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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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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.

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.

---

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.

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.

---

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 $$ \overline{X_1}\cup\ldots \cup \overline{X_n} $$ 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). The assignment $$ X_1\cup\ldots \cup X_n \mapsto \overline{X_1}\cup\ldots \cup \overline{X_n} $$ 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.

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.

---

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.

Recall that if two compact spaces $K_1$, $K_2$ are such that $C(K_1)\cong C(K_2)$ for, 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.

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.

---

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 $$ \overline{X_1}\cup\ldots \cup \overline{X_n} $$ 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). The assignment $$ X_1\cup\ldots \cup X_n \mapsto \overline{X_1}\cup\ldots \cup \overline{X_n} $$ 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.