Schauder basis for $c_0$

Question

So, I am trying to prove that $c_0$ has the dual space $\ell^1$ (I know this proof is out there). Except my professor told me that a Schauder basis for $c_0$ is $(e_k)$ where $ e_k = \delta_{j,k}$ has a 1 in the $k$th place and zeros otherwise is not correct (this is what every proof out there claims). He said that I need to think about this as a matrix to develop the basis.

This disagrees with everything I have found up until now. He said that I have $e_k$ depended upon $1$ variable while $\delta$ is depended upon $2$. Below I state the wording of the problem. Can anyone help me write a proper Schauder basis.

Represent $\ell^1$ as the space of all real functions $x$ on $S= \{(m,n): m\geq 1, n \geq 1\}$, such that $$ \|x\|_1 = \sum |x(m,n)| < \infty. $$ Let $c_0$ be the space of all real functions $\gamma$ on $S$ such that $y(m,n) \rightarrow 0$ as $m+n \rightarrow \infty$, with norm $\|y\|_\infty = \sup |y(m,n)|$. \ Let M be the subspace of $\ell^1$ consisting of all $x \in \ell^1$ that satisfy the equations $$ mx(m,1) = \sum_{n=2}^\infty x(m,n) \;\;\;\;\;\;\; (m = 1, 2, 3, \ldots) $$

Answer 1

I am not sure about this talk about matrices. $(e_n)_{n \geq 1} = ((\delta_{nj})_{j \geq 1})_{n \geq 1}$ is yes a Schauder basis for $c_0(\mathbb{N})$. I'll just prove that given $(a_n)_{n \geq 1}\in c_0(\mathbb{N})$, $\sum_{n \geq 1}a_ne_n = (a_n)_{n \geq 1}$.

This "$=$" means that $\sum_{k =1}^{n}a_ke_k \stackrel{\|\cdot\|_\infty}{\longrightarrow} (a_n)_{n \geq 1}$.

We have that $a_n \to 0$ because it is in $c_0(\mathbb{N})$. Let $\epsilon > 0$. There is $n_0 \in \mathbb{N}$ such that: $$|a_n| < \frac{\epsilon}{2}, \quad \forall\,n \geq n_0.$$

So: $$\sup_{n \geq n_0}|a_n| \leq \frac{\epsilon}{2} < \epsilon.$$

If $n \geq n_0$, we have: $$\left\|(a_n)_{n \geq 0} - \sum_{k=1}^{n}a_ke_k\right\|_\infty = \|(0,0,\cdots,0,a_{n+1},a_{n+2},\cdots)\|_\infty = \sup_{k \geq n+1}|a_k| \leq \sup_{n \geq n_0}|a_n| < \epsilon.$$Done.