Show that a given set is a closed subspace of $c_0$ Could someone check my proof of the following problem? I know that there might be some awkward argument. I would appreciate, if you have a better way to tackle this problem. You are welcome to share it.
Let $M=\{(x_n)_{n\in\mathbb{N}}\in c_0: \sum_{n=1}^{\infty}2^{-n}x_n=0\}\subset c_0\equiv \{(x_n)\in l^{\infty}: \lim_{n\rightarrow\infty}x_n=0\}$. Prove that $M$ is a closed subspace in $c_0$.
Let $\textbf{x}^{(k)}=(x^{(k)}_n)$ be any sequence in $M$ such that it converges to $\{x_n\}\subset c_0$. We want to show that $\textbf{x}=(x_n)\subset M$. Given $\varepsilon>0$. There exists $N\in\mathbb{N}$ such that $$|x^{(N)}_n-x_n|\leq \|\textbf{x}^{(N)}-\textbf{x}\|_{\infty}<\varepsilon.$$ Since $\{x^{(N)}_n\}\subseteq M$, which means $\lim_{n\rightarrow\infty}\frac{1}{2^n}x^{(N)}_n=0$ there is $J\in\mathbb{N}$ such that  $$\left|\frac{1}{2^n}x^{(N)}_n\right|<\varepsilon.$$  For each $n\geq J$ we have \begin{align*}\sum_{n}\frac{|x_n|}{2^n}&\leq \sum_{n}\frac{|x^{(N)}_n-x_n|}{2^n}+\sum_{n}\frac{|x^{(N)}_n|}{2^n}\\ &\leq \sum_{n}\frac{\varepsilon}{2^n}+\varepsilon\,\,\,\,(\text{since} \sum_{n\geq J}\frac{1}{2^n}<1)\\&\leq 2\varepsilon.\end{align*}Since $\varepsilon>0$ was arbitrary, we conclude that $\textbf{x}$ belongs to $M$. Hence $M$ is a closed subspace of $c_0.$ 
 A: There's a definite problem with this proof, but it's a little difficult to pinpoint precisely due to some muddled notation and missing details.
When you write

there is $J\in\mathbb{N}$ such that  $$\left|\frac{1}{2^n}x^{(N)}_n\right|<\varepsilon,$$

I'm guessing you mean

there is $J\in\mathbb{N}$ such that  $$n \ge J \implies \left|\frac{1}{2^n}x^{(N)}_n\right|<\varepsilon?$$

But, either which way, I can't follow the argument here:

\begin{align*}\sum_{n}\frac{|x_n|}{2^n}&\leq \sum_{n}\frac{|x^{(N)}_n-x_n|}{2^n}+\sum_{n}\frac{|x^{(N)}_n|}{2^n}\\ &\leq \sum_{n}\frac{\varepsilon}{2^n}+\color{red}{\varepsilon\,\,\,\,(\text{since} \sum_{n\geq J}\frac{1}{2^n}<1)}\\&\leq 2\varepsilon.\end{align*}

Firstly, you're using $\sum_n$, and my best guess is that $\sum_n = \sum_{n=1}^\infty$. However, I can't explain the highlighted bit in red. While you know that eventually the terms of the sum $\sum_{n}\frac{|x^{(N)}_n|}{2^n}$ are bounded above by $\varepsilon$, this does not even prove that the sum is convergent, let alone less than $\varepsilon$! This seems like a significant issue with the proof.

As an alternative, try working with the linear functional
$$\phi : c_0 \to \Bbb{F} : (x_n) \mapsto \sum_{n=1}^\infty \frac{1}{2^n}x_n,$$
(where $\Bbb{F} = \Bbb{R}$ or $\Bbb{C}$ is your field). Note that $M$ is the kernel of $\phi$, which establishes $M$ as a linear subspace of $c_0$, and is closed if and only if $\phi$ is bounded.
Suppose $\mathbf{x} = (x_n)$ and $\|\mathbf{x}\|_\infty \le 1$, i.e. $|x_n| \le 1$ for all $n$. Then,
$$|\phi(\mathbf{x})| = \left|\sum_{n=1}^\infty \frac{1}{2^n}x_n\right| \le \sum_{n=1}^\infty \frac{1}{2^n}|x_n| \le \sum_{n=1}^\infty \frac{1}{2^n} = 1,$$
and so $\phi$ is bounded. Thus, $M = \operatorname{ker} \phi$ is a closed subspace of $c_0$.
