Closed / open set in $\ell^\infty$ metric space.

Question

Let $F$ be the set of all $x$ in $\ell^\infty$ metric space with $x_n =0$ for all but finitely many $n$, then is $F$ closed? or open? or neither?

I know that $\ell^\infty$ is the space of all bounded sequences of real numbers $(x_n)$ with the sup norm. How to check if $F$ is closed or open or neither? Any hints are appreciated.

Answer 1

It is clearly not open: take for instance the all-zero sequence, $\mathbf{0}$, which clearly belongs to $F$.

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

It is also not closed. To show it, it's sufficient to show that its complement is not open either. Consider the sequence $u=(u_n)_n\notin F$ defined by $$ u_n = \frac{1}{n+1}. $$

For any $\varepsilon > 0$, the $\ell_\infty$ ball $B(u,\varepsilon)$ around $u$ will intersect $F$: indeed, letting $N\geq 0$ be the smallest integer such that $\frac{1}{N+1} < \varepsilon$, then the sequence $u^\prime$ which agrees with $u$ on the first $N$ terms and is identically $0$ afterwards belongs to $F$; and furthermore $\lVert u - u^\prime \rVert_\infty < \varepsilon$. Thus, $u^\prime\in B(u,\varepsilon)\cap F$, and so $B(u,\varepsilon)\not\subseteq F^c$.