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 &lt; \frac{\epsilon}{2} for all n &gt; n_0
for m,n &gt; n_0
\mid x^n - x^m \mid &lt; \mid x^n\mid + \mid x^m \mid &lt; \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 &lt; \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 &amp; : x=1 \\ 0 &amp; : 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

 

[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 &gt; 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 &lt; \frac{\epsilon}{2} for all n &gt; n_0
for m,n &gt; n_0
\mid x^n - x^m \mid &lt; \mid x^n\mid + \mid x^m \mid &lt; \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 &lt; \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 &amp; : x=1 \\ 0 &amp; : 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

<br /> <br /> 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.