dense subspace of $ c_0( \mathbb N)$ Prove that 
$$Y= \left\{ x=(x_n)_{n \in\mathbb{N}} \in c_{0}(\mathbb N )~ \Bigg | ~\sum_{n=1}^{\infty} x_n = 0 \right\}$$
is a dense linear subspace of $ c_0( \mathbb N)$.
where $ \displaystyle{c_0( \mathbb N) = \left\{  x=(x_n)_{n \in\mathbb{N}} \in \mathbb R ^{\mathbb N} : \lim_{n \to \infty} x_n =0 \right\}}$
I cannot prove that it is dense.
Any help?
Thank you in advance!
 A: The elegant proof is the following. Consider linear functional
$$
f:c_{00}(\mathbb{N})\to\mathbb{R}:x\mapsto\sum\limits_{n=1}^\infty x_n
$$
Then


*

*Show that $f$ is unbounded and $\mathrm{Ker}(f)\subset Y$. 

*Show that that kernel of each unbounded functional is dense in the domain space.

*Recall that $c_{00}(\mathbb{N})$ is dense in $c_0(\mathbb{N})$.

A: The fact that $Y$ is a subspace is quite clear. To see density, we can use a corollary of Hahn-Banach theorem: we just need to show that each linear continuous functional on $c_0(\Bbb N)$ which vanished on $Y$ vanishes on the whole space. 
Let $f$ such a functional. We have $f(e_n-e_m)=0$ if $m\neq n$, where $e_n$ is the sequence whose $n$-th term is $1$, the others $0$. So $f(e_k)=:K$. As $\left\lVert\sum_{k=0}^ne_k\right\rVert_{\infty}=1$, we show have $nK\leq \lVert f\rVert$ and $K=0$.
A: Taking on Norbert's answer: since the linear functional $\,f\,$ is obviously not the zero functional, we know $\,\ker f\,$ is a maximal subspace of $\,c_0(\Bbb N)\,$ , from which it follows that
$$c_0(\Bbb N)=\operatorname{Span}\{\ker f\,,\,v\,\}\;\;,\;\forall\;v\notin\ker f$$
Perhaps this now will make it simpler to find the solution ( hint: a subset $\,A\,$ of a topological space $\,X\,$is dense in it iff$\,A\cap Y\neq\emptyset\,$ for every non-empty open subset $\,Y\subset X\,$ )
