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Does there exist a linear independent and dense subset?

I am looking for an example of a countable dense subset of the Hilbert space $l^2(\mathbb{N})$ consisting of linearly independent vectors

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marked as duplicate by Jonas Meyer, Jennifer Dylan, sdcvvc, tomasz, Davide Giraudo Sep 7 '12 at 10:29

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You might want to begin by accepting answers. See why and how. –  Asaf Karagila Jul 8 '12 at 12:44
    
If $l^2$ are real sequences, is the set of square summable rational sequences not dense in it? –  Matt N. Jul 8 '12 at 13:24
    
I missed "linearly independent". –  Matt N. Jul 8 '12 at 13:27
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Related question: math.stackexchange.com/questions/60057/…. In my answer there, I show that any separable Banach space has such a subset. –  Nate Eldredge Jul 8 '12 at 16:19

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up vote 7 down vote accepted

Based on Davide's answer. Begin with the set $\{x_n\}$ of vectors with only finitely many nonzero coordinates, all rational. That is dense, but not linearly independent. Next choose a sequence (say $2^{-n}, n=1,2,\dots$) that goes to zero. Add $2^{-n}$ to coordinate $r_n$ of $x_n$, chosen so that $r_n > r_{n-1}$ and the $r_n$ coordinate of all $x_k, 1 \le k \le n$ is zero. This new sequence is still dense, but also linearly independent.

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That's a nice explicit example. –  Davide Giraudo Jul 8 '12 at 15:16

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