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My textbook said that

Suppose that $\{f_1,f_2,\dots\}$ is a linearly independent sequence in a Hilbert space $V$. There is a vector in $V$ which is not a finite linear combination of the $f_j$.

I am wondering if I have a Hamel basis $B$ for $V$. Then $B$ is linearly independent and by the above assertion, there is a vector $v\in V$ such that $v$ cannot be expressed as a finite linear combination, which contradicts with the fact that $B$ is a basis.

The proof on the textbook is

The Gram-Schmidt process produces an orthonormal sequence $\{e_1,e_2,\dots\}$ with $span\{e_1,e_2,\dots,e_n\}=span\{f_1,f_2,\dots,f_n\}$ for any $n$. Write down any sum $\sum_1^\infty c_je_j$ where infinitely many of $c_j$ are nonzero and $\sum_1^\infty |c_j|^2<\infty$. The sum then converges to a vector $g$. Then $g$ is not in the span of $\{e_1,e_2,\dots,e_n\}$ for any $n$, and thus not in the space of $\{f_1,f_2,\dots,f_n\}$ for any $n$.

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As a vector space, it must have a Hamel basis, and the theorem says that such a Hamel basis cannot be denumerable, so it must be uncountable.

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  • $\begingroup$ So an infinite-dimensional Hilbert space must have an uncountable Hamel basis? $\endgroup$ – SHBaoS Feb 4 '18 at 4:58
  • $\begingroup$ Yes, this is correct. $\endgroup$ – user284331 Feb 4 '18 at 4:58

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