Prove that $\{x=(x_n) \in c_0 \mid \sum_{n=1}^\infty x_n = 0\}$ is dense in $c_0$

Question

Let $(c_0,\|\cdot\|_{\infty})$ be the normed space of real sequences convergent to $0$, with the maximum norm. I need to prove that the subspace $M = \{x=(x_n) \in c_0 \mid \sum_{n=1}^\infty x_n = 0\}$ is dense in $c_0$.

I am trying to find an $x\in M$ such that, for a given $\epsilon>0$ and a given $y\in c_0$; $\|x-y\|_{\infty}<\epsilon$ , but I'm struggling with it. I have tried to construct this $x \in M$ but since $y \in c_0$ is arbitrary, I don't know how to deal with it. I suspect I may need to try it by contradiction, but so far I'm stuck.

Thanks in advance!

Answer 1

Since $\lim_{n\to\infty}y_n=0$, there is an $N$ such that $|y_n|<\epsilon/2$ for all $n\ge N$. Let $S=\sum_{n=1}^Ny_n$ and choose an integer $M$ such that $|S|/M<\epsilon/2$. Let $$ x=(y_1,\dots,y_N,\underbrace{-\frac{S}{M},\dots,-\frac{S}{M}}_{M\text{ times}},0,0,\dots) $$

Answer 2

For each $i=1,2\ldots$, let $e_{i}=(0,\ldots,0,1,0,\ldots)$ (i.e., the $i$-th entry is 1 and all other entries are 0). Clearly $e_{i}\in c_{0}$. Denote $A=\{x\in c_{0}\mid\sum_{i=1}^{\infty}x_{i}=0\}$. We show that for each $i\in\mathbb{N}$, $e_{i}\in\bar{A}$, where $\bar{A}$ denotes the closure of $A$ with respect to the $||\cdot||_{\infty}$-topology. Without loss of generality, we consider the case that $i=1$. Let $\epsilon\in(0,1)$ be arbitrary. Let $a=1-\epsilon/2$. Choose $n$ be sufficiently large such that $a/n<\frac{\epsilon}{2}$. Define $x=(a,-\frac{a}{n},\ldots,-\frac{a}{n},0,0,\ldots)$, where $-\frac{a}{n}$ appears $n$ times. Clearly $x\in A$ and $||x-e_{1}||_\infty<\epsilon$. This shows that $e_{i}\in\bar{A}$. Note that $A$ is a vector subspace of $c_{0}$ and hence $\bar{A}$ is also a vector subspace of $c_{0}$. Now $c_{c}=span\left\{ e_{i}\mid i\in\mathbb{N}\right\} $ and hence $c_{c}\subseteq\bar{A}$. Finally, $c_{c}$ is norm-dense in $c_{0}$, so $c_{0}=\overline{c_{c}}\subseteq\bar{A}$.