I am trying to understand the notion of extremally disconnected space
(in other words Stonean space),
i.e. a space in which any open set has an open closure.
Could you help me and give (reasonable) examples of such spaces except discrete space?
Thanks in advance.
Best Answer
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:
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