Closure of a subspace of $l^\infty$ Let $X$ be the following subspace of $l^\infty$:
$$
X=\mathrm{lin}\{e_n:n\in\mathbb{Z}^+\}
$$
where $e_j$ has zeroes everywhere except for one in the $j$-th entry. I want to know what the closure of $X$ is. The solution I was given claims that $\overline{X}$ is the space of sequences with limit zero, but I don't think that this is true. 
Let $(x_n)$ be any infinite convergent sequence in $\mathbb{R}$ with some limit x. Then we can construct a convergent sequence $(y_n)$ in $l^\infty$ by taking
$$
y_i=(x_1,x_2,...,x_i,0,0,...)\in X
$$
for every $i\in\mathbb{Z}^+$. Now, clearly, $\displaystyle\lim_{n\to\infty}y_n=(x_n)$ and if $x\neq0$ then $(x_n)$ is not in the space of sequences with limit zero. 
I understand that I haven't proved what the closure of $X$ actually is, but is my reasoning above correct?
 A: No, your argument is incorrect. The sequence $y_i$ converges poitwise to $(x_n)$, but not in the $\ell^\infty$ norm (i.e. uniformly) unless $x=0$. Take for instance $x_n=1$ for all $n$ and call $\bar x=(x_n)$. Then
$$
\|\bar x-y_i\|_\infty=1\quad\forall n.
$$
A: No, it’s not right, I’m afraid. Let $x$ be the target sequence, and let it converge to $\alpha\in\Bbb R$; then
$$\lim_{n\to\infty}\|y_n-x\|=|\alpha|\;,$$
so $\langle y_n:y\in\Bbb N\rangle\to x$ iff $\alpha=0$. Of course the $y_n$ converge pointwise to $x$, but that’s not enough to get $\ell^\infty$ convergence.
A: Following Jimmy R, we have
$$
||y_i-(x_n)||_{\infty}=\sup_{k \in N}|Pr_k(y_i)-Pr_k((x_n))|=\sup_{k >i}|Pr_k(y_i)-Pr_k((x_n))| \ge \sup_{k >i}|x_k|,
$$
where $Pr_k$ denotes $k$-th projection.
Since $\lim_{i\to \infty}|x_i|=|x| > 0$, we can choose such $n_{0}$ that  $|x_k| >\frac{|x|}{2}$ for $k \ge n_0$. Hence for $i\ge n_0$ we get 
$$
||y_i-(x_n)||_{\infty} \ge \sup_{k >i}|x_k| \ge \frac{|x|}{2}.
$$
Hence, 
$$
\lim_{i \to \infty} ||y_i-(x_n)||_{\infty}\ge \frac{|x|}{2}, 
$$
which means that $\lim_{i \to \infty}y_i \neq (x_n)$.
Note that $$
\lim_{i \to \infty} ||y_i-(x_n)||_{\infty}=0 
$$
holds if and only if $x=0$ which implies that  $\overline{X}$ is the space of sequences with limit zero.
