[Math] Showing the Universal Property of Stone-Čech Compactification Holds Using Ultrafilters

compactnessfiltersgeneral-topology

Let $X$ be a topological space, and $F$ an ultrafilter on $X$. We say that $x$ is an $F$-limit if every neighborhood of $x$ is in $F$. If $f: X \rightarrow Y$ is a map, then $f_{\ast}F = \{B \subseteq Y : f^{-1}(B) \in F\}$ is an ultrafilter on $Y$.

If $X$ is compact and Hausdorff, it's pretty straightforward to prove that the $F$-limit exists (compactness) for every ultrafilter $F$ and is unique (Hausdorff), so we'll denote it by $\lim(F)$. If the $f$ above is continuous, then it is not difficult to show if $x = \lim(F)$ then $f(x) = \lim(f_{\ast}F)$

Now, let $\beta X$ be the set of ultrafilters on $X$ with the topology induced by the basis consisting of the sets $\{ F \in U(X) : A \in F\}$ for each subset $A \subseteq X$. Showing $\beta X$ is compact and Hausdorff is pretty straightforward.

My question is, how do we show $\beta X$ is the Stone-Čech compactification of $X$ (with the discrete topology on $X$) with respect to the map $i: X \rightarrow \beta X$? Here $i$ sends an element $a$ to the principal ultrafilter with minimal element $\{a\}$. If $K$ is compact and Hausdorff with $f: X \rightarrow K$ continuous, I'm guessing we define $\bar{f}: \beta X \rightarrow K$ by $\bar{f}(F) = f(\lim(F))$, which is also equal to $\lim(f_{\ast}F)$ by the above.

I have already shown diagram commutativity and uniqueness. But why is $\bar{f}$ continuous? One way to show this would be to show that the map $\lim: \beta X \rightarrow X$ is continuous, since $\bar{f} = f \circ \lim$. I have been looking at several sources on the internet and they seem to give contradictory explanations or approaches, and I cannot see the connection between them.

Best Answer

Since $X$ has the discrete topology, $\lim(F)$ doesn’t exist unless $F$ is a principal (fixed) ultrafilter. However, $\lim(f_*F)$ does exist, since $f_*F$ is an ultrafilter on the compact space $K$, and you do indeed want to define $\bar f(F)=\lim(f_*F)$. Let $U$ be an open nbhd of $\bar f(F)$ in $K$; then $U\in f_*F$, so $f^{-1}[U]\in F$. Let $\mathscr{U}=\{G\in\beta X:f^{-1}[U]\in G\}$; $\mathscr{U}$ is an open nbhd of $F$ in $\beta X$, and it only remains to show that $\bar f[\mathscr{U}]\subseteq U$, i.e., that if $f^{-1}[U]\in G\in\beta X$, then $\bar f(G)\in U$. But $\bar f(G)\in U$ iff $U\in f_*G$ iff $f^{-1}[U]\in G$, so this is clear.