dense set in $c_o(\Bbb{Z})$ Let's consider the sequence space $c_o(\Bbb{Z}) = \{(a_n)_{n \in \Bbb{Z}} : lim_{n \rightarrow \infty} (a_n) = 0\}$ equipped with the usual norm $\lVert.\rVert_{l^{\infty}}$.
For any $k \in \Bbb{Z}$, let $e^{(k)} \in c_o(\Bbb{Z})$ be the sequence defined as $e^{(k)}_n := \delta_{nk}$ ($\forall n \in \Bbb{Z}$).
I read somewhere that $\{e^{(k)}\}_{k \in \Bbb{Z}}$ is a dense set in $c_o(\Bbb{Z})$ but it seems untrue. I might be missing something very obvious here but for example, I don't see why the sequence $a := (...,0 ,0 ,1, 1, 0, 0, 0,...) \in c_o(\Bbb{Z})$ would be arbitrarily close from one of the $e^{(k)}$. For me, the distance that this sequence has from $\{e^{(k)}\}_{k \in \Bbb{Z}}$ is just given by 1.
Could you please enlighten me ?
Thank you in advance for your help.

Your definition of $c_0(\mathbb Z)$ is not quite right: it is the set of all bilateral sequences $(a_k)_{k \in \mathbb Z}$ that "vanish at infinity", meaning that $\lim_{|k| \to \infty} a_k=0.$
What you are trying to show is not true. What is true however is that  $\operatorname{span}_{k\in \mathbb Z}\{e^{(k)}\}$ is dense in $c_0$. To see this, take $x=(x_k)_{k \in \mathbb Z} \in c_0(\mathbb Z)$ and let $ε>0$. Pick $n_0 \in \mathbb N$ such that for $|n| \ge n_0, \ |x_n|<\varepsilon.$ If we let
$$y= x_{-n_0+1} e^{(n_0-1) } + \dots + x_0 e^{(0)}+x_1 e^{(1)} + \dots + x_{n_0-1} e^{(n)}$$
then
$$\| y-x\|_{\infty} = \sup_{|n| \ge n_0} |x_n| \le \varepsilon. $$
