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I heard the claim in the title. ¿Is it true? If so, what is its name? Any references?

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 In fact, every Hilbert space is isometrically isomorphic to some $\ell^2$. Just take an orthonormal basis $B$ for your space $H$, and then the obvious map gives you an isometric iso $H \to \ell^2(B)$. – Faisal Apr 11 2011 at 23:55
To completely answer, one has to observe that $\ell^2(B)$ is the same thing as $L^2(B)$, if we endow the set $B$ with the counting measure. – Mariano Suárez-Alvarez Apr 12 2011 at 0:11
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 ... and what is the name? Would that be called the Riesz-Fischer Theorem? – Gerald Edgar Apr 12 2011 at 0:26
@Faisal, doesn't the existence of an orthonormal basis depend on the Hilbert space being separable? What is an ONB for $L^{2}(L^{2}(R))$? – BSteinhurst Apr 12 2011 at 3:14
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 @BSteinhurst: A Hilbert space is separable iff it has a countable ONB. You can have uncountable ONBs for nonseparable Hilbert spaces, though you probably can't write one down explicitly, because the axiom of choice is required (?) to prove the existence of ONBs for general Hilbert spaces. – Faisal Apr 12 2011 at 5:03

closed as not a real question by Bill Johnson, Zev Chonoles, Mariano Suárez-Alvarez, Anton Petrunin, Yemon Choi Apr 12 2011 at 0:57

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