$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?

  • $\begingroup$ 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

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.

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

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