From f94d019d5e3d646e22956dba4ef2dbb25669029d Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Fri, 24 Nov 2023 11:53:57 +0100 Subject: [PATCH] lecture 12 --- inputs/lecture_11.tex | 6 +- inputs/lecture_12.tex | 191 ++++++++++++++++++++++++++++++++++++++++++ logic.sty | 4 + logic3.tex | 1 + 4 files changed, 199 insertions(+), 3 deletions(-) create mode 100644 inputs/lecture_12.tex diff --git a/inputs/lecture_11.tex b/inputs/lecture_11.tex index f860080..91bf1a5 100644 --- a/inputs/lecture_11.tex +++ b/inputs/lecture_11.tex @@ -99,7 +99,7 @@ (continuous wrt.~to the topology of $X$) On the other hand \[ - X \hookrightarrow\cN \hookrightarrow[\text{continuous embedding}]\cC + X \hookrightarrow\cN \overset{\text{continuous embedding}}{\hookrightarrow}\cC \] \todo{second inclusion was on a homework sheet} For the first inclusion, @@ -132,7 +132,7 @@ \arrow["\subseteq", hook, from=1-2, to=2-3] \arrow["\subseteq", hook, from=2-3, to=1-4] \arrow["\subseteq", hook, from=2-3, to=3-4] -\end{tikzcd}\] +\end{tikzcd}\] \begin{definition} Let $X$ be a Polish space. @@ -174,7 +174,7 @@ Let $\Tr = \{T \in {2^{\N}}^{<\N} : T \text{ is a tree}\} \subseteq {2^{\N}}^{<\ \begin{observe} \[ \Tr \subseteq {2^{\N}}^{<\N} - \] + \] is closed (where we take the topology of the Cantor space). \end{observe} Indeed, for any $ s \in \N^{<\N}$ diff --git a/inputs/lecture_12.tex b/inputs/lecture_12.tex new file mode 100644 index 0000000..be9c892 --- /dev/null +++ b/inputs/lecture_12.tex @@ -0,0 +1,191 @@ +\lecture{12}{2023-11-24}{} + +\begin{definition} + A tree $T$ is \vocab{ill-founded} + if it has an infinite branch, + i.e. $[T] \neq \emptyset$ + Otherwise it is called \vocab{well-founded}. + Let + \[\IF \coloneqq \{T \in \Tr : T \text{ is ill-founded}\}\] + and + \[\WF \coloneqq \{T \in \Tr : T \text{ is well-founded}\}\] +\end{definition} + +\begin{proposition} + $\IF \in \Sigma^1_1(\Tr)$. +\end{proposition} +\begin{proof} + We have + \begin{IEEEeqnarray*}{rCl} + T \in \IF &\iff& \exists \beta \in \cN .~\forall n \in \N.~T(\beta\defon{n}) = 1. + \end{IEEEeqnarray*} + + Consider $\{(T, \beta) \in \Tr \times \cN : \forall n.~ T(\beta\defon{n}) = 1\}$. + Note that this set is closed in $\Tr \times \cN$, + since it is a countable intersection of clopen sets. + % TODO Why clopen? + Then $\IF = \proj_{\Tr}(D) \in \Sigma^1_1$. +\end{proof} + +\begin{definition} + An analytic set $B$ in some Polish space $Y$ + is \vocab{complete analytic} (\vocab{$\Sigma^1_1$-complete}) + iff for any analytic $A \in \Sigma^1_1(X)$ for some Polish space $X$, + there exists a Borel function $f\colon X\to Y$ + such that $x \in A \iff f(x) \in B$. + + Similarly, define \vocab{complete coanalytic} (\vocab{$\Pi^1_1$-complete}). +\end{definition} +\begin{observe} + \leavevmode + \begin{itemize} + \item Complements of $\Sigma^1_1$-complete sets are $\Pi^1_1$-complete. + \item $\Sigma^1_1$-complete sets are never Borel: + Suppose there is a $\Sigma^1_1$-complete set $B \in \cB(Y)$. + Take $A \in \Sigma^1_1(X) \setminus \cB(X)$ + and $f\colon X \to Y$ Borel. + But then $f^{-1}(B)$ is Borel. + \end{itemize} +\end{observe} + +\begin{theorem} + \label{thm:lec12:1} + Suppose that $A \subseteq \cN$ is analytic. + Then there is $f\colon \cN \to \Tr$ + such that $x \in A \iff f(x)$ is ill-founded. +\end{theorem} +For the proof we need some prerequisites: +\begin{enumerate}[1.] + \item Recall that for $S$ countable, + the pruned\footnote{no maximal elements, in particular this implies ill-founded if the tree is non empty.} trees + $T \subseteq S^{<\N}$ on $S$ correspond + to closed subsets of $S^{\N}$: + \begin{IEEEeqnarray*}{rCl} + T &\longmapsto & [T]\\ + \{\alpha\defon{n} : \alpha \in D, n \in \N\} &\longmapsfrom & D\\ + \end{IEEEeqnarray*} + \todo{Copy from exercises} + \item \leavevmode\begin{definition} + If $T$ is a tree on $\N \times \N$ + and $x \in \cN$, + then the \vocab{section at $x$} + %denoted $T(x)$, + is the following tree on $\N$ : + \[ + T(x) = \{s \in \N^{<\N} : (x\defon{|s|}, s) \in T\}. + \] + \end{definition} + \item \leavevmode + \begin{proposition} + \label{prop:lec12:2} + Let $A \subseteq \cN$. + The following are equivalent: + \begin{itemize} + \item $A$ is analytic. + \item There is a pruned tree on $\N \times \N$ + such that + \[A = \proj_1 ([T]) = \{x \in \cN : \exists y \in \cN.~ (x,y) \in [T]\}.\] + \end{itemize} + \end{proposition} + \begin{proof} + $A$ is analytic iff + there exists $F \overset{\text{closed}}{\subseteq} \N \times \N$ + such that $A = \proj_1(F)$. + But closed sets of $\N \times \N$ correspond to pruned trees, + by the first point. + \end{proof} +\end{enumerate} +\begin{refproof}{thm:lec12:1} + Take a tree $T$ on $\N \times \N$ + as in \autoref{prop:lec12:2}, i.e.~$A = \proj_1([T])$. + Consider + \begin{IEEEeqnarray*}{rCl} + f\colon \cN &\longrightarrow & \Tr \\ + x &\longmapsto & T(x). + \end{IEEEeqnarray*} + Clearly $x \in A \iff f(x)$ is ill-founded. + $f$ is continuous: + Let $x\defon{n} = y\defon{n}$ for some $n \in \N$. + Then for all $m \le n, s,t \in \N^{<\N}$ + such that $s = x\defon{m} = y \defon{m}$ and $|t| = |s|$, + we have + \begin{itemize} + \item $t \in T(x) \iff (s,t) \in T$, + \item $t \in T(y) \iff (s,t) \in T$. + \end{itemize} + + So if $x\defon{n} = y\defon{n}$, + then $t \in T(x) \iff t \in T(y)$ as long as $|t| \le n$.. +\end{refproof} + +\begin{corollary} + $\IF$ is $\Sigma^1_1$-complete. +\end{corollary} +\begin{proof} + Let $A \subseteq X$ is analytic + and $X$ Polish and uncountable, + then + % https://q.uiver.app/#q=WzAsNSxbMCwwLCJYIl0sWzEsMCwiXFxjTiJdLFsyLDAsIlxcVHIiXSxbMCwxLCJBIl0sWzEsMSwiYihBKSJdLFsxLDIsImYiXSxbMCwxLCJiIl0sWzMsMCwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbNCwxLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dXQ== +\[\begin{tikzcd} + X & \cN & \Tr \\ + A & {b(A)} + \arrow["f", from=1-2, to=1-3] + \arrow["b", from=1-1, to=1-2] + \arrow[hook, from=2-1, to=1-1] + \arrow[hook, from=2-2, to=1-2] +\end{tikzcd}\] + + If $X$ is Polish and countable and $A \subseteq X$ analytic, + just consider + \begin{IEEEeqnarray*}{rCl} + g \colon X &\longrightarrow & \Tr \\ + x &\longmapsto & \begin{cases} + a &: x \in A,\\ + b &: x \not\in A,\\ + \end{cases} + where $a \in \IF$ and $b \not\in \IF$ + are chosen arbitrarily. + \end{IEEEeqnarray*} +\end{proof} + +\subsection{Linear Orders} + +Let us consider the space +\[\LO \coloneqq \{x \in 2^{\N\times \N} : x \text{ is a linear order on $\N$}\},\] +where we code a linear order $(\N, <)$ +by $x \in 2^{\N \times \N}$ with $x(m,n) = 1 \iff m \le n$. + +Let +\[ +\WO \coloneqq \{x \in \LO: x \text{ is a well ordering}\}. +\] + +Recall that +\begin{itemize} + \item $(A,<)$ is a well ordering iff there are no infinite descending chains. + \item Every well ordering is isomorphic to an ordinal. + \item Any two well orderings are comparable, + i.e.~they are isomorphic, + or one is isomorphic to an initial segment of the other. + + Let $(A, <_A) \prec (B, <_B)$ denote that + $(A, <_A)$ is isomorphic to a proper initial segment of $(B, <_B)$. +\end{itemize} + +\begin{definition} + A \vocab{rank} on some set $C$ + is a function + \[ + \phi\colon C \to \Ord. + \] +\end{definition} +\begin{example} + Let $C = \WO$ + and + \begin{IEEEeqnarray*}{rCl} + \phi\colon \WO &\longrightarrow & \Ord \\ + \end{IEEEeqnarray*} + where $\phi((A,<_A))$ is the unique ordinal + isomorphic to $(A, <_A)$. +\end{example} + diff --git a/logic.sty b/logic.sty index d708fe0..30a9a8e 100644 --- a/logic.sty +++ b/logic.sty @@ -127,6 +127,10 @@ \DeclareSimpleMathOperator{tcl} \DeclareSimpleMathOperator{Tr} +\DeclareSimpleMathOperator{IF} +\DeclareSimpleMathOperator{WF} +\DeclareSimpleMathOperator{LO} % linear orders +\DeclareSimpleMathOperator{WO} % well orderings \newcommand{\concat}{\mathop{{}^{\scalebox{.7}{$\smallfrown$}}}} diff --git a/logic3.tex b/logic3.tex index 1615c2f..247b987 100644 --- a/logic3.tex +++ b/logic3.tex @@ -35,6 +35,7 @@ \input{inputs/lecture_09} \input{inputs/lecture_10} \input{inputs/lecture_11} +\input{inputs/lecture_12}