# Separable Hilbert space have a countable orthonormal basis

I want to show that every an infinite-dimensional separable (contains countable dense set) Hilbert space has a countable orthonormal basis.

I know that every orthogonal set in a separable Hilbert space is countable,it is help me with the proof?

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Hint: Take a countable dense subset $Q$ and build an orthonormal basis of $\text{span}(Q)$.

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Build by Gram–Schmidt process ? –  ali baba Mar 8 '13 at 9:20
How am I define Q ?: –  ali baba Mar 8 '13 at 9:26
@ali baba: $Q$ is any countable dense subset of $\mathcal{H}$, which exists because you have assumed that $\mathcal{H}$ is separable. –  Haskell Curry Mar 8 '13 at 9:32

You can use Zorn's lemma to show that there exists a maximal orthogonal set, show that it has to be a Schauder basis. Then use what you know about orthogonal sets.

(Remark: Zorn's full power is not needed here. It's an overkill, but it's easy and convenient.)

The above hint with more details: Every Hilbert space has an orthonomal basis - using Zorn's Lemma

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If I assume that Hilbert space contains a uncountable ortonormal basis and shows that I get a conflict, is that acceptable? –  ali baba Mar 9 '13 at 21:56
It isn't enough. Bases are not only orthonormal, but they are maximal. It might be the case that every orthonormal set is countable, but there is a point which is not in its closure. You need to show that there is a set which is not only orthonormal, but its span is dense. –  Asaf Karagila Mar 9 '13 at 21:59