Can I say that the normed linear space $(\Bbb{R}(\Bbb{Q}), \lvert\, \cdot\,\rvert)$ is an infinite dimensional, separable, Banach space and hence cannot have a Schauder basis?
My argument is based on the fact that an infinite dimensional Banach space cannot have a countable basis.
Best Answer
No. Schauder bases are not Hamel bases. Schauder bases allow representations as infinite linear combinations. Finding a separable Banach space without a Schauder basis is non-trivial. Per Enflo was the first to do so.
In fact, he constructed a separable Banach space without the Approximation property. A Banach space that has a Schauder basis also has the Approximation property; so his space also gives an example of a separable banach space without a basis. This was done in: Enflo, P., A counterexample to the approximation property in Banach spaces, Acta Math. 130, 309–317 (1973).
A link to this paper is here.