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@ -1,4 +1,5 @@
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\lecture{13}{2023-11-08}{}
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\gist{%
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% Recap
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$\LO = \{x \in 2^{\N\times \N} : x \text{ is a linear order}\} $.
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@ -124,7 +125,8 @@ with $(f^{-1}(\{1\}), <)$.
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and $f$ is continuous.
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\end{proof}
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% TODO: new section?
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\subsection{$\Pi_1^1$-ranks}
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\gist{%
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Recall that a \vocab{rank} on a set $C$
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is a map $\phi\colon C \to \Ord$.
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@ -1,47 +1,52 @@
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\lecture{14}{2023-12-01}{}
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% TODO ANKI-MARKER
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\begin{theorem}[Moschovakis]
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If $C$ is coanalytic,
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then there exists a $\Pi^1_1$-rank on $C$.
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\end{theorem}
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\begin{proof}
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\gist{%
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Pick a $\Pi^1_1$-complete set.
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It suffices to show that there is a rank on it.
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Then use the reduction to transfer
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it to any coanalytic set,
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i.e.~for $x,y \in C'$
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i.e.~for $x,y \in C'$
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let
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\[
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x \le^{\ast}_{C'} y :\iff f(x) \le^\ast_{C} f(y)
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\]
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\]
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and similarly for $<^\ast$.
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% https://q.uiver.app/#q=WzAsNSxbMCwwLCJZIl0sWzIsMCwiWCJdLFswLDEsIkMnIl0sWzIsMSwiQyJdLFsyLDIsIlxcUGlfMV4xLVxcdGV4dHtjb21wbGV0ZX0iXSxbMCwxLCJmIl0sWzIsMCwiXFxzdWJzZXRlcSIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzMsMSwiXFxzdWJzZXRlcSIsMix7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV1d
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\[\begin{tikzcd}
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Y && X \\
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{C'} && C \\
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&& {\Pi_1^1-\text{complete}}
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\arrow["f", from=1-1, to=1-3]
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\arrow["\subseteq", hook, from=2-1, to=1-1]
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\arrow["\subseteq"', hook, from=2-3, to=1-3]
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\end{tikzcd}\]
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% https://q.uiver.app/#q=WzAsNCxbMCwwLCJZIl0sWzIsMCwiWCJdLFswLDEsIkMnIl0sWzIsMSwiXFxiZWdpbnthcnJheX17Y31DXFxcXFxcUGlfMV4xLVxcdGV4dHtjb21wbGV0ZX1cXGVuZHthcnJheX0iXSxbMCwxLCJmIl0sWzIsMCwiXFxzdWJzZXRlcSIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzMsMSwiXFxzdWJzZXRlcSIsMix7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV1d
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\begin{tikzcd}
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Y && X \\
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{C'} && {\begin{array}{c}C\\\Pi_1^1-\text{complete}\end{array}}
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\arrow["f", from=1-1, to=1-3]
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\arrow["\subseteq", hook, from=2-1, to=1-1]
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\arrow["\subseteq"', hook, from=2-3, to=1-3]
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\end{tikzcd}
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Let $X = 2^{\Q} \supseteq \WO$.
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We have already show that $\WO$ is $\Pi^1_1$-complete.
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Set $\phi(x) \coloneqq \otp(x)$
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($\otp\colon \WO \to \Ord$ denotes the order type).
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We show that this is a $\Pi^1_1$-rank.
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}{%
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It suffices to show this for a $\Pi^1_1$-complete set.
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We show that $\phi \coloneqq \otp$ is a
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$\Pi^1_1$-rank for $\WO \subseteq 2^{\Q}$.
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}
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Define $E \subseteq \Q^{\Q} \times 2^{\Q} \times 2^{\Q}$
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by
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\begin{IEEEeqnarray*}{rCl}
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(f,x,y) \in E &:\iff& f \text{ order embeds $(x, \le_{\Q})$ to $(y,\le _{\Q})$}\\
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&\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))
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\begin{IEEEeqnarray*}{Cl}
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& (f,x,y) \in E\\
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:\iff& f \text{ order embeds $(x, \le_{\Q})$ to $(y,\le _{\Q})$}\\
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\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))
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\end{IEEEeqnarray*}
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$E$ is Borel as a countable intersection of clopen sets.
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$E$ is Borel\gist{ as a countable intersection of clopen sets}{}.
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Define
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$x <^\ast_{\phi}$
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$x <^\ast_{\phi} y$
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iff
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\begin{itemize}
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\item $(x, <_{\Q})$ is well ordered and
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@ -51,8 +56,9 @@
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This is equivalent to
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\begin{itemize}
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\item $x \in \WO$ and
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\item $\forall f \in \Q^\Q.~(f,y,x) \not\in E$.
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\item $\forall f \in \Q^\Q.~(f,y,x) \not\in E$,
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\end{itemize}
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so it is $\Pi^1_1$.
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Furthermore $x \le_\phi^\ast y \iff$
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either $x <^\ast_\phi y$ or
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@ -61,12 +67,11 @@
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i.e.~either $x<^\ast_\phi y$ or
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$x,y \in \WO$ and any order embedding of $(x,<_{\Q})$ to
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$(y, <_{\Q})$ is cofinal%
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\footnote{%
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\gist{\footnote{%
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Recall that $A \subseteq (x,<_{\Q})$
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is \vocab{cofinal} if $\forall t \in x.~\exists a \in A.~t\le _{\Q} a$.%
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}
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in $(y, <_\Q)$ is
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cofinal in $(y, <_{\Q})$ and vice versa.
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}}{}
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in $(y, <_\Q)$ and vice versa.
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Equivalently, either $(x <^\ast_\phi y)$
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or
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\begin{IEEEeqnarray*}{rCl}
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\end{IEEEeqnarray*}
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\end{proof}
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\begin{theorem}
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\label{thm:uniformization}
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Let $X$ be Polish and $R \subseteq X \times \N$ by $\Pi^1_1$
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\]
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We say that $R^\ast$ \vocab[uniformization]{uniformizes} $R$.%
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\footnote{Wikimedia has a \href{https://upload.wikimedia.org/wikipedia/commons/4/4c/Uniformization_ill.png}{nice picture.}}
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\end{theorem}
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\begin{proof}
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Let $\phi\colon R \to \Ord$
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and
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\[
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\bigcup_{n \in \N} C_n^\ast = \bigcup_{n \in \N} C_n.
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\]
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\]
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\end{corollary}
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\begin{proof}
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Define $R \subseteq X \times \N$ by setting
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$(x,n) \in R :\iff x \in C_n$
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$(x,n) \in R :\iff x \in C_n$
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and apply \yaref{thm:uniformization}.
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\end{proof}
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Let $X$ be a Polish space.
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If $(X, \prec)$ is well founded (i.e.~there are no infinite descending chains)
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then we define a rank $\rho_{y}\colon X > \Ord$
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as follows:
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If $(X, \prec)$ is well-founded%
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\gist{ (i.e.~there are no infinite descending chains)}{}
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then we define a rank $\rho_{y}\colon X \to \Ord$ as follows:
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For minimal elements the rank is $0$.
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Otherwise set $\rho_<(x) \coloneqq \sup \{\rho_<(y) + 1 : y \prec x\}$.
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Otherwise set $\rho_\prec(x) \coloneqq \sup \{\rho_\prec(y) + 1 : y \prec x\}$.
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Let $\rho(\prec) \coloneqq \sup \{\rho_{\prec}(x) + 1 : x \in X\}$.
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\begin{exercise}
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$\rho(\prec) \le |X|^+$ (successor cardinal).
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(for countable $<$)
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\todo{TODO}
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\end{exercise}
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% \begin{exercise}
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% $\rho(\prec) \le |X|^+$ (successor cardinal).
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% (for countable $<$)
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% \todo{TODO}
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% % TODO QUESTION
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% \end{exercise}
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\begin{theorem}[Kunen-Martin]
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\yalabel{Kunen-Martin}{Kunen-Martin}{thm:kunenmartin}
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If $(X, \prec)$ is wellfounded
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@ -140,80 +144,85 @@ Let $\rho(\prec) \coloneqq \sup \{\rho_{\prec}(x) + 1 : x \in X\}$.
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then $\rho(\prec) < \omega_1$.
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\end{theorem}
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\begin{proof}
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% TODO GIST
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% TODO QUESTION where did we use analytic?
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Wlog.~$X = \cN$.
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There is a tree $S$ on $\N \times \N \times \N$
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(i.e.~$S \subseteq \cN^3$)
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such that
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(i.e.~$S \subseteq (\N \times \N \times \N)^{<\N}$)
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such that
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\[
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\forall x, y \in \cN.~\left(x \succ y \iff \exists \alpha \in \N.~(x,\alpha,y) \in [S]\right).
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\]
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\forall x, y \in \cN.~\left(x \succ y \iff \exists \alpha \in \cN.~(x,\alpha,y) \in [S]\right).
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\]
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Let
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\[
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W \coloneqq \{w = (s_0,u_1,s_1,\ldots, u_n, s_n) : s_i, u_i \in \N^{n}
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\land (s_{i-1}, u_i, s_i) \in S\}.
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\]
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\]
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So $|W| \le \aleph_0$.
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Clearly $|W| \le \aleph_0$.
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Define $\prec^\ast$ on $W$
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by setting
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\[(s_0,u_1,s_1,\ldots, u_n,s_n) \succ (s_0',u_1', s_1', \ldots, u_m', s_m') :\iff\]
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\begin{itemize}
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\item $n < m$,
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\item $\forall i \le n.~s_i \subsetneq s_i'$ and
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\item $\forall i \le n.~u_i \subsetneq u_i'$.
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\item $n < m$ and
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\item $\forall i \le n.~s_i \subsetneq s_i' \land u_i \subsetneq u_i'$.
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%\todo{$\subseteq$ or $\subsetneq$?} % TODO QUESTION
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\end{itemize}
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\begin{claim}
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$\prec^\ast$ is well-founded.
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\end{claim}
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\begin{subproof}
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If $w_n = (s_0^n, u_1^n, \ldots, u_n^n, s_n^n)$
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was descending,
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then let
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\[
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x_i \coloneqq \bigcup s_i^n \in \cN
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\]
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and
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\[
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\alpha_i \coloneqq \bigcup_n u_i^n \cN.
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\]
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We get $(x_{i-1}, \alpha_i, x_i) \in [S]$,
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hence $x_{i-1} \succ x_i$ for all $i$,
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but this is an infinite descending chain
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in the original relation $\lightning$
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\gist{%
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If $w_n = (s_0^n, u_1^n, \ldots, u_n^n, s_n^n)$
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was descending,
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then let
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\[
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x_i \coloneqq \bigcup s_i^n \in \cN
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\]
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and
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\[
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\alpha_i \coloneqq \bigcup_n u_i^n \cN.
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\]
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We get $(x_{i-1}, \alpha_i, x_i) \in [S]$,
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hence $x_{i-1} \succ x_i$ for all $i$,
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but this is an infinite descending chain
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in the original relation $\lightning$
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}{%
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Use that $ \prec$ is well-founded.
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}
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\end{subproof}
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Hence $\rho(\prec^\ast) < |W|^+ \le \omega_1$.
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% TODO QUESTION
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We can turn $(X, \prec)$ into a tree $(T_\prec, \subsetneq)$
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with
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\begin{IEEEeqnarray*}{rCl}
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\rho(\prec) &=& \rho(T_{\prec})
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\end{IEEEeqnarray*}
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by setting $\emptyset \in T_{\prec}$
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and
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$(x_0,\ldots,x_n) \in T_\prec$,$x_i \in X =\cN$,
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iff $x_0 \succ x_1 \succ x_2 \succ \ldots \succ x_n$.
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\todo{Fix typos and end proof}
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% Hence $\rho(\prec^\ast) < \omega_1$.
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%
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% We can turn $(X, \prec)$ into a tree $(T_\prec, \subsetneq)$
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% with
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% \begin{IEEEeqnarray*}{rCl}
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% \rho(\prec) &=& \rho(T_{\prec})
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% \end{IEEEeqnarray*}
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% by setting $\emptyset \in T_{\prec}$
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% and
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% $(x_0,\ldots,x_n) \in T_\prec$,$x_i \in X =\cN$,
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% iff $x_0 \succ x_1 \succ x_2 \succ \ldots \succ x_n$.
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%
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% For all $x \succ y$
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% pick $\alpha_{x,y} \in \cN$
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% such that $(x, \alpha_{x,y}, y) \in [S]$
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% define
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% \begin{IEEEeqnarray*}{rCl}
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% \phi\colon T_{\prec} &\longrightarrow & W \\
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% \phi(x_0,x_1,\ldots,x_n) &\longmapsto & (x_0\defon{n}, \alpha_{x_0}, x_1\defon{n},\ldots,
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% \alpha_{x_{n-1}}, x_n\defon{n}).
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% \end{IEEEeqnarray*}
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% Then $\phi$ is a homomorphism of $\subsetneq$ to $<^\ast$
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% and
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% \[
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% \rho(<) = \rho(T_{\prec}, \subsetneq) \le \rho(<^\ast) < \omega_1.
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% \]
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\todo{Exercise}
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For all $x \succ y$
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pick $\alpha_{x,y} \in \cN$
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such that $(x, \alpha_{x,y}, y) \in [S]$
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define
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\begin{IEEEeqnarray*}{rCl}
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\phi\colon T_{\prec} \setminus \{\emptyset\} &\longrightarrow & W \\
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\phi(x_0,x_1,\ldots,x_n) &\longmapsto & (x_0\defon{n}, \alpha_{x_0}, x_1\defon{n},\ldots,
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\alpha_{x_{n-1}}, x_n\defon{n}).
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\end{IEEEeqnarray*}
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Then $\phi$ is a homomorphism of $\subsetneq$ to $\prec^\ast$
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so
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\[
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\rho(\prec)
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= \rho(T_{\prec} \setminus \{\emptyset\} , \subsetneq)
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\le \rho(\prec^\ast)
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< \omega_1.
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\]
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\end{proof}
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@ -1,5 +1,8 @@
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\lecture{15}{2023-12-05}{}
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% TODO ANKI-MARKER
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\begin{theorem}[Boundedness Theorem]
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\yalabel{Boundedness Theorem}{Boundedness}{thm:boundedness}
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