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?


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 $i \geq k$.

  • $\begingroup$ Could you please check again? I have edited the question. $\endgroup$ – Hitman Jan 6 at 0:19
  • $\begingroup$ Your last part is still not rigorous. There are two variables $i$ and $n$ and you have to state clearly that you first choose $n$ and then the final inequality becomes true for all $i$ sufficiently large. $\endgroup$ – Kabo Murphy Jan 6 at 0:38
  • $\begingroup$ I think now it is okay. $\endgroup$ – Hitman Jan 6 at 0:59
  • $\begingroup$ what do you say? $\endgroup$ – Hitman Jan 6 at 1:00
  • 1
    $\begingroup$ In the last part you did not specify $m$. If you take $m=N$ your proof will be correct. $\endgroup$ – Kabo Murphy Jan 6 at 4:47

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