Fix typos

iii/forcing/04_forcing_and_independence.tex

@@ -12,8 +12,8 @@ \subsection{Independence of the constructible universe}
 
     We show that \( M[G] \vDash \mathrm{V} \neq \mathrm{L} \).
     Therefore,
-    \[ \mathrm{L}_{\Ord \cap M} = \mathrm{L}^M \subseteq M \subsetneq M[G] \]
-    By the generic model theorem, \( \mathrm{Ord} \cap M = \mathrm{Ord} \cap M[G] \), so \( M[G] \neq \mathrm{L}_{\Ord \cap M[G]} = \mathrm{L}^{M[G]} \).
+    \[ \mathrm{L}_{\mathrm{Ord} \cap M} = \mathrm{L}^M \subseteq M \subsetneq M[G] \]
+    By the generic model theorem, \( \mathrm{Ord} \cap M = \mathrm{Ord} \cap M[G] \), so \( M[G] \neq \mathrm{L}_{\mathrm{Ord} \cap M[G]} = \mathrm{L}^{M[G]} \).
     In particular, we have \( (\mathrm{V} \neq \mathrm{L})^{M[G]} \).
 \end{proof}
 We will now discuss how to remove the assumption that we have a countable transitive model of \( \mathsf{ZFC} \).

iii/lc/02_measurable_cardinals.tex

@@ -810,7 +810,7 @@ \subsection{The fundamental theorem on measurable cardinals}
     Then for each \( \alpha \in C \), one can find an \( \alpha \)-complete nonprincipal ultrafilter on \( \alpha \) called \( U_\alpha \).
     Define
     \[ f(\alpha) = \begin{cases}
-        U_\alpha & \text{if } \alpha \in C
+        U_\alpha & \text{if } \alpha \in C \\
         \varnothing & \text{if } \alpha \notin C
     \end{cases} \]
     Thus the set of \( \alpha \) such that \( f(\alpha) \) is an \( \alpha \)-complete nonprincipal ultrafilter on \( \alpha \) is \( C \), so in \( U \).