Let $l_\infty$ be the set of all bounded real sequences.
$S:l_\infty\to l_\infty$ be the shift operator defined by $S(x_1,x_2,x_3,\dots)=(x_2,x_3,x_4,\dots)$ for all $x\in l_\infty$
Let $M=\{x-Sx:x\in l_\infty\}$. Then it is clear to me that $M$ is a linear subspace of $l_\infty$.
For a subset $P$ of $\mathbb N$ the natural/asymptotic density is defined by the limit (if it exists)$$\delta(P)=\lim\limits_{n\to \infty}\frac{|P\cap\{1,2,\dots,n\}|}{n}$$
We say a condition $\mathcal C$ is satisfied by $x_k\ a.a.k$ (for almost all k) iff $E=\{k\in\mathbb N:x_k$ does not satisfy condition $\mathcal C$ $\}\implies\delta(E)=0$.
Let $N=\{y\in l_\infty: y_k=x_k\ a.a.k,\ x\in M \}$. Then $N$ is a linear subspace of $l_\infty$. So, $M\subset N\subset l_\infty$.
My question : Is the subspace $N$ closed in $l_\infty$ or not?