$x=\lim_{i \to \infty}e_i$ is a sequence which is not convergent, so $x\notin c_0$, then why {$e_i$}$_{i=1}^\infty$ is a Schauder basis for $c_0$?

Question

Let $e_i$ denote the sequence whose $i$th term is 1 and rest all are zero. Then $x= lim_{i \to \infty}e_i$ is a sequence which is not convergent. So $x\notin c_0$ and so $x\notin l_p$ because $l_p\subset c_0$. Then this $x$ can't be a member of the Schauder basis for $c_0$ or $l_p$, then why {$e_i$}$_{i=1}^\infty$ is a Schauder basis for $C_0$ or $l_p$?

($c_0$ is the set of sequences converging to $0$ and is equipped with supremum norm, $l_p=${$(x_n):x_n\in R,\sum_{n=1}^\infty|x_n|^p<\infty $} equipped with the norm $||(x_n)||_p=$$[\sum_{n=1}^\infty|x_n|^p]^{1/p}$ )

Answer 1

First of all $x$ is not a sequence at all, it doesn't exist. The sequence $\{e_i\}$ is not convergent so $\lim e_i$ doesn't make any sense. So how can you talk about $x$ being a member or not of some set? This is illogical.

Secondly recall the definition of a Schauder basis:

A sequence $\{b_i\}\subseteq V$ is a Schauder basis for a Banach space $V$ if for every $v\in V$ there exists a unique sequence $\{\lambda_i\}$ of scalars such that $$v=\sum_{i=0}^{\infty}\lambda_i b_i$$

Note that it doesn't mean that for every sequence of scalars the sum above produces a vector from $V$. Generally this is not true.

And finally $\{e_i\}$ is a Schauder basis for $c_0$ or $l^p$, well, simply because it has been proven. I'm pretty sure you will find a proof in some book on functional analysis.

Answer 2

Did you mean $x=\sum_{i=1}^\infty e_i$ rather that $x = \lim_{i\to\infty} e_i$? If so, notice that $\sum_{i=1}^\infty e_i$ is not $(1, 1, 1, \ldots)$ if the convergence is understood in the $\|\cdot\|_\infty$ norm. The series $\sum_{i=1}^\infty e_i$ does not converge as it is not Cauchy. Take $m > n \in \mathbb{N}$:

$$\left\|\sum_{i=1}^m e_i - \sum_{i=1}^n e_i\right\|_\infty = \left\|\sum_{i=n+1}^m e_i\right\|_\infty = \|(\underbrace{0, \ldots, 0}_{n}, \underbrace{1, \ldots, 1}_{m-n}, 0, 0, \ldots)\|_\infty= 1$$

It is true that $(1, 1, 1, \ldots)$ cannot be obtained as $\sum_{i=1}^\infty \alpha_ie_i$ for some sequence of scalars $(\alpha_i)_{i=1}^\infty$ but this does not violate the definition of Schauder basis since $(1, 1, 1, \ldots) \notin c_0$.