# Spectral decomposition of compact self-adjoint operator

Consider the following fragment from Takesaki's book "Theory of operator algebra I":

I can't quite figure out rigorously why the boxed part of the proof is true. Note that I want to make sure that $$n \mapsto \xi_n$$ is injective (of course, for distinct $$n$$ the same $$\alpha_n$$ may occur).

I tried to write $$e_0 := 0$$. Then we have the convergence $$x = \sum_{n=1}^\infty x(e_n-e_{n-1})$$ in the norm-topology and $$xe_n-xe_{n-1}$$ is a linear combination of $$\alpha_n$$'s and $$t_{\xi_n, \xi_n}$$'s, so at best we can write something like $$x= \sum_{n=1}^\infty \sum_{k=1}^{z_n} \alpha_{k,n} t_{\xi_{k,n}, \xi_{k,n}}.$$ Of course, we still need to eliminate the second sum (depending on $$n$$) and somehow absorb it in the large sum and we also need to ensure that $$n \mapsto \xi_n$$ is injective in the end product. I can't get these technical details right. Any help will be greatly appreciated!

What he means is that you start with $$e_1$$ (assume that is nonzero, otherwise you start with a bigger index). Then you have an orthonormal basis $$\tag1 \xi_{1,1},\ldots,\xi_{1,m_1}$$ of $$e_1\mathfrak H$$. As $$e_2\geq e_1$$, you take an orthonormal basis of $$e_2\mathfrak H$$ formed by expading $$(1)$$ to an orthonormal basis. That is, $$\tag2 \xi_{2,1},\ldots,\xi_{2,m_2}=\xi_{1,1},\ldots,\xi_{1,m_1},\xi_{2,m_1+1},\ldots,\xi_{2,m_2}.$$ So in each step you are enlarging the orthonormal family, and when you consider all $$n$$ you denote it by $$\{\xi_n\}$$. For each $$n$$ you have $$xe_n=\sum_{k=1}^{m_n}\alpha_{n,k}\,t_{\xi_{n,k},\xi_{n,k}}.$$ The way the elements were constructed in $$(2)$$ guarantees that $$\xi_{n+r,k}=\xi_{n,k}$$ and $$\alpha_{n+r,k}=\alpha_{n,k}$$ if $$k\leq m_n$$. So, for $$n$$ big enough, the $$\xi_{n,k}$$ and the $$\alpha_{n,k}$$ do not depend on $$n$$. That, together with $$x=\lim_n xe_n$$, allow us to write $$x=\sum_k\alpha_k\,t_{\xi_k,\xi_k}.$$