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I know that Per Enflo gave examples of separable Banach spaces without Schauder basis, but... I have seen that:

1.- Every vector space has a basis.

2.- A normed space is separable if and only if it has a dense subspace of countably infinite dimension.

So it seems reasonable to me to take the Hamel/algebraic basis of that dense subspace as the Schauder basis, say {${u_n: n\in\mathbb{N}}$} as $X=\overline{L(\{u_n: n\in\mathbb{N}\})}$. What's wrong with that argument?

Also, why do we define Schauder basis just for Banach spaces?

Jorge Ortiz
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    There is a huge difference between Hamel bases and Schauder bases. In the fomer you use only finite sums. – Kavi Rama Murthy Feb 06 '21 at 08:53
  • yes @KaviRamaMurthy, so every element of that dense subspace (U) would be a finite sum of elements of the Hamel basis (B). Also every element of the normed space X would be a limit of elements of U, which are finite sums of elements of B, so it seems intuitive to me that every element of X is a series of elements of B, and I don't know the flaw – Jorge Ortiz Feb 06 '21 at 09:09
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    No, your thinking is techically wrong. For example any continuous function on $[0,1]$ is a uniform limit of linear combinatios sof $1,x,x^{2},...$ but it need not be a uniformly convergent infinite sum $\sum a_nx^{n}$ Only analytic functions are infinite sums of this type. – Kavi Rama Murthy Feb 06 '21 at 09:20
  • @Kavi, I deleted my answer after noticing that I misread the question. Your argument is clearly the right one! – Ruy Feb 06 '21 at 17:05
  • Note that the Baire category theorem implies that there cannot be a countable Hamel basis for an infinite dimensional Banach space since otherwise the Banach space could be written as countable union of closed subspaces with empty interior (take the linear span of the first $n$ elements of your basis). – Christian Feb 16 '21 at 14:52
  • that's right @Christian, good one – Jorge Ortiz Feb 16 '21 at 18:10

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