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

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}