I'd like to prove that if I consider $X$ Banach space with $\dim X =\infty$, then $X$ can't have a countable Hamel's basis. Someone can give me a hint? Even a counterexample is enough. I think this can be linked with Baire's theorem about Banach's space, but I'm not really sure... Thanks!
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See this. For just a hint: Otherwise $X=\cup_{n=1}^\infty F_i$ whee each $F_i$ is a finite dimensional subspace of $X$. Obtain a contradiction using Baire. – David Mitra Feb 01 '14 at 17:03
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Thanks, I searched for similar questions about this, but I didn't find this one! – rusca91 Feb 01 '14 at 17:09