Various typo fixes.

src/dg/complex/runge.tex

@@ -135,12 +135,12 @@
 
     where the convergence is uniform on $K$, so there exists a neighbourhood of $\infty$ that is contained in $A$. 
     
-    Since $P \cup \bracs{\infty} \subset A$, and $A$ is an open and closed subset of $C_\infty \setminus K$, $A$ is a union of connected components of $C_\infty \setminus K$ by \autoref{lemma:union-connected-components}. Given that $P$ intersects every connected component of $C_\infty \setminus K$, the lemma holds for all $a \in \complex \setminus K$. 
+    Since $P \cup \bracs{\infty} \subset A$, and $A$ is an open and closed subset of $C_\infty \setminus K$, $A$ is a union of connected components of $\complex_\infty \setminus K$ by \autoref{lemma:union-connected-components}. Given that $P$ intersects every connected component of $\complex_\infty \setminus K$, the lemma holds for all $a \in \complex \setminus K$. 
 \end{proof}
 
 \begin{theorem}[Runge]
 \label{theorem:runge}
-    Let $K \subset \complex$ be compact and $P \subset C_\infty \setminus K$ such that $P$ intersects every connected component of $C_\infty \setminus K$, then $\complex(x) \cap H(C_\infty \setminus P; \complex)$ is dense in $H(K; \complex)$ with respect to the uniform topology. 
+    Let $K \subset \complex$ be compact and $P \subset \complex_\infty \setminus K$ such that $P$ intersects every connected component of $\complex_\infty \setminus K$, then $\complex(x) \cap H(\complex_\infty \setminus P; \complex)$ is dense in $H(K; \complex)$ with respect to the uniform topology. 
 \end{theorem}
 \begin{proof}
     Let $U \in \cn_\complex(K)$, $f \in H(U; \complex)$, and $\eps > 0$. By \autoref{proposition:existence-curves}, there exists closed rectifiable curves $\seqf{\gamma_j}$ in $U \setminus V$ such that for each $z_0 \in K$, 

src/dg/derivative/taylor.tex

@@ -141,7 +141,7 @@
     [r(h)]_F \le \frac{1}{(n+1)!} \cdot \sup_{t \in [0, 1]}[D^{n+1}_\sigma f(x_0 + th)(h^{n+1})]
     \]
 \end{theorem}
-\begin{proof}{Proof, {{\cite[Section XIII.6]{Lang}}}. }
+\begin{proof}[Proof, {{\cite[Section XIII.6]{Lang}}}. ]
     Firstly, if $n = 0$, then by the \hyperref[Fundamental Theorem of Calculus]{theorem:ftc-riemann}, 
     \[
     f(x_0 + h) - f(x_0) = \int_0^1 D_\sigma f(x_0 + th)(h) dt

src/fa/norm/normed.tex

@@ -80,7 +80,7 @@
     then $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)} < \infty$.
 \end{theorem}
 \begin{proof}
-    By the \autoref{theorem:banach-steinhaus} theorem, $\mathcal{T}$ is equicontinuous. Therefore there exists $r > 0$ such that $\bigcup_{T \in \mathcal{T}}T[B_E(0, r)] \subset B_F(0, 1)$. In which case, $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)} \le r^{-1}$. 
+    By the \hyperref[Banach-Steinhaus Theorem]{theorem:banach-steinhaus}, $\mathcal{T}$ is equicontinuous. Therefore there exists $r > 0$ such that $\bigcup_{T \in \mathcal{T}}T[B_E(0, r)] \subset B_F(0, 1)$. In which case, $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)} \le r^{-1}$. 
 \end{proof}