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?
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:
topology. Must $K$ be extremelly disconnected? 


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 σstrong topology, it is a von Neumann algebra (that condition is equivalent to being closed in the σstrong* 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. 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. 

