13

inputs/lecture_12.tex

@@ -119,6 +119,7 @@ For the proof we need some prerequisites:
 \end{refproof}
 
 \begin{corollary}
+\label{cor:ifsi11c}
     $\IF$ is $\Sigma^1_1$-complete.
 \end{corollary}
 \begin{proof}

inputs/lecture_13.tex

@@ -0,0 +1,186 @@
+\lecture{13}{2023-11-08}{}
+
+% 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-fonuded iff
+    $(T, <_{KB}\defon{T})$ is well ordered.
+\end{proposition}
+\begin{proof}
+    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 $(a_0,a_1,a_2, \ldots) \in [T]$.
+\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
+    \[
+    x \in \WF \iff f(x) \in \WO
+    \]
+    (equivalently $x \in \IF \iff f(x) \in \LO \setminus \WO$).
+    This suffices, since $\IF \subseteq \Tr$ is $\Sigma^1_1$-complete
+    (see \yaref{cor:ifiss11c}).
+
+    Fix a bijection $b\colon \N \to \N^{<\N}$.
+
+    \begin{idea}
+        For $T \in \Tr$ consider
+        $<_{KB}\defon{T}$
+        % TODO?
+    \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?
+
+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 $\prec$ ($x \prec y \iff x \preceq y \land y \not\preceq x$) 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
+\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}
+
+
+
+
+

logic3.tex

@@ -36,6 +36,7 @@
 \input{inputs/lecture_10}
 \input{inputs/lecture_11}
 \input{inputs/lecture_12}
+\input{inputs/lecture_13}