\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}