# Prove that $C_0$ is Banach.

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 $$i \geq k$$.
• 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. – Kabo Murphy Jan 6 at 0:38
• In the last part you did not specify $m$. If you take $m=N$ your proof will be correct. – Kabo Murphy Jan 6 at 4:47