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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.

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    $\begingroup$ 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: Enflo, P.: A counterexample to the approximation property in Banach spaces. Acta Math. 130, 309–317(1973). $\endgroup$ – David Mitra Jan 20 '14 at 10:01
  • $\begingroup$ OK. Got it. Thank you. $\endgroup$ – Alexander Jan 20 '14 at 10:07
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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.

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