Maybe its to late for You but this might be helpful for someone else
A topological space $K$ is said to be extremally disconnected if $ \bar U$ is open for every open $U ⊆ K$; equivalently, in $K$ the following separation axiom is satisfied:
If $U, V ⊆ K$ are open and $U ∩ V = \emptyset $ then $\bar U ∩ \bar V = ∅$.
This separation is extremally (sic!) storng, and unusual, I can't think of any natural examples that do not come from Stone spaces of boolean algebras.
By an ultrafilter $ \mathcal F$ in boolean algebra $ \mathfrak A = (A,0,1, -, \vee, \wedge)$ we call a subset of $A$ such that:
- $1 \in \mathcal F $ and $ 0 \not \in \mathcal F$
- $a, b \in \mathcal F $ then $a \wedge b \in \mathcal F$
- $a \in \mathcal F$, $b \geq a$ then $ b \in \mathcal F$
- for every $a \in A$ either $a \in \mathcal F$ or $ -a \in \mathcal F$
Assuming AC every boolean algebra has an ultrafilter. A Stone space $K$ of boolean algebra $\mathfrak A$ is a set of all ultrafilters of $\mathfrak A$ called $\mathrm{Ult } \ \mathfrak A $ with topology generated by base set of a form $ \hat a$, $ a \in A$ where:
$$ \hat a = \{ \mathcal F \in \mathrm{Ult} \ \mathfrak A : \ a \in \mathcal F \}$$
Theorem The space $K$ is a compact, totally disconnected topological space. Observe that it from this disconnectedness and compactness with have zero-dimensionality.
The mapping $a \mapsto \hat a$ is boolean isomorphism between $ \mathfrak A$ and algebra of a clopen sets in $K$.
Theorem The algebra $\mathfrak A$ is complete (every subfamily has a least upperbound) iff $K$ is extremally disconnected.
Example Consider algebra $ \mathfrak A = \mathcal P (\mathbb N )$ with natural operations. It is complete, a subset $\mathcal S$ has an upper bound $ \bigcup \mathcal S$. Hence $K_{\mathfrak A} = \beta \mathbb N$ (wiki) is extremally disconnected.
Example Let $\mathfrak B$ be the the measure algebra, i.e. the family of all equivalence classes $[B], B ∈ Bor[0, 1]$ where:
$$[B] = \{ A ∈ Bor[0, 1] : λ(A \Delta B) = 0\}$$
This is again complete, by taking the sum and assuring that it is countable one by some approximation theorems. So its Stone space $K_{\mathfrak B}$ is extremally disconnected.
Remark One can check that $\ell^\infty$ is isometric to $C ( \beta \mathbb N)$ and $L^\infty [0, 1]$ is isometric to $C(K_{\mathfrak B})$.
Someone objected what this topic has to do with operator theory. But the notion of extremal disconnectedness arises naturally in functional analysis:
Theorem Space K is an extremally disconnected compact space iff $C(K)$
is an order complete Banach lattice, i.e. for every bounded from above family $Φ ⊆ C(K)$ there is an least upper bound in $C(K)$.
A Banach space $X$ is 1-injective if for every Banach space $F$ and its subspace $E$ every bounded operator $T : E → X$ has an extension to a bounded operator $ \bar T : F → X$ such that $ \| \bar T \| = \| T \|$.
Example Space $\mathbb R$ is 1-injective, which is exactly Hahn-Banach theorem.
Example Space $\ell^\infty$ is 1-injective by using Hahn-Banach for every coordinate.
Theorem [Kelly] A Banach space $X$ is 1-injective iff it is isometric to $C(K)$ space with $K$ extremally disconnected.
Corollary Space $L^\infty [0,1]$ is 1-injective.
Notion of 1-injectivity is important in isomorphic theory of seperable Banach spaces, and extremal disconectedness gives us some characterisation of this property.
Reference: Lecture notes on combinatorics in Banach spaces