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I would like to know:

Why is the Stone-Čech compactification of a discrete space the set of ultrafilters on that space?

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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:

  1. $ F(X) $ is a compact Hausdorff space;
  2. $ \tilde{f} $ extends $ f $ (where $ X $ is identified with its image under $ \iota $);
  3. $ \tilde{f} $ is continuous;
  4. $ \tilde{f} $ the unique extension satisfying (2) and (3).
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That is just one of many ways to define the Čech-Stone compactification $\beta D$ of a discrete space $D$. The Čech-Stone compactification of a Tikhonov space is characterized by the fact that it is the maximal compactification of that space, or by the fact that it has a certain universal property. Any construction that yields the maximal compactification of a space $X$, or that yields a compactification with that universal property, is a construction of $\beta X$, the Čech-Stone compactification of $X$. However, in the case of a discrete space $D$, the ultrafilter construction is in many ways the simplest.

And the answer to your question is that $\beta D$ is the set of ultrafilter on $D$ (equipped with the appropriate topology) precisely because the space that you get from that construction can be shown to have the universal property that characterizes $\beta D$.

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