# Prove that $C$ is Banach.

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

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?

• You will not get a faster answer by reposting, even less if you do not explain, say, your notations. – Mindlack Jan 5 at 23:09
• The second part seems flawed.. What is $u$ there? – Berci Jan 5 at 23:10
• 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. – user3482749 Jan 5 at 23:10
• @Berci Please check now. – Hitman Jan 5 at 23:30
• @Mindlack it is not a repost. – 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.