Proving density of a set I want to show that $A=\{(x_n)\in c_{00}:\sum\limits_{n=1}^{\infty}x_n=0\}$ is dense in $(c_0,\|.\|_{\infty})$. 
Let $x=(x_n)\in c$ and let $\epsilon >0$. Thus there exists $n_0\in \mathbb N$ such that $|x_n|<\epsilon $ for all $n\geq n_0$. Now the question is how to choose an element from $A$? I wanted to choose $y=(x_1-x_2,x_2-x_3,\ldots,x_{n_0-1}-x_1,0,0,\ldots)\in A$ but it didn't work as $\|x-y\|_{\infty}=\|x\|_{\infty}$ and we cann't say it is less than $\epsilon$. Now what to do?

We must show that $$\forall a_n\in c_0\quad,\quad\forall\epsilon>0\quad,\quad\exists b_n\in A\quad,\quad||a_n-b_n||_\infty<\epsilon$$also according to definition $$||a_n-b_n||_\infty<\epsilon\leftarrow\rightarrow \sup_{n}|a_n-b_n|<\epsilon\leftarrow\rightarrow|a_n-b_n|<\epsilon\quad,\quad\forall n$$we try to make such a sequence $b_n$ for any given $a_n\in c_0$ . Since $a_n$ tends to $0$ there exists some $M_\epsilon$ such that $$\forall n>M_\epsilon\quad,\quad|a_n|<\dfrac{\epsilon}{2}$$For $n\le M_\epsilon$ define $b_n$ such that $$a_n-\epsilon<b_n<a_n+\epsilon$$once again define $$\Large S=\sum_{n=1}^{M_\epsilon}b_n$$if $S\ge 0$ for $M_\epsilon+1\le n \le M_\epsilon+\left[\dfrac{2S}{\epsilon}\right]$ define $b_n=-\dfrac{\epsilon}{2}$ and $$\Large b_{M_\epsilon+\left[\dfrac{2S}{\epsilon}\right]+1}=-\sum_{n=1}^{M_\epsilon+\left[\dfrac{2S}{\epsilon}\right]}b_n$$
if $S< 0$ for $M_\epsilon+1\le n \le M_\epsilon+\left[\dfrac{2S}{\epsilon}\right]$ define $b_n=\dfrac{\epsilon}{2}$ and $$\Large b_{M_\epsilon+\left[\dfrac{2S}{\epsilon}\right]+1}=-\sum_{n=1}^{M_\epsilon+\left[\dfrac{2S}{\epsilon}\right]}b_n$$ and $$b_n=0\quad,\quad\forall n\ge M_\epsilon+\left[\dfrac{2S}{\epsilon}\right]+2$$using this approach we make sure that $$|a_n-b_n|<\epsilon\quad,\quad \forall n\in\Bbb N$$which completes our proof on existence of such sequence.
