Schauder Basis Confusion I'm learning out of the Kreyszig book for Introductory Functional Analysis and I'm having trouble understanding one of the questions. From section 2.3, question 10 reads: Show that if a normed space has a Schauder basis; it is separable.
I attempted it and compared my attempt to solutions by others.


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*Is the scalar field, $Q$, attached to any vector space always separable? I don't think the proof is possible without this assumption. Are scalar fields always separable? With how much structure they have, I would guess that they are, but I would like know where to begin to look or just the answer would be nice. 

*Couldn't the set $B=\{\sum\limits_{i=1}^m a_{n}e_{n}:a_{1}...a_{n}\in{Q},m\in{N}\}$ be considered an infinite Cartesian product of $Q$ if each element of the set was written as $(a_{1},a_{2},...,a_{m},0,0,...)$? From what I understand infinite Cartesian products are uncountable. I understand that what I wrote down only considers finite sums, but how does that make the set, $B$, countable? Is there some detail I'm missing about countable products?


Any help would be greatly appreciated.
Item 2 has a short answer. The infinite Cartesian product is not in the set. Adding that would definitely make this set uncountable.
 A: 1) Ordinarily in functional analysis the scalar field is $\mathbb R$ or $\mathbb C$.  So yes, it is separable.
2) I'm not quite sure what you're asking.  You want to take the coefficients from a countable subset of the scalar field.  The union of countably many countable sets is countable.
A: Let $X$ be a normed space and $(e_i)_{i=1}^{\infty}$ be a Schauder basis of $X$ and let $\|e_i\|=1$ consider $$B=\left\{\sum\limits_{i=1}^{n}q_ie_i: n\in \mathbf{ N }, q_i\in\mathbf{ Q } \right\}.$$ Let $x=\sum\limits_{i=1}^{\infty}x_ne_i$ so for all $\epsilon>0$  there exists $n_0\in \mathbf{N}$ such that $$\|x-\sum\limits_{i=1}^{n_0}x_ie_i\|<\frac{\epsilon}{2}$$ since $x_i\in \mathbf{R}$ and $\mathbf{Q}$ is dense in $\mathbf{R}$, $\exists y_i \in \mathbf{Q}$ such that $$|x_i-y_i|<\frac{\epsilon}{2^{n+1}}$$ from the last equations we have $$\|x-\sum\limits_{i=1}^{n_0}y_ie_i\|\leq\|x-\sum\limits_{i=1}^{n_0}x_ie_i\|+\|\sum\limits_{i=1}^{n_0}x_ie_i-\sum\limits_{i=1}^{n_0}y_ie_i\|<\epsilon$$. This shows that if $X$ has a Schauder basis, then it is separable. 
