I need to show:
Show that for a Hilbert Space $H\neq \{0\}$ there exists a total orthonormal set.
Case 1:$\dim H<\infty $
Let $B=\{b_1,b_2,\dots ,b_n\}$ be a basis of $H$.Since $B$ is a basis so the set $\{b_1,b_2,\dots ,b_n\}$ is linearly independent and hence by Gram-Schmidt process I can create an orthonormal set out of $B$ say $B_1$ such that $\text{span B}=\text{span} B_1=H$
Since $\text{span}(B_1)=H\implies B_1$ is a total orthonormal set of $H$.
Case 2:$\dim H=\infty $.
Can I use the same technique here also like the finite case since Gram-Schmidt process is applicable in infinite dimensional case also.
The problem is it is given that for $\dim H=\infty $ we should break it into two cases like $H$ is separable and not-separable.
Will someone please tell me why is it needed and if it is so how to do it?
Any help will be appreciated.
No, you cannot use the same proof, because a Hamel basis of an infinite-dimensional Hilbert space is necessarily uncountable.
The best way to prove this is by applying Zorn's lemma to the partially ordered set of all orthonormal sets in $H$ to obtain a maximal orthonormal set, and show that this maximal orthonormal set is total.