Let $F$ be the set of all $x\in l_{\infty}$ such that $x_n = 0$ for all but finitely many n. Is $F$ open? closed? neither?

Question

Let $F$ be the set of all $x\in l_{\infty}$ such that $x_n = 0$ for all but finitely many n. Is $F$ open? closed? neither?

$l_{\infty}$ is the collection of all bounded real sequences.

I know F is neither open nor closed.

This is someone's a part of proof.

For any $\varepsilon>0$, the $l_∞$ ball around $\mathbf{0}$ includes the constant sequence $(\varepsilon)_{n∈N}$, which does not belong to $F$.

$\mathbf{0}$ is all-zero sequence.

In this problem, we don't define distance. So, Why is the sequence $(\varepsilon)$ ball around $\mathbf{0}$.

Answer 1

The distance is defined. As soon as you say $\ell_\infty$, it comes equipped with the norm $$\|x\|_\infty=\sup\{|x_n|:\ n\in\mathbb N\},$$ and thus with the distance $$ d(x,y)=\|x-y\|_\infty. $$