Prove that subspace of $s^*\subset l_p$, $p\ge 1$ of real sequences with a finite support is not closed. Prove that subspace of $s^*\subset l_p$, $p\ge 1$ of real sequences with a finite support is not closed. I have to show, I suppose, that it doesn't contain all the limit points. I try to think of something but I get that as long as the number of zeros is infinite, it will always stay like that. No constructions of sequence  smart enough to change it seems to exist. Is there some concept or idea I am missing? I could really use some help here. 

For a sequence $x:\mathbb{N} \to \mathbb{R}$, define the truncation
$\tau_k(x)(n) = \begin{cases} x_n, & n \le k \\
0, & \text{otherwise}\end{cases}$.
It should be clear that for any sequence $x$, we have  $\tau_k(x) \in l_p$
for any $p \ge 1$ or $p=\infty$.
Note that $\tau_k(x)$ has finite support.
Let $x^*(n) = {1 \over n^2}$. Note that $x^* \in l_p$ for any $p \ge 1$ or $p=\infty$. Note that $x^*$ does not
have finite support.
For any $p \ge 1$ or $p=\infty$, check that 
 $\|x^*-\tau_k(x^*) \|_p \to 0$. 
