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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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6You might want to begin by accepting answers. See why and how. – Asaf Karagila Jul 08 '12 at 12:44
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If $l^2$ are real sequences, is the set of square summable rational sequences not dense in it? – Rudy the Reindeer Jul 08 '12 at 13:24
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I missed "linearly independent". – Rudy the Reindeer Jul 08 '12 at 13:27
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3Related question: http://math.stackexchange.com/questions/60057/does-there-exist-a-linear-independent-and-dense-subset/60109#60109. In my answer there, I show that any separable Banach space has such a subset. – Nate Eldredge Jul 08 '12 at 16:19
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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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