A simple question of density in $l^1$ (sequence space)

Question

I need to show that the linear span of $S=\{e_n|n≥1\}$, where $e_n = (\delta_{nj})_{j≥1}$ is dense in $l^1$.

I know I'm doing something wrong here, but why can't I make a sequence of sequences in sp{$S$} such as

$1,0,0,0,0,\dots$

$1,1,0,0,0, \dots$

$1,1,1,0,0,\dots$

etc.

whose limit is clearly not in $l^1$?

How would I then go about showing the answer to the question?

Answer 1

Take $x=(\xi_k)\in \ell_1$ and $\varepsilon > 0$. Then you will find $N$ such that $\sum_{k=n}^\infty |\xi_k| < \varepsilon$ for $n\geqslant N$. Let $y = (\xi_1, \ldots, \xi_N, 0,0,\ldots)$. Certainly, $y$ is in the span of $e_1, \ldots, e_N$. Moreover

$$\|x-y\|_{\ell_1} = \sum_{k=N+1}^\infty |\xi_k|<\varepsilon.$$

Answer 2

Why should this be a contradiction? In the same manner, $\mathbb{Q}$ is dense in $\mathbb{R}$, while the sequence $a_n = n$ of rational numbers does not converge in $\mathbb{R}$.

To solve the problem, you need to show that every sequence in $l^1$ can be approximated arbitrarily well by a sequence in $\langle S \rangle$. You just need to truncate the sequence you want to approximate after sufficiently many elements.

Answer 3

Let $x=(\alpha_k) \in l^1$. Then $x=\sum_{k=1}^{\infty}\alpha_ke_k$, since, with $x_n:=\sum_{k=1}^{n}\alpha_ke_k$, we have

$$||x_n-x||_{l^1} \to 0.$$

Thus $x \in \overline{S}$