From acb56df1c2ad130d87047567bf47c68014067e92 Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Wed, 24 Jan 2024 15:35:14 +0100 Subject: [PATCH] gist 12 --- inputs/lecture_09.tex | 1 + inputs/lecture_11.tex | 14 +-- inputs/lecture_12.tex | 240 ++++++++++++++++++++++++------------------ 3 files changed, 148 insertions(+), 107 deletions(-) diff --git a/inputs/lecture_09.tex b/inputs/lecture_09.tex index c5f67d5..f344eee 100644 --- a/inputs/lecture_09.tex +++ b/inputs/lecture_09.tex @@ -135,6 +135,7 @@ We will see later that $\Sigma^1_1(X) \cap \Pi^1_1(X) = \cB(X)$. \begin{theorem} + \label{thm:universals11} Let $X,Y$ be uncountable Polish spaces. There exists a $Y$-universal $\Sigma^1_1(X)$ set. \end{theorem} diff --git a/inputs/lecture_11.tex b/inputs/lecture_11.tex index 641f969..ebdd95f 100644 --- a/inputs/lecture_11.tex +++ b/inputs/lecture_11.tex @@ -153,13 +153,14 @@ We will not proof this in this lecture. \subsection{Ill-Founded Trees} - +\gist{% Recall that a \vocab{tree} on $\N$ is a subset of $\N^{<\N}$ closed under taking initial segments. We now identify trees with their characteristic functions, -i.e.~we want to associate a tree $T \subseteq \N^{<\N}$ +i.e.~we want to associate a tree $T \subseteq \N^{<\N}$}% +{We identify trees $T \subseteq \N^{<\N}$ with their characteristic functions:} \begin{IEEEeqnarray*}{rCl} \One_T\colon \omega^{<\omega} &\longrightarrow & \{0,1\} \\ x &\longmapsto & \begin{cases} @@ -167,9 +168,9 @@ i.e.~we want to associate a tree $T \subseteq \N^{<\N}$ 0 &: x \not\in T. \end{cases} \end{IEEEeqnarray*} -Note that $\One_T \in {\{0,1\}^\N}^{< \N}$. +\gist{Note that $\One_T \in {\{0,1\}^\N}^{< \N}$.}{} -Let $\Tr = \{T \in {2^{\N}}^{<\N} : T \text{ is a tree}\} \subseteq {2^{\N}}^{<\N}$. +Let \vocab{$\Tr$} $ \coloneqq \{T \in {2^{\N}}^{<\N} : T \text{ is a tree}\} \subseteq {2^{\N}}^{<\N}$. \begin{observe} \[ @@ -177,6 +178,7 @@ Let $\Tr = \{T \in {2^{\N}}^{<\N} : T \text{ is a tree}\} \subseteq {2^{\N}}^{<\ \] is closed (where we take the topology of the Cantor space). \end{observe} +\gist{% Indeed, for any $ s \in \N^{<\N}$ we have that $\{T \in {2^{\N}}^{<\N} : s \in T\}$ and $\{T \in {2^{\N}}^{<\N} : s\not\in T\}$ are clopen. @@ -185,6 +187,4 @@ In particular for $s$ fixed, we have that \[\{A \in {2^{\N}}^{<\N} : s \in A \text{ and } s' \in A \text{ for any initial segment $s' \subseteq s$}\}\] is clopen in ${2^{\N}}^{<\N}$. - - - +}{} diff --git a/inputs/lecture_12.tex b/inputs/lecture_12.tex index 5b9d551..feb3e2e 100644 --- a/inputs/lecture_12.tex +++ b/inputs/lecture_12.tex @@ -21,102 +21,136 @@ T \in \IF &\iff& \exists \beta \in \cN .~\forall n \in \N.~T(\beta\defon{n}) = 1. \end{IEEEeqnarray*} - Consider $D \coloneqq \{(T, \beta) \in \Tr \times \cN : \forall n.~ T(\beta\defon{n}) = 1\}$. - Note that this set is closed in $\Tr \times \cN$, - since it is a countable intersection of clopen sets. - % TODO Why clopen? - Then $\IF = \proj_{\Tr}(D) \in \Sigma^1_1$. + Consider + \[D \coloneqq \{(T, \beta) \in \Tr \times \cN : \forall n.~ T(\beta\defon{n}) = 1\}.\] + \gist{% + Note that this set is closed in $\Tr \times \cN$, + since it is a countable intersection of clopen sets. + Then $\IF = \proj_{\Tr}(D) \in \Sigma^1_1$. + }{$D \overset{\text{closed}}{\subseteq} \Tr \times \cN$ and $\IF = \proj_{\Tr}(D)$.} \end{proof} \begin{definition} An analytic set $B$ in some Polish space $Y$ is \vocab{complete analytic} (\vocab{$\Sigma^1_1$-complete}) + \gist{% 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 $x \in A \iff f(x) \in B$. + such that $x \in A \iff f(x) \in B$, + i.e.~$f^{-1}(B) = A$. + }{% + iff for any $A \in \Sigma^1_1(X)$, $X$ Polish + there exists $f\colon X \to Y$ Borel such that $f^{-1}(B) = A$.} - Similarly, define \vocab{complete coanalytic} (\vocab{$\Pi^1_1$-complete}). + \gist{% + Similarly, a conalytic set $B$ is called + \vocab{complete coanalytic} (\vocab{$\Pi^1_1$-complete}) + iff for any $A \subseteq \Pi^1_1(X)$ + there exists $f\colon X \to Y$ Borel such that $f^{-1}(B) = A$. + }{Similarly we 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. + \item $\Sigma^1_1$-complete sets are never Borel% + \gist{: + Suppose there is a $\Sigma^1_1$-complete set $B \in \cB(Y)$. + Take $A \in \Sigma^1_1(X) \setminus \cB(X)$% + \footnote{e.g.~\yaref{thm:universals11}} + and $f\colon X \to Y$ Borel. + But then we get that $f^{-1}(B)$ is Borel $\lightnig$. + }{.} \end{itemize} \end{observe} +% TODO ANKI-MARKER \begin{theorem} \label{thm:lec12:1} Suppose that $A \subseteq \cN$ is analytic. - Then there is $f\colon \cN \to \Tr$\todo{Borel?} - such that $x \in A \iff f(x)$ is ill-founded. + \gist{% + Then there is a continuous function $f\colon \cN \to \Tr$ + such that $x \in A \iff f(x)$ is ill-founded, + i.e.~$A = f^{-1}(\IF)$. + }{% + Then there exists $f\colon \cN \to \Tr$ continuous + such that $A = f^{-1}(\IF)$. + } \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} + +\gist{% + 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 $S^{\N}$:% + \footnote{cf.~\yaref{s3e1} (c)} + \begin{IEEEeqnarray*}{rCl} + T &\longmapsto & [T]\\ + \{\alpha\defon{n} : \alpha \in D, n \in \N\} &\longmapsfrom & D\\ + \end{IEEEeqnarray*} +}{% + For $S$ countable, + pruned trees on $S$ correspond to closed subsets of $S^{\N}$ + via $T \mapsto [T]$. +} +\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} +\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} + \gist{% + $A$ is analytic iff + there exists $F \overset{\text{closed}}{\subseteq} (\N \times \N)^{\N}$ + such that $A = \proj_1(F)$. + But closed sets of $\N^\N \times \N^{\N}$ correspond to pruned trees, + by the first point. + }{Closed subsets of $\N^\N \times \N^\N$ correspond to pruned trees.} +\end{proof} \begin{refproof}{thm:lec12:1} - Take a tree $T$ on $\N \times \N$ - as in \autoref{prop:lec12:2}, i.e.~$A = \proj_1([T])$. + \gist{% + Take a tree $T$ on $\N \times \N$ + as in \autoref{prop:lec12:2}, i.e.~$A = \proj_1([T])$. + }{Write $A = \proj_1([T])$ for a pruned tree $T$ on $\N \times \N$.} 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} + \gist{% + Clearly $x \in A \iff f(x) \in \IF$. + $f$ 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 $|t| \le n$.. + 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} @@ -125,7 +159,9 @@ For the proof we need some prerequisites: \end{corollary} \begin{proof} Let $X$ be Polish. - Suppose that $A \subseteq X$ is analytic and uncountable. + Suppose that $A \subseteq X$ is analytic and uncountable% + \gist{}{ (trivial for countable)}. + Then % https://q.uiver.app/#q=WzAsNSxbMCwwLCJYIl0sWzEsMCwiXFxjTiJdLFsyLDAsIlxcVHIiXSxbMCwxLCJBIl0sWzEsMSwiYihBKSJdLFsxLDIsImYiXSxbMCwxLCJiIl0sWzMsMCwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbNCwxLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dXQ== \[\begin{tikzcd} @@ -138,16 +174,18 @@ For the proof we need some prerequisites: \end{tikzcd}\] where $f$ is chosen as in \yaref{thm:lec12:1}. - 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} - \end{IEEEeqnarray*} - where $a \in \IF$ and $b \not\in \IF$ are chosen arbitrarily. + \gist{% + 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} + \end{IEEEeqnarray*} + where $a \in \IF$ and $b \not\in \IF$ are chosen arbitrarily. + }{} \end{proof} \subsection{Linear Orders} @@ -161,18 +199,19 @@ Let \[ \WO \coloneqq \{x \in \LO: x \text{ is a well ordering}\}. \] +\gist{% + 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. -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} + 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 $C$ @@ -181,13 +220,14 @@ Recall that \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} - +\gist{% + \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} +}{}