These are the completely ultrametrizable spaces.
Recall that a d:E2→[0,∞) is an ultrametric if
- d(x,y) = 0 ↔ x = y
- d(x,y) = d(y,x)
- d(x,z) ≤ max(d(x,y),d(y,z))
As usual, (E,d) is a complete ultrametric space if every Cauchy sequence converges.
Suppose E∞ is the inverse limit of the sequence En of discrete spaces, with fn:E∞→En being the limit maps. Then
d∞(x,y) = inf { 2-n : fn(x) = fn(y) }
is a complete ultrametric on E∞, which is compatible with the inverse limit topology.
Conversely, given a complete ultrametric space (E,d), the relation x ∼n y defined by d(x,y) ≤ 2-n is an equivalence relation. Let En be the quotient E/∼n, with the discrete topology. These spaces have obvious commuting maps between them, let E∞ be the inverse limit of this system. The map which sends each point of E to the sequence of its ∼n equivalence classes is a continuous map f:E→E∞. Because E is complete, this map f is a bijection. Moreover, a simple computation shows that this bijection is in fact a homeomorphism. Indeed, with d∞ defined as above, we have
d∞(f(x),f(y)) ≥ d(x,y) ≥ d∞(f(x),f(y))/2.
As Pete Clark pointed out in the comments, the above is an incomplete answer since the question does not assume that the inverse system is countable. However, the general case does admit a similar characterization in terms of uniformities. For the purposes of this answer, let us say that an ultrauniformity is a unformity with a fundamental system of entourages which consists of open (hence clopen) equivalence relations. The spaces in question are precisely the complete Hausdorff ultrauniform spaces.
Suppose E is the inverse limit of the discrete spaces Ei with limit maps fi:E→Ei. Without loss of generality, this is a directed system. Then the sets Ui = {(x,y): fi(x) = fi(y)} form a fundamental system of entourages for the topology on E, each of which is a clopen equivalence relation on E. The universal property of inverse limits guarantees that E is complete and Hausdorff. Indeed, every Cauchy filter on E defines a compatible sequence of points in the spaces Ei, which is the unique limit of this filter.
Conversely, suppose E is a complete Hausdorff ultrauniform space. If U is a fundamental entourage (so U is a clopen equivalence relation on E) then the quotient space E/U is a discrete space since the diagonal is clopen. In fact, E is the inverse limit of this directed system of quotients. It is a good exercise (for Pete's students) to show that completeness and Hausdorffness of E ensure that E satisfies the universal property of inverse limits.
No, the Stone-Cech compactification $\beta\mathbf{N}$ of $\mathbf{N}$ is extremally disconnected, but not the Stone-Cech boundary $\beta\mathbf{N}\smallsetminus\mathbf{N}$.
To see this, it is enough to find an increasing sequence $(F_n)$ of clopen subsets with no supremum (=least upper bound) in the Boolean algebra of clopen subsets, or equivalently of the Boolean algebra of subsets modulo symmetric difference finite subsets.
For this, it is more convenient to work with $\mathbf{N}^2$: then $F_n$ is just the boundary of $\mathbf{N}\times\{0,\dots,n\}$. An upper bound for this sequence is just a subset $Y$ intersecting each horizontal line in a cofinite subset. By given any such $Y$ one can remove one point in each horizontal layer and get another upper bound $Y'\subset Y$ with $Y\smallsetminus Y'$ infinite. So there is no least upper bound.
Best Answer
Here's an example, I think. Let $X=(\omega_1+1)\times(\omega_0+1)$ and let $Y$ denote the subset $\omega_1\times(\omega_0+1)$. Define a topology on $X$ by declaring every set of the form $Y\setminus C$ with $C$ countable open. So $Y$ will be an open subspace. We ensure that $Y$ has the co-countable topology by specifying local bases at the other points.
At $\langle\omega_1,n\rangle$ (with $n\in\omega_0$) the basic neighbourhoods are of the form $\{\langle\omega_1,n\rangle\}\cup(Y\setminus C)$ with $C$ countable.
At $\langle\omega_1,\omega_0\rangle$ the basic neighbourhoods are of the form $\{\{\omega_1\}\times(n,\omega_0]\cup(Y\setminus C)$ with $n\in\omega_0$ and $C$ countable.
If $A$ is a countable subset of $X$ then $A\cap Y$ is closed in $X$ and so if $A\cap Y$ is infinite then $A$ cannot form a convergent sequence under any enumeration. Hence if $A$ is an infinite convergent sequence then $A\cap Y$ must be finite and upon deleting that finite set we can assume that $A\subseteq \{\omega_1\}\times(\omega_0+1)$; the latter has the natural topology of $\omega_0+1$ so $A$ converges to $\omega_0$ (and no other point).
Finally: every nonempty open set is co-countable, and so every nonempty open set is dense and that makes $X$ extremally disconnected.