Fix a compact Hausdorff space $K$ and think about $C(K)$ as a C\*-algebra acting on a Hilbert space $H$. Suppose that $C(K)$ is closed in $\mathcal{B}(H)$ in:


* $\sigma$-strong
* $\sigma$-strong\*


topology. Must $K$ be extremelly disconnected? 



When C(K) is closed in sigma strong topology?
Recall that if two compact spaces $K\_1$, $K\_2$ are such that $C(K\_1)\cong C(K\_2)$, then $K\_1\cong K\_2$.
The space $K$ is called the spectrum of the abelian *C*\*-algebra $C(K)$.


Since $C(K)$ is closed in the σ-strong topology, it is a von Neumann algebra (that condition is equivalent to being closed in the σ-strong\* topology).


Now, the spectrum of an abelian von Neumann algebra is indeed an extremely disconnected space.
So yes: $K$ has to be extremely disconnected. This kind of space is also called *hyperstonean* space.  
  
  





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By the way,
here's one way to visualize the hyperstonean space associated to the von Neumann algebra $L^\infty ([0,1])$:  

For every measurable partition of $[0,1]$ into finitely many subsets $$[0,1]=X\_1\cup\ldots \cup X\_n$$
(where each $X\_i$ is well defined up to measure zero sets), we form the space
$$
\overline{X\_1}\sqcup\ldots \sqcup \overline{X\_n}
$$
where $\overline{X\_i}$ is the closure of $X\_i$ (more precisely, it is the intersection of all closures of sets that are equal to $X\_i$ up to a measure zero set) and $\sqcup$ denotes disjoint union. The assignment
$$
X\_1\cup\ldots \cup X\_n \mapsto \overline{X\_1}\sqcup\ldots \sqcup \overline{X\_n}
$$ 
is a functor from the poset of measurable partitions of $[0,1]$ to the category of compact topological spaces. The hyperstonean space associated to $L^\infty ([0,1])$ is the inverse limit of that functor.


