Added uniform structures for completely regular spaces. Added calculus lemma.

src/topology/main/compactify.tex

@@ -0,0 +1,75 @@
+\section{Compactifications}
+\label{section:compactifications}
+
+\begin{definition}[Compactification]
+\label{definition:compactification}
+    Let $X$ be a topological space, then a \textbf{compactification} of $X$ is a pair $(Y, f)$ where
+    \begin{enumerate}
+        \item $Y$ is a compact Hausdorff space. 
+        \item $f \in C(X; Y)$ is an embedding.
+        \item $f(X)$ is dense in $Y$. 
+    \end{enumerate}
+\end{definition}
+
+\begin{definition}[Stone-Čech Compactification]
+\label{definition:stone-cech}
+    Let $X$ be a completely regular space, then there exists a pair $(\beta X, e)$ such that:
+    \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}
+
+    Moreover, if $(\beta X, e)$ is \textit{any} pair that satisfies (1) and (U1), then
+    \begin{enumerate}
+        \item[(U2)] For any pair $(Y, \varphi)$ satisfying (1), there exists a unique $\beta \varphi \in C(\beta X; Y)$ such that the following diagram commutes:
+        \[
+        \xymatrix{
+        \beta X \ar@{->}[r]^{\beta \varphi} & Y \\
+        X \ar@{->}[u]^{e} \ar@{->}[ru]_{\varphi} & 
+        }
+        \]
+    \end{enumerate}
+
+    The pair $(\beta X, e)$ is the \textbf{Stone-Čech compactification} of $X$. 
+\end{definition}
+\begin{proof}
+    Let $e: X \to [0, 1]^{C(X; [0, 1])}$ be the embedding of $X$ into $[0, 1]^{C(X; [0, 1])}$ associated with $C(X; [0, 1])$ in \autoref{definition:embedding-in-cube}, and $\beta X = \ol{e(X)}$. 
+
+    (1): By \autoref{theorem:tychonoff} and \autoref{proposition:product-hausdorff}, $[0, 1]^{C(X; [0, 1])}$ is a compact Hausdorff space. By definition, $e(X)$ is dense in $\beta X$. 
+
+    (U1): For each $f \in C(X; [0, 1])$, $\pi_f \in C([0, 1]^{C(X; [0, 1])}; [0, 1])$ is an extension of $f$ to $e(x)$. 
+
+    (U2): Let $(Y, \varphi)$ be a compactification of $X$. For each $f \in C(Y; [0, 1])$, by (U1), there exists a unique $\beta(f \circ \varphi) \in C(X; [0, 1])$ such that the following diagram commutes: 
+    \[
+    \xymatrix{
+    X \ar@{->}[d]_{\varphi} \ar@{->}[r]^{e} & \beta X \ar@{->}[d]^{\beta (f \circ \varphi)} \\
+    Y \ar@{->}[r]_{f} & [0, 1]
+    }
+    \]
+
+    Let $e': Y \to [0, 1]^{C(Y; [0, 1])}$ be the embedding of $Y$ into $[0, 1]^{C(Y; [0, 1])}$ associated with $C(Y; [0, 1])$, then by (U) of the \hyperref[product topology]{definition:product-topology}, there exists $\beta(e' \circ \varphi) \in C(\beta X; [0, 1])$ such that the following diagram commutes:
+    \[
+    \xymatrix{
+    X \ar@{->}[d]_{\varphi} \ar@{->}[r]^{e} & \beta X \ar@{->}[d]^{\beta (e' \circ \varphi)} \\
+    Y \ar@{->}[r]_{e'} & [0, 1]^{C(Y; [0, 1])}
+    }
+    \]
+
+    Since $Y$ is a compact Hausdorff space, $e'(Y)$ is closed by \autoref{proposition:compact-extensions} and \autoref{proposition:compact-closed}. As $e'$ is an embedding, identify $Y$ as a subspace of $[0, 1]^{C(Y; [0, 1])}$. Given that $e(X)$ is dense in $\beta X$, the the image of $\beta (e' \circ \varphi)$ lies in $Y$ by \autoref{proposition:closure-of-image}. Therefore under the identification, the following diagram commutes:
+    \[
+    \xymatrix{
+    X \ar@{->}[d]_{\varphi} \ar@{->}[r]^{e} & \beta X \ar@{->}[ld]^{\beta (e' \circ \varphi)} \\
+    Y & 
+    }
+    \] 
+    
+\end{proof}
+
+
+

src/topology/main/cube.tex

@@ -0,0 +1,63 @@
+\section{Embeddings in Cubes}
+\label{section:embeddings-in-cubes}
+
+\begin{definition}[Completely Regular]
+\label{definition:completely-regular}
+    Let $X$ be a topological space, then $X$ is \textbf{completely regular} if for any $E \subset X$ closed and $x \in X \setminus E$, there exists $f \in C(X; [0, 1])$ such that $f(x) = 1$ and $f|_E = 0$. 
+\end{definition}
+
+\begin{definition}[Separation of Points and Closed Sets]
+\label{definition:separate-points-closed-sets}
+    Let $X$ be a topological space and $\cf \subset C(X; [0, 1])$, then $\cf$ \textbf{separates points and closed sets} if for any $E \subset X$ closed and $x \in X \setminus E$, there exists $f \in \cf$ such that $f(x) \not\in \ol{f(E)}$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:completely-regular-separate}
+    Let $X$ be a $T_1$ space, then the following are equivalent:
+    \begin{enumerate}
+        \item $X$ is completely regular. 
+        \item There exists $\cf \subset C(X; [0, 1])$ that separates points and closed sets.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (2) $\Rightarrow$ (1): Let $E \subset X$ closed and $x \in X \setminus E$, then there exists $f \in \cf$ such that $x \not\in \ol{f(E)}$. By \hyperref[Urysohn's lemma]{lemma:urysohn}, there exists $\phi \in C([0, 1]; [0, 1])$ such that $\phi(f(x)) = 1$ and $\phi(f(E)) = 0$. 
+\end{proof}
+
+\begin{definition}[Embedding in Cube]
+\label{definition:embedding-in-cube}
+    Let $X$ be a topological space, $\cf \subset C(X; [0, 1])$, and
+    \[
+    e: X \to [0, 1]^\cf \quad \pi_f(e(x)) = f(x)
+    \]
+
+    then:
+    \begin{enumerate}
+        \item $e \in C(X; [0, 1]^\cf)$. 
+        \item If $\cf$ separates points, then $e$ is injective.
+        \item If $X$ is $T_1$ and $\cf$ separates points and closed sets, then $e$ is an embedding. 
+    \end{enumerate}
+
+    The mapping $e$ is the \textbf{mapping of $X$ into the cube $[0, 1]^\cf$ associated with $\cf$}. 
+\end{definition}
+\begin{proof}[Proof, {{\cite[Proposition 4.53]{Folland}}}. ]
+    (1): By (U) of the \hyperref[product topology]{definition:product-topology}. 
+
+    (3): Since $X$ is $T_1$, $e$ is injective by (2). Let $x \in X$ and $U \in \cn_X^o(x)$, then there exists $f \in \cf$ such that $f(x) \not\in \ol{f(U^c)}$. In which case, there exists $V \in \cn_{[0, 1]}^o(f(x))$ such that $V \cap f(U^c) = \emptyset$. Thus for any $y \in X$ with $\pi_f(e(y)) \in V$, $f(y) \not\in f(U^c)$, so $y \in U$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:completely-regular-uniformisable}
+    Let $X$ be a $T_1$ space, then the following are equivalent:
+    \begin{enumerate}
+        \item $X$ is completely regular.
+        \item There exists a uniformity $\fU$ on $X$ that induces the topology on $X$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): By \autoref{definition:uniform-separated}. 
+
+    (2) $\Rightarrow$ (1): By \autoref{definition:embedding-in-cube}, $X$ embeds into $[0, 1]^{C(X; [0, 1])}$, which is a uniform space. The subspace uniformity on $X$ then induces the topology on $X$. 
+\end{proof}
+
+
+

src/topology/main/index.tex

@@ -24,3 +24,5 @@
 \input{./c0.tex}
 \input{./semicontinuity.tex}
 \input{./baire.tex}
+\input{./cube.tex}
+\input{./compactify.tex}

src/topology/uniform/compact.tex

@@ -71,4 +71,12 @@
     Let $V$ be an entourage of $Y$. For each $x \in X$, let $U_x$ be an entourage of $X$ such that $(f(y), f(z)) \in V$ for all $y, z \in (U_x \circ U_x)(x)$. Since $X$ is compact, there exists $\seqf{x_j} \subset X$ such that $X = \bigcup_{j = 1}^nU_{x_j}(x_j)$. 
     
     Let $U = \bigcap_{j = 1}^n U_{x_j}$, then for any $(x, y) \in U$, there exists $1 \le j \le n$ such that $x \in U_{x_j}(x_j)$. In which case, $x, y \in (U_{x_j} \circ U_{x_j})(x_j)$, so $(f(x), f(y)) \in V$. 
-\end{proof}
+\end{proof}
+
+\begin{proposition}
+\label{proposition:compact-uniform-structure}
+    Let $X$ be a compact Hausdorff space, then there exists a unique uniformity on $X$ that induces its topology. 
+\end{proposition}
+\begin{proof}
+    By \autoref{proposition:completely-regular-uniformisable}, there exists a uniformity on $X$ that induces its topology. By \autoref{proposition:uniform-continuous-compact}, $X$ admits a unique uniformity. 
+\end{proof}

src/topology/uniform/definition.tex

@@ -270,19 +270,16 @@ V = (\bracs{x} \times V)(x) = (U \cup (\bracs{x} \times V))(x) = W(x) \in \cn(x)
         \item $X$ is T1.
         \item $X$ is Hausdorff.
         \item $X$ is regular.
+        \item $X$ is completely regular. 
         \item $\Delta = \bigcap_{U \in \fU}U$.
     \end{enumerate}
 
     If the above holds, then $X$ is \textbf{separated}.
 \end{definition}
 \begin{proof}
-    $(1) \Rightarrow (5)$: Let $x, y \in X$ with $x \ne y$. Assume without loss of generality that there exists $U(x) \in \cn(x)$ such that $y \not\in U$. In which case, $(x, y) \not\in U$ and $\Delta \supset \bigcap_{U \in \fU}U$.
+    (1) $\Rightarrow$ (6): Let $x, y \in X$ with $x \ne y$. Assume without loss of generality that there exists $U(x) \in \cn(x)$ such that $y \not\in U$. In which case, $(x, y) \not\in U$ and $\Delta \supset \bigcap_{U \in \fU}U$.
 
-    $(5) \Rightarrow (2)$: By \autoref{proposition:goodentourages}, $\ol \Delta \subset \bigcap_{U \in \fU}\ol U = \Delta$, so $\ol \Delta$ is closed. By (6) of \autoref{definition:hausdorff}, $X$ is Hausdorff.
-
-    $(1) \Rightarrow (4)$: $X$ is T1 and satisfies (2) of \autoref{definition:regular} by \autoref{proposition:uniform-neighbourhoods}, so $X$ is regular.
-
-    $(4) \Rightarrow (3) \Rightarrow (2) \Rightarrow (1)$:  (T3) $\Rightarrow $ (T2) $\Rightarrow$ (T1) $\Rightarrow$ (T0).
+    (6) $\Rightarrow$ (5): Let $E \subset X$ be closed and $x \in X \setminus E$. Since $\Delta = \bigcap_{U \in \fU}U$, there exists $U \in \fU$ such that $U(x) \subset E^c$. By \autoref{theorem:uniform-pseudometric}, there exists a pseudometric $d: X \times X \to [0, \infty)$ such that $d(x, E) > 0$. Thus the function $y \mapsto d(x, y)$ is a continuous function that separates $x$ and $E$. 
 \end{proof}