Josia Pietsch bbb891594c
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% 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 $(f^{-1}(\{1\}), <)$.
Any countable wellorder embeds into $(\Q,<)$.
\begin{proof}\footnote{In the lecture this was only done for countable \emph{ordinals}.}
% TODO $\WF \subseteq 2^\Q$ is $\Sigma^1_1$-complete.
\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:
s = (s_0,\ldots,s_{m-1}), t = (t_0, \ldots, t_{n-1}).
We set $s < t$ iff
\item $(s \supsetneq t)$ or
\item $s_i < t_i$ for the minimal $i$ such that
$s_i \neq t_i$.
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.
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 $(a_0,a_1,a_2, \ldots) \in [T]$.
The set $\LO \setminus \WO$
(resp.~$2^{\Q} \setminus \WO$)
is $\Sigma_1^1$-complete.
We will find a continuous function
$f\colon \Tr \to \LO$ such that
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 $b\colon \N \to \N^{<\N}$.
For $T \in \Tr$ consider
Let $\alpha \in \Tr$.
For $m,n \in \N$ define $f(\alpha)(m,n) \coloneqq 1$
(i.e.~$m \le_{f(\alpha)} n$)
\item $\alpha(b(m)) = \alpha(b(n)) = 1$
and $b(m) \le_{KB} b(n)$
(recall that we identified $\Tr$
with a subset of ${2^{\N}}^{<\N}$),
\item $\alpha(b(m)) = 1$ and $\alpha(b(n)) = 0$ or
\item $\alpha(b(m)) = \alpha(b(n)) = 0$ and $m \le n$.
Then $\alpha \in \WF \iff f(\alpha) \in \WO$
and $f$ is continuous.
\subsection{\texorpdfstring{$\Pi_1^1$-ranks}{Coanalytic Ranks}}
Recall that a \vocab{rank} on a set $C$
is a map $\phi\colon C \to \Ord$.
\otp \colon \WO &\longrightarrow & \Ord \\
x &\longmapsto & \text{the unique $\alpha \in \Ord$ such that $x \cong \alpha$}.
A \vocab{prewellordering} $\preceq$
on a set $C$
is a binary relation that is
\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.
\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.
We have the following correspondence
between downwards-closed ranks and prewellorderings:
\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,
where $\phi_\preceq(x)$ is defined as
\phi_{\preceq}(x) &\coloneqq &0 \text{ if $x$ is minimal},\\
\phi_{\preceq}(x) &\coloneqq & \sup \{\phi_{\preceq}(y) + 1 : y \prec x\},
\phi_{\preceq}(x) = \otp\left(\faktor{\{y \in C : y \prec x\}}{\sim}\right).
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 $\le^\ast_\phi$ and $<^\ast_\phi$ are coanalytic subsets of $X \times X$,
$x \le^\ast_{\phi} y$
\item $y \in X \setminus C \land x \in C$ or
\item $x,y \in C \land \phi(x) \le \phi(y)$
and similarly for $<^\ast_{\phi}$.