Determining closure of $F=\{(z_n)_{n \in \mathbb{N}} \in l^\infty(\mathbb{N},\mathbb{C}):\exists N \in \mathbb{N}, z_n=0 \;\forall \;n\ge N \}$

Question

$l^\infty(\mathbb{N},\mathbb{C})$= Banach-space of all bounded sequences $\in \mathbb{C}$ with the norm $\|\cdot\|_\infty$, so $\|(z_n)_{n \in \mathbb{N}}\|_\infty=\sup_{n\in \mathbb{N}}|z_n|$

$F=\{(z_n)_{n \in \mathbb{N}} \in l^\infty(\mathbb{N},\mathbb{C}):\exists N \in \mathbb{N}, z_n=0 \;\forall \;n\ge N \}$

I need to show that the set $c_o(\mathbb{N}, \mathbb{C)}$of all sequences $\in \mathbb{C}$ converging to $0$ is a closed subspace of $l^\infty(\mathbb{N},\mathbb{C})$ which I managed but now I need to determine the closure of $F$ and I don't know how to do that.

My try:

Let $x_n \in c_0(\mathbb{N},\mathbb{C})$

$x_n^1=(x_1,0,...)$

$x_n^2=(x_1,x_2,0,...)$

$x_n^3=(x_1,x_2,x_3,0...)$ and so on

$c(F)\subset c(c_0(\mathbb{N},\mathbb{C}))=c_0(\mathbb{N},\mathbb{C})$ because $c_0$ is closed. But I guess that's not enough. How can I determine the closure of $F$?

Answer 1

Clearly $F \subseteq c_0(\mathbb{N}, \mathbb{C})$ so $\overline{F} \subseteq \overline{c_0(\mathbb{N}, \mathbb{C})} = c_0(\mathbb{N}, \mathbb{C})$ because $c_0(\mathbb{N}, \mathbb{C})$ is closed.

Conversely, if $(x_n)_n \in c_0(\mathbb{N}, \mathbb{C})$, then consider the sequence $((x_1, x_2, \ldots, x_n, 0, 0, \ldots))_n$ in $F$. We have $$\|(x_n)_n - (x_1, x_2, \ldots, x_n, 0, 0, \ldots)\|_\infty = \|(0, \ldots, 0, x_{n+1}, x_{n+2}, \ldots)\|_\infty = \sup_{k > n} |x_k| \xrightarrow{n\to\infty} 0$$

because $x_n \xrightarrow{n\to\infty} 0$. Therefore, $(x_1, x_2, \ldots, x_n, 0, 0, \ldots) \xrightarrow{n\to\infty} (x_n)_n$.

Hence every element of $c_0(\mathbb{N}, \mathbb{C})$ is a limit of elements of $F$, so we conclude $c_0(\mathbb{N}, \mathbb{C}) \subseteq \overline{F}$.

Therefore $$\overline{F} = c_0(\mathbb{N}, \mathbb{C}) $$