How to prove that $\ell_{\infty}$ has no Schauder basis by contradiction?

Question

I have a question about:schauder basis for $\ell_\infty$

I know that $\ell_\infty$ is not separable, therefore has no Schauder basis. However I cannot understand why the set $\{e_1, e_2, e_3, \dotsc \}$ where $e_1=(1,0,0,\dotsc), e_2=(0,1,0,0,\dotsc), \dotsc$ can not work as a Schauder basis.

One answer is that let $x=(1,1,1,\ldots)\in \ell_\infty$. The proof of that "$\ell_{\infty}$ has no Schauder basis by contradiction?

Question:

(1) The standard basis $\{e_n\}_{n\ge 1}$ is not Schauder basis

(2) What can we get from the results "$\sum_{I=1}^{\infty}x_ie_i$ does not converges to $x$"? I just read the proof in the second answer in schauder basis for $\ell_\infty$. I am confused about the answer.

Let $x=(1,1,1,\ldots)\in \ell_\infty$ if $x=\sum\limits_{i=1}^{\infty}x_ie_i,$ then for any $N\in\mathbb{N}$ we have $$\left\|x-\sum_{n=1}^{N}x_ne_n\right\|_{\infty}=\|(0,\ldots,0,1,1,1,\ldots)\|_{\infty}=1\not\to0,$$ so $\sum\limits_{i=1}^{\infty}x_ie_i$ does not converge to $x.$

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The proof of (1): Does it make sense?

Suppse that $\{e_n\}$ is a Schauder basis. Then for every $x\in \ell_{\infty}$ we have $x=\sum_{I=1}^{\infty} x_ie_i$. That means as $n \to\infty$, $$ 0\le \sup_{k\ge n+1}|x_k|=\|(0,0,\dots,0, x_{n+1}, x_{n+2},\dots)\|_{\infty}=\|x-\sum_{I=1}^n x_ie_i\|_{\infty}\to 0 $$

Thus, $x_k=0$ for $k\ge n+1$. So $$ x=\sum_{I=1}^n x_ie_i $$

But this is impossible. Does it a contradiction?

Answer 1

No, your proof isn't valid. You can't draw the conclusion that $\ x_k=0\ $ for all $\ k\ $ exceeding some integer $\ n\ $ from the mere fact that $\ \sup_\limits{k\ge n+1}\big|x_k\big|\rightarrow$$\,0\ $ as $\ n\rightarrow\infty\ $. The latter condition will hold if $\ x_k=\frac{1}{k}\ $, for instance, but then $\ x_k\ne0\ $ for any $\ k\ $.

Nor can you obtain a contradiction by choosing some arbitrary $\ x\in\ell_\infty\ $ and assuming that $\ \lim_\limits{n\rightarrow\infty}\sum_\limits{k=1}^nx_ke_k=x\ $, because there happen to be members $\ x\ $ of $\ \ell_\infty\ $ for which this latter condition does hold (namely, those $\ x\ $ for which $\ \lim_\limits{k\rightarrow\infty}x_k=0\ $), and an arbitrarily chosen $\ x\ $ could well be one of them unless you've specified otherwise.

For $\ \big\{e_n\big\}_{n\ge1}\ $ to be a Schauder basis, however, it must be true that for every $\ x\in\ell_\infty\ $ there exists a unique sequence $\ \big\{y_k\big\}_{k=1}^\infty\ $ such that $\ \lim_\limits{n\rightarrow\infty}\sum_\limits{k=1}^ny_ke_k=$$\ x\ $. Therefore, to show that it's not a Schauder basis you only need to find a single member of $\ \ell_\infty\ $ for which $\ \lim_\limits{n\rightarrow\infty}\sum_\limits{k=1}^ny_ke_k\ne x\ $ for any sequence $\ \big\{y_k\big\}_{k=1}^\infty\ $, and $\ x=(1,1,1,\dots)\ $ happens to be one such.

There's one relatively minor blemish in the proof you cite from here, which you also repeat in your attempted proof— namely, the assumption that if $\ x=\big(x_1,x_2,\dots\big)\in\ell_\infty\ $ and $\ x=\sum_\limits{n=1}^\infty y_ne_n\ $ then $\ y_i=x_i\ $ must necessarily hold for all $\ i\ $. Since I'm pretty sure I can see how to prove this hypothetical statement, and also that the proof won't be difficult, I might be prepared to let you get away with "It's easy to see that" it's true. However, I wouldn't be prepared to let you get away with "It's obvious" that it's true, especially because I know that in the case when $\ x=$$\,(1,1,\dots)\ $ no such sequence $\ \big\{y_k\big\}_{k=1}^\infty\ $ can exist.

Here's how you can modify your attempted proof so that it does work:

Suppose that $\ \big\{e_n\big\} $ is a Schauder basis. Then since $\ x=$$\,(1,1,\dots)\in$$\,\ell_\infty\ $ we have $\ x=$$\,\sum_\limits{n=1}^\infty y_ne_n\ $ for some sequence $\ \big\{y_k\big\}_{k=1}^\infty\ $ of complex numbers. This means that as $\ n\rightarrow\infty\ $, \begin{align} \left\|x-\sum_\limits{i=1}^n y_ie_i\right\|_\infty&=\big\|\big(1-y_1,1-y_2,\dots,1-y_n,1,1,\dots\big)\big\|_\infty\\ &\rightarrow0\ . \end{align} But \begin{align} \big\|\big(1-y_1,1-y_2,\dots,1-y_n,1,1,\dots\big)\big\|_\infty &=\sup\big(|1-y_1|,|1-y_2|,\dots,|1-y_n|,1,1,\dots\big)\\ &\ge1 \end{align} for all $\ n\ $, and it's impossible for a sequence that's never less than $\ 1\ $ to converge to $\ 0\ $.

Answer 2

Hint: By contradiction

1.Show that, $(\mathbb R^\infty,\|\cdot\|_{\infty})$ is a Banach space

2. For $n\in\mathbb N$ set $$\mathbb R_n = \{x=(x_1,x_2,...)\in\mathbb R^\infty :x_i =0 ~~\text{if}~~i>n~i.e~x=(x_1,x_2,\cdots,x_n,0,0,\cdots)\}$$
   Show that $\mathbb R_n$ is closed subspace of $(\mathbb R^\infty,\|\cdot\|_{\infty})$
3. Since $(e_1,e_2,..)$ is Schauder Basis, observe that, $$\mathbb R^\infty = \bigcup_{n\in\mathbb N}\mathbb R_n $$
4. From Baire Theorem's there exists $n_0$ such that, $\overset{\circ}{\mathbb R}_{n_0} \neq \emptyset$ that is there are $x_0\in \mathbb R^\infty$ and $r>0$ such that, $B(x_0,r)\subset\mathbb R_{n_0}$

5.Deduce from 4. that, $\mathbb R_{n_0}=\mathbb R^\infty$ contradiction.