# Stone-Čech compactification of a discrete space

I would like to know:

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

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).

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 $\beta D$.