Typo fixes.

src/fa/convex/legendre.tex

@@ -221,14 +221,9 @@
     E \times \real \setminus \ol{\text{Conv}}(\text{epi}(f)) \subset E \times \real \setminus \text{epi}(f)
     \]
     
-    To this end, let $A = \ol{\text{Conv}}(\text{epi}(f))$ and $(x, \alpha) \in E \times \real \setminus A$. By the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-2}, there exists $\phi \in F$ and $\mu \in \real$ such that
+    To this end, let $A = \ol{\text{Conv}}(\text{epi}(f))$ and $(x, \alpha) \in E \times \real \setminus A$. By the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-2}, there exists $\phi \in F$, $\mu \in \real$, and $\alpha_0 \in (\alpha, \infty)$ such that
     \[
-    \sup_{(y, \beta) \in A}\dpn{y, \phi}{\lambda} - \mu \beta < \dpn{x, \phi}{\lambda} - \mu \alpha
+    \sup_{(y, \beta) \in A}\dpn{y, \phi}{\lambda} - \mu \beta \le \dpn{x, \phi}{\lambda} - \mu \alpha_0 \le \dpn{x, \phi}{\lambda} - \mu \alpha
-    \]
-
-    Given that the inequality is strict, there exists $\alpha_0 \in (\alpha, \infty)$ such that
-    \[
-    \sup_{(y, \beta) \in A}\dpn{y, \phi}{\lambda} - \mu \beta \le \dpn{x, \phi}{\lambda} - \mu\alpha_0
     \]
 
     For any $(y, \beta) \in A$, $\beta$ may be arbitrarily large by \autoref{lemma:closed-convex-epigraph}, so $\mu \ge 0$. 

src/fa/duality/mackey.tex

@@ -52,7 +52,7 @@
     \bracsn{x \in E|\dpn{x, \psi}{E} \le \mu \forall \psi \in A} \subset \bracsn{x \in E|\dpn{x, \phi}{E} \le 1} = \bracs{\phi}^\circ
     \]
 
-    Since $\sigma$ is saturated, assume without loss of generality that $\mu = 1$ and that $A$ is convex, circled, and $\sigma(F, E)$-compact. In which case, let $A^\circ$ be the polar of $A$ with respect to $\dpn{E, E^*}{E}$, then
+    Since $\sigma$ is covering and saturated, assume without loss of generality that $\mu = 1$ and that $A$ is convex, circled, and $\sigma(F, E)$-compact with $0 \in A$. In which case, let $A^\circ$ be the polar of $A$ with respect to $\dpn{E, E^*}{E}$, then
     \[
     A^\circ = \bracsn{x \in E|\dpn{x, \psi}{E} \le 1 \forall \psi \in A} \subset \bracs{\phi}^\circ
     \]

src/fa/duality/polar.tex

@@ -113,7 +113,7 @@
 \begin{proof}[Proof, {{\cite[IV.1.5]{SchaeferWolff}}}. ]
     By \autoref{proposition:polar-properties}, $A^{\circ \circ}$ is a $\sigma(E, F)$-closed, convex set that contains $0$. Since $A^{\circ \circ} \supset A$, it is sufficient to show that $A^{\circ\circ} \subset \ol{\conv}(A \cup \bracs{0})$. 
 
-    Let $x_0 \in E \setminus \ol{\conv}(A \cup \bracs{0})$, then by the \hypreref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-2}, there exists $\phi: E \to \real$ such that:
+    Let $x_0 \in E \setminus \ol{\conv}(A \cup \bracs{0})$, then by the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-2}, there exists $\phi: E \to \real$ such that:
     \begin{enumerate}
         \item $\phi$ is $\sigma(E, F)$-continuous.
         \item $\phi(\ol{\conv}(A \cup \bracs{0})) \subset (-\infty, 1)$ and $\phi(x_0) > 1$.