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- Jan 17, 2013

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Thanks in advance !

- Thread starter ZaidAlyafey
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- Thread starter
- #1

- Jan 17, 2013

- 1,667

Thanks in advance !

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

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:

- Thread starter
- #3

- Jan 17, 2013

- 1,667

Wow very nice , thanks for the examplesTake 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

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