some changes

inputs/lecture_04.tex

@@ -109,7 +109,7 @@
             (If we intersect $A$ with an open $U$,
             then $A \cap U$ is not dense in $U$).
     \end{itemize}
-    \todo{Think about this}
+    %\todo{Think about this}
 
     A set $B \subseteq  X$ is \vocab{meager} 
     (or \vocab{first category}),
@@ -139,8 +139,6 @@
 Note that open sets and meager sets have the Baire property.
 
 
-
-
 \begin{example}
     \begin{itemize}
         \item $\Q \subseteq  \R$ is $F_\sigma$.

inputs/lecture_05.tex

@@ -91,7 +91,7 @@
     Then every comeager set of $X$ is dense in $X$.
 \end{theorem}
 \todo{Proof (copy from some other lecture)}
-\begin{proposition}
+\begin{theoremdef}
     Let $X$ be a topological space.
     The following are equivalent:
     \begin{enumerate}[(i)]
@@ -101,7 +101,9 @@
         \item The intersection of countable many
             open dense sets is dense.
     \end{enumerate}
-\end{proposition}
+    In this case $X$ is called a \vocab{Baire space}.
+    \footnote{see \yaref{s5e1}}
+\end{theoremdef}
 \begin{proof}
     \todo{Proof (short)}
 
@@ -118,7 +120,6 @@
 \end{proof}
 
 
-% TODO Fubini
 \begin{notation}
     Let $X ,Y$ be topological spaces,
     $A \subseteq  X \times Y$
@@ -139,10 +140,11 @@ The following similar to Fubini,
 but for meager sets:
 
 \begin{theorem}[Kuratowski-Ulam]
-    \label{thm:kuratowskiulam}
+    \yalabel{Kuratowski-Ulam}{Kuratowski-Ulam}{thm:kuratowskiulam}
     Let $X,Y$ be second countable topological spaces.
     Let $A \subseteq X \times Y$
-    be a set with the Baire property.
+    be a set with the Baire property.%
+    \footnote{It is important that $A$ has the Baire property (cf. \yaref{s5e4}).}
 
     Then
     \begin{enumerate}[(i)]

inputs/lecture_06.tex

@@ -203,6 +203,3 @@ i.e.~$\Delta^0_1$ is the set of clopen sets.
     Consider the cofinite topology on $\omega_1$.
     Then the non-empty open sets of this are not $F_\sigma$.
 \end{example}
-
-
-

inputs/lecture_09.tex

@@ -119,7 +119,7 @@ We will see that  not every analytic set is Borel.
             Take $f^+\colon X \times  Z \to Y \times Z, f^+ = f \times \id$.
             Then
             \[f^{-1}(B) = \underbrace{\proj_X(\overbrace{(\underbrace{f^+}_{\mathclap{\text{Borel}}})^{-1}(\underbrace{B_0}_{\mathclap{\text{Borel}}})}^{\text{Borel}})}_{\text{analytic}}.\]
-        \item See \yaref{ex:7.2}.
+        \item See \yaref{s7e2}.
     \end{enumerate}
 \end{proof}
 
@@ -167,5 +167,6 @@ We will see later that $\Sigma^1_1(X) \cap \Pi^1_1(X) = \cB(X)$.
 
 \begin{remark}+
     Showing that there exist sets that don't have the Baire property
-    requires the axiom of choice.\todo{Copy exercise from sheet 5}
+    requires the axiom of choice.
+    An example of such a set is constructed in \yaref{s5e4}.
 \end{remark}

inputs/lecture_11.tex

@@ -74,7 +74,7 @@
     Then $A$ and $B$ are Borel isomorphic.
 \end{theorem}
 \begin{proof}
-    \todo{Homework}
+    Cf.~\yaref{s7e4}.
 \end{proof}
 
 \begin{theorem}[\vocab{Isomorphism Theorem}]

inputs/tutorial_06.tex

@@ -5,11 +5,6 @@
 % Sheet 5 - 18.5 / 20
 
 \nr 1
-\todo{handwritten}
-Let $B \subseteq C$ be comeager.
-Then $B = B_1 \cup B_2$,
-where $B_1$ is dense $G_\delta$
-and $B_2$ is meager.
 
 \begin{fact}
      $X$ is Baire iff every non-empty open set is non-meager.
@@ -19,6 +14,60 @@ and $B_2$ is meager.
      is Baire.
 \end{fact}
 
+
+\begin{enumerate}[(a)]
+    \item Let $X$ be a non-empty Baire space
+        and let $A \subseteq X$.
+        Show that $A$ cannot be both meager and comeager.
+
+
+        Suppose that $A \subseteq X$ is meager
+        and comeager.
+        Then $A = \bigcup_{n < \omega} U_n$
+        and  $X \setminus A = \bigcup_{n < \omega} V_n$
+        for some nwd sets $U_n, V_n$.
+        Then $X = A \cup (X \setminus A)$ is meager.
+        Let $X = \bigcup_{n < \omega} W_n$ be a union of nwd
+        sets.
+        Wlog.~the $W_n$ are closed (otherwise replace them $\overline{W_n}$)
+        Then $\emptyset = \bigcap_{n < \omega} (X \setminus W_n)$
+        is a countable intersection of open, dense sets,
+        hence dense $\lightning$
+
+    \item Let $X$ be a topological space. The relation $=^\ast$ is transitive:
+
+        Suppose $A =^\ast B$ and $B =^\ast C$.
+        Then $A \symdif C \subseteq (A \symdif B \cup B \symdif C)$
+        is contained in a meager set.
+        Since a subset of a nwd set
+        is nwd, a subset of a meager set is meager.
+        Hence $A \symdif C$ is meager, thus $A =^\ast C$.
+
+    \item Let $X$ be a topological space.
+        Let $A \subseteq X$ be a set with the Baire property,
+        then at least one of the following hold:
+        \begin{enumerate}[(i)]
+            \item $A $ is meager,
+            \item there exists $\emptyset = U \overset{\text{open}}{\subseteq} X$
+                such that $A \cap U$ is comeager in $U$.
+        \end{enumerate}
+
+        Suppose there was $A \subseteq X$ such that (i)
+        does not hold.
+        Then there exists $U \overset{\text{open}}{\subseteq} X$
+        such that $A =^\ast U$.
+        In particular, $A \symdif U$ is meager,
+        hence  $U \cap (A \symdif U) = U \setminus A$ is meager.
+        Thus $A \cap U$ is comeager in $U$.
+
+        Now suppose that $X$ is a Baire space.
+        Suppose that for $A$ (i) and (ii) hold.
+        Let $\emptyset \neq U \overset{\text{open}}{\subseteq} X$
+        be such that $A \cap U$ is comeager in $U$.
+        Since $U$ is a Baire space,
+        this contradicts (a).
+\end{enumerate}
+
 \nr 2
 
 Let $(U_i)_{i < \omega}$ be a countable base of $Y$.
@@ -41,15 +90,34 @@ so $f\defon{A}^{-1}(U_i) = V_i \cap A$ is open.
 
 
 \nr 4
+\begin{lemma}
+    There exists a non-meager subset $A \subseteq  \R^2$
+    such that no three points of $A$ are collinear.
+\end{lemma}
+This requires the use of the axiom of choice.
+\begin{proof}
+    Enumerate the continuum-many $F_\sigma$ subsets of  $\R^2$
+    as $(F_i)_{i < \fc}$.
+    We will inductively construct a sequence $(a_i)_{i < \fc}$
+    of points of $\R^2$ such that for each $i < \fc$:
+    \begin{enumerate}[(i)]
+        \item $\{a_j | j \le i\} $ is not a subset of $F_i$ and
+        \item $\{a_j | j \le i\}$ does not contain any three collinear points.
+    \end{enumerate}
 \begin{enumerate}[(a)]
-    \item $|B| = \fc$, since $B$ contains a comeager
+    \item Let $B = \{x \in \R | (F_i)_x \text{ is meager}\}$.
+        Then $B$ is comeager in $\R$ and $|B| = \fc$.
+
+        We have $|B| = \fc$, since $B$ contains a comeager
         $G_\delta$ set, $B'$:
         $B'$ is Polish,
         hence $B' =  P \cup C$
         for $P$ perfect and $C$ countable,
         and $|P| \in \{\fc, 0\}$.
         But $B'$ can't contain isolated point.
-    \item To ensure that (a) holds, it suffices to chose
+    \item We use $B$ to find a suitable point $a_i$:
+
+        To ensure that (i) holds, it suffices to chose
         $a_i \not\in F_i$.
         Since $|B| = \fc$ and $|\{a_i | j < i\}| = |i| < \fc$,
         there exists some $x \in B \setminus \{\pi_1(a_j)| j <i\}$,
@@ -77,13 +145,16 @@ so $f\defon{A}^{-1}(U_i) = V_i \cap A$ is open.
     \item $A$ is by construction not a subset of any $F_\sigma$ meager set.
         Hence it is not meager, since any meager set is contained
         in an $F_\sigma$ meager set.
-    \item For every $x \in \R$ we have that $A_x$ contains at most
+\end{enumerate}
+\end{proof}
+\begin{enumerate}
+    \item[(d)] For every $x \in \R$ we have that $A_x$ contains at most
         two points, hence it is meager.
         In particular $\{x \in \R | A_x \text{ is meager}\} = \R$
         is comeager.
         However $A$ is not meager.
         Hence $A$ can not be a set with the Baire property
-        by the theorem.
+        by \yaref{thm:kuratowskiulam}.
         In particular, the assumption of the set having
         the BP is necessary.
 

inputs/tutorial_08.tex

@@ -1,5 +1,4 @@
 \subsection{Sheet 7}
-
 \tutorial{08}{2023-12-05}{}
 % 17 / 20
 \nr 1
@@ -45,7 +44,6 @@
 \end{itemize}
 
 \nr 2
-\yalabel{Exercise}{}{ex:7.2}
 
 
 Recall \autoref{thm:analytic}.

logic.sty

@@ -153,6 +153,6 @@
 
 \newcommand\lecture[3]{\hrule{\color{darkgray}\hfill{\tiny[Lecture #1, #2]}}}
 \newcommand\tutorial[3]{\hrule{\color{darkgray}\hfill{\tiny[Tutorial #1, #2]}}}
-\newcommand\nr[1]{\subsubsection{Exercise #1}}
+\newcommand\nr[1]{\subsubsection{Exercise #1}\yalabel{Sheet \arabic{subsection}, Exercise #1}{E \arabic{subsection}.#1}{s\arabic{subsection}e#1}}
 
 \usepackage[bibfile=bibliography/references.bib, imagefile=bibliography/images.bib]{mkessler-bibliography}