Fix typo

Signed-off-by: zeramorphic <[email protected]>

iii/forcing/04_forcing_and_independence.tex

@@ -561,7 +561,7 @@ \subsection{The mixing lemma}
 For example, in \( p \Vdash \exists x.\, (\dot a \in x \wedge \dot b \in x) \), we can find a name \( \dot x = \operatorname{op}(\dot a, \dot b) \) without needing to know \( G \).
 Informally, the mixing lemma says that this is always the case, as long as \( M \) has \( \mathsf{AC} \).
 \begin{theorem}[the mixing lemma]
-    (\mathsf{ZFC})
+    (\( \mathsf{ZFC} \))
     Suppose that \( (p \Vdash \exists x.\, \varphi(x))^M \).
     Then there is a name \( \dot x \in M^{\mathbb P} \) such that \( (p \Vdash \varphi(\dot x))^M \).
 \end{theorem}

iii/lc/02_measurable_cardinals.tex

@@ -545,7 +545,7 @@ \subsection{Ultrapowers of the universe}
         (c_\alpha) < (\id_\kappa) &\leftrightarrow \qty{\gamma \mid c_\alpha(\gamma) < \id_\kappa(\gamma)} \in U \\
         &\leftrightarrow \qty{\gamma \mid \alpha < \gamma} \in U
     \end{align*}
-    But by a size argument, \( \qty{\gamma \mid \gamma \leq \alpha} \notin U \) as \( U \) is nonprincipal, so we must have \( \alpha < (id) \).
+    But by a size argument, \( \qty{\gamma \mid \gamma \leq \alpha} \notin U \) as \( U \) is nonprincipal, so we must have \( \alpha < (\id) \).
     Also,
     \begin{align*}
         (\id_\kappa) < (c_\kappa) &\leftrightarrow \qty{\gamma \mid \id_\kappa(\gamma) < c_\kappa(\gamma)} \in U \\
@@ -795,7 +795,7 @@ \subsection{The fundamental theorem on measurable cardinals}
 \end{proposition}
 This proposition provides an alternative view of reflection.
 Suppose that the ultrafilter \( U \) on \( \kappa \) is normal.
-If \( M \vDash \Phi(\kappa) \), then \( M \vDash \Phi((id)) \).
+If \( M \vDash \Phi(\kappa) \), then \( M \vDash \Phi((\id)) \).
 By \L{}o\'s' theorem,
 \[ \qty{\alpha < \kappa \mid \Phi(\id(\alpha))} \in U \]
 So \( \Phi \) reflects not only on a set of size \( \kappa \), but on an ultrafilter set.
@@ -815,7 +815,7 @@ \subsection{The fundamental theorem on measurable cardinals}
     \end{cases} \]
     Thus the set of \( \alpha \) such that \( f(\alpha) \) is an \( \alpha \)-complete nonprincipal ultrafilter on \( \alpha \) is \( C \), so in \( U \).
     Equivalently, the set of \( \alpha \) such that \( f(\alpha) \) is an \( \id(\alpha) \)-complete nonprincipal ultrafilter on \( \id(\alpha) \) is in \( U \).
-    So by \L{}o\'s' theorem, \( M \) believes that \( (f) \) is an \( (id) \)-complete nonprincipal ultrafilter on \( (id) \).
+    So by \L{}o\'s' theorem, \( M \) believes that \( (f) \) is an \( (\id) \)-complete nonprincipal ultrafilter on \( (\id) \).
     So \( (f) \) witnesses that \( \kappa \) is measurable in \( M \).
 \end{proof}
 This shows that whether a cardinal \( \kappa \) is surviving depends only on \( \mathrm{V}_{\kappa + 2} \), and is therefore a 2-stable property.