lecture 12

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}$

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}
+

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}{$\smallfrown$}}}}
 

logic3.tex

@@ -35,6 +35,7 @@
 \input{inputs/lecture_09}
 \input{inputs/lecture_10}
 \input{inputs/lecture_11}
+\input{inputs/lecture_12}