From 14cc41cea592da7c342275073d66d2d6f7617ccc Mon Sep 17 00:00:00 2001
From: Josia Pietsch <git@jrpie.de>
Date: Thu, 18 Jan 2024 17:17:27 +0100
Subject: [PATCH] some small changes

---
 inputs/lecture_09.tex |  1 +
 inputs/lecture_10.tex | 11 ++++++-----
 inputs/lecture_11.tex |  4 ++--
 3 files changed, 9 insertions(+), 7 deletions(-)

diff --git a/inputs/lecture_09.tex b/inputs/lecture_09.tex
index ea1c2f6..c5f67d5 100644
--- a/inputs/lecture_09.tex
+++ b/inputs/lecture_09.tex
@@ -40,6 +40,7 @@ We will see that  not every analytic set is Borel.
     $f$ is continuous
     by the weaker assertion of $f$ being Borel.
     \todo{Copy exercise from sheet 5}
+    % TODO WHY?
 \end{remark}
 
 \begin{theorem}
diff --git a/inputs/lecture_10.tex b/inputs/lecture_10.tex
index d6191dc..a8fb5ea 100644
--- a/inputs/lecture_10.tex
+++ b/inputs/lecture_10.tex
@@ -1,5 +1,6 @@
+\subsection{The Lusin Separation Theorem}
+
 \lecture{10}{2023-11-17}{}
-\todo{Start a new subsection here?}
 \begin{theorem}[\vocab{Lusin separation theorem}]
     \yalabel{Lusin Separation Theorem}{Lusin Separation}{thm:lusinseparation}
     Let $X$ be Polish and $A,B \subseteq  X$ disjoint analytic.
@@ -55,7 +56,7 @@ we need the following definition:
     such that $A \cap B = \emptyset$
     Then there are continuous surjections
     $f\colon \cN \twoheadrightarrow  A \subseteq X$
-    and $g\colon \twoheadrightarrow B \subseteq X$.
+    and $g\colon \cN \twoheadrightarrow B \subseteq X$.
 
     Write $A_s \coloneqq  f(\cN_s)$ and $B_s \coloneqq g(\cN_s)$.
     Note that $A_s = \bigcup_m A_{s\concat m}$
@@ -105,8 +106,8 @@ we need the following definition:
     Then  $f(A)$ is Borel.
 \end{theorem}
 \begin{refproof}{thm:lusinsouslin}
-    W.l.o.g.~suppose that $f$ is continuous
-    $A$ is closed,\footnote{We might even assume that $A$ is clopen, but we only need closed.}
+    W.l.o.g.~suppose that $f$ is continuous,
+    $A$ is closed\footnote{We might even assume that $A$ is clopen, but we only need closed.}
     and $X = \cN$ by \yaref{thm:bairetopolish}:
 
 
@@ -139,7 +140,7 @@ we need the following definition:
     \end{itemize}
     The second point follows from injectivity of $f$
     and the fact that $\cN_{s \concat n} \cap \cN_{s\concat n'} = \emptyset$.
-    In particular, the $(B_s)$ for a Lusin scheme.
+    In particular, the $(B_s)$ form a Lusin scheme.
 
     Note that $f(A) = \bigcup_{s \in \omega^k} B_s$
     for every $k <\omega$,
diff --git a/inputs/lecture_11.tex b/inputs/lecture_11.tex
index 60a9e6f..641f969 100644
--- a/inputs/lecture_11.tex
+++ b/inputs/lecture_11.tex
@@ -118,7 +118,7 @@
     Hence \yaref{thm:bsb} can be applied.
 \end{proof}
 
-\subsection{Projective Hierarchy}
+\subsection{The Projective Hierarchy}
 
         
 % https://q.uiver.app/#q=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
@@ -151,7 +151,7 @@
 We will not proof this in this lecture.
 
 
-\subsection{Trees} % TODO section?
+\subsection{Ill-Founded Trees}
 
 
 Recall that a \vocab{tree} on $\N$ is a subset of