Added extremely disconnected spaces.

src/topology/main/index.tex

@@ -27,3 +27,4 @@
 \input{./cube.tex}
 \input{./compactify.tex}
 \input{./preimage.tex}
+\input{./stonean.tex}

src/topology/main/normal.tex

@@ -60,7 +60,7 @@
 
     is open. On the other hand, let $x \in X$, then $f(x) > \alpha$ if and only if there exists $q \in (\alpha, f(x)) \cap \rational$ such that $x \not\in U_q$. By (b) of (1), this is equivalent to the existence of $p \in (\alpha, f(x)) \cap \rational$ such that $x \not\in \ol{U_p}$. In which case,
     \[
-    f^{-1}((b, 1]) = \bigcup_{\substack{q \in \rational \cap [0, 1] \\ q > \alpha}}\overline{U_p}^c
+    f^{-1}((\alpha, 1]) = \bigcup_{\substack{q \in \rational \cap [0, 1] \\ q > \alpha}}\overline{U_p}^c
     \]
 
     is open.

src/topology/main/stonean.tex

@@ -0,0 +1,55 @@
+\section{Extremely Disconnected Spaces}
+\label{section:extremely-disconnected-space}
+
+\begin{definition}[Extremely Disconnected]
+\label{definition:extremely-disconnected}
+    Let $X$ be a topological space, then $X$ is \textbf{extremely disconnected} if for every $U \subset X$ open, $\ol U$ is also open. 
+\end{definition}
+
+
+\begin{theorem}[Stone-Nakano]
+\label{theorem:stone-nakano-extremely-disconnected}
+    Let $X$ be a topological space. If $X$ is extremely disconnected, then $C(X; \real)$ is order complete.  Conversely, if $X$ is ??? and $C(X; \real)$ is order complete, then $X$ is extremely disconnected. 
+\end{theorem}
+\begin{proof}
+    ($\Rightarrow$): Suppose that $X$ is extremely disconnected. Let $\cf \subset C(X; \real)$ and $F \in C(X; \real)$ be an upper bound of $\cf$. For each $q \in \real$, let
+    \[
+    U(q) = \overline{\bigcup_{f \in \cf}\bracs{f > q}} \subset \overline{\bracs{F > q}}
+    \]
+
+    then $U(q)$ is open. Now, define
+    \[
+    g: X \to [-\infty, \infty] \quad x \mapsto \sup\bracs{q \in \real|x \in U(q)}
+    \]
+
+    For each $x \in X$, $g(x) \ge \sup_{f \in \cf}f(x)$, so $g(x) > -\infty$. On the other hand, for any $q > F(x)$, $x\not\in \overline{\bracs{F > q}}$, so $g(x) < \infty$. Thus $g$ is defined as a real-valued function. 
+
+    Let $p \in \real$, then $\bracs{g > p} = \bigcup_{q \in \real, q > p}U(q)$ is open. On the other hand, 
+    \[
+    \bracs{g < p} = \bigcup_{\substack{q \in \real \\ q < p}}\overline{U(q)}^c = \bigcup_{\substack{q \in \real \\ q < p}}U(q)^c
+    \]
+
+    which is also open, so $g$ is continuous. 
+
+    Now, for each $p \in \real$, since $F$ is continuous, 
+    \begin{align*}
+        \bracs{g > p} &= \bigcup_{\substack{q \in \real \\ q > p}}U(q) = \bigcup_{\substack{q \in \real \\ q > p}}\overline{\bigcup_{f \in \cf}\bracs{f > q}} \\
+        &\subset \bigcup_{\substack{q \in \real \\ q > p}}\overline{\bracs{F > q}} \subset \bigcup_{\substack{q \in \real \\ q > p}}\bracs{F \ge q} = \bracs{F > p}
+    \end{align*}
+
+    so $g \le F$. As this holds for all upper bounds of $\cf$, $g$ is the supremum of $\cf$ in $C(X; \real)$. 
+    
+
+    ($\Leftarrow$, \cite[Theorem II.9.6]{Zhu}): Suppose that $X$ is completely regular and $C(X; \real)$ is order complete. Let $U \subset X$ be open and
+    \[
+    \cf = \bracs{f \in C(X; [0, 1])| 0 \le f \le \one_U}
+    \]
+
+    and $F$ be the supremum of $\cf$ in $C(X; \real)$. Since $X$ is completely regular, for each $x \in U$, there exists $f \in C(X; [0, 1])$ with $f(x) = 1$ and $f|_{U^c} = 0$. Thus $f \in \cf$ and $\one_{\bracs{x}} \le f \le F$. As this holds for all $x \in U$, $F \ge \one_U$.
+
+    On the other hand, for any $x \in \ol{U}^c$, there exists $g \in C(X;[0, 1])$ with $g(x) = 0$ and $g|_{\ol{U}} = 1$. In this case, $g \ge f$ for all $f \in \cf$, so $g \ge F$ as well. This yields that $F \le g \le \one_{X \setminus \bracs{x}}$. As this holds for all $x \in X \setminus \ol{U}$, $F \le \one_{\ol{U}}$. 
+
+    Given that $\one_U \le F \le \one_{\ol{U}}$ and $F$ is continuous, $\bracs{F = 1} \supset \ol{U}$, so $F = \one_{\ol{U}}$ is continuous. Hence $\ol{U} = F^{-1}(1)$ must be open. 
+\end{proof}
+
+