This post is closely related to [this one](https://mathoverflow.net/questions/24440/must-we-close-weakly-to-apply-the-spectral-theorem). (In fact I copied some of its content.)


Let $H$ be an infinite dimensional separable complex Hilbert space. All $C^{\star}$-subalgebras of $B(H)$ are assumed to be non-degenerate.
The spectral projections of a self-adjoint element $T$ of $B(H)$ lie in the weakly closed algebra generated by $T$. In the early 1970s Pedersen [proved](http://blms.oxfordjournals.org/cgi/reprint/4/2/171) that if a $C^{\star}$-subalgebra $A$ of $B(H)$ contains all of the spectral projections of each of its self-adjoint elements, then $A$ is weakly closed, i.e. $A=A''$. Now, concerning the consequences of this result, Pedersen says in his book:


"For any $C^\star$-subalgebra $A$ of $B(H)$ define $a(A)$ as the smallest $C^{\star}$-subalgebra of $B(H)$ containing all spectral projections of each self-adjoint element in $A$. [...] Note that [...] a transfinite (but countable) application of the operation $a$ will produce $A''$."



> 
> **Question 1:** What is the reasoning here? 
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I see that $\omega\_1$ applications of the operation $a$ produce $A''$ and also that each element of $A''$ appears at the $\alpha$-th application for some $\alpha < \omega\_1$ (This is due to the fact that the closure in the norm-topology is a sequential closure and hence, only countably many elements play a role.); but why is $a^{\alpha}(A)=A''$ for some $\alpha < \omega\_1$?


Maybe there is something deeper behind:



> 
> **Question 2:** Is there an $\omega\_1$-chain of unital subalgebras of $B(H)$?
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Here $\omega\_1$-chain means an $\omega\_1$-index family $(A\_{\alpha})\_{\alpha< \omega\_1}$ of subalgebras of $B(H)$ such that for all $\beta < \omega\_1$ we have 
$$\overline{\cup\_{\alpha< \beta}A\_{\alpha}}^{\|.\|} \subsetneq A\_{\beta}.$$



Transfinite induction, a theorem of Pedersen, and chains of subalgebras of $B(H)$
The answer to Question 2 is yes; such a chain can be found using only diagonal operators with respect to some fixed basis $(e\_n)\_{n\in{\mathbb N}}$ for $H$. As a starting point, you can find an $\omega\_1$-sequence of subsets $S\_\alpha\subset{\mathbb N}$ such that if $\alpha<\beta$, $S\_\beta\setminus S\_\alpha$ is finite and $S\_\alpha\setminus S\_\beta$ is infinite (that is, the sequence $(S\_\alpha)$ is strictly decreasing "modulo finite sets"). This is easy to do because by diagonalizing, given a countable decreasing sequence of infinite subsets of ${\mathbb N}$, you can always find another infinite set which is strictly contained in all of them modulo finite sets. Now let $A\_\alpha$ denote the algebra of diagonal operators such that the eigenvalues of the eigenvectors $e\_n$ for $n\in S\_\alpha$ form a convergent sequence. These are closed because a uniform limit of convergent sequences is convergent. These are nested because if $\alpha<\beta$, then all but finitely many elements of $S\_\beta$ are contained in $S\_\alpha$ so if the $S\_\alpha$-eigenvalues converge, so do the $S\_\beta$-eigenvalues.


Here's a more conceptual explanation of this construction. The algebra of diagonal operators is naturally isomorphic to the algebra $C\_b({\mathbb N})$ of bounded continuous functions on $\mathbb N$, which is in turn naturally isomorphic to the algebra $C(\beta{\mathbb N})$ of all continuous functions on the Stone-Cech compactification of $\mathbb N$. Each set $S\_\alpha\subset{\mathbb N}$ has a closure $\overline{S\_\alpha}\subset\beta{\mathbb N}$, and the statement that the $S\_\alpha$ are decreasing modulo finite sets says exactly that the sequence of closed sets $C\_\alpha=\overline{S\_\alpha}\setminus S\_\alpha$ is actually decreasing. The subalgebra $A\_\alpha$ is then exactly the set of continuous functions on $\beta{\mathbb N}$ which are constant on the set $C\_\alpha$.


