14

inputs/lecture_13.tex

@@ -1,4 +1,5 @@
 \lecture{13}{2023-11-08}{}
+
 \gist{%
 % Recap
 $\LO = \{x \in 2^{\N\times \N} : x \text{ is a linear order}\} $.
@@ -124,7 +125,8 @@ with $(f^{-1}(\{1\}), <)$.
     and $f$ is continuous.
 \end{proof}
 
-% TODO: new section?
+\subsection{$\Pi_1^1$-ranks}
+
 \gist{%
     Recall that a \vocab{rank} on a set $C$
     is a map $\phi\colon C \to \Ord$.

inputs/lecture_14.tex

@@ -1,11 +1,11 @@
 \lecture{14}{2023-12-01}{}
 
-% TODO ANKI-MARKER
 \begin{theorem}[Moschovakis]
     If $C$ is coanalytic,
     then there exists a $\Pi^1_1$-rank on $C$.
 \end{theorem}
 \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
@@ -16,32 +16,37 @@
     x \le^{\ast}_{C'} y :\iff f(x) \le^\ast_{C} f(y)
     \]
     and similarly for $<^\ast$.
-    % https://q.uiver.app/#q=WzAsNSxbMCwwLCJZIl0sWzIsMCwiWCJdLFswLDEsIkMnIl0sWzIsMSwiQyJdLFsyLDIsIlxcUGlfMV4xLVxcdGV4dHtjb21wbGV0ZX0iXSxbMCwxLCJmIl0sWzIsMCwiXFxzdWJzZXRlcSIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzMsMSwiXFxzdWJzZXRlcSIsMix7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV1d
-\[\begin{tikzcd}
-	Y && X \\
-	{C'} && C \\
-	&& {\Pi_1^1-\text{complete}}
-	\arrow["f", from=1-1, to=1-3]
-	\arrow["\subseteq", hook, from=2-1, to=1-1]
-	\arrow["\subseteq"', hook, from=2-3, to=1-3]
-\end{tikzcd}\]
-
+    % 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 show that $\WO$ is $\Pi^1_1$-complete.
+
     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*}{rCl}
-        (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))
+    \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 as a countable intersection of clopen sets.
+    $E$ is Borel\gist{ as a countable intersection of clopen sets}{}.
 
     Define
-    $x <^\ast_{\phi}$
+    $x <^\ast_{\phi} y$
     iff
     \begin{itemize}
         \item $(x, <_{\Q})$ is well ordered and
@@ -51,8 +56,9 @@
     This is equivalent to
     \begin{itemize}
         \item $x \in \WO$ and
-        \item $\forall f \in \Q^\Q.~(f,y,x) \not\in E$.
+        \item $\forall f \in \Q^\Q.~(f,y,x) \not\in E$,
     \end{itemize}
+    so it is $\Pi^1_1$.
 
     Furthermore $x \le_\phi^\ast y \iff$
     either $x <^\ast_\phi y$ or
@@ -61,12 +67,11 @@
     i.e.~either $x<^\ast_\phi y$ or
     $x,y \in \WO$ and any order embedding of $(x,<_{\Q})$ to
     $(y, <_{\Q})$ is cofinal%
-    \footnote{%
+    \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 $(y, <_\Q)$ is
-    cofinal in $(y, <_{\Q})$ and vice versa.
+    }}{}
+    in $(y, <_\Q)$ and vice versa.
     Equivalently, either $(x <^\ast_\phi y)$
     or
     \begin{IEEEeqnarray*}{rCl}
@@ -76,7 +81,6 @@
     \end{IEEEeqnarray*}
 \end{proof}
 
-
 \begin{theorem}
     \label{thm:uniformization}
     Let $X$ be Polish and $R \subseteq X \times \N$ by $\Pi^1_1$
@@ -88,7 +92,6 @@
     \]
     We say that $R^\ast$ \vocab[uniformization]{uniformizes} $R$.%
     \footnote{Wikimedia has a \href{https://upload.wikimedia.org/wikipedia/commons/4/4c/Uniformization_ill.png}{nice picture.}}
-
 \end{theorem}
 \begin{proof}
     Let $\phi\colon R \to  \Ord$
@@ -121,18 +124,19 @@
 \end{proof}
 
 Let $X$ be a Polish space.
-If $(X, \prec)$ is well founded (i.e.~there are no infinite descending chains)
-then we define a rank $\rho_{y}\colon X > \Ord$
-as follows:
+If $(X, \prec)$ is well-founded%
+\gist{ (i.e.~there are no infinite descending chains)}{}
+then we define a rank $\rho_{y}\colon X \to  \Ord$ as follows:
 For minimal elements the rank is $0$.
-Otherwise set $\rho_<(x) \coloneqq  \sup \{\rho_<(y) + 1 : y \prec x\}$.
+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{exercise}
-    $\rho(\prec) \le  |X|^+$ (successor cardinal).
-    (for countable $<$)
-    \todo{TODO}
-\end{exercise}
+% \begin{exercise}
+%     $\rho(\prec) \le  |X|^+$ (successor cardinal).
+%     (for countable $<$)
+%     \todo{TODO}
+%     % TODO QUESTION
+% \end{exercise}
 \begin{theorem}[Kunen-Martin]
     \yalabel{Kunen-Martin}{Kunen-Martin}{thm:kunenmartin}
     If $(X, \prec)$ is wellfounded
@@ -140,12 +144,14 @@ Let $\rho(\prec) \coloneqq  \sup \{\rho_{\prec}(x) + 1 : x \in X\}$.
     then $\rho(\prec) < \omega_1$.
 \end{theorem}
 \begin{proof}
+    % TODO GIST
+    % TODO QUESTION where did we use analytic?
     Wlog.~$X = \cN$.
     There is a tree $S$ on $\N \times \N \times \N$
-    (i.e.~$S \subseteq \cN^3$)
+    (i.e.~$S \subseteq (\N \times \N \times \N)^{<\N}$)
     such that
     \[
-    \forall x, y \in \cN.~\left(x \succ y \iff \exists \alpha \in \N.~(x,\alpha,y) \in [S]\right).
+    \forall x, y \in \cN.~\left(x \succ y \iff \exists \alpha \in \cN.~(x,\alpha,y) \in [S]\right).
     \]
 
     Let
@@ -154,66 +160,69 @@ Let $\rho(\prec) \coloneqq  \sup \{\rho_{\prec}(x) + 1 : x \in X\}$.
     \land (s_{i-1}, u_i, s_i) \in S\}.
     \]
 
-    So $|W| \le \aleph_0$.
+    Clearly $|W| \le \aleph_0$.
     Define $\prec^\ast$ on $W$
     by setting
     \[(s_0,u_1,s_1,\ldots, u_n,s_n) \succ (s_0',u_1', s_1', \ldots, u_m', s_m') :\iff\]
     \begin{itemize}
-        \item $n < m$,
-        \item $\forall i \le n.~s_i \subsetneq s_i'$ and
-        \item $\forall i \le n.~u_i \subsetneq u_i'$.
+        \item $n < m$ and
+        \item $\forall i \le n.~s_i \subsetneq s_i' \land u_i \subsetneq u_i'$.
+            %\todo{$\subseteq$ or $\subsetneq$?} % TODO QUESTION
     \end{itemize}
     \begin{claim}
         $\prec^\ast$ is well-founded.
     \end{claim}
     \begin{subproof}
-        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
-         \[
-        \alpha_i \coloneqq  \bigcup_n u_i^n \cN.
-        \] 
-        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 $\lightning$
+        \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
+             \[
+            \alpha_i \coloneqq  \bigcup_n u_i^n \cN.
+            \]
+            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 $\rho(\prec^\ast) < |W|^+ \le \omega_1$.
+    % TODO QUESTION
 
+    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 $x_0 \succ x_1 \succ x_2 \succ \ldots \succ x_n$.
 
-    \todo{Fix typos and end proof}
-% Hence $\rho(\prec^\ast) < \omega_1$.
-% 
-% 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 $x_0 \succ x_1 \succ x_2 \succ \ldots \succ x_n$.
-% 
-% 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} &\longrightarrow & W \\
-%     \phi(x_0,x_1,\ldots,x_n) &\longmapsto & (x_0\defon{n}, \alpha_{x_0}, x_1\defon{n},\ldots,
-%     \alpha_{x_{n-1}}, x_n\defon{n}).
-% \end{IEEEeqnarray*}
-% Then $\phi$ is a homomorphism of $\subsetneq$ to $<^\ast$
-% and
-% \[
-% \rho(<) = \rho(T_{\prec}, \subsetneq) \le \rho(<^\ast) < \omega_1.
-% \] 
-
-\todo{Exercise}
+    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 \\
+        \phi(x_0,x_1,\ldots,x_n) &\longmapsto & (x_0\defon{n}, \alpha_{x_0}, x_1\defon{n},\ldots,
+        \alpha_{x_{n-1}}, x_n\defon{n}).
+    \end{IEEEeqnarray*}
+    Then $\phi$ is a homomorphism of $\subsetneq$ to $\prec^\ast$
+    so
+    \[
+    \rho(\prec)
+    = \rho(T_{\prec} \setminus \{\emptyset\} , \subsetneq)
+    \le \rho(\prec^\ast)
+    < \omega_1.
+    \]
 \end{proof}
 
 

inputs/lecture_15.tex

@@ -1,5 +1,8 @@
 \lecture{15}{2023-12-05}{}
 
+% TODO ANKI-MARKER
+
+
 \begin{theorem}[Boundedness Theorem]
     \yalabel{Boundedness Theorem}{Boundedness}{thm:boundedness}