$\sigma_p(T)$ is at most countable for a normal operator

This is the part (b) of Rudin's Functional Analysis Chapter 12, question 18.

Show that if $H$ is a separable Hilbert space then the point spectrum (denoted by $\sigma_p(T)$) of a normal operator $T$ is at most countable.

I have done part (a) and (c)
Part (a) is to prove that, the residual spectrum $\sigma_r(T)$ of a normal $T \in \mathcal{B}(H)$ is empty. (perhaps this result is useful)

• (a) is to show the residual spectrum is empty if $T \in B(H)$ is normal. If $– reuns Nov 21 '16 at 5:39 • I edited your question to give a correct statement of part (a). The comment of @user1952009 agrees with what is written in the text. – Disintegrating By Parts Nov 21 '16 at 6:00 1 Answer If$\lambda_1,\lambda_2$are distinct complex numbers in the point spectrum$\sigma_p(T)$, and if$e_1,e_2$are unit vectors such that$Te_j=\lambda_j e_j$, then$\langle e_1,e_2\rangle = 0$. So, if you are working in a separable space, the point spectrum$\sigma_p(T)\$ must be at most countable.

• Does your method of proof use the Axiom of Choice? Is there a proof which does not? – Jeff Rubin Jul 25 '20 at 3:35