Added Hausdorff characterisation for uniform spaces.

src/topology/uniform/metric.tex

@@ -1,16 +1,21 @@
 \section{Pseudometrics}
 \label{section:pseudometric}
 
-The axioms of uniform spaces strongly resembles working in a metric space. In fact, any uniform space may arise from a family of uniformly continuous pseudometrics. This allows understanding uniform spaces in a more familiar languagge.
+The axioms of uniform spaces strongly resembles working in a metric space. In fact, any uniform space may arise from a family of uniformly continuous pseudometrics. This allows understanding uniform spaces in a more familiar language.
 
 \begin{definition}[Pseudometric]
 \label{definition:pseudometric}
-    Let $X$ be a set, a \textbf{pseudometric} is a mapping $d: X \times X \to [0, \infty)$ such that:
+    Let $X$ be a set, and $d: X \times X \to [0, \infty)$, then $d$ is a \textbf{pseudometric} on $X$ if
     \begin{enumerate}
         \item[(PM1)] For any $x \in X$, $d(x, x) = 0$.
         \item[(PM2)] For any $x, y \in X$, $d(x, y) = d(y, x)$.
         \item[(PM3)] For any $x, y, z \in X$, $d(x, z) \le d(x, y) + d(y, z)$.
     \end{enumerate}
+    If $d$ satisfies the above and
+    \begin{enumerate}
+        \item[(M)] For any $x, y \in X$ with $x \ne y$, $d(x, y)$
+    \end{enumerate}
+    then $d$ is a \textbf{metric}.
 \end{definition}
 
 \begin{lemma}
@@ -141,7 +146,7 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
 \end{remark}
 
 \begin{remark}
-	It may be tempting to construst the level sets of the pseudometric on the dyadic rational numbers by composing these sets, then proceed to construct the pseudometric as in Urysohn's lemma. However, this approach has a major shortcoming in that the composition of symmetric entourages are not necessarily symmetric. As such, it is difficult to construct symmetric level sets for the desired pseudometric.
+	It may be tempting to construct the level sets of the pseudometric on the dyadic rational numbers by composing these sets, then proceed to construct the pseudometric as in Urysohn's lemma. However, this approach has a major shortcoming in that the composition of symmetric entourages are not necessarily symmetric. As such, it is difficult to construct symmetric level sets for the desired pseudometric.
 \end{remark}
 
 
@@ -159,6 +164,30 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
     By \ref{definition:pseudometric-uniformity}, $d$ is a uniformly continuous pseudometric on $X$. Since $\bracs{(x, y) \in X \times X|d(x, y) < 1/4} \subset U_1$, $U_1 \in \mathfrak{V}$. Therefore $\fU = \mathfrak{V}$.
 \end{proof}
 
+\begin{proposition}
+\label{proposition:uniform-pseudometric-hausdorff}
+    Let $(X, \fU)$ be a uniform space and $\seqi{d}$ be a family of pseudometrics that induces the topology on $X$, then the following are equivalent:
+    \begin{enumerate}
+        \item $X$ is separated.
+        \item For any $x, y \in X$ with $x \ne y$, there exists $i \in I$ such that $d_i(x, y) > 0$.
+        \item For any $x, y \in X$ with $x \ne y$, there exists a uniformly continuous pseudometric $d$ on $X$ such that $d(x, y) > 0$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): By assumption, there exists $U \in \fU$ such that $(x, y) \not\in U$, so there exists $J \subset I$ finite and $r > 0$ such that
+    \[
+    \bracs{(x', y') \in X \times X| d_j(x', y') < r \forall j \in J} \subset U
+    \]
+    In which case, there must exist $j \in J$ such that $d_j(x, y) \ge r > 0$.
+
+    (3) $\Rightarrow$ (1): Let $x, y \in X$ with $x \ne y$ and $d$ be a continuous pseudometric on $X$ such that $r = d(x, y) > 0$. By \ref{theorem:uniform-pseudometric},
+    \[
+    \bracs{(x', y') \in X \times X| d(x', y') < r} \in \fU
+    \]
+    Therefore $\bigcup_{U \in \fU}U = \Delta$, and $X$ is separated by \ref{definition:uniform-separated}.
+\end{proof}
+
+
 \begin{definition}[Equivalent Pseudometrics]
 \label{definition:equivalent-pseudometrics}
     Let $X$ be a set and $\seqi{d}, \seqj{d}$ be pseudometrics on $X$, then $\seqi{d}$ and $\seqj{d}$ are \textbf{equivalent} if their induced uniformities coincide.