Not closed set in $l^2$ Let $H=l^2$ equipped with the inner product $$\lVert \{a_n\}\rVert_2=\left(\sum_{n=1}^\infty\lvert a_n \rvert ^2\right)^\frac{1}{2}$$ We consider the set $$V:=\left\{\{a_n\}\in l^2\;|\;a_n=0\quad\text{for all}\quad n>k;\;k\in\mathbb{N}\right\}$$

How can i prove that $V$ is not closed?

 A: One way would be to find a sequence of elements in $V$ which converges to something that does not belong to $V$. Hint: what do you think of the sequence $v_1 = (2^0,0,0,\ldots)$, $v_2 = (2^0,2^{-1},0,0\ldots)$, $v_3 = (2^0,2^{-1},2^{-2},0,0\ldots)$, etc.
A: $V\subset \ell^2$ is not closed as
$(x^{(n) })_{n\in\Bbb{N}}\subset V $ be a sequence where the $n$-th term is defined by $x^{(n)}=(x^{(n) }_j) {_{j\in \Bbb{N}}}$
$\begin{align}&x^{(1)}=\{1,0,\dots\}\\&x^{(2)}=\{1,\frac{1}{2},0,\ldots\}\\&x^{(3)}=\{1,\frac{1}{2},\frac{1}{3},\ldots\}\\&\vdots \\&x^{(n)}=\{1,\frac{1}{2},\frac{1}{3},\ldots,\frac{1}{n},0,\ldots\}\\&\vdots\end{align}$

Then for each fixed $j$ , the coordinate sequence $(x^{(n) }_j) \to \frac{1}{j}$.
Let $x_n=(\frac{1}{n})_{n\in\Bbb{N}}$
Check:  $(x^{(n)})_{n\in\Bbb{N}}\to (x_n)_{n\in\Bbb{N}} \space $ in $\ell^2$ but clearly $(x_n)\notin V$ as it is not eventually constant.

$V\subset \ell^2$ is proper as $(\frac{1}{n}) \in \ell^2$ but not in $V$ . Now we want to show show $V$ is dense in $\ell^2$  and it follows that $V$ is not closed ( any proper dense subset can't be closed)
Let $(x_n) \in \ell^2$ be any element.
Let us consider the sequence $(x^{(n) }) $ defined by
$\begin{align}&x^{(1)}=\{x_1,0,\dots\}\\&x^{(2)}=\{x_1,x_{2},0,\ldots\}\\&x^{(3)}=\{x_1,x_{2},x_{3},\ldots\}\\&\vdots \\&x^{(n)}=\{x_1,x_{2},x_{3},\ldots,x_{n},0,\ldots\}\\&\vdots\end{align}$
Now show that $(x^{(n) })\to (x_n) $ in $\ell_2$ space.
A: As I suggested in the comment every element of $\ell^2$ can be approximated by elements of $V,$ therefore $V$ cannot be closed, as $V\subsetneq \ell^2.$
There is another, more sophisticated explanation. We have
$$V=\bigcup_{k=1}^\infty V_k,\qquad V_k=\{x\in \ell^2\,:\, x_n=0,\ n\ge k\}$$
Hence $V$ is a set of first Baire category. Thus $V$ cannot be complete, therefore closed in $\ell^2.$
A: Another way :
Suppose $V\subset \ell^2$ is closed. As we know any closed subspace of a Banach space is Banach, $V$ is a Banach space.
$V$ has a countable infinite basis( Hamel basis) $\mathcal{B}=\{e_k :k\in \Bbb{N}\}$ where $e_k=(0, \ldots , 1,\ldots )$
$1$ in $k$-th place.
But a Banach space can't have countable infinite dimension. ( can be proved using Baire's theorem) . Hence $V$ can't be closed .
