\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}



