Canonical basis in $l^{\infty}$

Question

For an exercise, I need to show that the canonical basis is not a valid basis in $l^{\infty}$. Concretely, the exercise states :

Consider the Banach space $l^{\infty}$ of sequences $x = \{x_n\}_{n = 1}^{\infty} \subset \mathbb{C}$ with the usual norm $\left\| x \right\|_{\infty} = \sup_{n \in \mathbb{N}} |x_n| < \infty$. Then, we define the sequence $\{e_m\}_{m \in \mathbb{N}} \in l^{\infty}$ as

$e_m := \{0, ..., 1, 0, ...\}$ , $m \in \mathbb{N}$

i.e. with 1 at the $m$-th position and zeros everywhere else. Show that the system $\{e_m\}_{m \in \mathbb{N}}$ is not a basis for $l^{\infty}$.

My thoughts so far, please correct me if I am wrong!

1. We need to show that every $x \in l^{\infty}$ can be written as $x = \sum_{n = 1}^{\infty} c_n(x) e_n$
2. For the dense subset of all those sequences which have only finitely many non-zero entries, $l_c^{\infty} \subset l^{\infty}$, we have that $c_n(x) = x_n$. Then, the expression $x = \sum_{n = 1}^{\infty} c_n(x) e_n$ is unique.
3. We now need to show that for above choice of $c_n$ and for every $x \in l^{\infty}$ we have that $\lim_{N \rightarrow \infty} \left\| x - \sum_{n = 1}^{N} c_n(x) e_n \right\|_{l^{\infty}} = 0$
4. With the chosen $c_n(x) = x_n$ we have that $x - \sum_{n = 1}^{N} c_n(x) e_n = \{0, ..., 0, x_{N+1}, x_{N+2}, ...\}$
5. With $x \in l^{\infty}$ and $\left\| x \right\|_{\infty} = \sup_{n \in \mathbb{N}} |x_n| < \infty$, we have that $\left\| x - \sum_{n = 1}^{N} c_n(x) e_n \right\|_{l^{\infty}} = \sup_{n \in \mathbb{N}} \sum_{n = N + 1}^{\infty} |x_n|$
6. This however is not a Cauchy sequence, hence there is no convergence on this side s.t. $\lim_{N \rightarrow \infty} \left\| x - \sum_{n = 1}^{N} c_n(x) e_n \right\|_{l^{\infty}} = 0$ does not hold.

I am not sure whether my conclusion in 6) is right and suffices for this proof. However, I suppose that you can prove this in a much more compact way (and if so, how?).

Help is much appreciated :-)

Edit :

The easiest way to solve the problem is to show that $l^{\infty}$ is non-separable since the existence of a Schauder basis as defined in above problem implies that the underlying (sequence) space is separable. I refer to this post to show the non-separability : Why is $L^{\infty}$ not separable?

Thanks @AlvinL and the others!

Answer 1

The sequence $\sum _{n=1}^N e_n$ is eventually zero for any $N$. Thus, for the constant sequence $x:=(1,1,\ldots)$, $$ \left\| x - \sum _{n=1}^N e_n\right\|_\infty = 1 $$ for every $N\in\mathbb N$. What do we conclude?

The above also disproves the claim of $c_{00}$ being a dense subset of $\ell_\infty$.

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Alternatively, existence of Schauder basis implies separable. But $\ell_\infty$ is not separable.