# Example of a separable space without a Schauder basis.

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.

• 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). – David Mitra Jan 20 '14 at 10:01
• OK. Got it. Thank you. – Alexander Jan 20 '14 at 10:07