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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.

  • 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.

Martin Sleziak
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MathMigrant
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2 Answers2

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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.

Robert Israel
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  • I'm confused about the difference between a countable union of countable sets, which is countable since you can enumerable along diagonals, and the infinite Cartesian product, which is uncountable by Cantor's diagonalization.

    I would like to know why representing the linear combinations as $(a_{1},a_{2},...)$ does not result in an uncountable set by cantor's diagonalization.

     – MathMigrant Sep 14 '14 at 07:38
  • Cantor diagonalization would produce a "linear combination" with infinitely many nonzero coefficients. Here you only allow finitely many. – Robert Israel Sep 14 '14 at 07:50
  • I can't wrap my head around it. The first $m$ entries are non-zero coefficients, but $m$ is any natural number. For any "linear combination", there is an $m$ which allows the "linear combination" to be in the set. – MathMigrant Sep 14 '14 at 07:58
  • Sorry, I just realized my mistake. $(a_{1},a_{2},a_{3},...)$, the infinite Cartesian product, is not in the set. – MathMigrant Sep 14 '14 at 08:40
  • Heyy, really like your answer, if you could help with an exercise, would really appreciate it :) https://math.stackexchange.com/questions/3096022/normed-vector-space-schauder-basis-exercise – Homaniac Feb 02 '19 at 06:53
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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.

user62498
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  • I wasn't asking for the proof. I was asking about 2 details. This proof isn't complete. You need to count the basis. – MathMigrant Sep 14 '14 at 18:43