Stone-Cech.

src/topology/main/compactify.tex

@@ -71,6 +71,51 @@
     
 \end{proof}
 
+
+\begin{theorem}[Stone-Čech Compactification of Discrete Spaces]
+\label{theorem:stone-cech-ultrafilter}
+    Let $X$ be a discrete topological space and $\beta X$ be the set of all ultrafilters on $X$. For each $\emptyset \ne A \subset X$, let
+    \[
+    U_A = \bracsn{\fU \in \beta X| A \in \fU}
+    \]
+
+    Equip $\beta X$ with the topology generated by $\fU := \bracsn{U_A|\emptyset \ne A \subset X}$. For each $x \in X$, let $e(x)$ be the principal ultrafilter at $x$, then
+    \begin{enumerate}
+        \item $(\beta X, e)$ is a compactification of $X$. 
+        \item[(U1)] For any $f \in C(X; [0, 1])$, there exists a unique $\beta f \in C(\beta X; [0, 1])$ such that the following diagram commutes:
+        \[
+        \xymatrix{
+        \beta X \ar@{->}[r]^{\beta f} & [0, 1] \\
+        X \ar@{->}[u]^{e} \ar@{->}[ru]_{f} & 
+        }
+        \]
+    \end{enumerate}
+
+    In other words, $(\beta X, e)$ as constructed above is the Stone-Čech compactification of $X$. 
+\end{theorem}
+\begin{proof}
+    (1): Firstly, $\beta X = U_{X}$. For any $A, B \subset X$, $U_{A \cap B} = U_A \cap U_B$. Thus $\fU$ is a base for the topology that it generates. 
+
+    Let $\mathfrak{U}, \mathfrak{V} \in \beta X$ be distinct, then there exists $A \subset X$ with $A \in \mathfrak{U}$ and $A^c \in \mathfrak{V}$, so $\mathfrak{U} \in U_A$, $\mathfrak{V} \in U_{A^c}$, and $U_A \cap U_{A^c} = \emptyset$. Thus $\beta X$ is Hausdorff. 
+
+    Let $\mathcal{U} \subset 2^{\beta X}$ be an open cover of $X$. Since $\fU$ is a base for $\beta X$, assume without loss of generality that $\mathcal{U}$ is of the form $\bracsn{U_A|A \in \mathcal{S}}$, where $\mathcal{S} \subset 2^X \setminus \bracsn{\emptyset}$. Suppose for contradiction that $\mathcal{U}$ admits no finite subcover, then for any $F \subset \mathcal{S}$ finite,
+    \[
+    \bigcap_{A \in F}U_{A^c} = \bigcap_{A \in F}U_A^c = \braks{\bigcup_{A \in F}U_A}^c \ne \emptyset
+    \]
+    
+    In particular, $\bigcap_{A \in F}A^c \ne \emptyset$. As such, $\bracsn{A^c|A \in \mathcal{S}}$ generates an ultrafilter $\fU \in \beta X$. Since $\fU \not\in U_A$ for all $A \in \mathcal{S}$, $\mathcal{U}$ does not cover $X$, which is a contradiction. Thus $\beta X$ is compact. 
+
+    Finally, since $X$ is discrete, $e$ is automatically continuous. For any $\emptyset \ne A \subset X$, $e(x) \in U_A$ for all $x \in A$, so $U_A \cap e(X) \ne \emptyset$, and $e(X)$ is dense in $\beta X$. Therefore $e(X)$ is embedded as a dense subspace of $\beta X$. 
+
+    (U1): Let $f \in C(X; [0, 1])$, and let $\beta f: \beta X \to [0, 1]$ be defined by $\beta f(\fU) = \lim_{\fU}f$. Fix $\fU \in \beta X$ and $\eps > 0$, then there exists $A \in \fU$ such that $f(A) \subset B_{[0, 1]}(\lim_{\fU}f, \eps)$. Thus for any $\fV \in U_A$, 
+    \[
+    \beta f(\fV) = \lim_{\fV}f \in \ol{f(A)} \subset B_{[0, 1]}\paren{\lim_{\fU}f, \eps}
+    \]
+
+    and $|\beta f(\fU) - \beta f(\fV)| \le \eps$. Since $A \in \fU$, $U_A \in \cn_{\beta X}(\fU)$, so $\beta f$ is continuous at $\fU$. As this holds for all $\fU \in \beta X$, $\beta f$ is continuous. 
+\end{proof}
+
+
 \begin{lemma}
 \label{lemma:lch-compactification-open}
     Let $X$ be an LCH space and $(Y, \varphi)$ be a compactification of $X$, then $\varphi(X) \subset Y$ is open.