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},
@@ -19,6 +20,10 @@
     \[z \in A \iff z \in \cU_z \iff (z,z) \in \cU.\]
     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}
 
 
@@ -33,8 +38,10 @@
     f(B) = A.
     \]
 \end{definition}
+\gist{%
 Trivially, every Borel set is analytic.
 We will see that not every analytic set is Borel.
+}{}
 \begin{remark}
     In the definition we can replace the assertion that
     $f$ is continuous
@@ -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 $h(\cN) = A$.