lecture 09

inputs/lecture_08.tex

@@ -84,6 +84,7 @@ where $X$ is a metrizable, usually second countable space.
 \end{example}
 
 \begin{theorem}
+    \label{thm:cantoruniversal}
     Let $X$ be a separable, metrizable space.
     Then for every $\xi \ge 1$,
     there is a $2^{\omega}$-universal
@@ -141,3 +142,12 @@ where $X$ is a metrizable, usually second countable space.
     % TODO
     Furthermore $\cU \in \Sigma^0_{\xi}((2^\omega)^\omega \times X)$.
 \end{proof}
+\begin{remark}
+    Since $2^{\omega}$ embeds
+    into any uncountable polish space $Y$
+    such that the image is closed,
+    we can  $2^{\omega}$ by $Y$
+    in the statement of the theorem.%
+    \footnote{By definition of the subspace topology
+    and transfinite induction, $\Sigma^0_\xi(Y)\defon{2^\omega} = \Sigma^0_\xi(2^\omega)$.}
+\end{remark}

inputs/lecture_09.tex

@@ -0,0 +1,170 @@
+\lecture{09}{2023-11-14}{}
+
+\begin{theorem}
+    Let $X$ be an uncountable Polish space.
+    Then for all $\xi < \omega_1$,
+    we have that $\Sigma^0_\xi(X) \neq \Pi^0_\xi(X)$.
+\end{theorem}
+\begin{proof}
+    Fix $\xi < \omega_1$.
+    Towards a contradiction assume $\Sigma^0_\xi(X) = \Pi^0_\xi(X)$.
+    By \autoref{thm:cantoruniversal},
+    there is a $X$-universal set $\cU$ for $\Sigma^0_\xi(X)$.
+
+    Take $A \coloneqq  \{y \in X : (y,y) \not\in \cU\}$.
+    Then $A \in \Pi^0_\xi(X)$.\todo{Needs a small proof}
+    By assumption $A \in \Sigma^0_\xi(X)$,
+    i.e.~there exists some $z \in X$ such that $A = \cU_z$.
+    We have
+    \[z \in A \iff z \in \cU_z \iff (z,z) \in \cU.\]
+    But by the definition of $A$,
+    we have $z \in A \iff (z,z) \not\in \cU \lightning$.
+\end{proof}
+
+
+\begin{definition}
+    Let $X$ be a Polish space.
+    A set $A \subseteq X$
+    is called \vocab{analytic} 
+    if
+    \[
+    \exists Y \text{ Polish}.~\exists  B \in \cB(Y).~
+    \exists \underbrace{f\colon Y \to X}_{\text{continuous}}.~
+    f(B) = A.
+    \] 
+\end{definition}
+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
+    by the weaker assertion of $f$ being Borel.
+    \todo{Copy exercise from sheet 5}
+\end{remark}
+
+\begin{theorem}
+    Let $X$ be Polish,
+    $\emptyset \neq A \subseteq X$.
+    Then the following are equivalent:
+    \begin{enumerate}[(i)]
+        \item $A$ is analytic.
+        \item There exists a Polish space $Y$ 
+            and $f\colon Y \to X$
+            continuous\footnote{or Borel}
+            such that $A = f(Y)$.
+        \item There exists $h\colon \cN \to X$
+            continuous with $h(\cN) = A$.
+        \item There is $F \overset{\text{closed}}{\subseteq} X \times \cN$
+            such that $A = \proj_X(F)$.
+        \item There is a Borel set $B \subseteq X \times Y$
+            for some Polish space $Y$,
+            such that $A = \proj_X(B)$.
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    To show (i) $\implies$ (ii):
+    take $B \in \cB(Y')$
+    and $f\colon Y' \to X$
+    continuous with $f(B) = A$.
+    Take a finer Polish topology $\cT$ on $Y'$
+    adding no Borel sets,
+    such that $B$ is clopen with respect to the new topology.
+    Then let $g = f\defon{B}$
+    and $Y = (B, \cT\defon{B})$.
+
+    (ii) $\implies$ (iii):
+    Any Polish space is the continuous image of $\cN$.
+    Let $g_1: \cN \to  Y$
+    and $h \coloneqq g \circ g_1$.
+
+    (iii) $\implies$ (iv):
+    Let $h\colon \cN \to X$ with $h(\cN) = A$.
+    Let $G(h) \coloneqq \{(a,b) : h(a) = b\} \overset{\text{closed}}{\subseteq} \cN \times X$
+    be the \vocab{graph} of $h$.
+    Take $F \coloneqq  G(h)^{-1} \coloneqq  \{(c,d) | (d,c) \in G(h)\}$
+
+    Clearly (iv) $\implies$ (v).
+
+    (v) $\implies$ (i):
+    Take $f \coloneqq \proj_X$.
+\end{proof}
+
+
+\begin{theorem}
+    Let $X,Y$ be Polish spaces.
+    Let $f\colon X \to Y$ be \vocab{Borel}
+    (i.e.~preimages of open sets are Borel).
+    \begin{enumerate}[(a)]
+        \item The image of an analytic set is analytic.
+        \item The preimage of an analytic set is analytic.
+        \item Analytic sets are closed under countable unions
+            and countable intersections.
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    \begin{enumerate}[(a)]
+        \item Let $A \subseteq X$ analytic.
+            Then there exists $Z$ Polish and  $g\colon Z \to X$
+            continuous
+            with $g(Z) = A$.
+            We have that  $f(A) = (f \circ g)(Z)$
+            and $f \circ g$ is Borel.
+        \item Let $f\colon X \to Y$ be Borel
+            and $B \subseteq Y$ analytic.
+
+            Take $Z$ Polish
+            and $B_0 \subseteq Y \times Z$
+            such that $\proj_Y(B_0) = B$.
+            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 \todo{Exercise}
+    \end{enumerate}
+\end{proof}
+
+\begin{notation}
+    Let $X$ be Polish.
+    Let $\Sigma^1_1(X)$ denote the set of all analytic
+    subsets of $X$.
+    $\Pi^1_1(X) \coloneqq \{B \subseteq X : X \setminus B \in \Sigma^1_1(X)\}$
+    is the set of \vocab{coanalytic} sets.
+\end{notation}
+We will see later that $\Sigma^1_1(X) \cap \Pi^1_1(X) = \cB(X)$.
+
+
+\begin{theorem}
+    Let $X,Y$ be uncountable Polish spaces.
+    There exists a $Y$-universal $\Sigma^1_1(X)$ set.
+\end{theorem}
+\begin{proof}
+    Take $\cU \subseteq  Y \times X \times \cN$
+    which is $Y$-universal for $\Pi^0_1(X \times \cN)$.
+    Let $\cV \coloneqq  \proj_{Y \times Y}(\cU)$.
+    Then $\cV$ is $Y$-universal for $\Sigma^1_1(X)$:
+    \begin{itemize}
+        \item $\cV \in \Sigma^1_1(Y \times X)$
+            since $\cV$ is a projection of a closed set.
+        \item All sections of $\cV$ are analytic.
+            Let $A \in \Sigma^1_1(X)$.
+            Let $C \subseteq  X \times \cN$
+            be closed such that $\proj_X(C) = A$.
+            There is $y \in Y$
+            such that $\cU_y = C$,
+            hence $\cV_y = A$.
+    \end{itemize}
+\end{proof}
+\begin{remark}
+    In the same way that we proved
+    $\Sigma^0_\xi(X) \neq  \Pi^0_\xi(X)$ for $\xi < \omega_1$,
+    we obtain that $\Sigma^1_1(X) \neq \Pi^1_1(X)$.
+
+    In fact if $\cU$ is universal for $\Sigma^1_1(X)$,
+    then $\{y : (y,y) \in \cU\} \in \Sigma^1_1(X) \setminus \Pi^1_1(X)$.
+    In particular, this set is not Borel.
+\end{remark}
+
+
+\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}
+\end{remark}

logic3.tex

@@ -32,6 +32,7 @@
 \input{inputs/lecture_06}
 \input{inputs/lecture_07}
 \input{inputs/lecture_08}
+\input{inputs/lecture_09}