I'm currently studying the spectral theorem for bounded self-adjoint operators, and came across Lemma 2 in section VII.2 from the book "Functional Analysis" by Reed and Simon. It says:
Let $A$ be a self adjoint operator on a separable Hilbert space $H$. Then there is a direct sum decomposition $H=\bigoplus_{n=1}^N H_n$ with $N=1,2,\ldots,$ or $\infty$ so that
- $A$ leaves each $H_n$ invariant, that is $\psi\in H_n$ implies $A\psi\in H_n$
- For each $n$, there is a $\phi_n\in H_n$, which is cyclic for $A \upharpoonright H_n$ i.e. $H_n=\overline{\{f(A)\phi_n|f\in C(\sigma(A))\}}$
For the proof, the book says it is "a simple Zornication".
My question is why we need the lemma of Zorn to prove this theorem in a separable Hilbert space. For an arbitrary Hilbert space I have no trouble understanding why this is necessary, but in a separable Hilbert space, similar to the Hahn-Banach theorem, I think it is not necessary here. Most proofs I found online were similar to that of Reed and Simon, although sometimes in non-separable Hilbert space.
My proof without using the Lemma would be similar to the following:
Choose an orthonormal basis $(e_i)$ of $H$ and define $H_1$ to be the completion of $\{A^n e_1| n\in \mathbb{N}\}$. If $e_2\not \in H_1$ let $\tilde e_2$ be the projection of $e_2$ to $H_1^\perp$ and set $H_2 = \overline{\{A^n \tilde e_2| n\in \mathbb{N}\}}$. Since $\tilde e_2\perp H_1$, we have $(\tilde e_2,A^ne_1) = 0$ and then for all $n,m\in \mathbb{N}$ we have $(A^m\tilde e_2,A^n e_1) = (\tilde e_2,A^{m+n}e_1) = 0$ so $H_2\perp H_1$. If $e_2\in H_1$ continue with $e_3$. Similarly we construct spaces $H_3,H_4\ldots$ which are all orthogonal.
Do I make a mistake here or is it just more convenient to use Zorns lemma and state an unconstructive, more general proof of the theorem?