Closure of a subspace of $l^\infty$

Question

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?

Answer 1

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.

Answer 2

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 $$ \|\bar x-y_i\|_\infty=1\quad\forall n. $$