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Complete spaces and Cauchy sequences

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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

Active member
Mar 1, 2012
275
Take the set of all continuous functions at [0,1]
with
[tex]d(f(x),g(x)) = \int_{0}^{1} \mid f(x) - g(x) \mid dx [/tex]
Let
[tex]f_n(x) = x^n [/tex] , I want to show [tex] f_n [/tex] is Cauchy
Given [tex] \epsilon > 0 [/tex]
if x= 1 ,
[tex] f_n (x) = 1 [/tex] which is Cauchy
if x<1

[tex]x^n \rightarrow 0 [/tex]
There exist [tex]n_0 [/tex] such that
[tex] \mid x^n \mid < \frac{\epsilon}{2} [/tex] for all [tex] n > n_0 [/tex]
for [tex] m,n > n_0 [/tex]
[tex] \mid x^n - x^m \mid < \mid x^n\mid + \mid x^m \mid < \frac{\epsilon}{2} + \frac{\epsilon}{2}= \epsilon [/tex]

[tex] 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 [/tex]
Using the fact for f,g continuous functions with f<= g
[tex] \int_{a}^{b} f(x) \leq \int_{a}^{b} g(x) [/tex]

so it is Cauchy but
[tex] f_n(x) [/tex]
converges to
[tex] f(x) = \left\{ \begin{array}{11} 1 & : x=1 \\ 0 & : x\in [0,1) \end{array} \right. [/tex]
f not continuous so it is not in the Metric
so [tex]f_n(x) [/tex] is not converge in our metric
but it is Cauchy

Another Example
The metric on the Rational number set [tex]Q [/tex]
with [tex] \mid x - y \mid [/tex]
Let
[tex] P_n = \sum_{k=0}^{n} \frac{1}{k!} [/tex]
this sequence is Cauchy but it is converge to [tex] e [\tex] which is not rational number
Another sequence in the same metric is
[tex] P_n = \left( 1 + \frac{1}{n} \right)^n [/tex] which converge to e
 
Last edited:

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Take the set of all continuous functions at [0,1]
with
[tex]d(f(x),g(x)) = \int_{0}^{1} \mid f(x) - g(x) \mid dx [/tex]
Let
[tex]f_n(x) = x^n [/tex] , I want to show [tex] f_n [/tex] is Cauchy
Given [tex] \epsilon > 0 [/tex]
if x= 1 ,
[tex] f_n (x) = 1 [/tex] which is Cauchy
if x<1

[tex]x^n \rightarrow 0 [/tex]
There exist [tex]n_0 [/tex] such that
[tex] \mid x^n \mid < \frac{\epsilon}{2} [/tex] for all [tex] n > n_0 [/tex]
for [tex] m,n > n_0 [/tex]
[tex] \mid x^n - x^m \mid < \mid x^n\mid + \mid x^m \mid < \frac{\epsilon}{2} + \frac{\epsilon}{2}= \epsilon [/tex]

[tex] 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 [/tex]
Using the fact for f,g continuous functions with f<= g
[tex] \int_{a}^{b} f(x) \leq \int_{a}^{b} g(x) [/tex]

so it is Cauchy but
[tex] f_n(x) [/tex]
converges to
[tex] f(x) = \left\{ \begin{array}{11} 1 & : x=1 \\ 0 & : x\in [0,1) \end{array} \right. [/tex]
f not continuous so it is not in the Metric
so [tex]f_n(x) [/tex] is not converge in our metric
but it is Cauchy

Another Example
The metric on the Rational number set [tex]Q [/tex]
with [tex] \mid x - y \mid [/tex]
Let
[tex] P_n = \sum_{k=0}^{n} \frac{1}{k!} [/tex]
this sequence is Cauchy but it is converge to [tex] e [\tex] which is not rational number
Another sequence in the same metric is
[tex] P_n = \left( 1 + \frac{1}{n} \right)^n [/tex] which converge to e
Wow very nice , thanks for the examples :)
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
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

Well-known member
MHB Math Helper
Jan 29, 2012
661
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.