From a0c5bbad0b6074c1e66871a7a7d724d5d8ba4021 Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Tue, 28 Nov 2023 11:58:58 +0100 Subject: [PATCH] 13 --- inputs/lecture_12.tex | 1 + inputs/lecture_13.tex | 186 ++++++++++++++++++++++++++++++++++++++++++ logic3.tex | 1 + 3 files changed, 188 insertions(+) create mode 100644 inputs/lecture_13.tex diff --git a/inputs/lecture_12.tex b/inputs/lecture_12.tex index 7ebb1cb..6f88217 100644 --- a/inputs/lecture_12.tex +++ b/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} diff --git a/inputs/lecture_13.tex b/inputs/lecture_13.tex new file mode 100644 index 0000000..753e015 --- /dev/null +++ b/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} + + + + + diff --git a/logic3.tex b/logic3.tex index 247b987..300f8af 100644 --- a/logic3.tex +++ b/logic3.tex @@ -36,6 +36,7 @@ \input{inputs/lecture_10} \input{inputs/lecture_11} \input{inputs/lecture_12} +\input{inputs/lecture_13}