# schauder basis for $\ell_\infty$ [duplicate]

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.

Every sequence $x=(x_1,x_2,x_3,\dotsc)$ that belongs to $\ell_\infty$ can be written in one and only one way as $\sum_{i=1}^\infty x_i e_i$.

Could you explain why this is wrong? Thank you!

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• You wrote "Every sequence $x=(x_1,x_2,x_3,\dotsc)$ that belongs to $\ell_\infty$ can be written in one and only one way as $\sum_{i=1}^\infty x_i e_i$". It is NOT true. Please, remember that $\sum_{i=1}^\infty x_i e_i=\lim_{n \to +\infty}\sum_{i=1}^n x_i e_i$, and in this case the limit must be taken in the norm (topology) of $\ell_\infty$. As a simple example, $\sum_{i=1}^n e_i$ does NOT converge to $(1, 1, \cdots)$ in $\ell_\infty$. – Ramiro Aug 11 '15 at 11:44
That sum does not converge in the $\ell_\infty$ norm if the sequence $x_n$ does not converge to $0$.
• Your sequence $\{e_n\}$ is a Schauder basis of the closed subspace $c_0$ of sequences that converge to zero. – GEdgar Aug 11 '15 at 0:41
• @Dear, Robert Israel,That sum does not converge in the $\ell_\infty$ norm, you means that $\|x-\sum\limits_{i=1}^{n}x_ie_i\|$ does not converge to 0 in $\ell_\infty$? – user62498 Aug 11 '15 at 10:42
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 $x.$