Let $x^n \in C_0$ is Cauchy.
$\rightarrow$ For $\epsilon> 0$ there is N such that $n,m \ge N $ $$ \lVert x^m -x^n\rVert< \frac {\epsilon} {2} $$
we know that for every k $$\lvert x^m_k -x^n_k\rvert \le sup_{i\ge 1} \ \lvert x^m_i -x^n_i\rvert < \frac {\epsilon} {2} $$
So $x^n_k$ is Cauchy in $\mathbb R$ which is Banach so $$x^n_k \rightarrow x_k \in \mathbb R \ \ \ \ \ \ or \ \ \ \ \ \lvert x^n_k -x_k\rvert \le \frac {\epsilon} {2} $$
by this we can say that there is an N, $n\ge N$ such that $\lVert x^n -x\rVert = sup_{i\ge 1} \ \lvert x^n_i -x_i \lvert < \frac {\epsilon} {2} $ $\ \ \ \ \ \ \ $
which means $x^n \rightarrow x$
Since $x^m \in c_0$,
there is some $N'$ such that $|x_i^m| < {1 \over 2 } \epsilon$ for $i \ge N' \ \ \ \ \ \ \ \ \ \ \ (m=N)$
Now to show that $x\in C_0$
$$\lvert x_i\rvert \le \lvert x^m_i-x_i\rvert+ \lvert x^m_i\rvert <\frac {\epsilon} {2}+\frac {\epsilon} {2}=\epsilon \ \ for \ i\ge N' $$
This gives us $x_i\rightarrow 0$ for $i\ge N'$
So $x\in C_0$
Is this Correct?
Best Answer
The question has been modified based on this answer. The answer below is for the first version of the question. First correction: $|a_n|<\epsilon /2$ for all $n$ and $a_n \to a$ does not imply $|a|<\epsilon /2$. It implies $|a|\leq \epsilon /2$. Second correction: in the last part $|x|$ has no meaning for a sequence $x$. Use coordinates. You should write $|x_i| \leq |x_i^{n}|+|x_i^{n}-x_i|$. Then you should make the argument more rigorous as follows. First choose $n$ such that $|x_i^{n}-x_i|<\epsilon /2$ for all $i$. Fixing this $n$ choose $k$ such that $|x_i^{n}|<\epsilon /2$ for all $i \geq k$. Now you get $|x_i| <\epsilon$ for all $i \geq k$.