From 0612713145031c504e3acc7e1b377ea8d716ae6a Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Fri, 1 Dec 2023 11:58:27 +0100 Subject: [PATCH] lecture 14 --- inputs/lecture_13.tex | 5 - inputs/lecture_14.tex | 219 ++++++++++++++++++++++++++++++++++++++++++ logic3.tex | 1 + 3 files changed, 220 insertions(+), 5 deletions(-) create mode 100644 inputs/lecture_14.tex diff --git a/inputs/lecture_13.tex b/inputs/lecture_13.tex index 753e015..27ac96d 100644 --- a/inputs/lecture_13.tex +++ b/inputs/lecture_13.tex @@ -179,8 +179,3 @@ i.e. \end{itemize} and similarly for $<^\ast_{\phi}$. \end{definition} - - - - - diff --git a/inputs/lecture_14.tex b/inputs/lecture_14.tex new file mode 100644 index 0000000..baef8a4 --- /dev/null +++ b/inputs/lecture_14.tex @@ -0,0 +1,219 @@ +\lecture{14}{2023-12-01}{} + +\begin{theorem}[Moschovakis] + If $C$ is coanalytic, + then there exists a $\Pi^1_1$-rank on $C$. +\end{theorem} +\begin{proof} + Pick a $\Pi^1_1$-complete set. + It suffices to show that there is a rank on it. + Then use the reduction to transfer + it to any coanalytic set, + i.e.~for $x,y \in C'$ + let + \[ + x \le^{\ast}_{C'} y :\iff f(x) \le^\ast_{C} f(y) + \] + and similarly for $<^\ast$. + % https://q.uiver.app/#q=WzAsNSxbMCwwLCJZIl0sWzIsMCwiWCJdLFswLDEsIkMnIl0sWzIsMSwiQyJdLFsyLDIsIlxcUGlfMV4xLVxcdGV4dHtjb21wbGV0ZX0iXSxbMCwxLCJmIl0sWzIsMCwiXFxzdWJzZXRlcSIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzMsMSwiXFxzdWJzZXRlcSIsMix7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV1d +\[\begin{tikzcd} + Y && X \\ + {C'} && C \\ + && {\Pi_1^1-\text{complete}} + \arrow["f", from=1-1, to=1-3] + \arrow["\subseteq", hook, from=2-1, to=1-1] + \arrow["\subseteq"', hook, from=2-3, to=1-3] +\end{tikzcd}\] + + Let $X = 2^{\Q} \supseteq \WO$. + We have already show that $\WO$ is $\Pi^1_1$-complete. + Set $\phi(x) \coloneqq \otp(x)$ + ($\otp\colon \WO \to \Ord$ denotes the order type). + We show that this is a $\Pi^1_1$-rank. + + Define $E \subseteq \Q^{\Q} \times 2^{\Q} \times 2^{\Q}$ + by + \begin{IEEEeqnarray*}{rCl} + (f,x,y) \in E &:\iff& f \text{ order embeds $(x, \le_{\Q})$ to $(y,\le _{\Q})$}\\ + &\iff& \forall p,q \in \Q.~(p,q \in x \land p <_{\Q} q \implies f(p), f(q) \in y \land f(p) <_{\Q} f(q)) + \end{IEEEeqnarray*} + $E$ is Borel as a countable intersection of clopen sets. + + Define + $x <^\ast_{\phi}$ + iff + \begin{itemize} + \item $(x, <_{\Q})$ is well ordered and + \item $(y, <_{\Q})$ does not order embed into $(x, <_{\Q})$, + \end{itemize} + where we identify $2^\Q$ and the powerset of $\Q$. + This is equivalent to + \begin{itemize} + \item $x \in \WO$ and + \item $\forall f \in \Q^\Q.~(f,y,x) \not\in E$. + \end{itemize} + + Furthermore $x \le_\phi^\ast y \iff$ + either $x <^\ast_\phi y$ or + $(x, <_{\Q})$ and $(y, <_\Q)$ + are well ordered with the same order type, + i.e.~either $x<^\ast_\phi y$ or + $x,y \in \WO$ and any order embedding of $(x,<_{\Q})$ to + $(y, <_{\Q})$ is cofinal% + \footnote{% + Recall that $A \subseteq (x,<_{\Q})$ + is \vocab{cofinal} if $\forall t \in x.~\exists a \in A.~t\le _{\Q} a$.% + } + in $(y, <_\Q)$ is + cofinal in $(y, <_{\Q})$ and vice versa. + Equivalently, either $(x <^\ast_\phi y)$ + or + \begin{IEEEeqnarray*}{rCl} + & &x,y \in \WO\\ + &\land& \forall f \in \Q^\Q .~(E(f,x,y) \implies \forall p \in y.~\exists q \in x.~p \le f(q))\\ + &\land& \forall f \in \Q^\Q .~(E(f,y,x) \implies \forall p \in x.~\exists q \in y.~p \le f(q)) + \end{IEEEeqnarray*} +\end{proof} + + +\begin{theorem} + \label{thm:uniformization} + Let $X$ be Polish and $R \subseteq X \times \N$ by $\Pi^1_1$ + (we only need that $\N$ is countable). + Then there is $R^\ast \subseteq R$ coanalytic + such that + \[ + \forall x \in X.~(\exists n.~(x,n) \in R \iff \exists! n.~(x,n)\in R^\ast). + \] + We say that $R^\ast$ \vocab[uniformization]{uniformizes} $R$. + \todo{missing picture + \url{https://upload.wikimedia.org/wikipedia/commons/4/4c/Uniformization_ill.png}} + +\end{theorem} +\begin{proof} + Let $\phi\colon R \to \Ord$ + by a $\Pi^1_1$-rank. + Set + \begin{IEEEeqnarray*}{rCl} + (x,n) \in R^\ast &:\iff& (x,n) \in R\\ + &&\land \forall m.~(x,n) \le^\ast_\phi (x,m)\\ + &&\land \forall m.~\left( (x,n) <^\ast_\phi (x,m) \lor n \le m \right), + \end{IEEEeqnarray*} + i.e.~take the element with minimal rank + that has the minimal second coordinate among those elements. +\end{proof} +\begin{corollary}[Countable Reduction for $\Pi^1_1$ Sets] + Let $X$ be a Polish space + and $(C_n)_n$ a sequence of coanalytic subsets of $X$. + + Then there exists a sequence $(C_n^\ast)$ + of pairwise disjoint $\Pi^1_1$-sets + with $C_n^\ast \subseteq C_n$ + and + \[ + \bigcup_{n \in \N} C_n^\ast = \bigcup_{n \in \N} C_n. + \] +\end{corollary} +\begin{proof} + Define $R \subseteq X \times \N$ by setting + $(x,n) \in R :\iff x \in C_n$ + and apply \yaref{thm:uniformization}. +\end{proof} + +Let $X$ be a Polish space. +If $(X, \prec)$ is well founded (i.e.~there are no infinite descending chains) +then we define a rank $\rho_{y}\colon X > \Ord$ +as follows: +For minimal elements the rank is $0$. +Otherwise set $\rho_<(x) \coloneqq \sup \{\rho_<(y) + 1 : y \prec x\}$. +Let $\rho(\prec) \coloneqq \sup \{\rho_{\prec}(x) + 1 : x \in X\}$. + +\begin{exercise} + $\rho(\prec) \le |X|^+$ (successor cardinal). + (for countable $<$) + \todo{TODO} +\end{exercise} +\begin{theorem}[Kunen-Martin] + If $(X, \prec)$ is wellfounded + and $\prec \subseteq X^2$ is $\Sigma^1_1$ + then $\rho(\prec) < \omega_1$. +\end{theorem} +\begin{proof} + Wlog.~$X = \cN$. + There is a tree $S$ on $\N \times \N \times \N$ + (i.e.~$S \subseteq \cN^3$) + such that + \[ + \forall x, y \in \cN.~\left(x \succ y \iff \exists \alpha \in \N.~(x,\alpha,y) \in [S]\right). + \] + + Let + \[ + W \coloneqq \{w = (s_0,u_1,s_1,\ldots, u_n, s_n) : s_i, u_i \in \N^{n} + \land (s_{i-1}, u_i, s_i) \in S\}. + \] + + So $|W| \le \aleph_0$. + Define $\prec^\ast$ on $W$ + by setting + \[(s_0,u_1,s_1,\ldots, u_n,s_n) \succ (s_0',u_1', s_1', \ldots, u_m', s_m') :\iff\] + \begin{itemize} + \item $n < m$, + \item $\forall i \le n.~s_i \subsetneq s_i'$ and + \item $\forall i \le n.~u_i \subsetneq u_i'$. + \end{itemize} + \begin{claim} + $\prec^\ast$ is well-founded. + \end{claim} + \begin{subproof} + If $w_n = (s_0^n, u_1^n, \ldots, u_n^n, s_n^n)$ + was descending, + then let + \[ + x_i \coloneqq \bigcup s_i^n \in \cN + \] + and + \[ + \alpha_i \coloneqq \bigcup_n u_i^n \cN. + \] + We get $(x_{i-1}, \alpha_i, x_i) \in [S]$, + hence $x_{i-1} \succ x_i$ for all $i$, + but this is an infinite descending chain + in the original relation $\lightning$ + \end{subproof} + + + + \todo{Fix typos and end proof} +% Hence $\rho(\prec^\ast) < \omega_1$. +% +% We can turn $(X, \prec)$ into a tree $(T_\prec, \subsetneq)$ +% with +% \begin{IEEEeqnarray*}{rCl} +% \rho(\prec) &=& \rho(T_{\prec}) +% \end{IEEEeqnarray*} +% by setting $\emptyset \in T_{\prec}$ +% and +% $(x_0,\ldots,x_n) \in T_\prec$,$x_i \in X =\cN$, +% iff $x_0 \succ x_1 \succ x_2 \succ \ldots \succ x_n$. +% +% For all $x \succ y$ +% pick $\alpha_{x,y} \in \cN$ +% such that $(x, \alpha_{x,y}, y) \in [S]$ +% define +% \begin{IEEEeqnarray*}{rCl} +% \phi\colon T_{\prec} &\longrightarrow & W \\ +% \phi(x_0,x_1,\ldots,x_n) &\longmapsto & (x_0\defon{n}, \alpha_{x_0}, x_1\defon{n},\ldots, +% \alpha_{x_{n-1}}, x_n\defon{n}). +% \end{IEEEeqnarray*} +% Then $\phi$ is a homomorphism of $\subsetneq$ to $<^\ast$ +% and +% \[ +% \rho(<) = \rho(T_{\prec}, \subsetneq) \le \rho(<^\ast) < \omega_1. +% \] + +\todo{Exercise} +\end{proof} + + + diff --git a/logic3.tex b/logic3.tex index 300f8af..6e23238 100644 --- a/logic3.tex +++ b/logic3.tex @@ -37,6 +37,7 @@ \input{inputs/lecture_11} \input{inputs/lecture_12} \input{inputs/lecture_13} +\input{inputs/lecture_14}