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

Question

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.

Answer 1

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.

Answer 2

NOTE: I am working with a discrete topology on $\mathbb N$ and the Stone-Čech compactification of this space. But the same proof works for any discrete space.$\newcommand{\N}{\mathbb N}\newcommand{\Filt}[1]{\mathcal{#1}}\newcommand{\UA}[1]{{\widehat{#1}}}\newcommand{\prin}[1]{{{#1}^*}} \newcommand{\Obr}[2]{{#1}[#2]}\newcommand{\Invobr}[2]{{#1}^{-1}[#2]} \newcommand{\ol}[1]{\overline{#1}}\newcommand{\FF}{{\mathcal{F}}}\newcommand{\GG}{{\Filt{G}}} \newcommand{\Flim}{\operatorname{\mathcal{F}-lim}}\newcommand{\Glim}{\operatorname{\mathcal{G}-lim}} \newcommand{\Zobr}[3]{{#1}\colon{#2}\to{#3}}\newcommand{\sm}{\setminus} \newcommand{\emps}{\emptyset}$

Let $\beta\N$ be the set of all ultrafilters on $\N$. The set of free ultrafilters is denoted by $\beta\N^*$. The principal ultrafilter determined by $n$ will be denoted by $\prin n$. Notice that we have $$\begin{gather*} \UA {A\cap B}=\UA A \cap \UA B \\ \UA {A\cup B}=\UA A \cup \UA B \\ \UA {\N\sm A}=\beta\N\sm\UA{A} \end{gather*}$$

From the first property we see that the set $\{\UA A; A\subseteq\N\}$ is a base for a topology on $\beta\N$.

Theorem. The topology on $\beta\N$ generated by $\{\UA A; A\subseteq\N\}$ is compact and Hausdorff. This space is zero-dimensional.

The map $\Zobr e{\N}{\beta\N}$ which maps $n$ to $n^*$ is a dense embedding if $\N$ is endowed with the discrete topology. Moreover, for each sequence $\Zobr x{\N}K$ to a compact Hausdorff space $K$ there is a unique continuous extension $\Zobr{\ol x}{\beta\N}K$ such that $\ol x\circ e=x$. $($This extension can be expressed using $\FF$-limit as $\ol x(\FF)=\Flim x$.$)$

Proof can be found also e.g. in [HS,Chapter 3].

Proof.
$\beta\N$ is Hausdorff. If $\FF\ne\GG$ are two ultrafilters on $\N$, then there is a set $A$ such that $A\in\FF$ and $A\notin\GG$. Then $\UA{A}$ is a basic neighborhood of $\FF$, and $\UA{\N\sm A}$ is a basic neighborhood of $\GG$ and these two neighborhoods are disjoint.

$\beta\N$ is compact. Suppose that there is an open cover $\{\UA{A_i}; i\in I\}$ consisting of basic sets, which does not have a finite subcover. This means that for every finite set $F\subset I$ we have $\beta\N\sm\bigcup\limits_{i\in F}\UA{A_i}\ne\emps$. Since $\beta\N\sm\bigcup\limits_{i\in F}\UA{A_i}=\UA{\N\sm\bigcup\limits_{i\in F}A_i}$, we see that $\N\sm\bigcup\limits_{i\in F}A_i=\bigcap_{i\in F}(\N\sm A_i)\ne\emps$. Hence the system $\{\N\sm A_i; i\in I\}$ has finite intersection property, and therefore there exists an ultrafilter $\FF$ which contains this system. For this ultrafilter $\FF$ we have $\FF\notin\bigcup\limits_{i\in I}A_i$, which contradicts the assumption that $\{\UA{A_i}; i\in I\}$ is a cover of $\beta\N$.

$\beta\N$ is zero-dimensional. All basic set $\UA{A}$ are clopen, since $\UA{\N\sm A}=\beta\N\sm\UA{A}$.

The map $e$ is an embedding. The map $e$ is continuous, since $\N$ has discrete topology. For each $n\in\N$ we have $\Obr e{\{n\}}=\{n^*\}=\UA{\{n\}}$, therefore $e$ is open.

$\Obr e{\N}$ is dense. If $A\ne\emps$, then there is $n\in A$ and we have $n^*\in\UA A$.

Definition of $\ol{x}$. Let $K$ be a compact Hausdorff space. Then for every ultrafilter and any sequence $\Zobr x{\N}K$ there exists a unique limit $\Flim x_n$. For the proof see this question.

Continuity of $\Flim$. Let $\Zobr x{\N}K$, where $K$ is compact. For any $\FF$ we denote $\ol x(\FF)=\Flim x$. Then the map $\ol x$ is continuous. Let $\ol{x}(\FF)=\Flim x=L$. Choose a neighborhood $V$ of $L$ such that $\ol V\subseteq U$. Then the set $A=\Invobr xV$ belongs to $\FF$. We show that $\ol{x}(\GG)\in U$ for any $\GG\in\UA{A}$.

To see this, it suffices to notice that $\Glim x\in\ol{V}$ for each such ultrafilter $\GG$. Indeed, if $\Glim x=L'$ then for any neighborhood $U'$ of $L'$ we have $\Invobr x{U'}\in\GG$. Since $\Invobr xV\in\GG$, we get that $\Invobr x{U'\cap V}=\Invobr x{U'}\cap\Invobr xV\in\GG$. This implies that $U'\cap V\ne\emptyset$. Since every neighborhood of $L'$ intersects $V$, we see that $L'\in\ol{V}$.

Uniqueness of $\ol{x}$ follows from the fact that $\Obr e{\N}$ is dense in $\beta\N$. $\hspace{2cm}\square$

[HS] Hindman, Strauss: Algebra in the Stone-Čech compactification, Walter de Gruyter, Berlin-New York, 1998.

This proof is taken from my notes available here.