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.

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.
