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In a comment of this answer someone said that

if an inner product space satisfy that every closed subspace is complemented then it has an (countable) orthonormal basis

and this answer claim that

H has a countable orthonormal basis if and only if H has a countable dense subset (which mean H is separable).

I trying to get the first one:take $e_1$ from $H$ such that $\||e_1\||=1$,then $E_{1} = \langle e_{1} \rangle $ is closed,so $H=E_{1}\oplus E_{1}^{\bot}$ then take $e_2$ from $H\setminus \langle e_{1} \rangle$ such that $\||e_2\||=1$ ,$E_{2}=\langle e_{1},e_{2} \rangle$ is closed,so $H=E_{2}\oplus E_{2}^{\bot}$, and so on ...... but I don't know that is it a right way to get a countable orthonormal basis of $H$.

Thanks if you can give me any advice of any form.

anyon
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    You misunderstood the comment you link to – Jakobian Jun 12 '23 at 12:17
  • @Jakobian: So you mean that with the existence of orthogonal decompositions, the inner product space $H$ could has an orthonormal basis not necessarily countable? Could you tell me how to prove that? Thanks a lot. – anyon Jun 12 '23 at 12:50
  • For any Hilbert space $H$, each closed subspace $E\subseteq H$ is complemented. But $H$ doesn't necessarily have a countable orthonormal basis. The comment refers to if we can always find an orthonormal basis in a dense subspace of a Hilbert space. Turns out that the answer is yes in the separable case, and no in the non-separable case. This is what the comment answers. – Jakobian Jun 12 '23 at 12:56
  • @Jakobian: Sorry to bother you, I think I get what you mean, but what I'm actually confused about is why Eric says "$H$ no need to assume separability" under the comment of the first answer (in proving that $H$ is a Hilbert space). – anyon Jun 12 '23 at 13:27
  • Because the theorem given in the answer under the comment is true without assumption of separability. That is, if $K^\perp \neq{0}$ for any proper closed subspace $K\subseteq H$, then $H$ is a Hilbert space. – Jakobian Jun 12 '23 at 13:51

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