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Does there exist a linear independent and dense subset?

Please bring a hint for following problem:

Every separable infinite-dimensional normed space has a linearly independent countable dense subset.

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    Can you take the countable dense subset and apply Gram-Schmidt?2012-09-07
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    @Matt Gram-Schmidt doesn't make sense in general normed space. But incomplete basis theorem does.2012-09-07
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    True, we want to have an inner product.2012-09-07
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    What I can't see is how the infinite dimensional setting plays a role. The result is clearly false in a finite dimensional space.2012-09-07
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    There is a full answer in this discussion : http://math.stackexchange.com/questions/60057/does-there-exist-a-linear-independent-and-dense-subset/60109#601092012-09-07
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    @Matt I guess the two questions are almost equivalent and so this must be closed as the duplicate of the other one. Or maybe the fact that this questions explicitly asks for the theorem in a separable space can make it a different enough question in which case, we might let it remain?2012-09-07
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    See also [this thread](http://math.stackexchange.com/q/168229/5363) for a very neat explicit construction for $\ell^2(\mathbb{N})$ in GEdgar's answer.2012-09-07
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    @JayeshBadwaik I don't know.2012-09-07

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