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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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Ali Farzaneh
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  • Can you take the countable dense subset and apply Gram-Schmidt? – Rudy the Reindeer Sep 07 '12 at 07:05
  • @Matt Gram-Schmidt doesn't make sense in general normed space. But incomplete basis theorem does. – Ahriman Sep 07 '12 at 07:06
  • True, we want to have an inner product. – Rudy the Reindeer Sep 07 '12 at 07:07
  • What I can't see is how the infinite dimensional setting plays a role. The result is clearly false in a finite dimensional space. – Ahriman Sep 07 '12 at 07:11
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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#60109

     – Ahriman Sep 07 '12 at 07:18
  • @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? –  Sep 07 '12 at 08:33
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    See also this thread for a very neat explicit construction for $\ell^2(\mathbb{N})$ in GEdgar's answer. – t.b. Sep 07 '12 at 08:36
  • @JayeshBadwaik I don't know. – Rudy the Reindeer Sep 07 '12 at 09:32

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