diff --git a/inputs/lecture_05.tex b/inputs/lecture_05.tex index 36b1cf0..289847a 100644 --- a/inputs/lecture_05.tex +++ b/inputs/lecture_05.tex @@ -103,7 +103,9 @@ Let $X$ be a completely metrizable space. Then every comeager set of $X$ is dense in $X$. \end{theorem} -\todo{Proof (copy from some other lecture)} +\gist{% + \todo{Proof (copy from some other lecture)} +}{Not proved in the lecture.} \begin{theoremdef} Let $X$ be a topological space. The following are equivalent: @@ -118,7 +120,21 @@ \footnote{cf.~\yaref{s5e1}} \end{theoremdef} \begin{proof} - \todo{Proof (short)} + (i) $\implies$ (ii) + \gist{% + Consider a comeager set $A$. + Let $U\neq \emptyset$ be any open set. Since $U$ is + non-meager, we have $A \cap U \neq \emptyset$. + }{The intersection of a comeager and a non-meager set is nonempty.} + + (ii) $\implies$ (iii) + The complement of an open dense set is nwd. + \gist{% + Hence the intersection of countable + many open dense sets is comeager. + }{} + + (iii) $\implies$ (i) Let us first show that $X$ is non-meager. diff --git a/inputs/lecture_07.tex b/inputs/lecture_07.tex index 22a2026..dafa238 100644 --- a/inputs/lecture_07.tex +++ b/inputs/lecture_07.tex @@ -77,7 +77,7 @@ sets that are not clopen)}{}. \end{example} -\subsection{Turning Borels Sets into Clopens} +\subsection{Turning Borel Sets into Clopens} \begin{theorem}% \gist{% @@ -109,55 +109,74 @@ into $B$. \end{corollary} \begin{proof} - Pick $\cT_B \supset \cT$ - such that $(X, \cT_B)$ is Polish, - $B$ is clopen in $\cT_B$ and - $\cB(X,\cT) = \cB(X, \cT_B)$. + \gist{% + Pick $\cT_B \supset \cT$ + such that $(X, \cT_B)$ is Polish, + $B$ is clopen in $\cT_B$ and + $\cB(X,\cT) = \cB(X, \cT_B)$. - Therefore $(\cB, \cT_B\defon{B})$ is Polish. - We know that there is an embedding - $f\colon 2^{\omega} \to (B, \cT_{B}\defon{B})$. + Therefore $(\cB, \cT_B\defon{B})$ is Polish. + We know that there is an embedding + $f\colon 2^{\omega} \to (B, \cT_{B}\defon{B})$. - Consider $f\colon 2^{\omega} \to B \subseteq (X, \cT)$. - This is still continuous as $\cT \subseteq \cT_B$. - Since $2^{\omega}$ is compact, $f$ is an embedding. - %\todo{Think about this} + Consider $f\colon 2^{\omega} \to B \subseteq (X, \cT)$. + This is still continuous as $\cT \subseteq \cT_B$. + Since $2^{\omega}$ is compact, $f$ is an embedding. + }{% + Clopenize $B$. + We can embed $2^{ \omega}$ into Polish spaces. + Clopenization makes the topology finer, + so this is still continuous wrt.~the original topology. + $2^{\omega}$ is compact, so this is an embedding. + } \end{proof} \begin{refproof}{thm:clopenize} - We show that - \begin{IEEEeqnarray*}{rCl} - A \coloneqq \{B \subseteq \cB(X, \cT)&:\exists & \cT_B \supseteq \cT .\\ - && (X, \cT_B) \text{ is Polish},\\ - && \cB(X, \cT) = \cB(X, \cT_B)\\ - && B \text{ is clopen in $\cT_B$}\\ - \} - \end{IEEEeqnarray*} - is equal to the set of Borel sets. - - The proof rests on two lemmata: + \gist{% + We show that + \begin{IEEEeqnarray*}{rCl} + A \coloneqq \{B \subseteq \cB(X, \cT)&:\exists & \cT_B \supseteq \cT .\\ + && (X, \cT_B) \text{ is Polish},\\ + && \cB(X, \cT) = \cB(X, \cT_B)\\ + && B \text{ is clopen in $\cT_B$}\\ + \} + \end{IEEEeqnarray*} + is equal to the set of Borel sets. + }{% + Let $A$ be the set of clopenizable sets. + We show that $A = \cB(X)$. + } + \gist{The proof rests on two lemmata:}{} \begin{lemma} \label{thm:clopenize:l1} - Let $(X,\cT)$ be a Polish space. - Then for any $F \overset{\text{closed}}{\subseteq} X$ (wrt. $\cT$) - there is $\cT_F \supseteq \cT$ - such that $\cT_F$ is Polish, - $\cB(\cT) = \cB(\cT_F)$ - and $F$ is clopen in $\cT_F$. + \gist{% + Let $(X,\cT)$ be a Polish space. + Then for any $F \overset{\text{closed}}{\subseteq} X$ (wrt. $\cT$) + there is $\cT_F \supseteq \cT$ + such that $\cT_F$ is Polish, + $\cB(\cT) = \cB(\cT_F)$ + and $F$ is clopen in $\cT_F$. + }{% + Closed sets can be clopenized. + } \end{lemma} \begin{proof} - Consider $(F, \cT\defon{F})$ and $(X \setminus F, \cT\defon{X \setminus F})$. - Both are Polish spaces. - Take the coproduct% - \footnote{In the lecture, this was called the \vocab{topological sum}.} - $F \oplus (X \setminus F)$ of these spaces. - This space is Polish, - and the topology is generated by $\cT \cup \{F\}$, - hence we do not get any new Borel sets. + \gist{% + Consider $(F, \cT\defon{F})$ and $(X \setminus F, \cT\defon{X \setminus F})$. + Both are Polish spaces. + Take the coproduct% + \footnote{In the lecture, this was called the \vocab{topological sum}.} + $F \oplus (X \setminus F)$ of these spaces. + This space is Polish, + and the topology is generated by $\cT \cup \{F\}$, + hence we do not get any new Borel sets. + }{Consider $(F, \cT\defon{F}) \oplus (X \setminus F, \cT\defon{X \setminus F})$.} \end{proof} - So all closed sets are in $A$. - Furthermore $A$ is closed under complements, - since complements of clopen sets are clopen. + \gist{% + So all closed sets are in $A$. + Furthermore $A$ is closed under complements, + since complements of clopen sets are clopen. + }{So $\Sigma^0_1(X), \Pi^0_1(X) \subseteq A$.} \begin{lemma} \label{thm:clopenize:l2} @@ -170,22 +189,23 @@ is Polish and $\cB(\cT_\infty) = \cB(T)$. \end{lemma} - \begin{proof} - We have that $\cT_\infty$ is the smallest - topology containing all $\cT_n$. - To get $\cT_\infty$ - consider - \[ - \cF \coloneqq \{A_1 \cap A_2 \cap \ldots \cap A_n : A_i \in \cT_i\}. - \] - Then - \[ - \cT_\infty = \{\bigcup_{i<\omega} B_i : B_i \in \cF\}. - \] - (It suffices to take countable unions, - since we may assume that the $A_1, \ldots, A_n$ in the - definition of $\cF$ belong to - a countable basis of the respective $\cT_n$). + \begin{refproof}{thm:clopenize:l2} + \gist{% + We have that $\cT_\infty$ is the smallest + topology containing all $\cT_n$. + To get $\cT_\infty$ consider + \[ + \cF \coloneqq \{A_1 \cap A_2 \cap \ldots \cap A_n : A_i \in \cT_i\}. + \] + Then + \[ + \cT_\infty = \{\bigcup_{i<\omega} B_i : B_i \in \cF\}. + \] + (It suffices to take countable unions, + since we may assume that the $A_1, \ldots, A_n$ in the + definition of $\cF$ belong to + a countable basis of the respective $\cT_n$). + }{} % Proof was finished in lecture 8 Let $Y = \prod_{n \in \N} (X, \cT_n)$. @@ -195,6 +215,7 @@ \begin{claim} $\delta$ is a homeomorphism. \end{claim} + \gist{% \begin{subproof} Clearly $\delta$ is a bijection. We need to show that it is continuous and open. @@ -219,28 +240,34 @@ \delta(V) = D \cap (X \times X \times \ldots \times U_{n_1} \times \ldots \times U_{n_2} \times \ldots \times U_{n_u} \times X \times \ldots) \overset{\text{open}}{\subseteq} D. \] \end{subproof} + }{} - This will finish the proof since - \[ - D = \{(x,x,\ldots) \in Y : x \in X\} \overset{\text{closed}}{\subseteq} Y - \] - Why? Let $(x_n) \in Y \setminus D$. - Then there are $i < j$ such that $x_i \neq x_j$. - Take disjoint open $x_i \in U$, $x_j \in V$. - Then - \[(x_n) \in X \times X \times \ldots \times U \times \ldots \times X \times \ldots \times V \times X \times \ldots\] - is open in $Y\setminus D$. - Hence $Y \setminus D$ is open, thus $D$ is closed. - It follows that $D$ is Polish. - \end{proof} + \begin{claim} + $D = \{(x,x,\ldots) \in Y : x \in X\} \overset{\text{closed}}{\subseteq} Y.$ + \end{claim} + \gist{% + \begin{subproof} + Let $(x_n) \in Y \setminus D$. + Then there are $i < j$ such that $x_i \neq x_j$. + Take disjoint open $x_i \in U$, $x_j \in V$. + Then + \[(x_n) \in X \times X \times \ldots \times U \times \ldots \times X \times \ldots \times V \times X \times \ldots\] + is open in $Y\setminus D$. + Hence $Y \setminus D$ is open, thus $D$ is closed. + \end{subproof} + It follows that $D$ is Polish. + }{} + \end{refproof} - We need to show that $A$ is closed under countable unions. - By \yaref{thm:clopenize:l2} there exists a topology - $\cT_\infty$ such that $A = \bigcup_{n < \omega} A_n$ is open in $\cT_\infty$ - and $\cB(\cT_\infty) = \cB(\cT)$. - Applying \yaref{thm:clopenize:l1} - yields a topology $\cT_\infty'$ such that - $(X, \cT_\infty')$ is Polish, - $\cB(\cT_\infty') = \cB(\cT)$ - and $A $ is clopen in $\cT_{\infty}'$. + \gist{% + We need to show that $A$ is closed under countable unions. + By \yaref{thm:clopenize:l2} there exists a topology + $\cT_\infty$ such that $A = \bigcup_{n < \omega} A_n$ is open in $\cT_\infty$ + and $\cB(\cT_\infty) = \cB(\cT)$. + Applying \yaref{thm:clopenize:l1} + yields a topology $\cT_\infty'$ such that + $(X, \cT_\infty')$ is Polish, + $\cB(\cT_\infty') = \cB(\cT)$ + and $A $ is clopen in $\cT_{\infty}'$. + }{} \end{refproof} diff --git a/inputs/lecture_08.tex b/inputs/lecture_08.tex index 62298a0..bca267f 100644 --- a/inputs/lecture_08.tex +++ b/inputs/lecture_08.tex @@ -1,8 +1,9 @@ -\lecture{08}{2023-11-10}{}\footnote{% +\lecture{08}{2023-11-10}{}% +\gist{\footnote{% In the beginning of the lecture, we finished the proof of \yaref{thm:clopenize:l2}. This has been moved to the notes on lecture 7.% -} +}}{} \subsection{Parametrizations} %\todo{choose better title} @@ -22,13 +23,14 @@ where $X$ is a metrizable, usually second countable space. \item $\{U_y : y \in Y\} = \Gamma(X)$. \end{itemize} \end{definition} - -\begin{example} - Let $X = \omega^\omega$, $Y = 2^{\omega}$ - and consider $\Gamma = \Sigma^0_{\omega+5}(\omega^\omega)$. - We will show that there is a $2^{\omega}$-universal - set for $\Gamma$. -\end{example} +\gist{% + \begin{example} + Let $X = \omega^\omega$, $Y = 2^{\omega}$ + and consider $\Gamma = \Sigma^0_{\omega+5}(\omega^\omega)$. + We will show that there is a $2^{\omega}$-universal + set for $\Gamma$. + \end{example} +}{} \begin{theorem} \label{thm:cantoruniversal} diff --git a/inputs/lecture_11.tex b/inputs/lecture_11.tex index 53d2c7e..5174717 100644 --- a/inputs/lecture_11.tex +++ b/inputs/lecture_11.tex @@ -99,9 +99,10 @@ (continuous wrt.~to the topology of $X$) On the other hand \[ - X \hookrightarrow\cN \overset{\text{continuous embedding}}{\hookrightarrow}\cC + X \hookrightarrow\cN \overset{\text{continuous embedding\footnotemark}}{\hookrightarrow}\cC \] - \todo{second inclusion was on a homework sheet} + \footnotetext{cf.~\yaref{s2e4}} + For the first inclusion, recall that there is a continuous bijection $b\colon D \to X$, where $D \overset{\text{closed}}{\subseteq} \cN$. diff --git a/inputs/lecture_13.tex b/inputs/lecture_13.tex index f68c640..c534efa 100644 --- a/inputs/lecture_13.tex +++ b/inputs/lecture_13.tex @@ -79,8 +79,6 @@ with $(f^{-1}(\{1\}), <)$. }{easy} \end{proof} -% TODO ANKI-MARKER - \begin{theorem}[Lusin-Sierpinski] The set $\LO \setminus \WO$ (resp.~$2^{\Q} \setminus \WO$) @@ -89,19 +87,23 @@ with $(f^{-1}(\{1\}), <)$. \begin{proof} We will find a continuous function $f\colon \Tr \to \LO$ such that + \gist{% \[ x \in \WF \iff f(x) \in \WO \] (equivalently $x \in \IF \iff f(x) \in \LO \setminus \WO$). This suffices, since $\IF \subseteq \Tr$ is $\Sigma^1_1$-complete + }{ + $f^{-1}(\LO \setminus \WO) = \IF$. + This suffices + } (see \yaref{cor:ifs11c}). Fix a bijection $b\colon \N \to \N^{<\N}$. \begin{idea} For $T \in \Tr$ consider - $<_{KB}\defon{T}$ - % TODO? + $<_{KB}\defon{T}$. \end{idea} Let $\alpha \in \Tr$. @@ -109,7 +111,7 @@ with $(f^{-1}(\{1\}), <)$. (i.e.~$m \le_{f(\alpha)} n$) iff \begin{itemize} - \item $(\alpha(b(m)) = \alpha(b(n)) = 1$ + \item $\alpha(b(m)) = \alpha(b(n)) = 1$ and $b(m) \le_{KB} b(n)$ (recall that we identified $\Tr$ with a subset of ${2^{\N}}^{<\N}$), @@ -123,15 +125,16 @@ with $(f^{-1}(\{1\}), <)$. \end{proof} % TODO: new section? - -Recall that a \vocab{rank} on a set $C$ -is a map $\phi\colon C \to \Ord$. -\begin{example} - \begin{IEEEeqnarray*}{rCl} - \otp \colon \WO &\longrightarrow & \Ord \\ - x &\longmapsto & \text{the unique $\alpha \in \Ord$ such that $x \cong \alpha$}. - \end{IEEEeqnarray*} -\end{example} +\gist{% + Recall that a \vocab{rank} on a set $C$ + is a map $\phi\colon C \to \Ord$. + \begin{example} + \begin{IEEEeqnarray*}{rCl} + \otp \colon \WO &\longrightarrow & \Ord \\ + x &\longmapsto & \text{the unique $\alpha \in \Ord$ such that $x \cong \alpha$}. + \end{IEEEeqnarray*} + \end{example} +}{} \begin{definition} A \vocab{prewellordering} $\preceq$ @@ -141,7 +144,7 @@ is a map $\phi\colon C \to \Ord$. \item reflexive, \item transitive, \item total (any two $x,y$ are comparable), - \item $\prec$ ($x \prec y \iff x \preceq y \land y \not\preceq x$) is well-founded, + \item $\prec$ \gist{($x \prec y \iff x \preceq y \land y \not\preceq x$)}{} is well-founded, in the sense that there are no descending infinite chains. \end{itemize} \end{definition} @@ -149,7 +152,7 @@ is a map $\phi\colon C \to \Ord$. \begin{itemize} \item A prewellordering may not be a linear order since it is not necessarily antisymmetric. - \item The linearly ordered wellfounded sets are exactly the wellordered sets. + %\item The linearly ordered wellfounded sets are exactly the wellordered sets. \item Modding out $x \sim y :\iff x \preceq y \land y \preceq x$ turns a prewellordering into a wellordering. \end{itemize} @@ -163,11 +166,11 @@ between downwards-closed ranks and prewellorderings: \phi_{\preceq}&\longmapsfrom& \preceq, \end{IEEEeqnarray*} where $\phi_\preceq(x)$ is defined as -\begin{IEEEeqnarray*}{rCl} +\gist{\begin{IEEEeqnarray*}{rCl} \phi_{\preceq}(x) &\coloneqq &0 \text{ if $x$ is minimal},\\ \phi_{\preceq}(x) &\coloneqq & \sup \{\phi_{\preceq}(y) + 1 : y \prec x\}, \end{IEEEeqnarray*} -i.e. +i.e.}{} \[ \phi_{\preceq}(x) = \otp\left(\faktor{\{y \in C : y \prec x\}}{\sim}\right). \] diff --git a/inputs/lecture_14.tex b/inputs/lecture_14.tex index 975e4a5..3a52e99 100644 --- a/inputs/lecture_14.tex +++ b/inputs/lecture_14.tex @@ -1,5 +1,6 @@ \lecture{14}{2023-12-01}{} +% TODO ANKI-MARKER \begin{theorem}[Moschovakis] If $C$ is coanalytic, then there exists a $\Pi^1_1$-rank on $C$.