Examples of Dense\Not Dense sets Let the set of eventually zero sequences $c_{00} = \{x = (x_1, x_2, . . .) : x_n = 0 \ \text{for all but finitely many} \ n \}$ where $x_i$ are real numbers.
(a) Prove that $c_{00}$ is dense in $l^p, 1 \le p < \infty$.
(b) Prove that $c_{00}$ is not dense in $l^\infty$.
$\text{My attempt}$:
a) let $\epsilon>0$ and pick any $c_{00} \ni x= (x_1,\dots,x_n,0,0,\dots) , x_i\in\mathbb{R}$. Define the set $M=\{l^p \ni y=(y_i)_{i\ge 1} : (y_i)_{i=1}^n \in \mathbb{Q}, (y_i)_{i=n+1}^\infty=0 \}$, then the set $M$ is countable and $M\subset l^p$. then , since rationals are dense in reals;
$$\|x-y\|_{l^p}^p = \sum_{i=1}^\infty|x_i-y_i|^p = \sum_{i=1}^n|x_i-y_i|^p < \epsilon^p$$ $$\|x-y\|_{l^p}<\epsilon \ \ \ ,\text{for some}\ \ y\in M$$
b) similarly let $\epsilon>0$ and pick any $c_{00} \ni x= (x_1,\dots,x_n,0,0,\dots) , x_i\in\mathbb{R}$. Also pick $l^\infty \ni y=(y_i)_{i\ge1}$s.t. $(y_i)_{i=1}^n \in \mathbb{Q}, (y_i)_{i=n+1}^\infty=L$. Where $\sum_{i=1}^n|x_i-y_i|^p<\frac{\epsilon^p}{2}$ and $L<\infty$ is sufficiently large. Then ;
$$\|x-y\|_{l^p}^p = \sum_{i=1}^\infty|x_i-y_i|^p = \sum_{i=1}^n|x_i-y_i|^p + \sum_{i=n+1}^\infty|y_i|^p< \frac{\epsilon^p}{2} + \sum_{i=n+1}^\infty|y_i|^p$$
So $L$ can be chosen sufficiently large s.t. $\sum_{i=n+1}^\infty|y_i|^p > \frac{\epsilon^p}{2}$. So $c_{00}$ is not dense in $l^\infty.$
 A: For (a), your argument does not prove that $c_{00}$ is dense in $\ell^p$, rather it proves that the set of elements of $c_{00}$ having rational coordinates is dense in $c_{00}$. To prove that $c_{00}$ is dense in $\ell^p$, you need to start with an arbitrary element of $\ell^p$ and show that the $\varepsilon$ ball centered at this element contains an element of $c_{00}$ for any $\varepsilon>0$.
For (b), in order to show that $c_{00}$ is not dense in $\ell^{\infty}$ you need to show that there is some $x\in\ell^{\infty}$ and $\varepsilon>0$ such that no element of $c_{00}$ is contained in the $\varepsilon$ ball centered at $x$. So while you are on the right track, it doesn't really make sense to start with an element of $c_{00}$.
A: This is edited solution I added. 
a) let $y\in l^p$ be any arbitrary point , then $\sum_{i=1}^\infty |y_i|^p< \infty$. As this series converges, we can write that for any $\epsilon>0$, there exists $n$ such that $\sum_{i=n+1}^\infty |y_i|^p< \epsilon^p$ for all $i\ge n$. Then there exists $c_{00}\ni x=(x_1,⋯,x_n,0,⋯)$, s.t. $y_i=x_i \ \ \forall i\le n$ and $\|y-x\|_{l^p} < \epsilon \ \ .$
b)choose $\epsilon >0$ and let $y\in l^\infty$ be any arbitrary point s.t. $l^\infty\ni y=(y_1,⋯,y_n,L,L,,⋯)$. Then for any $x\in c_{00}$, not matter of how close it is to $y$ (even if $x_i=y_i=L$) we can choose $L$ large enough s.t.
$$\|x-y\|_{l^\infty} = \sup_{i}|x_i-y_i|  > \epsilon$$
