Is this subspace of $\ell_\infty$ a Banach space?

Question

Define $T: \ell_\infty \rightarrow \ell_\infty$ the continuous linear operator defined by $$T(x_1,x_2,x_3,\dots) = (x_2,x_3,\dots).$$

Consider the subspace $M$ of $\ell_\infty$ defined by $$M = \{ x- T(x) : x \in \ell_\infty\}.$$

Is it true that $M$ is a closed subspace of $\ell_\infty$? I want to know if $M$ is a Banach space, but I couldn't prove or disprove that.

Answer 1

The sequence $(x_n)_{n\in\mathbb N}$ where $x_n=1/n$ is not in $M$.

In the other hand, the sequences $(x_n^m)_{n\in \mathbb N}$ where $x_n^m = x_n$ for $n\leq m$ and $x_n^m = 0$ else, are in $M$.

The convergence of $ (x_n^m)_{N \in \mathbb N}$ to $(x_n)_{n\in\mathbb N}$ in $\ell^\infty$ is clear.

In other words, $M$ is not closed.