False proof that infinite dimensional Banach space has no countable basis

Question

Let $B$ be a Banach space. If $B$ is infinite-dimensional, then $B$ has no countable basis.

I suspect the following proof is wrong :

Let $(e_i)_i$ be a countable linearly independent set of elements of $B$. Let $S=<(e_i)_i>$ and let $x=\Sigma_i \lambda_ie_i$ be a convergent series. Such a series exists. Indeed, $B$ is Banach so it suffices to choose the $\lambda_i$ coefficients such that it converges normally. Since you can't just ignore the terms beyond a certain rank, it seems intuitively acceptable that if an infinite number of $\lambda_i $'s are nonzero, then $x$ cannot not be expressed as a finite linear combination of the $e_i$'s, them being independent. Therefore $x \notin S$ from where it follows that $(e_i)_i$ is not a basis of $B$.

But I'm not quite sure how to convince myself that the "$x \notin S$" statement is wrong. Is there an example of a convergent series with linearly independent terms that converges to a linear combination (i.e. finite linear combination) of its terms ?

Why is it clear that the sum of the series is not in the span of the vectors? — Mariano Suárez-Álvarez, Feb 20 '18 at 00:37
Whenever you have doubts about an argument, look for the first apparence of the word "clear" or "obvious" and expand it into a complete argument. Repeat until you have found the error. — Mariano Suárez-Álvarez, Feb 20 '18 at 00:38
(in any case, every correct argument goes thru Bairer's category theorem, like https://math.stackexchange.com/questions/217516/let-x-be-an-infinite-dimensional-banach-space-prove-that-every-hamel-basis-of) — Mariano Suárez-Álvarez, Feb 20 '18 at 00:40
I can't manage to prove that $x \notin F$ but at first glance doesn't it seem reasonable to think so ? — user493048, Feb 20 '18 at 00:47
"it doesn't seem reasonable to think so" is hand waving at Olympic level... — Mariano Suárez-Álvarez, Feb 20 '18 at 00:48
At the very least, you need some conditions on the coefficients for that to even have a hope of being true. As an extreme example, all lambdas could be zero. Notice you have not excluded that possibility: with that choice the series converges normally. — Mariano Suárez-Álvarez, Feb 20 '18 at 00:49
In that case the terms of the series would actually not be linearly independent. But that needs editing too. — user493048, Feb 20 '18 at 00:51

score 4 · Accepted Answer · 2018-02-20T02:07:31.867

Consider the usual Hilbert basis $( e_i)_{i\in\Bbb N}$ of $\ell^2$. The vector $e_{-1}=\sum_{k\in\Bbb N}2^{-k}e_k$ is not in $\operatorname{span}( e_i)_{i\in\Bbb N}$, so $( e_i)_{i\in\Bbb N\cup\{-1\}}$ is a linearly independent subset which can be extended to a Hamel basis of $\ell^2$ - call it $( e_i)_{i\in I}$. Well, now $x=e_{-1}$ is a counterexample to your "intuitively acceptable" claim, because $$e_{-1}=\sum_{k=-1}^\infty \lambda_k e_k=\sum_{k=-1}^\infty \mu_k e_k$$ for both $\lambda_k=\begin{cases}2^{-k}&\text{if }k\ge 0\\ 0&\text{if }k=-1\end{cases}$ and $\mu_k=\begin{cases}0&\text{if }k\ge 0\\ 1&\text{if }k=-1\end{cases}$. And this is by no means an exotism: it's a material difference between finite sum and series.

score 2 · Answer 2 · answered Feb 20 '18 at 03:35

The erroneous assumption is that $v_n=\sum_{j=1}^n\lambda_je_j$ cannot converge to $0$ as $n\to \infty$ unless every $\lambda_j=0.$

It is an easy exercise to find a linearly independent subset $S=\{e_j:j\in \Bbb N\}$ of a separable infinite-dimensional Banach space $B$ such that for every $v\in B$ there is a sequence $(\lambda_j)_{j\in \Bbb N}$ such that $v=\sum_{j\in \Bbb N}\lambda_je_j.$

If, for any given $v,$ the sequence $(\lambda_j)_j$ is unique, then $S$ is called a Schauder basis for $B.$ You cannot prove that $B$ has a Schauder basis. It was proven (fairly recently) that separable Banach spaces without Schauder bases do exist

False proof that infinite dimensional Banach space has no countable basis

2 Answers2