lecture 08

inputs/lecture_07.tex

@@ -0,0 +1,192 @@
+\lecture{07}{2023-11-07}{}
+
+\begin{proposition}
+    Let $X$ be second countable.
+    Then $|\cB(X)| \le \fc$.
+    % $\fc := 2^{\aleph_0}$
+\end{proposition}
+\begin{proof}
+    We use strong induction on  $\xi < \omega_1$.
+    We have $\Sigma^0_1(X) \le \fc$
+    (for every element of the basis, we can decide
+    whether to use it in the union or not).
+
+    Suppose that $\forall \xi' < \xi.~|\Sigma^0_{\xi'}(X)| \le \fc$.
+    Then $|\Pi^0_{\xi'}(X)| \le  \fc$.
+    We have that
+    \[
+        \Sigma^0_\xi(X) = \{\bigcup_{n} A_n : n \in \N, A_n \in \Pi^0_{\xi_n}(X), \xi_n < \xi\}.
+    \] 
+    Hence $|\Sigma^0_\xi(X)| \le (\overbrace{\aleph_0}^{\mathclap{\{\xi': \xi' < \xi\} \text{ is countable~ ~ ~ ~}}} \cdot \underbrace{\fc}_{\mathclap{\text{inductive assumption}}})^{\overbrace{\aleph_0}^{\mathclap{\text{countable unions}}}}$.
+
+    We have
+    \[
+    \cB(X) = \bigcup_{\xi < \omega_1}  \Sigma^0_\xi(X).
+    \] 
+    Hence
+    \[
+    |\cB(X)| \le  \omega_1 \cdot  \fc = \fc.
+    \] 
+\end{proof}
+
+\begin{proposition}[Closure properties]
+    Suppose that $X$ is metrizable.
+    Let $1 \le  \xi < \omega_1$.
+    Then
+    \begin{enumerate}[(a)]
+        \item \begin{itemize}
+                \item $\Sigma^0_\xi(X)$ is closed under countable unions.
+                \item  $\Pi^0_\xi(X)$ is closed under countable intersections.
+                \item  $\Delta^0_\xi(X)$ is closed under complements,
+                    countable unions and
+                    countable intersections.
+            \end{itemize}
+        \item \begin{itemize}
+                \item $\Sigma^0_\xi(X)$ is closed under \emph{finite} intersections.
+                \item $\Pi^0_\xi(X)$ is closed under \emph{finite} unions.
+            \end{itemize}
+
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    \begin{enumerate}[(a)]
+        \item This follows directly from the definition. 
+            Note that a countable intersection can be written
+            as a complement of the countable union of complements:
+            \[
+            \bigcap_{n} B_n = \left( \bigcup_{n} B_n^{c} \right)^{c}.
+            \] 
+        \item If suffices to check this for $\Sigma^0_{\xi}(X)$.
+            Let $A = \bigcup_{n}  A_n$ for $A_n \in \Pi^0_{\xi_n}(X)$
+            and $B = \bigcup_{m} B_m$ for $B_m \in \Pi^0_{\xi'_m}(X)$.
+            Then
+            \[
+            A \cap B = \bigcup_{n,m}  \left( A_n \cap B_m \right)
+            \] 
+            and $A_n \cap B_m \in \Pi^{0}_{\max(\xi_n, \xi'_m)}(X)$.
+
+    \end{enumerate}
+\end{proof}
+\begin{example}
+    Consider the cantor space $2^{\omega}$.
+    We have that $\Delta^0_1(2^{\omega})$
+    is not closed under countable unions
+    (countable unions yield all open sets, but there are open
+    sets that are not clopen).
+\end{example}
+
+\subsection{Turning Borels Sets into Clopens}
+
+\begin{theorem}
+    \label{thm:clopenize}
+    Let $(X, \cT)$ be a Polish space.
+    For any  Borel set $A \subseteq  X$,
+    there is a finer Polish topology,
+    \footnote{i.e.~$\cT_A \supseteq \cT$ and $(X, \cT_A)$ is Polish}
+    such that
+    \begin{itemize}
+        \item $A$ is clopen in $\cT_A$,
+        \item the Borel sets do not change,
+            i.e.~$\cB(X, \cT) = \cB(X, \cT_A)$.
+    \end{itemize}
+\end{theorem}
+\begin{corollary}[Perfect set property]
+    Let $(X, \cT)$ be Polish,
+    and let $B \subseteq X$ be Borel and uncountable.
+    Then there is an embedding
+    of the cantor space $2^{\omega}$
+    into $B$.
+\end{corollary}
+\begin{proof}
+    Pick $\cT_B \supset \cT$
+    such that $(X, \cT_B)$ is Polish,
+    $B$ is clopen in $\cT_B$ and
+    $\cB(X,\cT) = \cB(X, \cT_B)$.
+
+    Therefore $(\cB, \cT_B\defon{B})$ is Polish.
+    We know that there is an embedding
+    $f\colon 2^{\omega} \to (B, \cT_{B}\defon{B})$.
+
+    Consider $f\colon 2^{\omega} \to B \subseteq (X, \cT)$.
+    This is still continuous as $\cT \subseteq \cT_B$.
+    Since $2^{\omega}$ is compact, $f$ is an embedding.
+    %\todo{Think about this}
+\end{proof}
+
+\begin{refproof}{thm:clopenize}
+    We show that
+    \begin{IEEEeqnarray*}{rCl}
+        A \coloneqq  \{B \subseteq \cB(X, \cT)&:\exists & \cT_B \supseteq \cT .\\
+                                 && (X, \cT_B)  \text{ is Polish},\\
+                                 && \cB(X, \cT) = \cB(X, \cT_B)\\
+                                 && B \text{ is clopen in $\cT_B$}\\
+        \}
+    \end{IEEEeqnarray*}
+    is equal to the set of Borel sets.
+
+    The proof rests on two lemmata:
+    \begin{lemma}
+        \label{thm:clopenize:l1}
+        Let $(X,\cT)$ be a Polish space.
+        Then for any $F \overset{\text{closed}}{\subseteq} X$ (wrt. $\cT$)
+        there is $\cT_F \supseteq \cT$
+        such that $\cT_F$ is Polish,
+        $\cB(\cT) = \cB(\cT_F)$
+        and  $F$ is clopen in $\cT_F$.
+    \end{lemma}
+    \begin{proof}
+        Consider $(F, \cT\defon{F})$ and $(X \setminus F, \cT\defon{X \setminus F})$.
+        Both are Polish spaces.
+        Take the coproduct\footnote{topological sum} $F \oplus (X \setminus F)$
+        of these spaces.
+        This space is Polish,
+        and the topology is generated by $\cT \cup \{F\}$,
+        hence we do not get any new Borel sets.
+    \end{proof}
+    So all closed sets are in $A$.
+    Furthermore $A$ is closed under complements,
+    since complements of clopen sets are clopen.
+
+    \begin{lemma}
+        \label{thm:clopenize:l2}
+        Let $(X, \cT)$ be Polish.
+        Let $\{\cT_n\}_{n < \omega}$
+        be Polish topologies
+        such that $\cT_n \supseteq \cT$
+        and $\cB(\cT_n) = \cB(\cT)$.
+        Then the topology $\cT_\infty$ generated by  $\bigcup_{n} \cT_n$
+        is still Polish
+        and $\cB(\cT_\infty) = \cB(T)$.
+    \end{lemma}
+    \begin{proof}
+        We have that $\cT_\infty$ is the smallest
+        topology containing all $\cT_n$.
+        To get $\cT_\infty$
+        consider
+        \[
+        \cF \coloneqq  \{A_1 \cap A_2 \cap \ldots \cap A_n : A_i \in \cT_i\}.
+        \]
+        Then
+        \[
+        \cT_\infty = \{\bigcup_{i<\omega} B_i : B_i \in \cF\}.
+        \]
+        (It suffices to take countable unions,
+        since we may assume that the $A_1, \ldots, A_n$ in the
+        definition of $\cF$ belong to
+        a countable basis of the respective $\cT_n$).
+
+        \todo{This proof will be finished in the next lecture}
+    \end{proof}
+
+    We need to show that $A$ is closed under countable unions.
+    By \yaref{thm:clopenize:l2} there exists a topology
+    $\cT_\infty$ such that $A = \bigcup_{n < \omega}  A_n$ is open in $\cT_\infty$
+    and $\cB(\cT_\infty) = \cB(\cT)$.
+    Applying \yaref{thm:clopenize:l1}
+    yields a topology $\cT_\infty'$ such that
+    $(X, \cT_\infty')$ is Polish,
+    $\cB(\cT_\infty') = \cB(\cT)$
+    and $A $ is clopen in $\cT_{\infty}'$.
+\end{refproof}
+
+

inputs/lecture_08.tex

@@ -0,0 +1,143 @@
+\lecture{08}{2023-11-10}{}
+
+\todo{put this lemma in the right place}
+\begin{lemma}[Lemma 2]
+    Let $(X, \cT)$ be a Polish space.
+    Let $\cT_n \supseteq \cT$ be Polish
+    with $\cB(X, \cT_n) = \cB(X, \cT)$.
+    Let  $\cT_\infty$ be the topology generated
+    by $\bigcup_n \cT_n$.
+    Then $(X, \cT_\infty)$ is Polish
+    and $\cB(X, \cT_\infty) = \cB(X, \cT)$.
+\end{lemma}
+\begin{proof}
+    Let $Y = \prod_{n \in \N} (X, \cT_n)$.
+    Then $Y$ is Polish.
+    Let $\delta\colon (X, \cT_\infty) \to Y$
+    defined by $\delta(x) = (x,x,x,\ldots)$.
+    \begin{claim}
+        $\delta$ is a homeomorphism.
+    \end{claim}
+    \begin{subproof}
+        Clearly $\delta$ is a bijection.
+        We need to show that it is continuous and open.
+
+        Let $U \in \cT_i$.
+        Then
+        \[
+        \delta^{-1}(D \cap \left( X \times  X \times \ldots\times  U \times  \ldots) \right)) = U \in \cT_i \subseteq \cT_\infty,
+        \]
+        hence $\delta$ is continuous.
+        Let $U \in \cT_\infty$.
+        Then $U$ is the union of sets of the form
+        \[
+        V = U_{n_1} \cap U_{n_2} \cap \ldots \cap U_{nu}
+        \]
+        for some $n_1 < n_2 < \ldots < n_u$
+        and $U_{n_i} \in \cT_i$.
+
+        Thus is suffices to consider sets of this form.
+        We have that
+        \[
+        \delta(V) = D \cap (X \times  X \times  \ldots \times  U_{n_1} \times  \ldots \times  U_{n_2} \times  \ldots \times  U_{n_u} \times  X \times  \ldots) \overset{\text{open}}{\subseteq} D.
+        \]
+    \end{subproof}
+
+    This will finish the proof since
+    \[
+    D = \{(x,x,\ldots) \in Y : x \in X\} \overset{\text{closed}}{\subseteq} Y
+    \]
+    Why? Let  $(x_n) \in Y \setminus D$.
+    Then there are $i < j$ such that $x_i \neq x_j$.
+    Take disjoint open $x_i \in U$, $x_j \in V$.
+    Then
+    \[(x_n) \in  X \times  X \times  \ldots \times  U \times  \ldots \times  X \times  \ldots \times V \times  X \times  \ldots\]
+    is open in $Y\setminus D$.
+    Hence $Y \setminus D$ is open, thus $D$ is closed.
+    It follows that $D$ is Polish.
+\end{proof}
+
+\subsection{Parametrizations}
+\todo{choose better title}
+
+
+Let $\Gamma$ denote a collection of sets in some space.
+For us $\Gamma$ will be one of $\Sigma^0_\xi(X), \Pi^0_\xi(X), \Delta^0_\xi(X), \cB(X)$,
+where $X$ is a metrizable, usually second countable space.
+
+\begin{definition}
+    We say that $\cU \subseteq Y \times X$
+    is \vocab{$Y$-universal} for $\Gamma(X)$ /
+    $\cU$ \vocab{parametrizes}  $\Gamma(X)$
+    iff:
+    \begin{itemize}
+        \item $\cU \in \Gamma$,
+        \item $\{U_y : y \in Y\} = \Gamma(X)$.
+    \end{itemize}
+\end{definition}
+
+\begin{example}
+    Let $X = \omega^\omega$, $Y = 2^{\omega}$
+    and consider $\Gamma = \Sigma^0_{\omega+5}(\omega^\omega)$.
+    We will show that there is a $2^{\omega}$-universal
+    set for $\Gamma$.
+\end{example}
+
+\begin{theorem}
+    Let $X$ be a separable, metrizable space.
+    Then for every $\xi \ge 1$,
+    there is a $2^{\omega}$-universal
+    set for $\Sigma^0_\xi(X)$ and
+    similarly for $\Pi^0_\xi(X)$.
+\end{theorem}
+\begin{proof}
+    Note that if $\cU$ is $2^{\omega}$ universal for
+    $\Sigma^0_\xi(X)$, then $(2^{\omega} \times X) \setminus \cU$
+    is $2^{\omega}$-universal for $\Pi^0_\xi(X)$.
+    Thus it suffices to consider $\Sigma^0_\xi(X)$.
+
+    First let $\xi = 1$.
+    We construct $\cU \overset{\text{open}}{\subseteq} 2^{\omega} \times X$
+    such that
+    \[
+    \{U_y : y \in 2^\omega\}  = \Sigma^0_1(X).
+    \]
+
+    Let $(V_n)$ be a basis of open sets of $X$.
+    For all $y \in 2^\omega$ and $x \in X$
+    put $(y,x) \in \cU$ iff
+    $x \in \bigcup \{V_n : y_n = 1\}$.
+    $\cU$ is open.
+    Let $V = \bigcup \{V_n : V_n \subseteq V\}$.
+    Pick $y \in 2^\omega$
+    and let $y_n = 1$ iff $V_n \subseteq  V$.
+    Then $\cU_y = V$.
+
+
+    Now suppose that there exists a
+    $2^{\omega}$-universal set for $\Sigma^0_{\eta}(X)$
+    for all $\eta < \xi$.
+    Fix  $\xi_0 \le  \xi_1 \le \ldots < \xi$
+    such that $\xi_n \to \xi$ if $\xi$ is a limit,
+    or $\xi_n = \xi'$ if $\xi = \xi' +1$ is a successor.
+
+    Recall that $\eta_1 \le  \eta_2 \implies \Pi^0_{\eta_1}(X) \subseteq \Pi^0_{\eta_2}(X)$.
+
+    Note that if $A = \bigcup_n A_n$,  with $A_n \in \Pi^0_{\eta_n}(X)$
+    some $\eta_n < \xi$,
+    we also have
+    $A = \bigcup_n A_n'$ with  $A'_n \in \Pi^0_{\xi_n}(X)$.
+
+    We construct a $(2^\omega)^\omega \cong 2^\omega$-universal set
+    for $\Sigma^0_\xi(X)$.
+    For $(y_n) \in (2^\omega)^\omega$
+    and $x \in X$
+    we set $((y_n), x) \in \cU$
+    iff $\exists n.~(y_n, x) \in U_{\xi_n}$,
+    i.e.~iff $\exists n.~x \in (U_{\xi_n})_{y_n}$.
+
+    Let $A \in \Sigma^0_\xi(X)$.
+    Then $A = \bigcup_{n} B_n$ for some $B_n \in \Pi^0_{\xi_n}(X)$.
+    % TODO
+    Furthermore $\cU \in \Sigma^0_{\xi}((2^\omega)^\omega \times X)$.
+\end{proof}

logic.sty

@@ -134,5 +134,6 @@
 \DeclareMathOperator{\hght}{height}
 \DeclareMathOperator{\symdif}{\triangle}
 \DeclareSimpleMathOperator{proj}
+\newcommand{\fc}{\mathfrak{c}}
 
 \newcommand\lecture[3]{\hrule{\color{darkgray}\hfill{\tiny[Lecture #1, #2]}}}

logic3.tex

@@ -30,6 +30,8 @@
 \input{inputs/lecture_04}
 \input{inputs/lecture_05}
 \input{inputs/lecture_06}
+\input{inputs/lecture_07}
+\input{inputs/lecture_08}