Josia Pietsch 458dd9ab1f
Some checks failed
Build latex and deploy / checkout (push) Failing after 19m2s
fixed some typesetting problems
2024-02-09 20:23:05 +01:00

261 lines
9.4 KiB

If $C$ is coanalytic,
then there exists a $\Pi^1_1$-rank on $C$.
% TODO show that WO sse 2^QQ is Pi_1^1 complete
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 $x,y \in C'$
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 $E \subseteq \Q^{\Q} \times 2^{\Q} \times 2^{\Q}$
& (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))
$E$ is Borel\gist{ as a countable intersection of clopen sets}{}.
$x <^\ast_{\phi} y$
\item $(x, <_{\Q})$ is well ordered and
\item $(y, <_{\Q})$ does not order embed into $(x, <_{\Q})$,
where we identify $2^\Q$ and the powerset of $\Q$.
This is equivalent to
\item $x \in \WO$ and
\item $\forall f \in \Q^\Q.~(f,y,x) \not\in E$,
so it is $\Pi^1_1$.%
(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%
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 $(x <^\ast_\phi y)$
& &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))
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}.}
Let $\phi\colon R \to \Ord$
be a $\Pi^1_1$-rank.
(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),
i.e.~take}{Take} the element with minimal rank
that has the minimal second coordinate among those elements.
Uniformization also works for $R \in \Pi^1_1(X \times Y)$
for arbitrary Polish spaces $X,Y$,
\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 $C_n^\ast \subseteq C_n$
\bigcup_{n \in \N} C_n^\ast = \bigcup_{n \in \N} C_n.
Define $R \subseteq X \times \N$ by setting
$(x,n) \in R :\iff x \in C_n$
and apply \yaref{thm:uniformization}.
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 $\rho(\prec) < |X|^+$.%
\gist{\footnote{Here, $|X|^+$ denotes the successor cardinal.}}{}
\begin{theorem}[{Kunen-Martin, \cite[(31.1)]{kechris}}]
If $(X, \prec)$ is well-founded
and $\prec \subseteq X^2$ is $\Sigma^1_1$
then $\rho(\prec) < \omega_1$.
Wlog.~$X = \cN$.
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).%
\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 $S' = [S]$.
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
\[(s_0,u_1,s_1,\ldots, u_n,s_n) \succ^\ast (s_0',u_1', s_1', \ldots, u_m', s_m')\]
\item $n < m$ and
\item $\forall i \le n.~s_i \subsetneq s_i' \land u_i \subsetneq u_i'$.
% \footnote{sic! (there was a typo in the official notes)}
$\prec^\ast$ is well-founded.
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
\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.
Hence $\rho(\prec^\ast) < |W|^+ \le \omega_1$.
We can turn $(X, \prec)$ into a tree $(T_\prec, \subsetneq)$
\rho(\prec) &=& \rho(T_{\prec})
by setting $\emptyset \in T_{\prec}$
$(x_0,\ldots,x_n) \in T_\prec$,$x_i \in X =\cN$,
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 $(x, \alpha_{x,y}, y) \in [S]$.
\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}).
Then $\phi$ is a homomorphism of $\supsetneq$ to $\prec^\ast$
= \rho(T_{\prec} \setminus \{\emptyset\} , \supsetneq)
\le \rho(\prec^\ast)
< \omega_1.