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

@@ -21,63 +21,92 @@
         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
+    \[D \coloneqq \{(T, \beta) \in \Tr \times  \cN : \forall n.~ T(\beta\defon{n}) = 1\}.\]
+    \gist{%
         Note that this set is closed in $\Tr \times \cN$,
         since it is a countable intersection of clopen sets.
-    % TODO Why clopen?
         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%
+            \gist{:
                 Suppose there is a $\Sigma^1_1$-complete set $B \in \cB(Y)$.
-            Take $A \in \Sigma^1_1(X) \setminus \cB(X)$
+                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 $f^{-1}(B)$ is 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 $S^{\N}$:
+
+\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*}
-        \todo{Copy from exercises}
-    \item \leavevmode\begin{definition}
+}{%
+    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 $T(x)$,
+    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}
+\end{definition}
+\begin{proposition}
     \label{prop:lec12:2}
     Let $A \subseteq  \cN$.
     The following are equivalent:
@@ -87,24 +116,28 @@ For the proof we need some prerequisites:
             such that
             \[A = \proj_1 ([T]) = \{x \in \cN : \exists  y \in \cN.~ (x,y) \in [T]\}.\]
     \end{itemize}
-        \end{proposition}
-        \begin{proof}
+\end{proposition}
+\begin{proof}
+    \gist{%
         $A$ is analytic iff
-            there exists $F \overset{\text{closed}}{\subseteq} \N \times \N$
+        there exists $F \overset{\text{closed}}{\subseteq} (\N \times \N)^{\N}$
         such that $A = \proj_1(F)$.
-            But closed sets of $\N \times \N$ correspond to pruned trees,
+        But closed sets of $\N^\N \times \N^{\N}$ correspond to pruned trees,
         by the first point.
-        \end{proof}
-\end{enumerate}
+    }{Closed subsets of $\N^\N \times \N^\N$ correspond to pruned trees.}
+\end{proof}
 \begin{refproof}{thm:lec12:1}
+    \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.
+    \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 $m \le n, s,t \in \N^{<\N}$
@@ -116,7 +149,8 @@ For the proof we need some prerequisites:
         \end{itemize}
 
         So if $x\defon{n} = y\defon{n}$,
-    then $t \in T(x) \iff t \in  T(y)$ as long as $|t| \le n$..
+        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==
 \[\begin{tikzcd}
@@ -138,6 +174,7 @@ For the proof we need some prerequisites:
 \end{tikzcd}\]
     where $f$ is chosen as in \yaref{thm:lec12:1}.
 
+    \gist{%
         If $X$ is Polish and countable and $A \subseteq  X$ analytic,
         just consider
         \begin{IEEEeqnarray*}{rCl}
@@ -148,6 +185,7 @@ For the proof we need some prerequisites:
             \end{cases}
         \end{IEEEeqnarray*}
         where $a \in \IF$ and $b \not\in \IF$ are chosen arbitrarily.
+    }{}
 \end{proof}
 
 \subsection{Linear Orders}
@@ -161,9 +199,9 @@ Let
 \[
 \WO \coloneqq  \{x \in \LO: x \text{ is a well ordering}\}.
 \]
-
-Recall that
-\begin{itemize}
+\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,
@@ -172,7 +210,8 @@ Recall that
 
             Let $(A, <_A) \prec (B, <_B)$ denote that
             $(A, <_A)$ is isomorphic to a proper initial segment of $(B, <_B)$.
-\end{itemize}
+    \end{itemize}
+}{}
 
 \begin{definition}
     A \vocab{rank} on some set $C$
@@ -181,7 +220,8 @@ Recall that
     \phi\colon  C \to  \Ord.
     \]
 \end{definition}
-\begin{example}
+\gist{%
+    \begin{example}
         Let $C = \WO$
         and
         \begin{IEEEeqnarray*}{rCl}
@@ -189,5 +229,5 @@ Recall that
         \end{IEEEeqnarray*}
         where $\phi((A,<_A))$ is the unique ordinal
         isomorphic to $(A, <_A)$.
-\end{example}
-
+    \end{example}
+}{}