From 0d6d392d5d7dc348c79ee3771bd56e9618de9502 Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Tue, 14 Nov 2023 11:53:30 +0100 Subject: [PATCH] lecture 09 --- inputs/lecture_08.tex | 10 +++ inputs/lecture_09.tex | 170 ++++++++++++++++++++++++++++++++++++++++++ logic3.tex | 1 + 3 files changed, 181 insertions(+) create mode 100644 inputs/lecture_09.tex diff --git a/inputs/lecture_08.tex b/inputs/lecture_08.tex index 162b50c..09efe3f 100644 --- a/inputs/lecture_08.tex +++ b/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} diff --git a/inputs/lecture_09.tex b/inputs/lecture_09.tex new file mode 100644 index 0000000..f79c2aa --- /dev/null +++ b/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} diff --git a/logic3.tex b/logic3.tex index 3c1360b..cecc7a0 100644 --- a/logic3.tex +++ b/logic3.tex @@ -32,6 +32,7 @@ \input{inputs/lecture_06} \input{inputs/lecture_07} \input{inputs/lecture_08} +\input{inputs/lecture_09}