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
Suppose $X$ is extremally disconnected (this implies the following property: if $U$ and $V$ are disjoint open sets then $\overline{U}$ and $\overline{V}$ are also disjoint (and open)), Hausdorff and cwH as well.
Suppose that $(x_n)$ is a sequence in $X$ of all distinct points, such that $x_n \to x \in X$, where $x \notin \{x_n: n \in \omega\}$. Then as $X$ is cwH, and the set $\{x_n: n \in \omega\}$ is discrete (this uses that $X$ is a Hausdorff space), we can find open sets $U_n$ such that $x_n \in U_n$ and such that the $U_n$ are pairwise disjoint.
But then $U = \bigcup_{n} U_{2n}$ and $V = \bigcup_{n} U_{2n+1}$ are open and disjoint and $x \in \overline{U} \cap \overline{V} \neq \emptyset$ and this contradicts $X$ being extremally disconnected.
It follows that $X$ can only have trivial convergent sequences: ones that are eventually constant, in essence, or we could extract a sequence as above from it.
So if $X$ were also first countable or a sequential space (more generally), if $X$ were non-discrete, there would be a non-sequentially closed set $A$, and we'd have $x_n$, all from $A$, converging to some $x \notin A$, and we could apply the above argument to get a contradiction. But I don't think "$X$ cwH and e.d. implies $X$ discrete" holds in full generality. A counterexample is $\beta \omega$, the Čech-Stone compactification of the countable discrete $\omega$, which is even collectionwise normal.
Best Answer
Here is a proof for T2:
$X$ is a topological space.
Lemma. Let X be T2, $(x_n)_{n \in \mathbb N}$ a sequence in $X$ of pairwise distinct elements, converging to $x \in X, x \neq x_n$ for all $n \in \mathbb N$. Then there exists a pairwise disjoint family $(U_n)_{n \in \mathbb N}$ of open sets, such that $x_n \in U_n$ for all $n \in \mathbb N$.
PROOF. Step 1: For $n \in \mathbb N$ there is an open set $U$ containing $x_n$, such that $x_m \notin \overline{U}$ for all $m > n$. [Consider disjoint neighborhoods of $x_n$ and $x$. Then only finitely many $x_m$ are not in the latter, which can easily be taken care of by T2.]
Step 2: By induction, using step 1, it is now easy to define $(U_n)_{n \in \mathbb N}$ as required.
Now let $X$ be extremally disconnected, T2. Then each converging sequence is eventually constant:
Assume not. Then it is easy to see that by T1, there is a sequence $(x_n)_{n \in \mathbb N}$ in $X$ of pairwise distinct elements, converging to $x \in X, x \neq x_n$ for all $n \in \mathbb N$. Let $(U_n)_{n \in \mathbb N}$ be as in the lemma. Define $U := \bigcup_{n \in \mathbb N} U_{2n}$ and $V := \bigcup_{n \in \mathbb N} U_{2n+1}$. Then $U, V$ are open and disjoint, $x \in \overline{U} \cap \overline{V}$ contradicting extremal disconnectness of $X$.