A countable set in $\ell^\infty$ which is not dense

Question

Consider the set $S$ of all rational sequences which are eventually zero. I know $S \subset c_0 \subset \ell^\infty$ where $c_0$ means space of all sequences converges to zero and $\ell^\infty$ is the space of all bounded sequences.

Since $S$ has a bijection with $\Bbb{Q}^N$ for some $N$, it is a countable subset of $c_0$ as well as $\ell^\infty$

Also $S$ is a subset of $\ell^p$ , where $1 \leq p < \infty$.

I know $S$ is a countable dense subset of $\ell^p$ and $c_0$, so these two spaces are separable

My question is: Why this $S$ is not dense in $\ell^\infty$ ?

Any help?

Answer 1

Let $x=(1,1,...)$. If $S$ is dense then there would be a sequence $r=(r_n)$ which is eventiually $0$ such that $\|x-r\|< \frac 1 2$. Taking $n$ large enough we get $|x_n-r_n|=1-0<\frac 1 2 $ which is a contradiction.

Answer 2

Recall that $c_0$ is closed in $\ell^\infty$. Taking the closure in $S \subseteq c_0$ gives

$$\overline{S} \subseteq \overline{c_0} = c_0 \subsetneq \ell^\infty$$

so $S$ is not dense in $\ell^\infty$.