We can show this by checking the following universal property for the Stone-Čech compactification:
If $ X $ is a topological space, then the Stone-Čech compactification of $ X $ consists of a compact Hausdorff space $ \beta X $, together with a continuous map $ \iota: X \to \beta X $, such that given any other compact Hausdorff space $ Y $ and a continuous map $ f : X \to Y $, there exists a unique continuous map $ \tilde{f} : \beta X \to Y $ such that $ \tilde{f} \circ \iota = f $.
So let $ X $ be a discrete space. Note that an arbitrary function from $ X $ to any other space is continuous, so we don't have to worry about checking continuity for maps coming out of $ X $.
Let $ F(X) $ be the set of ultrafilters of $ X $. We claim that $ F(X) $ is the Stone-Čech compactification of $ X $. The topology on $ F(X) $ is defined as follows. For any $ S \subseteq X $, let $ {F_{S}}(X) $ be the set of ultrafilters containing $ S $. Then the sets $ {F_{S}}(X) $ are closed under finite union: $ {F_{S}}(X) \cup {F_{T}}(X) = {F_{S \cup T}}(X)$. (This is equivalent to the following property of ultrafilters: An ultrafilter contains $ S \cup T $ iff it contains either $ S $ or $ T $.) Thus the sets $ {F_{S}}(X) $ form a closed basis of a topology (closed sets are defined to be arbitrary intersections of sets of the form $ {F_{S}}(X) $).
The map $ \iota: X \to F(X) $ is defined by sending $ x $ to the principal ultrafilter associated to $ x $.
Now let $ Y $ be any compact Hausdorff space, and consider a map $ f: X \to Y $. How will we extend $ f $ to $ F(X) $? If $ U $ is an ultrafilter, then for any $ S \in U $, consider the subset of $ Y $ defined by $ f(S) \stackrel{\text{df}}{=} \{ f(s) \mid s \in S \} $. Now the properties of ultrafilters imply that for any finite number of elements $ S_{1},\ldots,S_{n} \in U $, the intersection $ \bigcap_{i} S_{i} $ is nonempty. Hence $ \bigcap_{i} f[S_{i}] $ is nonempty as well. So any finite number of $ f[S] $ have nonempty intersection. By compactness of $ Y $, the intersection of all the closures, $ \bigcap_{S \in U} \overline{f[S]} $, is nonempty.
We claim this intersection cannot have more than one element. If $ y_{1},y_{2} \in Y $ and $ y_{1} \neq y_{2} $, then the preimages (under $ f $) of $ y_{1} $ and $ y_{2} $ are disjoint subsets of $ X $, and so every ultrafilter of $ X $ contains an element meeting one of the preimages but not the other. Hence the intersection $ \bigcap_{S \in U} f[S] $ (before taking closures) does not contain more than one element. The intersection of closures $ \bigcap_{S \in U} \overline{f[S]} $ also cannot contain more than one element, since if $ y_{1} $ and $ y_{2} $ were distinct elements in $ \bigcap_{S \in U} \overline{f[S]} $, we use the Hausdorff property to find disjoint open subsets $Y_{1},Y_{2} $ of $ Y $ separating $ y_{1} $ and $ y_{2} $, and reach a similar contradiction by considering the disjoint preimages of $ Y_{1} $ and $ Y_{2} $ under $ f $.
Hence, for every ultrafilter $ U $, the intersection $ \bigcap_{S \in U} \overline{f[S]} $ contains a single element $ y \in Y $. We define $ \tilde{f}(U) \stackrel{\text{df}}{=} y $.
Now I will leave it to you to check all of the following:
- $ F(X) $ is a compact Hausdorff space;
- $ \tilde{f} $ extends $ f $ (where $ X $ is identified with its image under $ \iota $);
- $ \tilde{f} $ is continuous;
- $ \tilde{f} $ the unique extension satisfying (2) and (3).
Yes, it is consistent with ZF that all compact Hausdorff extremally disconnected spaces are finite. In Paul Howard and Jean E. Rubin's Consequences of the Axiom of Choice this is given as Form 371
FORM 371. There is an infinite, compact, Hausdorff, extremally disconnected topological space.
The reference they give for this is
- Marianne Morillon; Les compacts extrêmement discontinus sont finis!; Séminaire d’Analyse; Université Blaise Pascal, Clermont-Ferrand II; 9, Année 1993-1994, Exp. No. 11, 13 p. (1994).
I have been unable to find this paper, but the zbMath review of the paper by Alan Dow gives an overview of its contents:
A space is extremally disconnected if the closure of each open set is again open. Each compact extremally disconnected space is ‘equal’ to the Stone space of some complete Boolean algebra (and conversely). We must be careful that this is a consequence of just ZF (without the Axiom of Choice) since the topic of this paper is to show that ZF does not imply that there is an infinite compact extremally disconnected space. The approach is clever: work in a model of Definable Choice (DC) plus there is no ultrafilter on the integers. It is easy to see that DC implies that each infinite complete Boolean algebra contains the power set of the integers as a subalgebra. Therefore, from DC, if any infinite extremally disconnected space is compact, there are many ultrafilters on the integers. Finally the author shows that in a model constructed by A. Blass [Bull. Acad. Pol. Sci., Sér. Sci. Math. Astron. Phys. 25, No. 4, 329-331 (1977; Zbl 0365.02054)], in which DC, and even BC (compact spaces are Baire) fail, every compact extremally disconnected space is again finite. Of course this establishes that BCED (compact extremally disconnected spaces are Baire) does not imply BC.
Best Answer
If $O$ is open in $\beta D$ (where $D$ is discrete), let $f: D \to \{0,1\}$ be the function that sends points in $O \cap D$ to $0$ and all other points to $1$; this is a continuous function as $D$ is discrete so the universal property tells us that there is a (unique) $\beta f: \beta D \to \{0,1\}$ that extends $f$. As $O$ is open and $D$ is dense, $\overline{O \cap D} = \overline{O}$ and as $f[O \cap D]=\{0\}$ we get that $\beta f [\overline{O}]= \{0\}$ as well, and we can show that $\overline{O}=(\beta f)^{-1}[\{0\}]$, a clopen set.