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Can we take as a dense subset the collection of all the linear combinations of the vectors of the Schauder basis using the rationals as scalars (or the complex numbers with rational real and imaginary parts for that matter)?

What can we say about the converse?

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2 Answers

Yes, you have the right idea.

Let $\mathbb K$ be either $\mathbb R$ or $\mathbb C$ as appropriate. Suppose that the Banach space $X$ has a Schauder basis $\{x_n\}_{n=1}^\infty$, i.e., for every $x \in X$, there exists a unique sequence of scalars $\alpha_n \in \mathbb K$ such that $$ x = \sum_{n=1}^\infty \alpha_n x_n, $$ where this sum converges in the norm topology. We can renormalize so that $\Vert x_n \Vert =1$ for all $n$.

We fix such an $x \in X$ and show how to approximate it by elements from a countable set. Given $\varepsilon >0$, there exists $N \in \mathbb N$ such that $$ \Vert x - \sum_{n=1}^N \alpha_n x_n \Vert < \varepsilon / 2. $$

For each $\alpha_n \in \mathbb K$, we can find $\beta_n $ (in $\mathbb Q$ or $\mathbb Q + \mathbb Q i$ as appropriate) such that $| \alpha_n - \beta_n | < \varepsilon/ 2^{n+1}$.

Then, by the triangle inequality, $$ \Vert x - \sum_{n=1}^N \beta_n x_n \Vert < \Vert x - \sum_{n=1}^N \alpha_n x_n \Vert + \Vert \sum_{n=1}^N \alpha_n x_n - \sum_{n=1}^N \beta_n x_n \Vert < \varepsilon/2 + \sum_{n=1}^N \varepsilon/2^{n+1} < \varepsilon. $$ Thus every element in $X$ can be approximated by finite linear combinations of the elements of the Schauder basis, where the scalars come from $\mathbb Q$ or $\mathbb Q + \mathbb Q i$, as appropriate.

As for your second question: no, not every separable Banach space has a Schauder basis. This was a longstanding problem in the field, which was solved by Per Enflo in 1972 (for which he was awarded a live goose!).

This result can be found in:

Enflo, Per (July 1973). "A counterexample to the approximation problem in Banach spaces". Acta Mathematica 130 (1): 309–317. doi:10.1007/BF02392270

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It's quite impossible to resist the temptation to mention the goose :-) // Notice that OP does not assume the space to be Banach, but of course the same argument for separability works for incomplete spaces. –  Martin Feb 8 '13 at 9:36
    
@Martin Just to mention that later Enflo gave an example of a Banach space with AP but without a basis, answering in the negative the question whether the two notions are equivalent. –  Theo Feb 8 '13 at 17:31
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For your first question:

Let $X$ be a norm space and $(e_i)$ be a Schauder basis of $X$ (suppose $||e_i||=1$). Consider the countable set $$Q= \{ \sum\limits_{i=0}^n q_ie_i : n \in \mathbb{N}, q_i \in \mathbb{Q} \}$$ Let $\displaystyle x= \sum\limits_{i \geq 0} x_ie_i \in X$ and $\epsilon >0$. There exists $n \geq 0$ such that $\displaystyle \left\|x- \sum\limits_{i=0}^n x_ie_i \right\| \leq \epsilon$, and for $0 \leq i \leq n$, there exists $q_i \in \mathbb{Q}$ such that $\displaystyle |q_i-x_i| \leq \frac{\epsilon}{n+1}$ because $\mathbb{Q}$ is dense in $\mathbb{R}$. Then $\displaystyle y= \sum\limits_{i=0}^n q_ix_i \in Q$ and $$||x-y|| \leq ||x- \sum\limits_{i=0}^n x_ie_i||+ \sum\limits_{i=0}^n |x_i-q_i| \cdot ||e_i|| \leq 2 \epsilon$$ If $X$ is a norm space over $\mathbb{C}$, you can consider $Q+iQ$ rather than $Q$.

For your second question, the converse is not true; see here. However, Mazur gave a partial result: Any infinite dimensional Banach space has a subspace containing a Schauder basis.

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I feel that it's not enough. A vector space is defined on a field for the scalar multiplication. Then that field is not needed to be $R$ or $C$. In other case, we can't take $Q$ into our proof. Can you clarify more to make it clearer... –  le duc quang Nov 7 '13 at 14:41
    
My answer works when the field is $\mathbb{R}$ or $\mathbb{C}$, but if the field is $\mathbb{Q}$ then $Q$ is still dense in the normed vector space: any element $x$ can be written as $\sum_{i=1}^{+ \infty} q_ie_i$ and by definition $\sum_{i=1}^n q_ie_i \to x$, so $Q$ is dense. –  Seirios Nov 7 '13 at 16:51
    
Oh, what I means is in the theorem, the field is not restrict to be $R$,$Q$, or $C$, but a general field. In that general case, are the problem still right? Can you prove that? –  le duc quang Nov 7 '13 at 17:16
    
The fiel has to be restricted, notably because of the definition of a normed space: an absolute value is needed. –  Seirios Nov 7 '13 at 20:27
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