small changes

inputs/lecture_09.tex

@@ -6,6 +6,7 @@
     we have that $\Sigma^0_\xi(X) \neq \Pi^0_\xi(X)$.
 \end{theorem}
 \begin{proof}
+    \gist{%
     Fix $\xi < \omega_1$.
     Towards a contradiction assume $\Sigma^0_\xi(X) = \Pi^0_\xi(X)$.
     By \autoref{thm:cantoruniversal},
@@ -19,24 +20,30 @@
     \[z \in A \iff z \in \cU_z \iff (z,z) \in \cU.\]
     But by the definition of $A$,
     we have $z \in A \iff (z,z) \not\in \cU \lightning$.
+    }{%
+        Let $\cU$ be $X$-universal for $\Sigma^0_\xi(X)$.
+        Consider $\{y \in X : (y,y) \not\in \cU\} \in \Pi^0_\xi(X) \setminus \Sigma^0_\xi(X)$.
+    }
 \end{proof}
 
 
 \begin{definition}
     Let $X$ be a Polish space.
     A set $A \subseteq X$
-    is called \vocab{analytic} 
+    is called \vocab{analytic}
     iff
     \[
     \exists Y \text{ Polish}.~\exists  B \in \cB(Y).~
     \exists \underbrace{f\colon Y \to X}_{\text{continuous}}.~
     f(B) = A.
-    \] 
+    \]
 \end{definition}
+\gist{%
 Trivially, every Borel set is analytic.
-We will see that  not every analytic set is Borel.
+We will see that not every analytic set is Borel.
+}{}
 \begin{remark}
-    In the definition we can replace the assertion that 
+    In the definition we can replace the assertion that
     $f$ is continuous
     by the weaker assertion of $f$ being Borel.
     \todo{Copy exercise from sheet 5}
@@ -50,7 +57,7 @@ We will see that  not every analytic set is Borel.
     Then the following are equivalent:
     \begin{enumerate}[(i)]
         \item $A$ is analytic.
-        \item There exists a Polish space $Y$ 
+        \item There exists a Polish space $Y$
             and $f\colon Y \to X$
             continuous\footnote{or Borel}
             such that $A = f(Y)$.
@@ -64,6 +71,7 @@ We will see that  not every analytic set is Borel.
     \end{enumerate}
 \end{theorem}
 \begin{proof}
+    \gist{%
     To show (i) $\implies$ (ii):
     take $B \in \cB(Y')$
     and $f\colon Y' \to X$
@@ -73,11 +81,16 @@ We will see that  not every analytic set is Borel.
     such that $B$ is clopen with respect to the new topology.
     Then let $g = f\defon{B}$
     and $Y = (B, \cT\defon{B})$.
+    }{(i) $\implies$ (ii):
+        Clopenize the Borel set, then restrict.
+    }
 
     (ii) $\implies$ (iii):
     Any Polish space is the continuous image of $\cN$.
+    \gist{%
     Let $g_1: \cN \to  Y$
     and $h \coloneqq g \circ g_1$.
+    }{}
 
     (iii) $\implies$ (iv):
     Let $h\colon \cN \to X$ with $h(\cN) = A$.