diff --git a/bibliography/references.bib b/bibliography/references.bib index 2f44ca0..1958dc8 100644 --- a/bibliography/references.bib +++ b/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}, +} + diff --git a/inputs/lecture_02.tex b/inputs/lecture_02.tex index 01655a2..403d9c7 100644 --- a/inputs/lecture_02.tex +++ b/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 $\diam_{d_Y}(U \cap Y) \le \frac{1}{n}$. + \item \gist{$\diam_d(U) \le \frac{1}{n}$,% + \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 $Y = \bigcap_{n \in \N} V_n$. diff --git a/inputs/lecture_07.tex b/inputs/lecture_07.tex index dafa238..d350805 100644 --- a/inputs/lecture_07.tex +++ b/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 $U_{n_i} \in \cT_i$. diff --git a/inputs/lecture_08.tex b/inputs/lecture_08.tex index bca267f..898ceb0 100644 --- a/inputs/lecture_08.tex +++ b/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} diff --git a/inputs/lecture_13.tex b/inputs/lecture_13.tex index 2a1cdb7..a495a93 100644 --- a/inputs/lecture_13.tex +++ b/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 diff --git a/inputs/lecture_14.tex b/inputs/lecture_14.tex index 258aaca..71739c1 100644 --- a/inputs/lecture_14.tex +++ b/inputs/lecture_14.tex @@ -58,7 +58,19 @@ \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 @@ -105,6 +117,13 @@ 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 $X$. @@ -126,33 +145,43 @@ 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} -% $\rho(\prec) \le |X|^+$ (successor cardinal). -% (for countable $<$) -% \todo{TODO} -% % TODO QUESTION -% \end{exercise} -\begin{theorem}[Kunen-Martin] +\begin{fact}[{\cite[Appendix B]{kechris}}] + Since $\rho_\prec\colon X \to \rho(\prec)$ is surjective, + we have that $\rho(\prec) \le |X|^+$.% + \gist{\footnote{Here, $|X|^+$ denotes the successor cardinal.}}{} +\end{fact} +\begin{theorem}[{Kunen-Martin, \cite[(31.1)]{kechris}}] \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 $\N \times \N \times \N$ - (i.e.~$S \subseteq (\N \times \N \times \N)^{<\N}$) - such that - \[ - \forall x, y \in \cN.~\left(x \succ y \iff \exists \alpha \in \cN.~(x,\alpha,y) \in [S]\right). - \] + \gist{% + There is a tree $S$ on $\N \times \N \times \N$ + (i.e.~$S \subseteq (\N \times \N \times \N)^{<\N}$) + such that + \[ + \forall x, y \in \cN.~\left(x \succ y + \iff \exists \alpha \in \cN.~(x,\alpha,y) \in [S]\right).% + \footnotemark + \] + \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 \[ @@ -163,11 +192,11 @@ 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 $\forall i \le n.~s_i \subsetneq s_i' \land u_i \subsetneq u_i'$. - %\todo{$\subseteq$ or $\subsetneq$?} % TODO QUESTION + \item $\forall i \le n.~s_i \subsetneq s_i' \land u_i \subsetneq u_i'$.% + \footnote{sic! (there is a typo in the official notes)} \end{itemize} \begin{claim} $\prec^\ast$ is well-founded. @@ -182,7 +211,7 @@ 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 $i$, @@ -192,38 +221,38 @@ 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 - - We can turn $(X, \prec)$ into a tree $(T_\prec, \subsetneq)$ - with - \begin{IEEEeqnarray*}{rCl} - \rho(\prec) &=& \rho(T_{\prec}) - \end{IEEEeqnarray*} - by setting $\emptyset \in T_{\prec}$ - and - $(x_0,\ldots,x_n) \in T_\prec$,$x_i \in X =\cN$, - iff $x_0 \succ x_1 \succ x_2 \succ \ldots \succ x_n$. + \gist{% + We can turn $(X, \prec)$ into a tree $(T_\prec, \subsetneq)$ + with + \begin{IEEEeqnarray*}{rCl} + \rho(\prec) &=& \rho(T_{\prec}) + \end{IEEEeqnarray*} + by setting $\emptyset \in T_{\prec}$ + and + $(x_0,\ldots,x_n) \in T_\prec$,$x_i \in X =\cN$, + iff $x_0 \succ x_1 \succ x_2 \succ \ldots \succ x_n$. + }{% + 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 $(x, \alpha_{x,y}, y) \in [S]$ - define + such that $(x, \alpha_{x,y}, y) \in [S]$. + 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, - \alpha_{x_{n-1}}, x_n\defon{n}). + (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} , 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} - - - diff --git a/inputs/lecture_15.tex b/inputs/lecture_15.tex index c8022f4..2db81f9 100644 --- a/inputs/lecture_15.tex +++ b/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} - - Let $X$ be Polish, $C \subseteq X$ coanalytic, - $\phi\colon C \to \omega_1$ a coanalytic rank on $C$, - $A \subseteq C$ analytic, i.e.~$A \in \Sigma^1_1(X)$. + \gist{% + Let $X$ be Polish, $C \subseteq X$ coanalytic, + $\phi\colon C \to \omega_1$ a coanalytic rank on $C$, + $A \subseteq C$ 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 $g_n \pi(x_1) \to \pi(z)$ @@ -253,9 +254,10 @@ 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 maps. Then $(X, T)$ is the \vocab{limit} of $\Sigma$ iff diff --git a/inputs/tutorial_08.tex b/inputs/tutorial_08.tex index 0f2a140..c6ccca4 100644 --- a/inputs/tutorial_08.tex +++ b/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 $Y_\omega \coloneqq \bigcap Y_i$.