gist 12

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}

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 ${2^{\N}}^{<\N}$.
-
+}{}
-
-

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 $D \coloneqq \{(T, \beta) \in \Tr \times  \cN : \forall n.~ T(\beta\defon{n}) = 1\}$.
+    Consider
-    Note that this set is closed in $\Tr \times \cN$,
+    \[D \coloneqq \{(T, \beta) \in \Tr \times  \cN : \forall n.~ T(\beta\defon{n}) = 1\}.\]
-    since it is a countable intersection of clopen sets.
+    \gist{%
-    % TODO Why clopen?
+        Note that this set is closed in $\Tr \times \cN$,
-    Then $\IF = \proj_{\Tr}(D) \in \Sigma^1_1$.
+        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:
+        \item $\Sigma^1_1$-complete sets are never Borel%
-            Suppose there is a $\Sigma^1_1$-complete set $B \in \cB(Y)$.
+            \gist{:
-            Take $A \in \Sigma^1_1(X) \setminus \cB(X)$
+                Suppose there is a $\Sigma^1_1$-complete set $B \in \cB(Y)$.
-            and $f\colon X \to  Y$ Borel.
+                Take $A \in \Sigma^1_1(X) \setminus \cB(X)$%
-            But then $f^{-1}(B)$ is Borel.
+                \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?}
+    \gist{%
-    such that $x \in A \iff f(x)$ is ill-founded.
+        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,
+\gist{%
-        the pruned\footnote{no maximal elements, in particular this implies ill-founded if the tree is non empty.} trees
+    Recall that for $S$ countable,
-        $T \subseteq S^{<\N}$ on $S$ correspond
+    the pruned%
-        to closed subsets of $S^{\N}$:
+    \footnote{no maximal elements,
-        \begin{IEEEeqnarray*}{rCl}
+        in particular this implies ill-founded if the tree is non empty.
-            T &\longmapsto & [T]\\
+    } trees $T \subseteq S^{<\N}$ on $S$ correspond
-            \{\alpha\defon{n} : \alpha \in D, n \in \N\} &\longmapsfrom & D\\
+    to closed subsets of $S^{\N}$:%
-        \end{IEEEeqnarray*}
+    \footnote{cf.~\yaref{s3e1} (c)}
-        \todo{Copy from exercises}
+    \begin{IEEEeqnarray*}{rCl}
-    \item \leavevmode\begin{definition}
+        T &\longmapsto & [T]\\
-        If $T$  is a tree on $\N \times \N$
+        \{\alpha\defon{n} : \alpha \in D, n \in \N\} &\longmapsfrom & D\\
-        and $x \in \cN$,
+    \end{IEEEeqnarray*}
-        then the \vocab{section at $x$}
+}{%
-        %denoted $T(x)$,
+    For $S$ countable,
-        is the following tree on $\N$ :
+    pruned trees on $S$ correspond to closed subsets of $S^{\N}$
-        \[
+    via $T \mapsto [T]$.
-        T(x) = \{s \in \N^{<\N} : (x\defon{|s|}, s) \in T\}.
+}
-        \]
+\begin{definition}
-    \end{definition}
+    If $T$  is a tree on $\N \times \N$
-    \item \leavevmode
+    and $x \in \cN$,
-        \begin{proposition}
+    then the \vocab{section at $x$}
-            \label{prop:lec12:2}
+    denoted $T(x)$,
-            Let $A \subseteq  \cN$.
+    is the following tree on $\N$ :
-            The following are equivalent:
+    \[
-            \begin{itemize}
+    T(x) = \{s \in \N^{<\N} : (x\defon{|s|}, s) \in T\}.
-                \item   $A$ is analytic.
+    \]
-                \item There is a pruned tree on $\N \times \N$
+\end{definition}
-                    such that
+\begin{proposition}
-                    \[A = \proj_1 ([T]) = \{x \in \cN : \exists  y \in \cN.~ (x,y) \in [T]\}.\]
+    \label{prop:lec12:2}
-            \end{itemize}
+    Let $A \subseteq  \cN$.
-        \end{proposition}
+    The following are equivalent:
-        \begin{proof}
+    \begin{itemize}
-            $A$ is analytic iff
+        \item   $A$ is analytic.
-            there exists $F \overset{\text{closed}}{\subseteq} \N \times \N$
+        \item There is a pruned tree on $\N \times \N$
-            such that $A = \proj_1(F)$.
+            such that
-            But closed sets of $\N \times \N$ correspond to pruned trees,
+            \[A = \proj_1 ([T]) = \{x \in \cN : \exists  y \in \cN.~ (x,y) \in [T]\}.\]
-            by the first point.
+    \end{itemize}
-        \end{proof}
+\end{proposition}
-\end{enumerate}
+\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 $\N \times \N$
+    \gist{%
-    as in \autoref{prop:lec12:2}, i.e.~$A = \proj_1([T])$.
+        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.
+    \gist{%
-    $f$ is continuous:
+        Clearly $x \in A \iff f(x) \in \IF$.
-    Let $x\defon{n} = y\defon{n}$ for some $n \in \N$.
+        $f$ is continuous:
-    Then for all $m \le n, s,t \in \N^{<\N}$
+        Let $x\defon{n} = y\defon{n}$ for some $n \in \N$.
-    such that $s = x\defon{m} = y \defon{m}$ and $|t| = |s|$,
+        Then for all $m \le n, s,t \in \N^{<\N}$
-    we have
+        such that $s = x\defon{m} = y \defon{m}$ and $|t| = |s|$,
-    \begin{itemize}
+        we have
-        \item $t \in T(x) \iff (s,t) \in T$,
+        \begin{itemize}
-        \item $t \in T(y) \iff (s,t) \in T$.
+            \item $t \in T(x) \iff (s,t) \in T$,
-    \end{itemize}
+            \item $t \in T(y) \iff (s,t) \in T$.
+        \end{itemize}
 
-    So if $x\defon{n} = y\defon{n}$,
+        So if $x\defon{n} = y\defon{n}$,
-    then $t \in T(x) \iff t \in  T(y)$ as long as $|t| \le n$..
+        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,
+    \gist{%
-    just consider
+        If $X$ is Polish and countable and $A \subseteq  X$ analytic,
-    \begin{IEEEeqnarray*}{rCl}
+        just consider
-        g \colon X &\longrightarrow & \Tr \\
+        \begin{IEEEeqnarray*}{rCl}
-        x &\longmapsto & \begin{cases}
+            g \colon X &\longrightarrow & \Tr \\
-            a &: x \in A,\\
+            x &\longmapsto & \begin{cases}
-            b &: x \not\in A,\\
+                a &: x \in A,\\
-        \end{cases}
+                b &: x \not\in A,\\
-    \end{IEEEeqnarray*}
+            \end{cases}
-    where $a \in \IF$ and $b \not\in \IF$ are chosen arbitrarily.
+        \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
+            Let $(A, <_A) \prec (B, <_B)$ denote that
-\begin{itemize}
+            $(A, <_A)$ is isomorphic to a proper initial segment of $(B, <_B)$.
-    \item $(A,<)$ is a well ordering iff there are no infinite descending chains.
+    \end{itemize}
-    \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 $C$
@@ -181,13 +220,14 @@ Recall that
     \phi\colon  C \to  \Ord.
     \]
 \end{definition}
-\begin{example}
+\gist{%
-    Let $C = \WO$
+    \begin{example}
-    and
+        Let $C = \WO$
-    \begin{IEEEeqnarray*}{rCl}
+        and
-        \phi\colon \WO &\longrightarrow & \Ord \\
+        \begin{IEEEeqnarray*}{rCl}
-    \end{IEEEeqnarray*}
+            \phi\colon \WO &\longrightarrow & \Ord \\
-    where $\phi((A,<_A))$ is the unique ordinal
+        \end{IEEEeqnarray*}
-    isomorphic to $(A, <_A)$.
+        where $\phi((A,<_A))$ is the unique ordinal
-\end{example}
+        isomorphic to $(A, <_A)$.
-
+    \end{example}
+}{}