Is the set of sequences for which $\sum_{n=2}^\infty x_n=2x_1$ closed in $\ell_1$? In $\ell_1$ (with the $\ell_1$-norm), consider the subspace $M=\{(x_n)_{n=1}^\infty:\displaystyle{\sum_{n=2}^\infty x_n=2x_1}\}$.
I am trying to determine whether this space is closed or not and if its codimension is 1.
I have tried to find a sequence contained in $M$ that does not converge on $M$. However, everything I have come up with is convergent on $M$. I started to suspect $M$ may actually be closed, but I have been unable to to prove it.
Any hints would be very helpful. Thanks in advance.
 A: Let $(x^{(n)})$ be a sequence in $M$ converging to some $x\in \ell^1$. We know that $2x_1=\lim_{n\to\infty} 2 x^{(n)}_1$, and that $\lim_{n\to\infty}\sum_{k=2}^\infty x_k^{(n)}=\sum_{k=2}^\infty x_k$. As $2 x^{(n)}_1=\sum_{k=2}^\infty x_k^{(n)}$ for every $n\in \mathbb N$, we must have $x\in M$.
EDIT:
$M$ does has codimension $1$. My previous argument was incorrect. Indeed as pointed out by KeeperofSecrets $M$ is just the kernel of a continuous linear functional, and thus must have codimension $1$.
This is actually intuitively clear too, as any element in $M$ can be constructed by freely varying all the coordinates except the first one, and then fixing the first coordinate to be $-2$ multiplied by the sum of all the other coordinates, so the "degrees of freedom" of $M$ is only one less than for all of $\ell^1$.
Rigorously, we can notice that $e_1=[(1,0,0,\dots)]$ is a basis vector for $\ell^1/M$. Indeed for any $(x_n)\in \ell^1$ we note that
$$(x_n)-\left(x_1-\frac{1}{2}\sum_{n=2}^\infty x_n,0,0,\dots\right)\in M,$$
so $[(x_n)]\in\operatorname{span}([e_1])$.
