\lecture{13}{2023-11-08}{}
\gist{%
% Recap
$\LO = \{x \in 2^{\N\times \N} : x \text{ is a linear order}\} $.
$\LO \subseteq 2^{\N \times \N}$ is closed
and $\WO = \{x \in  \LO: x \text{ is a wellordering}\} $
 is coanalytic in $\LO$.
% End Recap
}{}

Another way to code linear orders:

Consider $(\Q, <)$, the rationals with the usual order.
We can view $2^{\Q}$ as the space of linear orders
embeddable into $\Q$,
by associating a function $f\colon \Q \to \{0,1\}$
with $(f^{-1}(\{1\}), <)$.

\begin{lemma}
    Any countable ordinal embeds into $(\Q,<)$.
\end{lemma}
\begin{proof}[sketch]
    Use transfinite induction.
    Suppose we already have $\alpha \hookrightarrow (\Q, <)$,
    we need to show that $\alpha +1 \hookrightarrow (\Q, <)$.
    Since $(0,1) \cap \Q \cong \Q$,
    we may assume $\alpha \hookrightarrow ((0,1), <)$
    and can just set $\alpha \mapsto 2$.

    For a limit $\alpha$
    take a countable cofinal subsequence $\alpha_1 < \alpha_2 < \ldots \to \alpha$.
    Then map $[0,\alpha_1)$ to $(0,1)$
    and  $[\alpha_i, \alpha_{i+1})$ to $(i,i+1)$.
\end{proof}


\begin{definition}[\vocab{Kleene-Brouwer ordering}]
    Let $(A,<)$ be a linear order and $A$ countable.
    We define the linear order $<_{KB}$ on $A^{<\N}$
    as follows:
    Let
    \[
    s = (s_0,\ldots,s_{m-1}), t = (t_0, \ldots, t_{n-1}).
    \]
    We set $s < t$ iff
    \begin{itemize}
        \item $(s \supsetneq t)$ or
        \item $s_i < t_i$ for the minimal $i$  such that
            $s_i \neq t_i$.
    \end{itemize}
\end{definition}

\begin{proposition}
    Suppose that $(A, <)$ is a countable well ordering.
    Then for a tree $T \subseteq A^{<\N}$ on $A$,
    Then $T$ is well-founded iff
    $(T, <_{KB}\defon{T})$ is well ordered.
\end{proposition}
\begin{proof}
    \gist{%
        If $T$ is ill-founded and $x \in [T]$,
        then for all $n$, we have $x\defon{n+1} <_{KB} x\defon{n}$.
        Thus $(T, <_{KB}\defon{T})$ is not well ordered.

        Conversely, let $<\defon{KB}$ be not a well-ordering
        on $T$.
        Let $s_0 >_{KB} s_1 >_{KB} s_2 >_{KB} \ldots$
        be an infinite descending chain.
        We have that $s_0(0) \ge s_1(0) \ge s_2(0) \ge \ldots$
        stabilizes for $n > n_0$.
        Let $a_0 \coloneqq  s_{n_0}(0)$.
        Now for $n \ge  n_0$ we have that $s_n(0)$ is constant,
        hence for $n > n_0$ the value $s_{n}(1)$ must be defined.
        Thus there is $n_1 \ge n_0$ such that $s_n(1)$
        is constant for all $n \ge n_1$.
        Let $a_1 \coloneqq s_{n_1}(1)$
        and so on.
        Then $(a_0,a_1,a_2, \ldots) \in [T]$.
    }{easy}
\end{proof}

\begin{theorem}[Lusin-Sierpinski]
    The set $\LO \setminus \WO$
    (resp.~$2^{\Q} \setminus \WO$)
    is $\Sigma_1^1$-complete.
\end{theorem}
\begin{proof}
    We will find a continuous function
    $f\colon \Tr \to \LO$ such that
    \gist{%
    \[
    x \in \WF \iff f(x) \in \WO
    \]
    (equivalently $x \in \IF \iff f(x) \in \LO \setminus \WO$).
    This suffices, since $\IF \subseteq \Tr$ is $\Sigma^1_1$-complete
    }{
         $f^{-1}(\LO \setminus \WO) = \IF$.
         This suffices
    }
    (see \yaref{cor:ifs11c}).

    Fix a bijection $b\colon \N \to \N^{<\N}$.

    \begin{idea}
        For $T \in \Tr$ consider
        $<_{KB}\defon{T}$.
    \end{idea}

    Let $\alpha \in \Tr$.
    For $m,n \in \N$ define $f(\alpha)(m,n) \coloneqq 1$
    (i.e.~$m \le_{f(\alpha)} n$)
    iff
    \begin{itemize}
        \item $\alpha(b(m)) = \alpha(b(n)) = 1$
            and $b(m) \le_{KB} b(n)$
            (recall that we identified $\Tr$
            with a subset of ${2^{\N}}^{<\N}$),
            or
        \item $\alpha(b(m)) = 1$ and $\alpha(b(n)) = 0$ or
        \item $\alpha(b(m)) = \alpha(b(n)) = 0$ and $m \le n$.
    \end{itemize}

    Then $\alpha \in \WF \iff f(\alpha) \in \WO$
    and $f$ is continuous.
\end{proof}

% TODO: new section?
\gist{%
    Recall that a \vocab{rank} on a set $C$
    is a map $\phi\colon C \to \Ord$.
    \begin{example}
        \begin{IEEEeqnarray*}{rCl}
            \otp \colon \WO &\longrightarrow & \Ord \\
             x &\longmapsto & \text{the unique $\alpha \in \Ord$ such that $x \cong \alpha$}.
        \end{IEEEeqnarray*}
    \end{example}
}{}

\begin{definition}
    A \vocab{prewellordering} $\preceq$
    on  a set $C$
    is a binary relation that is
    \begin{itemize}
        \item reflexive,
        \item transitive,
        \item total (any two $x,y$ are comparable),
        \item $\prec$ \gist{($x \prec y \iff x \preceq y \land y \not\preceq x$)}{} is well-founded,
            in the sense that there are no descending infinite chains.
    \end{itemize}
\end{definition}
\begin{remark}
    \begin{itemize}
        \item A prewellordering may not be a linear order since
            it is not necessarily antisymmetric.
        %\item The linearly ordered wellfounded sets are exactly the wellordered sets.
        \item Modding out $x \sim y :\iff x \preceq y \land y \preceq x$
            turns a prewellordering into a wellordering.
    \end{itemize}
\end{remark}

We have the following correspondence
between downwards-closed ranks and prewellorderings:
\begin{IEEEeqnarray*}{rCl}
    \text{ranks}&\longrightarrow & \text{prewellorderings} \\
    (\phi\colon C \to \Ord) &\longmapsto & (x \le_{\phi} y :\iff \phi(x) \le \phi(y), x,y \in C)\\
                            \phi_{\preceq}&\longmapsfrom& \preceq,
\end{IEEEeqnarray*}
where $\phi_\preceq(x)$ is defined as
\gist{\begin{IEEEeqnarray*}{rCl}
    \phi_{\preceq}(x) &\coloneqq &0 \text{ if $x$ is minimal},\\
    \phi_{\preceq}(x) &\coloneqq & \sup \{\phi_{\preceq}(y) + 1 : y \prec x\},
\end{IEEEeqnarray*}
i.e.}{}
\[
    \phi_{\preceq}(x) = \otp\left(\faktor{\{y \in C : y \prec x\}}{\sim}\right).
\]

\begin{definition}
    Let $X$ be Polish and $C \subseteq  X$ coanalytic.
    Then $\phi\colon  C \to \Ord$
    is a  \vocab[Rank!$\Pi^1_1$-rank]{$\Pi^1_1$-rank}
    provided that $\le^\ast$ and $<^\ast$ are coanalytic,
    where
    $x \le^\ast_{\phi} y$
    iff
    \begin{itemize}
        \item $y \in  X \setminus C \land x \in C$ or
        \item $x,y \in C \land \phi(x) \le  \phi(y)$
    \end{itemize}
    and similarly for $<^\ast_{\phi}$.
\end{definition}