $V$ of the complex sequences $x=(x_n)$ with finitely many non zero terms. The map $V\times V\to \mathbb{C},$ $\langle x,y\rangle=\sum x_n\overline{y_n}$ is an inner product, and the corresponding metric space is not complete.