Typo fixes.

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
     \[
     A^\circ = \bracsn{x \in E|\dpn{x, \psi}{E} \le 1 \forall \psi \in A} \subset \bracs{\phi}^\circ
     \]