0
$\begingroup$

$C=\{x_n \lvert x_n \ converges \} $

Let $x^n \in C$ is Cauchy.

$\rightarrow$ For $\epsilon> 0$ there is N such that $n,m >N $ $$ \lVert x^n -x^m\rVert< \frac {\epsilon} {3} $$ we know that for every k $$\lvert u^n -u^m\rvert \le sup_{i\ge 1} \ \lvert x^n_i -x^m_i\rvert < \frac {\epsilon} {3} $$ So $u^n$ is Cauchy in $\mathbb R$ which is Banach so $$u^n \rightarrow u \in \mathbb R \ \ \ \ \ \ or \ \ \ \ \ \lvert u^n -u\rvert < \frac {\epsilon} {3} $$ by this we can say that $\lVert x^n -x\rVert = sup \ \lvert x^n_i -x_i \rvert < \frac {\epsilon} {3} $ $\ \ \ \ \ \ \ $ ($\epsilon>0, n\ge N\in \mathbb N$)

which means $x_n \rightarrow x$

Now to show that $x\in C$

$$\lvert x-u\rvert \le \lvert x^n-x\rvert+ \lvert x^n-u^n\rvert +\lvert u^n-u\rvert <\frac {\epsilon} {3}+\frac {\epsilon} {3}+\frac {\epsilon} {3}=\epsilon$$

This gives us $x\rightarrow u$

So $x\in C$

Is this Correct?

$\endgroup$
  • $\begingroup$ You will not get a faster answer by reposting, even less if you do not explain, say, your notations. $\endgroup$ – Mindlack Jan 5 at 23:09
  • 1
    $\begingroup$ The second part seems flawed.. What is $u$ there? $\endgroup$ – Berci Jan 5 at 23:10
  • $\begingroup$ I have no idea what $C$ is, what your hypotheses on it are, what $u$ is, what $x$ is, what you mean by $x_n \to x$, if $x$ is not necessarily in $C$, or what these $x^n_k$ are. So no, it isn't correct. $\endgroup$ – user3482749 Jan 5 at 23:10
  • $\begingroup$ @Berci Please check now. $\endgroup$ – Hitman Jan 5 at 23:30
  • $\begingroup$ @Mindlack it is not a repost. $\endgroup$ – Hitman Jan 5 at 23:30
0
$\begingroup$

The second part is not correct: $x$ should converge to a real number.

A hint for that: show that the real sequence $(x^n_n)$ is Cauchy, and show that $x=(\lim x^n)$ converges to its limit.

$\endgroup$
  • $\begingroup$ Could you please check again. I have edited the question. $\endgroup$ – Hitman Jan 5 at 23:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.