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In every Hilbert space $H \neq \{0 \}$, there exists a total orthonormal set.

I think I've understood the proof given by Erwin Kreyszig in Introductory Functional Analysis With Applications.

The following questions arise in my mind:

Is there a total orthonormal set in \emph{every} inner product space?

Is there a total proper subset in every normed (or Banach) space?

A subset $M$ of a normed space $X$ is said to be total in $X$ if span of $M$ is dense in $X$.

If $X$ is an inner product space and if $M (\neq \emptyset) \subset X$ is total in $X$, then $$M^\perp \colon= \{ \ x \in X \ \colon \ \langle x, v \rangle = 0 \ \mbox{ for all } \ v \in M \ \} = \{0 \}.$$

If $X$ is a Hilbert space and if $M^\perp = \{0 \}$, then $M$ is total in $X$.

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  • $\begingroup$ In the Hilbert space $\{0\}$, $\varnothing$ is a total orthonormal subset. $\endgroup$ – Jonas Meyer Jun 3 '15 at 17:27
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In his answer to this question (A complete orthonormal system contained in a dense sub-space.), GEdgar constructs a (non-separable) Hilbert space $H$ and a dense subspace $V\leq H$, such that $V$ contains no orthonormal basis for $H$.

Thus, $V$ contains no total orthonormal family for $V$, since otherwise the span of this family would be dense in $V$ and thus in $H$, so that the total orthonormal set would be an orthonormal basis for $H$.

For the more general question (total proper subset), simply take $M=X\setminus \{0\}$.

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