Added Urysohn Metrisation theorem and compactness theorems.

src/dg/complex/derivative.tex

@@ -212,7 +212,7 @@
     \]
 \end{corollary}
 \begin{proof}
-    By \autoref[Cauchy's Integral Formula]{theorem:cauchy-formula} and \autoref{proposition:rs-bound}, 
+    By \hyperref[Cauchy's Integral Formula]{theorem:cauchy-formula} and \autoref{proposition:rs-bound}, 
     \begin{align*}
         D^kf(z_0) &= \frac{k!}{2\pi i}\int_{\omega_{z_0, r}} \frac{f(z)}{(z - z_0)^{k+1}}dz \\
         [D^kf(z_0)]_E &\le \frac{k!}{2\pi i}\int_0^{2\pi}\frac{[f(z)]_E}{|z - z_0|^{k+1}}dz \\

src/dg/complex/space.tex

@@ -11,7 +11,7 @@
     Let $E$ be a complete separated locally convex space over $\complex$ and $U \subset \complex$ be open, then $H(U; E)$ is complete. 
 \end{proposition}
 \begin{proof}
-    By \hyperref[corollary:cauchy-estimate]{Cauchy's estimate}, uniform convergence on compact sets is equivalent to uniform convergence of derivatives of all orders on compact sets. Since $U$ is locally compact, uniform limits of holomorphic functions are holomorphic by \autoref{theorem:differentiable-uniform-limit}.
+    By \hyperref[Cauchy's estimate]{corollary:cauchy-estimate}, uniform convergence on compact sets is equivalent to uniform convergence of derivatives of all orders on compact sets. Since $U$ is locally compact, uniform limits of holomorphic functions are holomorphic by \autoref{theorem:differentiable-uniform-limit}.
 \end{proof}
 
 \begin{theorem}[Montel]