Merge branch 'main' of https://git.abstractnonsen.se/josia-notes/w23-logic-3

inputs/lecture_04.tex

@@ -119,7 +119,6 @@
 
     The complement of a meager set is called
     \vocab{comeager}.
-    
 \end{definition}
 \begin{example}
     $\Q \subseteq \R$ is meager.
@@ -128,7 +127,7 @@
     Let $A, B \subseteq  X$.
     We write $A =^\ast B$ 
     iff the \vocab{symmetric difference},
-    $A \mathop{\triangle} B \coloneqq  (A\setminus B) \cup (B \setminus A)$,
+    $A \symdif B \coloneqq  (A\setminus B) \cup (B \setminus A)$,
     is meager.
 \end{notation}
 \begin{remark}
@@ -144,13 +143,14 @@ Note that open sets and meager sets have the Baire property.
 
 
 
-% \begin{example}
+\begin{example}
-%     $\Q \subseteq  \R$ is $F_\sigma$.
+    \begin{itemize}
-% 
+        \item $\Q \subseteq  \R$ is $F_\sigma$.
-%     $\R \setminus \Q \subseteq \R$ is $G_\delta$.
+        \item $\R \setminus \Q \subseteq \R$ is $G_\delta$.
-% 
+        \item $\Q \subseteq \R$ is not $G_{\delta}$.
-%     $\Q \subseteq \R$ is not $G_{\delta}$.
+            (It is dense and meager,
-%     (It is dense and meager,
+            hence it can not be $G_\delta$
-%     hence it can not be $G_\delta$,
+            by the Baire category theorem).
-%     by the Baire category theorem).
+    \end{itemize}
-% \end{example}
+
+\end{example}

inputs/lecture_05.tex

@@ -0,0 +1,271 @@
+\lecture{05}{2023-10-31}{}
+
+\begin{fact}
+    A set $A$ is nwd iff $\overline{A}$ is nwd.
+
+    If $F$ is closed then 
+    $F$ is nwd iff $X \setminus F$ is open and dense.
+
+    Any meager set $B$ is contained in
+    a meager $F_{\sigma}$-set.
+\end{fact}
+
+\begin{definition}
+    A  \vocab{$\sigma$-algebra} on a set $X$
+    is a collection of subsets of $X$
+    such that:
+    \begin{itemize}
+        \item $\emptyset, X \in \cA$,
+        \item $ A \in \cA \implies X \setminus A \in \cA$,
+        \item  $(A_i)_{i < \omega}, A_i \in \cA \implies \bigcup_{i < \omega} A_i \in \cA$.
+    \end{itemize}
+\end{definition}
+\begin{fact}
+    Since $\bigcap_{i < \omega} A_i = \left( \bigcup_{i < \omega} A_i^c \right)^c$
+    we have that $\sigma$-algebras are closed under countable intersections.
+\end{fact}
+
+\begin{theorem}
+    \label{thm:bairesigma}
+    Let $X$ be a topological space.
+    Then the collection of sets with the Baire property
+    is a $\sigma$-algebra on $X$.
+
+    It is the smallest $\sigma$-algebra
+    containing all meager and open sets.
+\end{theorem}
+\begin{refproof}{thm:bairesigma}
+    Let $\cA$ be the collection of sets with the Baire property.
+    Since open sets have the Baire property,
+    we have $\emptyset, X \in \cA$.
+
+
+    Let $A_n \in \cA$ for all $n < \omega$.
+    Take $U_n$ such that $A_n \symdif U_n$ is meager.
+    Then
+    \[
+        \left( \bigcup_{n < \omega} A_n \right) \symdif \left( \bigcup_{n < \omega} U_n \right)
+    \] 
+    is meager,\todo{small exercise}
+    hence $\bigcup_{n < \omega} A_n \in \cA$.
+
+    Let $A \in \cA$.
+    Take some open $U$ such that $U \symdif A$ is meager.
+    We have $(X \setminus U) \symdif (X \setminus A) = U \symdif A$.
+
+    \begin{claim}
+        \label{thm:bairesigma:c1}
+        If $F$ is closed,
+        then $F \setminus \inter(F)$
+        is nwd.
+        In particular, $F \symdif \inter(F)$ is nwd.
+    \end{claim}
+    \begin{refproof}{thm:bairesigma:c1}
+        \todo{TODO}
+    \end{refproof}
+
+    From the claim we get that
+    $X \setminus A =^\ast X \setminus U =^\ast \inter(X \setminus U)$.
+    Hence $X \setminus A \in \cA$.
+
+
+    It is clear that all meager sets have the Baire property.
+
+    Let $A \in \cA$. Then $A = (A \setminus U) \cup (A \cap U)$
+    for some open $U$
+    such that $A \setminus U$ is meager.
+    We have $A \cap U = U \setminus (U \setminus A)$.
+    Thus we get that $\cA$ is the minimal $\sigma$-algebra
+    containing all meager and all open sets.
+\end{refproof}
+
+%\begin{example}
+%    Nwd set of positive measure.
+%    TODO
+%    remove open intervals such that their length does not add to 0
+%       
+%\end{example}
+
+\begin{theorem}[Baire Category theorem]
+    Let $X$ be a completely metrizable space.
+    Then every comeager set of $X$ is dense in $X$.
+\end{theorem}
+\todo{Proof (copy from some other lecture)}
+\begin{proposition}
+    Let $X$ be a topological space.
+    The following are equivalent:
+    \begin{enumerate}[(i)]
+        \item Every nonempty open set
+            is non-meager in $X$.
+        \item Every comeager set is dense.
+        \item The intersection of countable many
+            open dense sets is dense.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    \todo{Proof (short)}
+
+    (iii) $\implies$ (i)
+    Let us first show that $X$ is non-meager.
+    Suppose that $X$ is meager. Then $X = \bigcup_{n} A_n = \bigcup_{n} \overline{A_n}$
+    is the countable union of nwd sets.
+    We have that
+    \[
+    \emptyset = \bigcap_{n} (X \setminus \overline{A_n})
+    \] 
+    is dense by (iii).
+    This proof can be adapted to other open sets $X$.
+\end{proof}
+
+
+% TODO Fubini
+\begin{notation}
+    Let $X ,Y$ be topological spaces,
+    $A \subseteq  X \times Y$,
+    $x \in X, y \in Y$.
+
+    Let
+    \[
+    A_x \coloneqq \{y \in Y : (x,y) \in A\} 
+    \] 
+    and
+    \[
+    A^y \coloneqq \{x \in X : (x,y) \in A\} .
+    \] 
+\end{notation}
+
+The following similar to Fubini,
+but for meager sets:
+
+\begin{theorem}[Kuratowski-Ulam]
+    \label{thm:kuratowskiulam}
+    Let $X,Y$ be second countable topological spaces.
+    Let $A \subseteq X \times Y$
+    be a set with the Baire property.
+
+    Then
+    \begin{enumerate}[(i)]
+        \item $\{x \in X : A_x \text{ has the BP }\}$
+            is comeager\footnote{Note that not necessarily all sections
+                have the BP. For example $\{x\} \times Y$ is meager in $X \times Y$}
+            and similarly for $y$.
+        \item $A$ is meager
+            \begin{IEEEeqnarray*}{rll}
+                \iff &\{x \in X : A_x \text{ is meager }\}&\text{ is comeager}\\
+                \iff &\{y \in Y : A^y \text{ is meager }\}& \text{ is comeager}.
+            \end{IEEEeqnarray*}
+        \item $A$ is comeager
+            \begin{IEEEeqnarray*}{rll}
+                \iff & \{x \in X: A_x \text{ is comeager }\} &\text{ is comeager}\\
+                \iff & \{y \in Y: A^y \text{ is comeager}\} & \text{ is comeager}.
+            \end{IEEEeqnarray*}
+    \end{enumerate}
+\end{theorem}
+\begin{refproof}{thm:kuratowskiulam}
+    (ii) and (iii) are equivalent by passing to the complement.
+
+    \begin{claim}%[1a]
+        \label{thm:kuratowskiulam:c1a}
+
+        If $F \overset{\text{closed}}{\subseteq}  X \times Y$
+        is nwd,
+        then
+        \[
+        \{x \in X : F_x \text{is nwd}\} 
+        \] 
+        is comeager.
+    \end{claim}
+    \begin{refproof}{thm:kuratowskiulam:c1a}
+        Put $W = F^c$.
+        This is open and dense in $X \times Y$.
+        It suffices to show that $\{x \in X : W_x \text{ is dense}\}$
+        is comeager.
+        Note that $W_x$ is open for all $x$.
+
+        Fix a countable basis $(V_n)$ of $Y$
+        with $V_n$ non-empty.
+        We want to show that
+        \[
+        \{x \in X: \forall n.~ (W_x \cap V_n) \neq \emptyset\}
+        \] 
+        is a comeager set.
+        This is equivalent to
+        \[
+        \{x \in X : (W_x \cap V_n) \neq \emptyset\}
+        \] 
+        being comeager for all $n$,
+        because the intersection
+        of countably many comeager sets is comeager.
+
+        Fix $n$ and let $U_n \coloneqq \{x \in X: (W_x \cap V_n) = \emptyset\}$.
+        We will show that $U_n$ is open and dense,
+        hence it is comeager.
+
+        $U_n = \proj_x(W \cap (X \times V_n))$ is open
+        since projections of open sets are open.
+
+        Let $U \subseteq X$ be nonempty and open.
+        We need to show that $U \cap U_n \neq \emptyset$.
+        It is
+        \[
+        U \cap U_n = \proj_x(W \cap (U \times V_n))
+        \] 
+        nonempty since $W$ is dense.
+        
+    \end{refproof}
+
+    \begin{claim} % [1a']
+        \label{thm:kuratowskiulam:c1ap}
+        If $F \subseteq  X \times Y$
+        is nwd,
+        then
+        \[
+        \{x \in X : F_x \text{is nwd}\} 
+        \] 
+        is comeager.
+        
+    \end{claim}
+    \begin{refproof}{thm:kuratowskiulam:c1ap}
+        We have that $\overline{F}$ is nwd.
+        Hence by \yaref{thm:kuratowskiulam:c1a}
+        the set
+         \[
+        \{x \in X: \overline{F_x} \text{ is nwd}\} \subseteq
+        \{x \in X: F_x \text{ is nwd}\} 
+        \] 
+        is comeager.
+    \end{refproof}
+
+    \begin{claim}% [1b]
+        \label{thm:kuratowskiulam:c1b}
+
+        If $M \subseteq  X \times Y$ is meager,
+        then
+        \[
+        \{x \in X : M_x \text{ is meager}\} 
+        \] 
+        is comeager.
+    \end{claim}
+    \begin{refproof}{thm:kuratowskiulam:c1b}
+        This follows from \yaref{thm:kuratowskiulam:c1ap}:
+        Let $M = \bigcup_{n < \omega} F_n$
+        where the $F_n$ are nwd.
+        Apply \yaref{thm:kuratowskiulam:c1ap}
+        to each $F_n$.
+        We get that
+        $M_x$ is comeager
+        as a countable intersection of comeager sets.
+    \end{refproof}
+
+    \todo{Finish proof}
+    \phantom\qedhere
+\end{refproof}
+
+\begin{remark}
+    Suppose that $A$ has the BP.
+    Then there is an open $U$ such that
+    $A \symdif U \mathbin{\text{\reflectbox{$\coloneqq$}}} M$ is meager.
+    Then $A = U \symdif M$.
+\end{remark}
+
+

inputs/tutorial_02.tex

@@ -0,0 +1,61 @@
+\tutorial{02}{2023-10-24}{}
+
+% Points: 15 / 16
+
+\subsubsection{Exercise 4}
+
+\begin{fact}
+    Let $X $ be a compact Hausdorffspace.
+    Then the following are equivalent:
+    \begin{enumerate}[(i)]
+        \item $X$ is Polish,
+        \item $X$ is metrisable,
+        \item $X$ is second countable.
+    \end{enumerate}
+\end{fact}
+\begin{proof}
+    (i) $\implies$ (ii) clear
+
+    (i) $\implies$ (iii) clear
+
+    (ii) $\implies$ (i) Consider the cover $\{B_{\epsilon}(x) | x \in X\}$
+    for every $\epsilon \in \Q$
+    and chose a finite subcover.
+    Then the midpoints of the balls from the cover
+    form a countable dense subset.
+
+    The metric is complete as $X$ is compact.
+    (For metric spaces: compact $\iff$ seq.~compact $\iff$ complete and totally bounded)
+
+    (iii) $\implies$ (ii)
+    Use Urysohn's metrisation theorem and the fact that compact
+    Hausdorff spaces are normal
+\end{proof}
+
+Let $X$ be compact Polish (compact metrisable $\implies$ compact Polish)
+and $Y $ Polish.
+Let $\cC(X,Y)$ be the set of continuous functions $X \to Y$.
+Consider the metric $d_u(f,g) \coloneqq  \sup_{x \in X} |d(f(x), g(x))|$.
+Clearly $d_u$ is a metric.
+
+\begin{claim}
+    $d_u$ is complete.
+\end{claim}
+\begin{subproof}
+    Let $(f_n)$ be a Cauchy sequence in $\cC(X,Y)$.
+    A $Y$ is complete,
+    there exists a pointwise limit $f$.
+
+    $f_n$ converges uniformly to $f$:
+
+    \[
+    d(f_n(x), f(x)) \le  \overbrace{d(f_n(x), f_m(x))}^{\mathclap{\text{$(f_n)$ is Cauchy}}}
+    + \underbrace{d(f_m(x), f(x))}_{\mathclap{\text{small for appropriate $m$}}}.
+    \]
+
+    $f$ is continuous by the uniform convergence theorem.
+\end{subproof}
+
+\begin{claim}
+    There exists a countable dense subset.
+\end{claim}

logic.sty

@@ -131,5 +131,7 @@
 %\mathbin{\raisebox{1ex}{\scalebox{.7}{$\frown$}}}%
     
 \DeclareMathOperator{\hght}{height}
+\DeclareMathOperator{\symdif}{\triangle}
+\DeclareSimpleMathOperator{proj}
 
 \newcommand\lecture[3]{\hrule{\color{darkgray}\hfill{\tiny[Lecture #1, #2]}}}

logic3.tex

@@ -28,6 +28,7 @@
 \input{inputs/lecture_02}
 \input{inputs/lecture_03}
 \input{inputs/lecture_04}
+\input{inputs/lecture_05}