merge

bibliography/references.bib

@@ -31,3 +31,17 @@ volume = {85},
 year = {1963},
 title = {The Structure of Distal Flows},
 }
+@book{kechris,
+abstract = {Descriptive set theory has been one of the main areas of research in set theory for almost a century. This text attempts to present a largely balanced approach, which combines many elements of the different traditions of the subject. It includes a wide variety of examples, exercises (over 400), and applications, in order to illustrate the general concepts and results of the theory. This text provides a first basic course in classical descriptive set theory and covers material with which mathematicians interested in the subject for its own sake or those that wish to use it in their field should be familiar. Over the years, researchers in diverse areas of mathematics, such as logic and set theory, analysis, topology, probability theory, etc., have brought to the subject of descriptive set theory their own intuitions, concepts, terminology and notation.},
+author = {Kechris, Alexander},
+address = {Netherlands},
+isbn = {9781461241904},
+keywords = {Mathematics. General principles ; Set theory},
+language = {eng},
+publisher = {Springer Nature},
+series = {Graduate Texts in Mathematics},
+title = {Classical Descriptive Set Theory},
+volume = {156},
+year = {2012},
+}
+

inputs/lecture_02.tex

@@ -130,8 +130,14 @@
         of all open sets $U \subseteq  X$ such that
         \begin{enumerate}[(i)]
             \item $U \cap Y \neq  \emptyset$,
-            \item $\diam_d(U) \le \frac{1}{n}$,
+            \item \gist{$\diam_d(U) \le \frac{1}{n}$,%
-            \item $\diam_{d_Y}(U \cap Y) \le \frac{1}{n}$.
+                \footnote{The proof gets a little easier if we bound by $\frac{1}{2^n}$ instead of $\frac{1}{n}$,
+                    as that allows to simply take $U'_n \coloneqq \bigcup_{m > n} U_m$,
+                    but both bounds work.}
+                }{$\diam_d(U) \le \frac{1}{2^n}$.}
+            \item \gist{%
+                $\diam_{d_Y}(U \cap Y) \le \frac{1}{n}$.}{%
+                $\diam_{d_Y}(U \cap Y) \le \frac{1}{2^n}$.}
         \end{enumerate}
         \gist{%
             We want to show that $Y = \bigcap_{n \in \N} V_n$.

inputs/lecture_07.tex

@@ -229,7 +229,7 @@
             Let $U \in \cT_\infty$.
             Then $U$ is the union of sets of the form
             \[
-            V = U_{n_1} \cap U_{n_2} \cap \ldots \cap U_{nu}
+            V = U_{n_1} \cap U_{n_2} \cap \ldots \cap U_{n_u}
             \]
             for some $n_1 < n_2 < \ldots < n_u$
             and $U_{n_i} \in \cT_i$.

inputs/lecture_08.tex

@@ -19,7 +19,7 @@ where $X$ is a metrizable, usually second countable space.
     $\cU$ \vocab{parametrizes}  $\Gamma(X)$
     iff:
     \begin{itemize}
-        \item $\cU \in \Gamma$,
+        \item $\cU \in \Gamma(Y \times X)$,
         \item $\{U_y : y \in Y\} = \Gamma(X)$.
     \end{itemize}
 \end{definition}

inputs/lecture_13.tex

@@ -34,6 +34,8 @@ with $(f^{-1}(\{1\}), <)$.
     and  $[\alpha_i, \alpha_{i+1})$ to $(i,i+1)$.
 \end{proof}
 
+% TODO $\WF \subseteq 2^\Q$ is $\Sigma^1_1$-complete.
+
 
 \begin{definition}[\vocab{Kleene-Brouwer ordering}]
     Let $(A,<)$ be a linear order and $A$ countable.
@@ -181,7 +183,7 @@ i.e.}{}
     Let $X$ be Polish and $C \subseteq  X$ coanalytic.
     Then $\phi\colon  C \to \Ord$
     is a  \vocab[Rank!$\Pi^1_1$-rank]{$\Pi^1_1$-rank}
-    provided that $\le^\ast$ and $<^\ast$ are coanalytic,
+    provided that $\le^\ast$ and $<^\ast$ are coanalytic subsets of $X \times X$,
     where
     $x \le^\ast_{\phi} y$
     iff

inputs/lecture_14.tex

@@ -59,7 +59,19 @@
         \item $x \in \WO$ and
         \item $\forall f \in \Q^\Q.~(f,y,x) \not\in E$,
     \end{itemize}
-    so it is $\Pi^1_1$.
+    so it is $\Pi^1_1$.%
+    \gist{\footnote{%
+            (very informal)
+            Note that $\Sigma^1_1$-sets work well with comprehensions using ``$\exists$'':
+            Writing $A \in \Sigma^1_1(X)$ as $A = \proj_X(B)$
+            for some Borel set $B \subseteq X \times Y$,
+            the second coordinate can be thought of
+            as being a witness for a statement.
+            Likewise, being complements of $\Sigma^1_1$-sets,
+            $\Pi_1^1$-sets can capture that a witness does not exist,
+            i.e.~they interact nicely with ``$\forall$''.%
+    }}{}
+
 
     Furthermore $x \le_\phi^\ast y \iff$
     either $x <^\ast_\phi y$ or
@@ -106,6 +118,13 @@
     i.e.~take the element with minimal rank
     that has the minimal second coordinate among those elements.
 \end{proof}
+\gist{
+    \begin{remark}
+        Uniformization also works for $R \in \Pi^1_1(X \times Y)$
+        for arbitrary Polish spaces $X,Y$,
+        cf.~\cite[(36.12)]{kechris}.
+    \end{remark}
+}{}
 \begin{corollary}[Countable Reduction for $\Pi^1_1$ Sets]
     Let $X$ be a Polish space
     and $(C_n)_n$ a sequence of coanalytic subsets of $X$.
@@ -127,33 +146,43 @@
 Let $X$ be a Polish space.
 If $(X, \prec)$ is well-founded%
 \gist{ (i.e.~there are no infinite descending chains)}{}
-then we define a rank $\rho_{y}\colon X \to  \Ord$ as follows:
+then we define a rank $\rho_{\prec}\colon X \to  \Ord$ as follows:
 For minimal elements the rank is $0$.
 Otherwise set $\rho_\prec(x) \coloneqq  \sup \{\rho_\prec(y) + 1 : y \prec x\}$.
 Let $\rho(\prec) \coloneqq  \sup \{\rho_{\prec}(x) + 1 : x \in X\}$.
 
-% \begin{exercise}
+\begin{fact}[{\cite[Appendix B]{kechris}}]
-%     $\rho(\prec) \le  |X|^+$ (successor cardinal).
+    Since $\rho_\prec\colon X \to \rho(\prec)$ is surjective,
-%     (for countable $<$)
+    we have that $\rho(\prec) \le  |X|^+$.%
-%     \todo{TODO}
+    \gist{\footnote{Here, $|X|^+$ denotes the successor cardinal.}}{}
-%     % TODO QUESTION
+\end{fact}
-% \end{exercise}
+\begin{theorem}[{Kunen-Martin, \cite[(31.1)]{kechris}}]
-\begin{theorem}[Kunen-Martin]
     \yalabel{Kunen-Martin}{Kunen-Martin}{thm:kunenmartin}
-    If $(X, \prec)$ is wellfounded
+    If $(X, \prec)$ is well-founded
     and $\prec \subseteq X^2$ is $\Sigma^1_1$
     then $\rho(\prec) < \omega_1$.
 \end{theorem}
 \begin{proof}
-    % TODO GIST
-    % TODO QUESTION where did we use analytic?
     Wlog.~$X = \cN$.
-    There is a tree $S$ on $\N \times \N \times \N$
+    \gist{%
-    (i.e.~$S \subseteq (\N \times \N \times \N)^{<\N}$)
+        There is a tree $S$ on $\N \times \N \times \N$
-    such that
+        (i.e.~$S \subseteq (\N \times \N \times \N)^{<\N}$)
-    \[
+        such that
-    \forall x, y \in \cN.~\left(x \succ y \iff \exists \alpha \in \cN.~(x,\alpha,y) \in [S]\right).
+        \[
-    \]
+        \forall x, y \in \cN.~\left(x \succ y
+            \iff \exists \alpha \in \cN.~(x,\alpha,y) \in [S]\right).%
+            \footnotemark
+        \]
+        \footnotetext{Here we use that $\prec$ is analytic,
+        i.e.~$\prec$ can be written as the projection of a closed
+        subset of $(\cN \times \cN) \times \cN$ and closed subsets
+        correspond to pruned trees.}
+    }{%
+        Take $S' \overset{\text{closed}}{\subseteq} \cN \times \cN \times \cN$
+        such that $\prec = \proj_{1,3}(S')$
+        and $S$ a tree on $\N \times  \N \times \N$
+        such that $S' = [S]$.
+    }
 
     Let
     \[
@@ -164,11 +193,11 @@ Let $\rho(\prec) \coloneqq  \sup \{\rho_{\prec}(x) + 1 : x \in X\}$.
     Clearly $|W| \le \aleph_0$.
     Define $\prec^\ast$ on $W$
     by setting
-    \[(s_0,u_1,s_1,\ldots, u_n,s_n) \succ (s_0',u_1', s_1', \ldots, u_m', s_m') :\iff\]
+    \[(s_0,u_1,s_1,\ldots, u_n,s_n) \succ^\ast (s_0',u_1', s_1', \ldots, u_m', s_m') :\iff\]
     \begin{itemize}
         \item $n < m$ and
-        \item $\forall i \le n.~s_i \subsetneq s_i' \land u_i \subsetneq u_i'$.
+        \item $\forall i \le n.~s_i \subsetneq s_i' \land u_i \subsetneq u_i'$.%
-            %\todo{$\subseteq$ or $\subsetneq$?} % TODO QUESTION
+            \footnote{sic! (there is a typo in the official notes)}
     \end{itemize}
     \begin{claim}
         $\prec^\ast$ is well-founded.
@@ -183,7 +212,7 @@ Let $\rho(\prec) \coloneqq  \sup \{\rho_{\prec}(x) + 1 : x \in X\}$.
             \]
             and
              \[
-            \alpha_i \coloneqq  \bigcup_n u_i^n \cN.
+             \alpha_i \coloneqq  \bigcup_n u_i^n \in\cN.
             \]
             We get $(x_{i-1}, \alpha_i, x_i) \in [S]$,
             hence $x_{i-1} \succ x_i$ for all $i$,
@@ -193,38 +222,38 @@ Let $\rho(\prec) \coloneqq  \sup \{\rho_{\prec}(x) + 1 : x \in X\}$.
             Use that $ \prec$ is well-founded.
         }
     \end{subproof}
-
     Hence $\rho(\prec^\ast) < |W|^+ \le \omega_1$.
-    % TODO QUESTION
+    \gist{%
-
+        We can turn $(X, \prec)$ into a tree $(T_\prec, \subsetneq)$
-    We can turn $(X, \prec)$ into a tree $(T_\prec, \subsetneq)$
+        with
-    with
+        \begin{IEEEeqnarray*}{rCl}
-     \begin{IEEEeqnarray*}{rCl}
+            \rho(\prec) &=& \rho(T_{\prec})
-         \rho(\prec) &=& \rho(T_{\prec})
+        \end{IEEEeqnarray*}
-    \end{IEEEeqnarray*}
+        by setting $\emptyset \in T_{\prec}$
-    by setting $\emptyset \in T_{\prec}$
+        and
-    and
+        $(x_0,\ldots,x_n) \in T_\prec$,$x_i \in X =\cN$,
-    $(x_0,\ldots,x_n) \in T_\prec$,$x_i \in X =\cN$,
+        iff $x_0 \succ x_1 \succ x_2 \succ \ldots \succ x_n$.
-    iff $x_0 \succ x_1 \succ x_2 \succ \ldots \succ x_n$.
+    }{%
+        Turn $(X,\prec)$ into a tree $(T_\prec, \subsetneq)$
+        with $\rho(\prec) = \rho(T_\prec)$,
+         $(x_0,\ldots,x_n) \in T_\prec$ iff $x_0 \succ \ldots \succ x_n$.
+    }
 
     For all $x \succ y$
     pick $\alpha_{x,y} \in \cN$
-    such that $(x, \alpha_{x,y}, y) \in [S]$
+    such that $(x, \alpha_{x,y}, y) \in [S]$.
-    define
+    Define
     \begin{IEEEeqnarray*}{rCl}
         \phi\colon T_{\prec} \setminus \{\emptyset\}  &\longrightarrow & W \\
-        \phi(x_0,x_1,\ldots,x_n) &\longmapsto & (x_0\defon{n}, \alpha_{x_0}, x_1\defon{n},\ldots,
+        (x_0,x_1,\ldots,x_n) &\longmapsto & (x_0\defon{n}, \alpha_{x_0,x_1}\defon{n} , x_1\defon{n},\ldots,
-        \alpha_{x_{n-1}}, x_n\defon{n}).
+        \alpha_{x_{n-1},x_n}\defon{n} , x_n\defon{n}).
     \end{IEEEeqnarray*}
-    Then $\phi$ is a homomorphism of $\subsetneq$ to $\prec^\ast$
+    Then $\phi$ is a homomorphism of $\supsetneq$ to $\prec^\ast$
     so
     \[
     \rho(\prec)
-    = \rho(T_{\prec} \setminus \{\emptyset\} , \subsetneq)
+    = \rho(T_{\prec} \setminus \{\emptyset\} , \supsetneq)
     \le \rho(\prec^\ast)
     < \omega_1.
     \]
 \end{proof}
-
-
-

inputs/lecture_15.tex

@@ -1,14 +1,15 @@
 \lecture{15}{2023-12-05}{}
 
-% TODO ANKI-MARKER
-
-
 \begin{theorem}[Boundedness Theorem]
     \yalabel{Boundedness Theorem}{Boundedness}{thm:boundedness}
-
+    \gist{%
-    Let $X$ be Polish, $C \subseteq X$ coanalytic,
+        Let $X$ be Polish, $C \subseteq X$ coanalytic,
-    $\phi\colon  C \to \omega_1$ a coanalytic rank on $C$,
+        $\phi\colon  C \to \omega_1$ a coanalytic rank on $C$,
-    $A \subseteq C$ analytic, i.e.~$A \in \Sigma^1_1(X)$.
+        $A \subseteq C$ analytic, i.e.~$A \in \Sigma^1_1(X)$.
+    }{%
+        Let $X$ Polish, $C \in \Pi^1_1(X)$, $A \in \Sigma^1_1(X)$,
+        $A \subseteq C$, $\phi\colon C \to \omega_1$ a coanalytic rank.
+    }
     Then $\sup \{\phi(x) : x \in A\} < \omega_1$.
 
     Moreover for all $\xi < \omega_1$,
@@ -36,7 +37,7 @@
     Since $D_\xi = \bigcup_{\eta < \xi} E_\xi$,
     it suffices to check $E_\xi \in \Sigma_1^1(X)$.
     Let $\alpha \coloneqq \sup \{\phi(x) : x \in C\}$.
-    Then $E_\xi = E_\alpha$ for all $\alpha < \xi < \omega_1$.
+    Then $E_\xi = E_\alpha$ for all $\alpha \le \xi < \omega_1$.
 
     Consider $\xi \le \alpha$.
     \begin{itemize}
@@ -165,6 +166,10 @@ Recall:
     A flow is \vocab{distal} iff
     it has no proximal pair.
 \end{definition}
+\begin{remark}
+    Note that a flow is minimal iff it has no proper subflows.
+\end{remark}
+
 
 \begin{definition}+
     Let $(T,X)$ and $(T,Y)$ be flows.
@@ -187,9 +192,6 @@ Recall:
 
 
 
-\begin{remark}
-    Note that a flow is minimal iff it has no proper subflows.
-\end{remark}
 \begin{example}
     Recall that $S_1 = \{z \in \C : |z| = 1\}$.
     Let $X = S_1$, $T = S_1$
@@ -226,10 +228,9 @@ Recall:
     A flow is isometric iff it is an isometric extension
     of the trivial flow,
     i.e.~the flow acting on a singleton.
-    Indeed maps $\rho\colon X\times_\star X = X^2 \to 2$
+    Indeed maps $\rho\colon X\times_\star X = X^2 \to \R$
     as in \yaref{def:isometricextension}
     correspond to metrics witnessing that the flow is isometric.
-    % TODO THINK ABOUT THIS!
 \end{remark}
 \begin{proposition}
     An isometric extension of a distal flow is distal.
@@ -238,7 +239,7 @@ Recall:
     Let $\pi\colon  X\to Y$ be an isometric extension.
     Towards a contradiction,
     suppose that $x_1,x_2 \in X$ are proximal.
-    Take $z \in X$ and $(g_n) \in T^{\omega}$
+    Take $z \in X$ and a sequence $(g_n)_{n < \omega}$ in $T$
     such that $g_n x_1 \to z$ and $g_n x_2 \to z$.
 
     Then $g_n \pi(x_1) \to  \pi(z)$
@@ -253,18 +254,18 @@ Recall:
     Hence $x_1 = x_2$ $\lightning$.
 \end{proof}
 
+% TODO ANKI-MARKER
 \begin{definition}
     Let $\Sigma = \{(X_i, T) : i \in I\} $
-    be a collection of factors of $(X,T)$. % TODO State precise definition of a factor
+    be a collection of factors of $(X,T)$.
     Let $\pi_i\colon (X,T) \to (X_i, T)$ denote the factor map.
     Then $(X, T)$ is a \vocab{limit}%
-    \footnote{This does not seem to be a limit in the category theory sense.}
+    \footnote{This is not a limit in the category theory sense and not uniquely determined.}
     of $\Sigma$
     iff
     \[
     \forall x_1,x_2 \in X.~\exists i \in I.~\pi_i(x_1) \neq \pi_i(x_2).
     \]
-    % TODO think about abstract nonsense
 \end{definition}
 
 \begin{proposition}

inputs/lecture_17.tex

@@ -169,6 +169,7 @@ In particular,
 for $x \in X$ and $\overline{Tx} = Y$
 we have that $(Y,T)$ is a flow.
 However if we pick $y \in Y$, $Ty$ might not be dense.
+% TODO: question!
 % TODO: think about this!
 % We want to a minimal subflow in a nice way:
 

inputs/tutorial_08.tex

@@ -259,7 +259,7 @@ for some $B_i \in \cB(Y_i)$.
 Proof of Schröder-Bernstein:
 
 Let $X_0 \coloneqq  X$, $Y_0 \coloneqq Y$
-and define $X_{i+1} \coloneqq  g(Y_i)$, $Y_{i+1 } \coloneqq g(X_i)$.
+and define $X_{i+1} \coloneqq  g(Y_i)$, $Y_{i+1 } \coloneqq f(X_i)$.
 We have $X_{i+1} \subseteq X_i$ and similarly for $Y_i$.
 $f$ and $g$ are bijections between
 $X_\omega \coloneqq \bigcap X_i$ and $Y_\omega \coloneqq \bigcap Y_i$.