w23-logic-3/inputs/lecture_14.tex

\lecture{14}{2023-12-01}{}

\begin{theorem}[Moschovakis]
    If $C$ is coanalytic,
    then there exists a $\Pi^1_1$-rank on $C$.
\end{theorem}
% TODO show that WO sse 2^QQ is Pi_1^1 complete
\begin{proof}
    \gist{%
    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=WzAsNCxbMCwwLCJZIl0sWzIsMCwiWCJdLFswLDEsIkMnIl0sWzIsMSwiXFxiZWdpbnthcnJheX17Y31DXFxcXFxcUGlfMV4xLVxcdGV4dHtjb21wbGV0ZX1cXGVuZHthcnJheX0iXSxbMCwxLCJmIl0sWzIsMCwiXFxzdWJzZXRlcSIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzMsMSwiXFxzdWJzZXRlcSIsMix7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV1d
    % \begin{tikzcd}
    % 	Y && X \\
    % 	{C'} && {\begin{array}{c}C\\\Pi_1^1-\text{complete}\end{array}}
    % 	\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 shown that $\WO$ is $\Pi^1_1$-complete.% TODO REF

    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.

    }{%
        It suffices to show this for a $\Pi^1_1$-complete set.
        We show that $\phi \coloneqq \otp$ is a
        $\Pi^1_1$-rank for $\WO \subseteq 2^{\Q}$.
    }
    Define $E \subseteq \Q^{\Q} \times 2^{\Q} \times  2^{\Q}$
    by
    \begin{IEEEeqnarray*}{Cl}
        & (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\gist{ as a countable intersection of clopen sets}{}.

    Define
    $x <^\ast_{\phi} y$
    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}
    so it is $\Pi^1_1$.%
    \gist{\footnote{%
            (very informal)
            Note that $\Sigma^1_1$-sets work well with comprehensions using ``$\exists$'':
            Writing $A \in \Sigma^1_1(X)$ as $A = \proj_X(B)$
            for some Borel set $B \subseteq X \times Y$,
            the second coordinate can be thought of
            as being a witness for a statement.
            Likewise, being complements of $\Sigma^1_1$-sets,
            $\Pi_1^1$-sets can capture that a witness does not exist,
            i.e.~they interact nicely with ``$\forall$''.%
    }}{}


    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%
    \gist{\footnote{%
        Recall that $A \subseteq (x,<_{\Q})$
        is \vocab{cofinal}  if $\forall  t \in x.~\exists a \in A.~t\le _{\Q} a$.%
    }}{}
    in $(y, <_\Q)$ 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} $R$.%
    \footnote{Wikimedia has a nice \href{https://upload.wikimedia.org/wikipedia/commons/4/4c/Uniformization_ill.png}{picture}.}
\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}
\gist{
    \begin{remark}
        Uniformization also works for $R \in \Pi^1_1(X \times Y)$
        for arbitrary Polish spaces $X,Y$,
        cf.~\cite[(36.12)]{kechris}.
    \end{remark}
}{}
\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%
\gist{ (i.e.~there are no infinite descending chains)}{}
then we define a rank $\rho_{\prec}\colon X \to  \Ord$ as follows:
For minimal elements the rank is $0$.
Otherwise set $\rho_\prec(x) \coloneqq  \sup \{\rho_\prec(y) + 1 : y \prec x\}$.
Let $\rho(\prec) \coloneqq  \sup \{\rho_{\prec}(x) + 1 : x \in X\}$.

\begin{fact}[{\cite[Appendix B]{kechris}}]
    Since $\rho_\prec\colon X \to \rho(\prec)$ is surjective,
    we have that $\rho(\prec) \le  |X|^+$.%
    \gist{\footnote{Here, $|X|^+$ denotes the successor cardinal.}}{}
\end{fact}
\begin{theorem}[{Kunen-Martin, \cite[(31.1)]{kechris}}]
    \yalabel{Kunen-Martin}{Kunen-Martin}{thm:kunenmartin}
    If $(X, \prec)$ is well-founded
    and $\prec \subseteq X^2$ is $\Sigma^1_1$
    then $\rho(\prec) < \omega_1$.
\end{theorem}
\begin{proof}
    Wlog.~$X = \cN$.
    \gist{%
        There is a tree $S$ on $\N \times \N \times \N$
        (i.e.~$S \subseteq (\N \times \N \times \N)^{<\N}$)
        such that
        \[
        \forall x, y \in \cN.~\left(x \succ y
            \iff \exists \alpha \in \cN.~(x,\alpha,y) \in [S]\right).%
            \footnotemark
        \]
        \footnotetext{Here we use that $\prec$ is analytic,
        i.e.~$\prec$ can be written as the projection of a closed
        subset of $(\cN \times \cN) \times \cN$ and closed subsets
        correspond to pruned trees.}
    }{%
        Take $S' \overset{\text{closed}}{\subseteq} \cN \times \cN \times \cN$
        such that $\prec = \proj_{1,3}(S')$
        and $S$ a tree on $\N \times  \N \times \N$
        such that $S' = [S]$.
    }

    Let
    \[
    W \coloneqq \{w = (s_0,u_1,s_1,\ldots, u_n, s_n) : s_i, u_i \in \N^{n}
    \land (s_{i-1}, u_i, s_i) \in S\}.
    \]

    Clearly $|W| \le \aleph_0$.
    Define $\prec^\ast$ on $W$
    by setting
    \[(s_0,u_1,s_1,\ldots, u_n,s_n) \succ^\ast (s_0',u_1', s_1', \ldots, u_m', s_m') :\iff\]
    \begin{itemize}
        \item $n < m$ and
        \item $\forall i \le n.~s_i \subsetneq s_i' \land u_i \subsetneq u_i'$.%
            \footnote{sic! (there is a typo in the official notes)}
    \end{itemize}
    \begin{claim}
        $\prec^\ast$ is well-founded.
    \end{claim}
    \begin{subproof}
        \gist{%
            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
             \[
             \alpha_i \coloneqq  \bigcup_n u_i^n \in\cN.
            \]
            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 $\lightning$
        }{%
            Use that $ \prec$ is well-founded.
        }
    \end{subproof}
    Hence $\rho(\prec^\ast) < |W|^+ \le \omega_1$.
    \gist{%
        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 $x_0 \succ x_1 \succ x_2 \succ \ldots \succ x_n$.
    }{%
        Turn $(X,\prec)$ into a tree $(T_\prec, \subsetneq)$
        with $\rho(\prec) = \rho(T_\prec)$,
         $(x_0,\ldots,x_n) \in T_\prec$ iff $x_0 \succ \ldots \succ x_n$.
    }

    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} \setminus \{\emptyset\}  &\longrightarrow & W \\
        (x_0,x_1,\ldots,x_n) &\longmapsto & (x_0\defon{n}, \alpha_{x_0,x_1}\defon{n} , x_1\defon{n},\ldots,
        \alpha_{x_{n-1},x_n}\defon{n} , x_n\defon{n}).
    \end{IEEEeqnarray*}
    Then $\phi$ is a homomorphism of $\supsetneq$ to $\prec^\ast$
    so
    \[
    \rho(\prec)
    = \rho(T_{\prec} \setminus \{\emptyset\} , \supsetneq)
    \le \rho(\prec^\ast)
    < \omega_1.
    \]
\end{proof}