From 4d3e3c2a49c806f1a97a4b62aeccddb7d58f26b0 Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Tue, 23 Jan 2024 14:29:27 +0100 Subject: [PATCH 1/9] tutorial 13 --- inputs/lecture_05.tex | 1 + inputs/tutorial_09.tex | 18 +---- inputs/tutorial_13.tex | 145 +++++++++++++++++++++++++++++++++++++++++ jrpie-gist.sty | 3 + logic3.tex | 1 + 5 files changed, 152 insertions(+), 16 deletions(-) create mode 100644 inputs/tutorial_13.tex diff --git a/inputs/lecture_05.tex b/inputs/lecture_05.tex index 10250ae..e2c10a1 100644 --- a/inputs/lecture_05.tex +++ b/inputs/lecture_05.tex @@ -97,6 +97,7 @@ %\end{example} \begin{theorem}[Baire Category theorem] + \yalabel{Baire Category Theorem}{Baire Category Theorem}{thm:bct} Let $X$ be a completely metrizable space. Then every comeager set of $X$ is dense in $X$. \end{theorem} diff --git a/inputs/tutorial_09.tex b/inputs/tutorial_09.tex index 8b5b5ee..49682d1 100644 --- a/inputs/tutorial_09.tex +++ b/inputs/tutorial_09.tex @@ -140,16 +140,7 @@ amounts to a finite number of conditions on the preimage. \end{pmatrix*}&\longmapsfrom & \beta \in \Homeo(X). \end{IEEEeqnarray*} Clearly this has the desired properties. - \item We have - \begin{IEEEeqnarray*}{Cl} - & \Z \circlearrowright X \text{ has a dense orbit}\\ - \iff& \exists x \in X.~ \overline{\Z\cdot x} = X\\ - \iff& \exists x \in X.~\forall U\overset{\text{open}}{\subseteq} X.~\exists z \in \Z.~ - z \cdot x \in U\\ - \iff&\exists x \in X.~\forall U \overset{\text{open}}{\subseteq} X.~ - \exists z \in \Z.~f^z(x) \in U. - \end{IEEEeqnarray*} - \item Let $X$ be a compact Polish space. + \item Let $X$ be a compact Polish space. What is the Borel complexity of $\Homeo(X)$ inside $\cC(X,X)$? Recall that $\cC(X,X)$ is a Polish space with the uniform topology. @@ -160,14 +151,9 @@ amounts to a finite number of conditions on the preimage. \end{IEEEeqnarray*} by the general fact \begin{fact} - Let $X$ be comapct and $Y$ Hausdorff, + Let $X$ be compact and $Y$ Hausdorff, $f\colon X \to Y$ a continuous bijection. Then $f$ is a homeomorphism. \end{fact} - \item It suffices to check the condition from part (b) - for open sets $U$ of a countable basis - and points $x \in X$ belonging to a countable dense subset. - Replacing quantifiers by unions resp.~intersections - gives that $D$ is Borel. \end{itemize} diff --git a/inputs/tutorial_13.tex b/inputs/tutorial_13.tex new file mode 100644 index 0000000..30e78c4 --- /dev/null +++ b/inputs/tutorial_13.tex @@ -0,0 +1,145 @@ +\tutorial{13}{2024-01-23}{} + + +Continuation of sheet 8, exercise 4. + +\begin{definition} + Let $X$ be a compact metric space. + For $K \subseteq X$ compact and $U \overset{\text{open}}{\subseteq} X$ + let + \[ + S_{K,U} \coloneqq \{f \in \cC(X,X): f(K) \subseteq U\}. + \] + The \vocab{compact open topology} on $\cC(X,X)$ + is the topology that has $S_{K,U}$ as a subbase. +\end{definition} +\begin{fact} + If $X$ is compact, + then the compact open topology + is the topology induced by the uniform metric $d_\infty$. +\end{fact} +\begin{proof} + Take some $S_{K,U}$. We need to show that this can be written + as a union of open $d_{\infty}$-balls. + Let $f_0 \in S_{K,U}$. + Consider the continuous function $d(-, U^c)$. + Since $f_0(K)$ is compact, + there exists $\epsilon \coloneqq \min d(f_0(K), U^c)$ + and $B_{\epsilon}(f_0) \subseteq S_{K,U}$. + + + On the other hand, consider $B_{\epsilon}(f_0)$ for some $\epsilon > 0$ + and $f_0 \in \cC(X,X)$. + + As $f_0$ is uniformly continuous, + there exists $\delta > 0$ such that $d(x,x') < \delta \implies d(f_0(x), f_0(x')) < \frac{\epsilon}{3}$. + Cover $X$ with finitely many $\delta$-balls $B_\delta(a_1), \ldots, B_{\delta}(a_k)$. + Then + \[f_0(\overline{B_{\delta}(a_i)}) \subseteq \overline{f_0(B_{\delta}(a_i)} \subseteq \overline{B_{\frac{\epsilon}{3}}(f_0(a_i))} \subseteq B_{\frac{\epsilon}{2}}(f_0(a_i)).\] + + For $i \le k$, let $S_i \coloneqq S_{\overline{B_{\delta}(a_i)}, B_{\frac{\epsilon}{2}}(f_0(a_i))}$. + Take $\bigcap_{i \le k} S_i$. This is open + in the compact open topology and + $B_{\epsilon}(f_0) \subseteq \bigcap_{i \le k} S_i$. +\end{proof} + +\begin{claim} + $f \in \cC(X,X)$ is surjective + iff for all basic open $\emptyset\neq U \subseteq X$ + there exists a basic open $\emptyset \neq V \subseteq X$ + with $f(\overline{V}) \subseteq U$. + + Note that we can write this as a $G_\delta$-condition. + +\end{claim} +\begin{subproof} + Take $B_\epsilon(f(x_0))\subseteq U$. + Then there exists $\delta > 0$ + such that $f(B_{\delta}(x_0)) \subseteq B_{\frac{\epsilon}{2}}(f(x_0))$ + hence $f(\overline{B_{\delta}(x_0)}) \subseteq B_\epsilon(f(x_0))$. + + + For the other direction take $y \in X$. + We want to find a preimage. + For every $B_{\frac{1}{n}}(y)$, + there exists a basic open set $V_n$ with $f(\overline{V}) \subseteq B_{\frac{1}{n}}(y)$. + Take $x_n \in V_n$. + Since $X$ is compact, it is sequentially compact, + so there exists a converging subsequence. + Wlog.~$x_n \to x$, + so $f(x_n) \to f(x) = y$. +\end{subproof} + +\begin{claim} + $f \in \cC(X,X)$ is injective iff + for all basic open $U$,$V$ + with $\overline{U} \cap \overline{V} = \emptyset$ + we have $f(\overline{U}) \cap f(\overline{V}) = \emptyset$. + + This is a $G_\delta$-condition, + since we can write it as + \[ + \bigcap_{U,V} S_{\overline{U}, f(\overline{V})^c}. + \] +\end{claim} +\begin{subproof} + $\implies$ is trivial. + + $\impliedby$ follows since for all pairs $x,y \in X$, + we can find $x \in U$, $y \in V$ such that $\overline{U} \cap \overline{V} = \emptyset$. +\end{subproof} + +Hence $\Homeo(X,X)$ is $G_\delta$. +In particular it is a Polish space. + + + +Let $D$ be the set of $\Z$-flows with dense orbit. +\begin{claim} + $f \in D$ $\iff$ + for all basic open $U,V \subseteq X$, + there exists $n \in \Z$ + such that $f^n(U) \cap V \neq \emptyset$. + + \end{claim} +\begin{subproof} + Suppose that the orbit of $x_0 \in X$ is dense. + Then there exist $k,l \in \Z$ + such that $f^k(x_0)\in U$ and $f^l(x_0) \in V$, + so $f^{l-k} U \cap V \neq \emptyset$. + + + For basic open sets $V$ + let + \[ + A_V \coloneqq \{ x \in X: \exists n.~ f^n(x) \in V\}. + \] + By assumption, all the $A_V$ are dense. + Hence $\bigcap_{V}A_V$ is dense by the \yaref{thm:bct}. + + $A_V = \bigcup_{n \in \Z} f^n(V)$ is open. +\end{subproof} + +\begin{claim} +The condition can be written as a $G_\delta$ set. +\end{claim} +\begin{subproof} + + For $f_0(U) \cap V \neq \emptyset$ + take $u \in U$ such that $f_0(u) \in V$. + Then there exists $\epsilon > 0$ such that $B_{\epsilon}(f_0(u)) \subseteq U$, + hence $B_{\epsilon}(f_0)$ is an open neighbourhood contained + in $\{f : f(U) \cap V \neq \emptyset \} $. + + For $n = 2$ note that + $d(f^2(u), f^2_0(u) \le d(f(f(u)), f_0(f(u))) + d(f_0(f(u)), f_0(f_0(u)))$. + The first part can be bounded by $d(f,f_0)$. + For the second part, + note that there exists $\delta$ such that + \[d(a,b) < \delta \implies d(f_0(a), f_0(b)) < \frac{\epsilon}{2}.\] + Let $\eta \coloneqq \min \{\delta, \frac{\epsilon}{2}\}$ + and consider $d_\infty(f,f_0) < \epsilon$. + + For other $n$ it is some more work, which is left as an exercise. +\end{subproof} + diff --git a/jrpie-gist.sty b/jrpie-gist.sty index 5bde104..ea7caa3 100644 --- a/jrpie-gist.sty +++ b/jrpie-gist.sty @@ -1,6 +1,9 @@ \NeedsTeXFormat{LaTeX2e} \ProvidesPackage{jrpie-gist}[2023/01/22 - gist version for lecture notes] +% TODO gist info +% TODO link to long version (provide link to main document) + \newcommand{\gist}[2]{% \ifcsname EnableGist\endcsname% #2% diff --git a/logic3.tex b/logic3.tex index b721b4c..6127c23 100644 --- a/logic3.tex +++ b/logic3.tex @@ -65,6 +65,7 @@ \input{inputs/tutorial_07} \input{inputs/tutorial_08} \input{inputs/tutorial_09} +\input{inputs/tutorial_13} % sic! \input{inputs/tutorial_10} \input{inputs/tutorial_11} \input{inputs/tutorial_12b} From 0a14244eb3c2a896b8d90f7fa40cea7197674334 Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Tue, 23 Jan 2024 20:56:14 +0100 Subject: [PATCH 2/9] gist intro --- inputs/intro.tex | 16 +++++++++++++++- inputs/tutorial_13.tex | 1 + 2 files changed, 16 insertions(+), 1 deletion(-) diff --git a/inputs/intro.tex b/inputs/intro.tex index 9b141c8..b661dfb 100644 --- a/inputs/intro.tex +++ b/inputs/intro.tex @@ -1,4 +1,6 @@ -These are my notes on the lecture Probability Theory, +\gist{% +These are my notes on the lecture +Logic 3: Abstract Topological Dynamics and Descriptive Set Theory taught by \textsc{Aleksandra Kwiatkowska} in the summer term 2023 at the University Münster. @@ -18,3 +20,15 @@ I could not attend! This notes follow the way the material was presented in the lecture rather closely. Additions (e.g.~from exercise sheets) and slight modifications have been marked with $\dagger$. +}{ + This document aims to give a very brief summary of + my \href{https://josia-notes.users.abstractnonsen.se/w23-logic-3/logic3.pdf}{notes on the course Logic 3}. + I try to omit most technical details and only summarize the most important + ideas. + + Note that this is currently work in progress. + Currently the differences to the original document are only + minor (this is still mostly a technical test), + but this document will get shorter as I work through + and summarize it. +} diff --git a/inputs/tutorial_13.tex b/inputs/tutorial_13.tex index 30e78c4..1160d1c 100644 --- a/inputs/tutorial_13.tex +++ b/inputs/tutorial_13.tex @@ -2,6 +2,7 @@ Continuation of sheet 8, exercise 4. + % Whiteboard https://wbo.ophir.dev/boards/Dphc7eylJJcIA0WbQsn7jzec1domqyx51gXb5qe6rzw-#263,0,0.5 \begin{definition} Let $X$ be a compact metric space. From c986475c775646dfdddcca81f63cf6a42ea3523f Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Tue, 23 Jan 2024 21:52:45 +0100 Subject: [PATCH 3/9] gist for lectures 1-4 --- inputs/lecture_01.tex | 29 +++++----- inputs/lecture_02.tex | 117 ++++++++++++++++++++++------------------- inputs/lecture_03.tex | 86 +++++++++++++++++------------- inputs/lecture_04.tex | 74 ++++++++++++++++---------- inputs/lecture_05.tex | 14 +++-- inputs/lecture_06.tex | 7 ++- inputs/lecture_07.tex | 78 +++++++++++++++++++++------ inputs/lecture_08.tex | 63 ++-------------------- inputs/tutorial_01.tex | 13 ++++- 9 files changed, 264 insertions(+), 217 deletions(-) diff --git a/inputs/lecture_01.tex b/inputs/lecture_01.tex index 7aebc22..df7eb8c 100644 --- a/inputs/lecture_01.tex +++ b/inputs/lecture_01.tex @@ -58,7 +58,7 @@ However the converse of this does not hold. Take $x_0 \in X$ and consider the topology given by \[ \tau = \{U \subseteq X | U \ni x_0\} \cup \{\emptyset\}. - \] + \] Then $\{x_0\}$ is dense in $X$, but $X$ is not second countable. \end{example} \begin{example}[Sorgenfrey line] @@ -76,7 +76,7 @@ However the converse of this does not hold. \end{itemize} \end{fact} \begin{fact} - Compact\footnote{It is not clear whether compact means compact and Hausdorff in this lecture.} Hausdorff spaces are \vocab{normal} (T4) + Compact Hausdorff spaces are \vocab{normal} (T4) i.e.~two disjoint closed subsets can be separated by open sets. \end{fact} @@ -114,7 +114,7 @@ However the converse of this does not hold. \end{absolutelynopagebreak} \subsection{Some facts about polish spaces} - +\gist{% \begin{fact} Let $(X, \tau)$ be a topological space. Let $d$ be a metric on $X$. @@ -130,9 +130,10 @@ To show that $\tau_d = \tau_{d'}$ for two metrics $d, d'$, suffices to show that open balls in one metric are unions of open balls in the other. \end{fact} +}{} \begin{notation} - We sometimes denote $\min(a,b)$ by $a \wedge b$. + We sometimes\footnote{only in this subsection?} denote $\min(a,b)$ by $a \wedge b$. \end{notation} \begin{proposition} @@ -142,18 +143,20 @@ suffices to show that open balls in one metric are unions of open balls in the o Then $d' \coloneqq \min(d,1)$ is also a metric compatible with $\tau$. \end{proposition} +\gist{% \begin{proof} To check the triangle inequality: \begin{IEEEeqnarray*}{rCl} d(x,y) \wedge 1 &\le & \left( d(x,z) + d(y,z) \right) \wedge 1\\ &\le & \left( d(x,z) \wedge 1 \right) + \left( d(y,z) \wedge 1 \right). \end{IEEEeqnarray*} - + For $\epsilon \le 1$ we have $B_\epsilon'(x) = B_\epsilon(x)$ and for $\epsilon > 1$, $B'_\epsilon(x) = X$. - + Since $d$ is complete, we have that $d'$ is complete. \end{proof} +}{} \begin{proposition} Let $A$ be a Polish space. Then $A^{\omega}$ Polish. @@ -164,11 +167,11 @@ suffices to show that open balls in one metric are unions of open balls in the o (Consider the basic open sets of the product topology). Let $d \le 1$ be a complete metric on $A$. - Define $D$ on $A^\omega$ by + Define $D$ on $A^\omega$ by \[ - D\left( (x_n), (y_n) \right) \coloneqq + D\left( (x_n), (y_n) \right) \coloneqq \sum_{n< \omega} 2^{-(n+1)} d(x_n, y_n). - \] + \] Clearly $D \le 1$. It is also clear, that $D$ is a metric. @@ -194,7 +197,7 @@ suffices to show that open balls in one metric are unions of open balls in the o Let $X$ be a separable, metrisable topological space\footnote{e.g.~Polish, but we don't need completeness.}. Then $X$ topologically embeds into the \vocab{Hilbert cube}, - i.e. there is an injective $f: X \hookrightarrow [0,1]^{\omega}$ + i.e. there is an injective $f: X \hookrightarrow [0,1]^{\omega}$ such that $f: X \to f(X)$ is a homeomorphism. \end{proposition} \begin{proof} @@ -252,15 +255,15 @@ suffices to show that open balls in one metric are unions of open balls in the o \begin{proposition} Closed subspaces of Polish spaces are Polish. \end{proposition} -\gist{}{ +\gist{% \begin{proof} Let $X$ be Polish and $V \subseteq X$ closed. Let $d$ be a complete metric on $X$. Then $d\defon{V}$ is complete. Subspaces of second countable spaces are second countable. -\end{proof} -} +\end{proof}% +}{} \begin{definition} Let $X$ be a topological space. diff --git a/inputs/lecture_02.tex b/inputs/lecture_02.tex index 6faa1f3..01655a2 100644 --- a/inputs/lecture_02.tex +++ b/inputs/lecture_02.tex @@ -72,41 +72,51 @@ \[d_1((x_1,y_1), (x_2, y_2)) \coloneqq d(x_1,x_2) + |y_1 - y_2|\] metric is complete. - $f_U$ is an embedding of $U$ into $X \times \R$\gist{: - \begin{itemize} - \item It is injective because of the first coordinate. - \item It is continuous since $d(x, U^c)$ is continuous - and only takes strictly positive values. % TODO - \item The inverse is continuous because projections - are continuous. - \end{itemize} - }{.} + $f_U$ is an embedding of $U$ into $X \times \R$% + \gist{: + \begin{itemize} + \item It is injective because of the first coordinate. + \item It is continuous since $d(x, U^c)$ is continuous + and only takes strictly positive values. % TODO + \item The inverse is continuous because projections + are continuous. + \end{itemize} + }{.} - So we have shown that $U$ and - the graph of $\tilde{f_U}\colon x \mapsto \frac{1}{d(x, U^c)}$ - are homeomorphic. - The graph is closed \gist{in $U \times \R$, - because $\tilde{f_U}$ is continuous. - It is closed}{} in $X \times \R$ \gist{because - $\tilde{f_U} \to \infty$ for $d(x, U^c) \to 0$}{}. - \todo{Make this precise} + \gist{% + So we have shown that $U$ and + the graph of $\tilde{f_U}\colon x \mapsto \frac{1}{d(x, U^c)}$ + are homeomorphic. + The graph is closed \gist{in $U \times \R$, + because $\tilde{f_U}$ is continuous. + It is closed}{} in $X \times \R$ \gist{because + $\tilde{f_U} \to \infty$ for $d(x, U^c) \to 0$}{}. + \todo{Make this precise} + + Therefore we identified $U$ with a closed subspace of + the Polish space $(X \times \R, d_1)$. + }{% + So $U \cong \mathop{Graph}(x \mapsto \frac{1}{d(x, U^c)})$ + and the RHS is a close subspace of the Polish space + $(X \times \R, d_1)$. + } - Therefore we identified $U$ with a closed subspace of - the Polish space $(X \times \R, d_1)$. \end{refproof} Let $Y = \bigcap_{n \in \N} U_n$ be $G_{\delta}$. - Take + Consider \begin{IEEEeqnarray*}{rCl} f_Y\colon Y &\longrightarrow & X \times \R^{\N} \\ x &\longmapsto & \left(x, \left( \frac{1}{\delta(x,U_n^c)} \right)_{n \in \N}\right) \end{IEEEeqnarray*} - As for an open $U$, $f_Y$ is an embedding. - Since $X \times \R^{\N}$ - is completely metrizable, - so is the closed set $f_Y(Y) \subseteq X \times \R^\N$. + \gist{ + As for an open $U$, $f_Y$ is an embedding. + Since $X \times \R^{\N}$ + is completely metrizable, + so is the closed set $f_Y(Y) \subseteq X \times \R^\N$. + }{} \begin{claim} \label{psubspacegdelta:c2} @@ -123,36 +133,35 @@ \item $\diam_d(U) \le \frac{1}{n}$, \item $\diam_{d_Y}(U \cap Y) \le \frac{1}{n}$. \end{enumerate} - \gist{ - We want to show that $Y = \bigcap_{n \in \N} V_n$. - For $x \in Y$, $n \in \N$ we have $x \in V_n$, - as we can choose two neighbourhoods - $U_1$ (open in $Y)$ and $U_2$ (open in $X$ ) of $x$, - such that $\diam_{d_Y}(U) < \frac{1}{n}$ - and $U_2 \cap Y = U_1$. - Additionally choose $x \in U_3$ open in $X$ - with $\diam_{d}(U_3) < \frac{1}{n}$. - Then consider $U_2 \cap U_3 \subseteq V_n$. - Hence $Y \subseteq \bigcap_{n \in \N} V_n$. + \gist{% + We want to show that $Y = \bigcap_{n \in \N} V_n$. + For $x \in Y$, $n \in \N$ we have $x \in V_n$, + as we can choose two neighbourhoods + $U_1$ (open in $Y)$ and $U_2$ (open in $X$ ) of $x$, + such that $\diam_{d_Y}(U) < \frac{1}{n}$ + and $U_2 \cap Y = U_1$. + Additionally choose $x \in U_3$ open in $X$ + with $\diam_{d}(U_3) < \frac{1}{n}$. + Then consider $U_2 \cap U_3 \subseteq V_n$. + Hence $Y \subseteq \bigcap_{n \in \N} V_n$. - Now let $x \in \bigcap_{n \in \N} V_n$. - For each $n$ pick $x \in U_n \subseteq X$ open - satisfying (i), (ii), (iii). - From (i) and (ii) it follows that $x \in \overline{Y}$, - since we can consider a sequence of points $y_n \in U_n \cap Y$ - and get $y_n \xrightarrow{d} x$. - For all $n$ we have that $U_n' \coloneqq U_1 \cap \ldots \cap U_n$ - is an open set containing $x$, - hence $U_n' \cap Y \neq \emptyset$. - Thus we may assume that the $U_i$ form a decreasing sequence. - We have that $\diam_{d_Y}(U_n \cap Y) \le \frac{1}{n}$. - If follows that the $y_n$ form a Cauchy sequence with respect to $d_Y$, - since $\diam(U_n \cap Y) \xrightarrow{d_Y} 0$ - and thus $\diam(\overline{U_n \cap Y}) \xrightarrow{d_Y} 0$. - The sequence $y_n$ converges to the unique point in - $\bigcap_{n} \overline{U_n \cap Y}$. - Since the topologies agree, this point is $x$. - }{Then $Y = \bigcap_n U_n$.} + Now let $x \in \bigcap_{n \in \N} V_n$. + For each $n$ pick $x \in U_n \subseteq X$ open + satisfying (i), (ii), (iii). + From (i) and (ii) it follows that $x \in \overline{Y}$, + since we can consider a sequence of points $y_n \in U_n \cap Y$ + and get $y_n \xrightarrow{d} x$. + For all $n$ we have that $U_n' \coloneqq U_1 \cap \ldots \cap U_n$ + is an open set containing $x$, + hence $U_n' \cap Y \neq \emptyset$. + Thus we may assume that the $U_i$ form a decreasing sequence. + We have that $\diam_{d_Y}(U_n \cap Y) \le \frac{1}{n}$. + If follows that the $y_n$ form a Cauchy sequence with respect to $d_Y$, + since $\diam(U_n \cap Y) \xrightarrow{d_Y} 0$ + and thus $\diam(\overline{U_n \cap Y}) \xrightarrow{d_Y} 0$. + The sequence $y_n$ converges to the unique point in + $\bigcap_{n} \overline{U_n \cap Y}$. + Since the topologies agree, this point is $x$. + }{Then $Y = \bigcap_n U_n$.} \end{refproof} \end{refproof} - diff --git a/inputs/lecture_03.tex b/inputs/lecture_03.tex index e376cbe..d173a04 100644 --- a/inputs/lecture_03.tex +++ b/inputs/lecture_03.tex @@ -2,7 +2,7 @@ \subsection{Trees} - +\gist{% \begin{notation} Let $A \neq \emptyset$, $n \in \N$. Then @@ -34,7 +34,7 @@ $s\defon{m} \coloneqq (s_0,\ldots,s_{m-1})$. Let $s,t \in A^{<\N}$. - We say that $s$ is an \vocab{initial segment} + We say that $s$ is an \vocab{initial segment} of $t$ (or $t$ is an \vocab{extension} of $s$) if there exists an $n$ such that $s = t\defon{|s|}$. We write this as $s \subseteq t$. @@ -44,8 +44,8 @@ Otherwise the are \vocab{incompatible}, we denote that as $s \perp t$. - The \vocab{concatenation} - of $s = (s_0,\ldots, s_{n-1})$ + The \vocab{concatenation} + of $s = (s_0,\ldots, s_{n-1})$ and $t = (t_0,\ldots, t_{m-1})$ is the sequence $s\concat t \coloneqq (s_0,\ldots,s_{n-1}, t_0,\ldots, t_{n-1})$ @@ -59,10 +59,11 @@ define extension, initial segments and concatenation of a finite sequence with an infinite one. \end{notation} +}{} \begin{definition} - A \vocab{tree} - on a set $A$ is a subset $T \subseteq A^{<\N}$ + A \vocab{tree} + on a set $A$ is a subset $T \subseteq A^{<\N}$ closed under initial segments, i.e.~if $t \in T, s \subseteq t \implies s \in T$. Elements of trees are called \vocab{nodes}. @@ -70,27 +71,27 @@ A \vocab{leave} is an element of $T$, that has no extension in $t$. - An \vocab{infinite branch} of a tree $T$ - is $x \in A^{\N}$ + An \vocab{infinite branch} of a tree $T$ + is $x \in A^{\N}$ such that $\forall n.~x\defon{n} \in T$. The \vocab{body} of $T$ is the set of all infinite branches: \[ [T] \coloneqq\{x \in A^{\N}: \forall n. x\defon{n} \in T\}. - \] + \] We say that $T$ is \vocab{pruned}, iff \[ \forall t\in T.\exists s \supsetneq t.~s \in T. - \] + \] \end{definition} \begin{definition} A \vocab{Cantor scheme} - on a set $X$ is a family - $(A_s)_{s \in 2^{< \N}}$ + on a set $X$ is a family + $(A_s)_{s \in 2^{< \N}}$ of subsets of $X$ such that \begin{enumerate}[i)] \item $\forall s \in 2^{<\N}.~A_{s \concat 0} \cap A_{s \concat 1} = \emptyset$. @@ -99,7 +100,7 @@ \end{definition} \begin{definition} - A topological space + A topological space is \vocab{perfect} if it has no isolated points, i.e.~for any $U \neq \emptyset$ open, @@ -108,15 +109,15 @@ \begin{theorem} \label{thm:cantortopolish} - Let $X \neq \emptyset$ + Let $X \neq \emptyset$ be a perfect Polish space. Then there is an embedding - of the Cantor space $2^{\N}$ + of the Cantor space $2^{\N}$ into $X$. \end{theorem} \begin{proof} We will define a Cantor scheme - $(U_s)_{s \in 2^{<\N}}$ + $(U_s)_{s \in 2^{<\N}}$ such that $\forall s \in 2^{< \N}$. \begin{enumerate}[(i)] \item $U_s \neq \emptyset$ and open, @@ -127,39 +128,46 @@ We define $U_s$ inductively on the length of $s$. - For $U_{\emptyset}$ take any non-empty open set - with small enough diameter. + \gist{% + For $U_{\emptyset}$ take any non-empty open set + with small enough diameter. - Given $U_s$, pick $x \neq y \in U_s$ - and let $U_{s \concat 0} \ni x$, - $U_{s \concat 1} \ni y$ - be disjoint, open, - of diameter $\le \frac{1}{2^{|s| +1}}$ - and such that $\overline{U_{s\concat 0}}, \overline{U_{S \concat 1}} \subseteq U_s$. + Given $U_s$, pick $x \neq y \in U_s$ + and let $U_{s \concat 0} \ni x$, + $U_{s \concat 1} \ni y$ + be disjoint, open, + of diameter $\le \frac{1}{2^{|s| +1}}$ + and such that $\overline{U_{s\concat 0}}, \overline{U_{S \concat 1}} \subseteq U_s$. + }{} + \gist{% Let $x \in 2^{\N}$. - Then let $f(x)$ be the unique point in $X$ + Then let $f(x)$ be the unique point in $X$ such that \[ \{f(x)\} = \bigcap_{n} U_{x \defon n} = \bigcap_{n} \overline{U_{x \defon n}}. - \] + \] (This is nonempty as $X$ is a completely metrizable space.) It is clear that $f$ is injective and continuous. % TODO: more details $2^{\N}$ is compact, hence $f^{-1}$ is also continuous. + }{Consider $f\colon 2^{\N} \hookrightarrow X, x \mapsto y$, where $\{y\} = \bigcap_n U_{x\defon n}$. + By compactness of $2^{\N}$, we get that $f^{-1}$ is continuous.} \end{proof} \begin{corollary} \label{cor:perfectpolishcard} Every nonempty perfect Polish - space $X$ has cardinality $\fc = 2^{\aleph_0}$ - % TODO: eulerscript C ? + space $X$ has cardinality $\fc = 2^{\aleph_0}$ \end{corollary} \begin{proof} - Since the cantor space embeds into $X$, - we get the lower bound. - Since $X$ is second countable and Hausdorff, - we get the upper bound. + \gist{% + Since the cantor space embeds into $X$, + we get the lower bound. + Since $X$ is second countable and Hausdorff, + we get the upper bound.% + }{Lower bound: $2^{\N} \hookrightarrow X$, + upper bound: \nth{2} countable and Hausdorff. \end{proof} \begin{theorem} @@ -173,13 +181,13 @@ \begin{definition} A \vocab{Lusin scheme} on a set $X$ - is a family $(A_s)_{s \in \N^{<\N}}$ + is a family $(A_s)_{s \in \N^{<\N}}$ of subsets of $X $ such that \begin{enumerate}[(i)] \item $A_{s \concat i} \cap A_{s \concat j} = \emptyset$ for all $j \neq i \in \N$, $s \in \N^{<\N}$. - \item $A_{s \concat i} \subseteq A_s$ + \item $A_{s \concat i} \subseteq A_s$ for all $i \in \N, s \in \N^{<\N}$. \end{enumerate} \end{definition} @@ -191,11 +199,11 @@ \[ D \subseteq \N^\N \text{\reflectbox{$\coloneqq$}} \cN % TODO correct N for the Baire space? - \] + \] and a continuous bijection from $D$ onto $X$ (the inverse does not need to be continuous). - Moreover there is a continuous surjection $g: \cN \to X$ + Moreover there is a continuous surjection $g: \cN \to X$ extending $f$. \end{theorem} \begin{definition} @@ -203,12 +211,14 @@ countable union of closed sets, i.e.~the complement of a $G_\delta$ set. \end{definition} +\gist{% \begin{observe} \begin{itemize} \item Any open set is $F {\sigma}$. \item In metric spaces the intersection of an open and closed set is $F_\sigma$. \end{itemize} \end{observe} +}{} \begin{refproof}{thm:bairetopolish} Let $d$ be a complete metric on $X$. W.l.o.g.~$\diam(X) \le 1$. @@ -220,7 +230,7 @@ \item $F_\emptyset = X$, \item $F_s$ is $F_\sigma$ for all $s$. \item The $F_{s \concat i}$ partition $F_s$, - i.e.~$F_{s} = \bigsqcup_i F_{s \concat i}$. % TODO change notation? + i.e.~$F_{s} = \bigsqcup_i F_{s \concat i}$. Furthermore we want that $\overline{F_{s \concat i}} \subseteq F_s$ @@ -228,6 +238,7 @@ \item $\diam(F_s) \le 2^{-|s|}$. \end{enumerate} + \gist{% Suppose we already have $F_s \text{\reflectbox{$\coloneqq$}} F$. We need to construct a partition $(F_i)_{i \in \N}$ of $F$ with $\overline{F_i} \subseteq F$ @@ -252,6 +263,7 @@ The sets $D_i \coloneqq F_i^0 \cap B_i \setminus (B_1 \cup \ldots \cup B_{i-1})$ are $F_\sigma$, disjoint and $F_i^0 = \bigcup_{j} D_j$. +}{Induction.} diff --git a/inputs/lecture_04.tex b/inputs/lecture_04.tex index ff9fbb8..4c16ba2 100644 --- a/inputs/lecture_04.tex +++ b/inputs/lecture_04.tex @@ -5,6 +5,7 @@ \end{remark} \begin{refproof}{thm:bairetopolish} + \gist{% Take \[D = \{x \in \cN : \bigcap_{n} F_{x\defon{n}} \neq \emptyset\}.\] @@ -12,16 +13,19 @@ we have \[ \bigcap_{n} F_{x\defon{n}} = \bigcap_{n} \overline{F_{x\defon{n}}}. - \] + \] + }{} $f\colon D \to X$ is determined by \[ \{f(x)\} = \bigcap_{n} F_{x\defon{n}} - \] + \] - $f$ is injective and continuous. - The proof of this is exactly the same as in - \yaref{thm:cantortopolish}. + \gist{% + $f$ is injective and continuous. + The proof of this is exactly the same as in + \yaref{thm:cantortopolish}. + }{} \begin{claim} \label{thm:bairetopolish:c1} @@ -47,7 +51,7 @@ there exists $y = \lim_n f(x_n)$. Since for all $m \ge M$, $f(x_m) \in F_{x\defon{N}}$, we get that $y \in \overline{F_{x\defon{N}}}$. - + Note that for $N' > N$ by the same argument we get $y \in \overline{F_{x\defon{N'}}}$. Hence @@ -55,20 +59,20 @@ i.e.~$y \in D$ and $y = f(x)$. \end{refproof} - We extend $f$ to $g\colon\cN \to X$ + We extend $f$ to $g\colon\cN \to X$ in the following way: Take $S \coloneqq \{s \in \N^{<\N}: \exists x \in D, n \in \N.~x=s\defon{n}\}$. Clearly $S$ is a pruned tree. - Moreover, since $D$ is closed, we have that\todo{Proof this (homework?)} + Moreover, since $D$ is closed, we have that (cf.~\yaref{s3e1}) \[ D = [S] = \{x \in \N^\N : \forall n \in \N.~x\defon{n} \in S\}. - \] - We construct a \vocab{retraction} $r\colon\cN \to D$ + \] + We construct a \vocab{retraction} $r\colon\cN \to D$ (i.e.~$r = \id$ on $D$ and $r$ is a continuous surjection). Then $g \coloneqq f \circ r$. - - To construct $r$, we will define + + To construct $r$, we will define $\phi\colon \N^{<\N} \to S$ by induction on the length such that \begin{itemize} @@ -76,21 +80,27 @@ \item $|s| = \phi(|s|)$, \item if $s \in S$, then $\phi(s) = s$. \end{itemize} - Let $\phi(\emptyset) = \emptyset$. - Suppose that $\phi(t)$ is defined. - If $t\concat a \in S$, then set - $\phi(t\concat a) \coloneqq t\concat a$. - Otherwise take some $b$ such that - $t\concat b \in S$ and define - $\phi(t\concat a) \coloneqq \phi(t)\concat b$. + \gist{% + Let $\phi(\emptyset) = \emptyset$. + Suppose that $\phi(t)$ is defined. + If $t\concat a \in S$, then set + $\phi(t\concat a) \coloneqq t\concat a$. + Otherwise take some $b$ such that + $t\concat b \in S$ and define + $\phi(t\concat a) \coloneqq \phi(t)\concat b$.% + }{}% This is possible since $S$ is pruned. - Let $r\colon \cN = [\N^{<\N}] \to [S] = D$ - be the function defined by $r(x) = \bigcup_n f(x\defon{n})$. + \gist{% + Let $r\colon \cN = [\N^{<\N}] \to [S] = D$ + be the function defined by $r(x) = \bigcup_n f(x\defon{n})$. + }{} $r$ is continuous, since $d_{\cN}(r(x), r(y)) \le d_{\cN}(x,y)$. % Lipschitz - It is immediate that $r$ is a retraction. + \gist{% + It is immediate that $r$ is a retraction. + }{} \end{refproof} \section{Meager and Comeager Sets} @@ -110,42 +120,48 @@ then $A \cap U$ is not dense in $U$). \end{itemize} - A set $B \subseteq X$ is \vocab{meager} + A set $B \subseteq X$ is \vocab{meager} (or \vocab{first category}), iff it is a countable union of nwd sets. The complement of a meager set is called \vocab{comeager}. \end{definition} +\gist{% \begin{example} $\Q \subseteq \R$ is meager. \end{example} +}{} \begin{notation} Let $A, B \subseteq X$. - We write $A =^\ast B$ + We write $A =^\ast B$ iff the \vocab{symmetric difference}, $A \symdif B \coloneqq (A\setminus B) \cup (B \setminus A)$, is meager. \end{notation} +\gist{% \begin{remark} $=^\ast$ is an equivalence relation. \end{remark} +}{} \begin{definition} A set $A \subseteq X$ has the \vocab{Baire property} (\vocab{BP}) if $A =^\ast U$ for some $U \overset{\text{open}}{\subseteq} X$. \end{definition} +\gist{% Note that open sets and meager sets have the Baire property. +}{} - +\gist{% \begin{example} \begin{itemize} \item $\Q \subseteq \R$ is $F_\sigma$. \item $\R \setminus \Q \subseteq \R$ is $G_\delta$. - \item $\Q \subseteq \R$ is not $G_{\delta}$. - (It is dense and meager, + \item $\Q \subseteq \R$ is not $G_{\delta}$: + It is dense and meager, hence it can not be $G_\delta$ - by the Baire category theorem). + by the \yaref{thm:bct}. \end{itemize} - \end{example} +} diff --git a/inputs/lecture_05.tex b/inputs/lecture_05.tex index e2c10a1..748b6a1 100644 --- a/inputs/lecture_05.tex +++ b/inputs/lecture_05.tex @@ -6,10 +6,9 @@ \item If $F$ is closed then $F$ is nwd iff $X \setminus F$ is open and dense. \item Any meager set $B$ is contained in a meager $F_{\sigma}$-set. \end{itemize} - - \end{fact} -\begin{proof} % remove? +\gist{% +\begin{proof} \begin{itemize} \item This follows from the definition as $\overline{\overline{A}} = \overline{A}$. \item Trivial. @@ -17,7 +16,9 @@ Then $B \subseteq \bigcup_{n < \omega} \overline{B_n}$. \end{itemize} \end{proof} +}{} +\gist{% \begin{definition} A \vocab{$\sigma$-algebra} on a set $X$ is a collection of subsets of $X$ @@ -32,14 +33,15 @@ Since $\bigcap_{i < \omega} A_i = \left( \bigcup_{i < \omega} A_i^c \right)^c$ we have that $\sigma$-algebras are closed under countable intersections. \end{fact} +}{} \begin{theorem} \label{thm:bairesigma} Let $X$ be a topological space. Then the collection of sets with the Baire property - is a $\sigma$-algebra on $X$. + is \gist{a $\sigma$-algebra on $X$. - It is the smallest $\sigma$-algebra + It is}{} the smallest $\sigma$-algebra containing all meager and open sets. \end{theorem} \begin{refproof}{thm:bairesigma} @@ -274,9 +276,11 @@ but for meager sets: % \end{refproof} % TODO fix claim numbers +\gist{% \begin{remark} Suppose that $A$ has the BP. Then there is an open $U$ such that $A \symdif U \mathbin{\text{\reflectbox{$\coloneqq$}}} M$ is meager. Then $A = U \symdif M$. \end{remark} +}{} diff --git a/inputs/lecture_06.tex b/inputs/lecture_06.tex index 1f74848..f3bb7e0 100644 --- a/inputs/lecture_06.tex +++ b/inputs/lecture_06.tex @@ -98,14 +98,13 @@ Let $X$ be a topological space. Then define \[\Sigma^0_1(X) \coloneqq \{U \overset{\text{open}}{\subseteq} X\},\] \[ -\Pi^0_\alpha(X) \coloneqq \lnot \Sigma^0_\alpha(X) \coloneqq +\Pi^0_\alpha(X) \coloneqq \lnot \Sigma^0_\alpha(X) \coloneqq \{X \setminus A | A \in \Sigma^0_\alpha(X)\}, -\] -% \todo{Define $\lnot$ (element-wise complement)} +\] and for $\alpha > 1$ \[ \Sigma^0_\alpha \coloneqq \{\bigcup_{n < \omega} A_n : A_n \in \Pi^0_{\alpha_n}(X) \text{ for some $\alpha_n < \alpha$}\}. -\] +\] Note that $\Pi_1^0$ is the set of closed sets, $\Sigma^0_2 = F_\sigma$, diff --git a/inputs/lecture_07.tex b/inputs/lecture_07.tex index 687e009..22a2026 100644 --- a/inputs/lecture_07.tex +++ b/inputs/lecture_07.tex @@ -16,17 +16,17 @@ We have that \[ \Sigma^0_\xi(X) = \{\bigcup_{n} A_n : n \in \N, A_n \in \Pi^0_{\xi_n}(X), \xi_n < \xi\}. - \] + \] Hence $|\Sigma^0_\xi(X)| \le (\overbrace{\aleph_0}^{\mathclap{\{\xi': \xi' < \xi\} \text{ is countable~ ~ ~ ~}}} \cdot \underbrace{\fc}_{\mathclap{\text{inductive assumption}}})^{\overbrace{\aleph_0}^{\mathclap{\text{countable unions}}}}$. We have \[ \cB(X) = \bigcup_{\xi < \omega_1} \Sigma^0_\xi(X). - \] + \] Hence \[ |\cB(X)| \le \omega_1 \cdot \fc = \fc. - \] + \] \end{proof} \begin{proposition}[Closure properties] @@ -37,54 +37,58 @@ \item \begin{itemize} \item $\Sigma^0_\xi(X)$ is closed under countable unions. \item $\Pi^0_\xi(X)$ is closed under countable intersections. - \item $\Delta^0_\xi(X)$ is closed under complements, - countable unions and - countable intersections. + \item $\Delta^0_\xi(X)$ is closed under complements. \end{itemize} \item \begin{itemize} \item $\Sigma^0_\xi(X)$ is closed under \emph{finite} intersections. \item $\Pi^0_\xi(X)$ is closed under \emph{finite} unions. + \item $\Delta^0_\xi(X)$ is closed under finite unions and + finite intersections. \end{itemize} \end{enumerate} \end{proposition} +\gist{% \begin{proof} \begin{enumerate}[(a)] - \item This follows directly from the definition. + \item This follows directly from the definition. Note that a countable intersection can be written as a complement of the countable union of complements: \[ \bigcap_{n} B_n = \left( \bigcup_{n} B_n^{c} \right)^{c}. - \] + \] \item If suffices to check this for $\Sigma^0_{\xi}(X)$. Let $A = \bigcup_{n} A_n$ for $A_n \in \Pi^0_{\xi_n}(X)$ and $B = \bigcup_{m} B_m$ for $B_m \in \Pi^0_{\xi'_m}(X)$. Then \[ A \cap B = \bigcup_{n,m} \left( A_n \cap B_m \right) - \] + \] and $A_n \cap B_m \in \Pi^{0}_{\max(\xi_n, \xi'_m)}(X)$. \end{enumerate} \end{proof} +}{} \begin{example} Consider the cantor space $2^{\omega}$. We have that $\Delta^0_1(2^{\omega})$ - is not closed under countable unions - (countable unions yield all open sets, but there are open - sets that are not clopen). + is not closed under countable unions% + \gist{ (countable unions yield all open sets, but there are open + sets that are not clopen)}{}. \end{example} \subsection{Turning Borels Sets into Clopens} \begin{theorem}% + \gist{% \footnote{Whilst strikingly concise the verb ``\vocab[Clopenization™]{to clopenize}'' unfortunately seems to be non-standard vocabulary. Our tutor repeatedly advised against using it in the final exam. Contrary to popular belief the very same tutor was \textit{not} the one first to introduce it, as it would certainly be spelled ``to clopenise'' if that were the case. - } + }% + }{}% \label{thm:clopenize} Let $(X, \cT)$ be a Polish space. For any Borel set $A \subseteq X$, @@ -163,7 +167,7 @@ such that $\cT_n \supseteq \cT$ and $\cB(\cT_n) = \cB(\cT)$. Then the topology $\cT_\infty$ generated by $\bigcup_{n} \cT_n$ - is still Polish + is Polish and $\cB(\cT_\infty) = \cB(T)$. \end{lemma} \begin{proof} @@ -183,7 +187,51 @@ definition of $\cF$ belong to a countable basis of the respective $\cT_n$). - \todo{This proof will be finished in the next lecture} + % Proof was finished in lecture 8 + Let $Y = \prod_{n \in \N} (X, \cT_n)$. + Then $Y$ is Polish. + Let $\delta\colon (X, \cT_\infty) \to Y$ + defined by $\delta(x) = (x,x,x,\ldots)$. + \begin{claim} + $\delta$ is a homeomorphism. + \end{claim} + \begin{subproof} + Clearly $\delta$ is a bijection. + We need to show that it is continuous and open. + + Let $U \in \cT_i$. + Then + \[ + \delta^{-1}(D \cap \left( X \times X \times \ldots\times U \times \ldots) \right)) = U \in \cT_i \subseteq \cT_\infty, + \] + hence $\delta$ is continuous. + Let $U \in \cT_\infty$. + Then $U$ is the union of sets of the form + \[ + V = U_{n_1} \cap U_{n_2} \cap \ldots \cap U_{nu} + \] + for some $n_1 < n_2 < \ldots < n_u$ + and $U_{n_i} \in \cT_i$. + + Thus is suffices to consider sets of this form. + We have that + \[ + \delta(V) = D \cap (X \times X \times \ldots \times U_{n_1} \times \ldots \times U_{n_2} \times \ldots \times U_{n_u} \times X \times \ldots) \overset{\text{open}}{\subseteq} D. + \] + \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} We need to show that $A$ is closed under countable unions. diff --git a/inputs/lecture_08.tex b/inputs/lecture_08.tex index 4288b7f..62298a0 100644 --- a/inputs/lecture_08.tex +++ b/inputs/lecture_08.tex @@ -1,61 +1,8 @@ -\lecture{08}{2023-11-10}{} - -\todo{put this lemma in the right place} -\begin{lemma}[Lemma 2] - Let $(X, \cT)$ be a Polish space. - Let $\cT_n \supseteq \cT$ be Polish - with $\cB(X, \cT_n) = \cB(X, \cT)$. - Let $\cT_\infty$ be the topology generated - by $\bigcup_n \cT_n$. - Then $(X, \cT_\infty)$ is Polish - and $\cB(X, \cT_\infty) = \cB(X, \cT)$. -\end{lemma} -\begin{proof} - Let $Y = \prod_{n \in \N} (X, \cT_n)$. - Then $Y$ is Polish. - Let $\delta\colon (X, \cT_\infty) \to Y$ - defined by $\delta(x) = (x,x,x,\ldots)$. - \begin{claim} - $\delta$ is a homeomorphism. - \end{claim} - \begin{subproof} - Clearly $\delta$ is a bijection. - We need to show that it is continuous and open. - - Let $U \in \cT_i$. - Then - \[ - \delta^{-1}(D \cap \left( X \times X \times \ldots\times U \times \ldots) \right)) = U \in \cT_i \subseteq \cT_\infty, - \] - hence $\delta$ is continuous. - Let $U \in \cT_\infty$. - Then $U$ is the union of sets of the form - \[ - V = U_{n_1} \cap U_{n_2} \cap \ldots \cap U_{nu} - \] - for some $n_1 < n_2 < \ldots < n_u$ - and $U_{n_i} \in \cT_i$. - - Thus is suffices to consider sets of this form. - We have that - \[ - \delta(V) = D \cap (X \times X \times \ldots \times U_{n_1} \times \ldots \times U_{n_2} \times \ldots \times U_{n_u} \times X \times \ldots) \overset{\text{open}}{\subseteq} D. - \] - \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} +\lecture{08}{2023-11-10}{}\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} diff --git a/inputs/tutorial_01.tex b/inputs/tutorial_01.tex index 2ef9ca6..3eb6d60 100644 --- a/inputs/tutorial_01.tex +++ b/inputs/tutorial_01.tex @@ -27,8 +27,10 @@ \begin{definition} A topological space is \vocab{Lindelöf} - if every open cover has a countable subcover. + iff every open cover has a countable subcover. \end{definition} + + \begin{fact} Let $X$ be a metric space. If $X$ is Lindelöf, @@ -64,5 +66,12 @@ and Lindelöf coincide. In arbitrary topological spaces, - Lindelöf is the strongest of these notions. + Lindelöf is the weakest of these notions. \end{remark} + +\begin{definition}+ + A metric space $X$ is \vocab{totally bounded} + iff for every $\epsilon > 0$ there exists + a finite set of points $x_1,\ldots,x_n$ + such that $X = \bigcup_{i=1}^n B_{\epsilon}(x_i)$. +\end{definition} From 68a7e9d4286e7c58876e5e560e4120946b3d1b60 Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Tue, 23 Jan 2024 21:55:41 +0100 Subject: [PATCH 4/9] fixed typo --- inputs/lecture_03.tex | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/inputs/lecture_03.tex b/inputs/lecture_03.tex index d173a04..dce6608 100644 --- a/inputs/lecture_03.tex +++ b/inputs/lecture_03.tex @@ -167,7 +167,7 @@ Since $X$ is second countable and Hausdorff, we get the upper bound.% }{Lower bound: $2^{\N} \hookrightarrow X$, - upper bound: \nth{2} countable and Hausdorff. + upper bound: \nth{2} countable and Hausdorff.} \end{proof} \begin{theorem} From 0cceab0af98b892bebc25f8e8ccfe44b76c8f335 Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Tue, 23 Jan 2024 22:07:45 +0100 Subject: [PATCH 5/9] some small changes --- inputs/lecture_04.tex | 2 +- inputs/lecture_05.tex | 15 +++++++-------- inputs/lecture_16.tex | 11 ++++++----- 3 files changed, 14 insertions(+), 14 deletions(-) diff --git a/inputs/lecture_04.tex b/inputs/lecture_04.tex index 4c16ba2..21ebbd1 100644 --- a/inputs/lecture_04.tex +++ b/inputs/lecture_04.tex @@ -164,4 +164,4 @@ Note that open sets and meager sets have the Baire property. by the \yaref{thm:bct}. \end{itemize} \end{example} -} +}{} diff --git a/inputs/lecture_05.tex b/inputs/lecture_05.tex index 748b6a1..36b1cf0 100644 --- a/inputs/lecture_05.tex +++ b/inputs/lecture_05.tex @@ -95,7 +95,7 @@ % Nwd set of positive measure. % TODO % remove open intervals such that their length does not add to 0 -% +% %\end{example} \begin{theorem}[Baire Category theorem] @@ -114,8 +114,8 @@ \item The intersection of countable many open dense sets is dense. \end{enumerate} - In this case $X$ is called a \vocab{Baire space}. - \footnote{see \yaref{s5e1}} + In this case $X$ is called a \vocab{Baire space}.% + \footnote{cf.~\yaref{s5e1}} \end{theoremdef} \begin{proof} \todo{Proof (short)} @@ -127,12 +127,11 @@ We have that \[ \emptyset = \bigcap_{n} (X \setminus \overline{A_n}) - \] + \] is dense by (iii). This proof can be adapted to other open sets $X$. \end{proof} - \begin{notation} Let $X ,Y$ be topological spaces, $A \subseteq X \times Y$ @@ -141,12 +140,12 @@ Let \[ - A_x \coloneqq \{y \in Y : (x,y) \in A\} - \] + A_x \coloneqq \{y \in Y : (x,y) \in A\} + \] and \[ A^y \coloneqq \{x \in X : (x,y) \in A\} . - \] + \] \end{notation} The following similar to Fubini, diff --git a/inputs/lecture_16.tex b/inputs/lecture_16.tex index 05e5f71..adf914f 100644 --- a/inputs/lecture_16.tex +++ b/inputs/lecture_16.tex @@ -175,9 +175,10 @@ By Zorn's lemma, this will follow from \end{IEEEeqnarray*} The topology induced by the metric - is given by basic open subsets\footnote{see Exercise Sheet 9% TODO REF - } - $[U_0; U_1,\ldots, U_n]$, $U_0,\ldots, U_n \overset{\text{open}}{\subseteq} X$, + is given by basic open subsets\footnote{cf.~\yaref{s9e2}} + of the form + $[U_0; U_1,\ldots, U_n]$, + for $U_0,\ldots, U_n \overset{\text{open}}{\subseteq} X$, where \[ [U_0; U_1,\ldots,U_n] \coloneqq @@ -188,8 +189,8 @@ By Zorn's lemma, this will follow from We want to view flows as a metric space. For a fixed compact metric space $X$, we can view the flows $(X,\Z)$ as a subset of $\cC(X,X)$. -Note that $\cC(X,X)$ is Polish. -Then the minimal flows on $X$ are a Borel subset of $\cC(X,X)$. +Note that $\cC(X,X)$ is Polish.\footnote{cf.~\yaref{s1e4}} +Then the minimal flows on $X$ are a Borel subset of $\cC(X,X)$.\footnote{Exercise} % TODO However we do not want to consider only flows on a fixed space $X$, but we want to look all flows at the same time. From 3917993d930e4cf8cb8cfb3df556833765cca732 Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Wed, 24 Jan 2024 00:13:59 +0100 Subject: [PATCH 6/9] some small changes --- inputs/lecture_14.tex | 9 ++++----- 1 file changed, 4 insertions(+), 5 deletions(-) diff --git a/inputs/lecture_14.tex b/inputs/lecture_14.tex index 3976165..975e4a5 100644 --- a/inputs/lecture_14.tex +++ b/inputs/lecture_14.tex @@ -66,7 +66,7 @@ } in $(y, <_\Q)$ is cofinal in $(y, <_{\Q})$ and vice versa. - Equivalently, either $(x <^\ast_\phi y)$ + Equivalently, either $(x <^\ast_\phi y)$ or \begin{IEEEeqnarray*}{rCl} & &x,y \in \WO\\ @@ -84,10 +84,9 @@ such that \[ \forall x \in X.~(\exists n.~(x,n) \in R \iff \exists! n.~(x,n)\in R^\ast). - \] - We say that $R^\ast$ \vocab[uniformization]{uniformizes} $R$. - \todo{missing picture - \url{https://upload.wikimedia.org/wikipedia/commons/4/4c/Uniformization_ill.png}} + \] + We say that $R^\ast$ \vocab[uniformization]{uniformizes} $R$.% + \footnote{Wikimedia has a \href{https://upload.wikimedia.org/wikipedia/commons/4/4c/Uniformization_ill.png}{nice picture.}} \end{theorem} \begin{proof} From acb56df1c2ad130d87047567bf47c68014067e92 Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Wed, 24 Jan 2024 15:35:14 +0100 Subject: [PATCH 7/9] gist 12 --- inputs/lecture_09.tex | 1 + inputs/lecture_11.tex | 14 +-- inputs/lecture_12.tex | 240 ++++++++++++++++++++++++------------------ 3 files changed, 148 insertions(+), 107 deletions(-) diff --git a/inputs/lecture_09.tex b/inputs/lecture_09.tex index c5f67d5..f344eee 100644 --- a/inputs/lecture_09.tex +++ b/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} diff --git a/inputs/lecture_11.tex b/inputs/lecture_11.tex index 641f969..ebdd95f 100644 --- a/inputs/lecture_11.tex +++ b/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 ${2^{\N}}^{<\N}$. - - - +}{} diff --git a/inputs/lecture_12.tex b/inputs/lecture_12.tex index 5b9d551..feb3e2e 100644 --- a/inputs/lecture_12.tex +++ b/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 $D \coloneqq \{(T, \beta) \in \Tr \times \cN : \forall n.~ T(\beta\defon{n}) = 1\}$. - Note that this set is closed in $\Tr \times \cN$, - since it is a countable intersection of clopen sets. - % TODO Why clopen? - Then $\IF = \proj_{\Tr}(D) \in \Sigma^1_1$. + 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. + 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: - Suppose there is a $\Sigma^1_1$-complete set $B \in \cB(Y)$. - Take $A \in \Sigma^1_1(X) \setminus \cB(X)$ - and $f\colon X \to Y$ Borel. - But then $f^{-1}(B)$ is 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 $A \in \Sigma^1_1(X) \setminus \cB(X)$% + \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?} - 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 $S^{\N}$: - \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} - If $T$ is a tree on $\N \times \N$ - and $x \in \cN$, - then the \vocab{section at $x$} - %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} - \label{prop:lec12:2} - Let $A \subseteq \cN$. - The following are equivalent: - \begin{itemize} - \item $A$ is analytic. - \item There is a pruned tree on $\N \times \N$ - such that - \[A = \proj_1 ([T]) = \{x \in \cN : \exists y \in \cN.~ (x,y) \in [T]\}.\] - \end{itemize} - \end{proposition} - \begin{proof} - $A$ is analytic iff - there exists $F \overset{\text{closed}}{\subseteq} \N \times \N$ - such that $A = \proj_1(F)$. - But closed sets of $\N \times \N$ correspond to pruned trees, - by the first point. - \end{proof} -\end{enumerate} + +\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*} +}{% + 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 $T(x)$, + is the following tree on $\N$ : + \[ + T(x) = \{s \in \N^{<\N} : (x\defon{|s|}, s) \in T\}. + \] +\end{definition} +\begin{proposition} + \label{prop:lec12:2} + Let $A \subseteq \cN$. + The following are equivalent: + \begin{itemize} + \item $A$ is analytic. + \item There is a pruned tree on $\N \times \N$ + such that + \[A = \proj_1 ([T]) = \{x \in \cN : \exists y \in \cN.~ (x,y) \in [T]\}.\] + \end{itemize} +\end{proposition} +\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 $\N \times \N$ - as in \autoref{prop:lec12:2}, i.e.~$A = \proj_1([T])$. + \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. - $f$ is continuous: - Let $x\defon{n} = y\defon{n}$ for some $n \in \N$. - Then for all $m \le n, s,t \in \N^{<\N}$ - such that $s = x\defon{m} = y \defon{m}$ and $|t| = |s|$, - we have - \begin{itemize} - \item $t \in T(x) \iff (s,t) \in T$, - \item $t \in T(y) \iff (s,t) \in T$. - \end{itemize} + \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 $m \le n, s,t \in \N^{<\N}$ + such that $s = x\defon{m} = y \defon{m}$ and $|t| = |s|$, + we have + \begin{itemize} + \item $t \in T(x) \iff (s,t) \in T$, + \item $t \in T(y) \iff (s,t) \in T$. + \end{itemize} - So if $x\defon{n} = y\defon{n}$, - then $t \in T(x) \iff t \in T(y)$ as long as $|t| \le n$.. + So if $x\defon{n} = y\defon{n}$, + 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, - just consider - \begin{IEEEeqnarray*}{rCl} - g \colon X &\longrightarrow & \Tr \\ - x &\longmapsto & \begin{cases} - a &: x \in A,\\ - b &: x \not\in A,\\ - \end{cases} - \end{IEEEeqnarray*} - where $a \in \IF$ and $b \not\in \IF$ are chosen arbitrarily. + \gist{% + If $X$ is Polish and countable and $A \subseteq X$ analytic, + just consider + \begin{IEEEeqnarray*}{rCl} + g \colon X &\longrightarrow & \Tr \\ + x &\longmapsto & \begin{cases} + a &: x \in A,\\ + b &: x \not\in A,\\ + \end{cases} + \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 -\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. - - Let $(A, <_A) \prec (B, <_B)$ denote that - $(A, <_A)$ is isomorphic to a proper initial segment of $(B, <_B)$. -\end{itemize} + 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} - Let $C = \WO$ - and - \begin{IEEEeqnarray*}{rCl} - \phi\colon \WO &\longrightarrow & \Ord \\ - \end{IEEEeqnarray*} - where $\phi((A,<_A))$ is the unique ordinal - isomorphic to $(A, <_A)$. -\end{example} - +\gist{% + \begin{example} + Let $C = \WO$ + and + \begin{IEEEeqnarray*}{rCl} + \phi\colon \WO &\longrightarrow & \Ord \\ + \end{IEEEeqnarray*} + where $\phi((A,<_A))$ is the unique ordinal + isomorphic to $(A, <_A)$. + \end{example} +}{} From e59e97ca03d78f35d27d1a1ba0cb575585480070 Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Wed, 24 Jan 2024 16:06:42 +0100 Subject: [PATCH 8/9] fix --- inputs/lecture_12.tex | 3 +-- inputs/lecture_13.tex | 5 ++++- 2 files changed, 5 insertions(+), 3 deletions(-) diff --git a/inputs/lecture_12.tex b/inputs/lecture_12.tex index feb3e2e..4f62638 100644 --- a/inputs/lecture_12.tex +++ b/inputs/lecture_12.tex @@ -59,12 +59,11 @@ 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 we get that $f^{-1}(B)$ is Borel $\lightnig$. + But then we get that $f^{-1}(B)$ is Borel $\lightning$. }{.} \end{itemize} \end{observe} -% TODO ANKI-MARKER \begin{theorem} \label{thm:lec12:1} Suppose that $A \subseteq \cN$ is analytic. diff --git a/inputs/lecture_13.tex b/inputs/lecture_13.tex index ab38e1e..5d9cd9c 100644 --- a/inputs/lecture_13.tex +++ b/inputs/lecture_13.tex @@ -1,11 +1,12 @@ \lecture{13}{2023-11-08}{} - +\gist{% % Recap $\LO = \{x \in 2^{\N\times \N} : x \text{ is a linear order}\} $. $\LO \subseteq 2^{\N \times \N}$ is closed and $\WO = \{x \in \LO: x \text{ is a wellordering}\} $ is coanalytic in $\LO$. % End Recap +}{} Another way to code linear orders: @@ -32,6 +33,8 @@ with $(f^{-1}(\{1\}), <)$. and $[\alpha_i, \alpha_{i+1})$ to $(i,i+1)$. \end{proof} +% TODO ANKI-MARKER + \begin{definition}[\vocab{Kleene-Brouwer ordering}] Let $(A,<)$ be a linear order and $A$ countable. We define the linear order $<_{KB}$ on $A^{<\N}$ From f7f9d7d638953d666059fc48f0d6fc799670acc2 Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Wed, 24 Jan 2024 23:13:56 +0100 Subject: [PATCH 9/9] some changes --- inputs/lecture_11.tex | 6 ++---- inputs/lecture_13.tex | 40 ++++++++++++++++++++++------------------ jrpie-gist.sty | 1 + 3 files changed, 25 insertions(+), 22 deletions(-) diff --git a/inputs/lecture_11.tex b/inputs/lecture_11.tex index ebdd95f..53d2c7e 100644 --- a/inputs/lecture_11.tex +++ b/inputs/lecture_11.tex @@ -173,10 +173,8 @@ i.e.~we want to associate a tree $T \subseteq \N^{<\N}$}% Let \vocab{$\Tr$} $ \coloneqq \{T \in {2^{\N}}^{<\N} : T \text{ is a tree}\} \subseteq {2^{\N}}^{<\N}$. \begin{observe} - \[ - \Tr \subseteq {2^{\N}}^{<\N} - \] - is closed (where we take the topology of the Cantor space). + $\Tr \subseteq {2^{\N}}^{<\N}$ is closed + (where we take the topology of the Cantor space). \end{observe} \gist{% Indeed, for any $ s \in \N^{<\N}$ diff --git a/inputs/lecture_13.tex b/inputs/lecture_13.tex index 5d9cd9c..f68c640 100644 --- a/inputs/lecture_13.tex +++ b/inputs/lecture_13.tex @@ -33,7 +33,6 @@ with $(f^{-1}(\{1\}), <)$. and $[\alpha_i, \alpha_{i+1})$ to $(i,i+1)$. \end{proof} -% TODO ANKI-MARKER \begin{definition}[\vocab{Kleene-Brouwer ordering}] Let $(A,<)$ be a linear order and $A$ countable. @@ -58,25 +57,30 @@ with $(f^{-1}(\{1\}), <)$. $(T, <_{KB}\defon{T})$ is well ordered. \end{proposition} \begin{proof} - If $T$ is ill-founded and $x \in [T]$, - then for all $n$, we have $x\defon{n+1} <_{KB} x\defon{n}$. - Thus $(T, <_{KB}\defon{T})$ is not well ordered. + \gist{% + If $T$ is ill-founded and $x \in [T]$, + then for all $n$, we have $x\defon{n+1} <_{KB} x\defon{n}$. + Thus $(T, <_{KB}\defon{T})$ is not well ordered. - Conversely, let $<\defon{KB}$ be not a well-ordering - on $T$. - Let $s_0 >_{KB} s_1 >_{KB} s_2 >_{KB} \ldots$ - be an infinite descending chain. - We have that $s_0(0) \ge s_1(0) \ge s_2(0) \ge \ldots$ - stabilizes for $n > n_0$. - Let $a_0 \coloneqq s_{n_0}(0)$. - Now for $n \ge n_0$ we have that $s_n(0)$ is constant, - hence for $n > n_0$ the value $s_{n}(1)$ must be defined. - Thus there is $n_1 \ge n_0$ such that $s_n(1)$ - is constant for all $n \ge n_1$. - Let $a_1 \coloneqq s_{n_1}(1)$ - and so on. - Then $(a_0,a_1,a_2, \ldots) \in [T]$. + Conversely, let $<\defon{KB}$ be not a well-ordering + on $T$. + Let $s_0 >_{KB} s_1 >_{KB} s_2 >_{KB} \ldots$ + be an infinite descending chain. + We have that $s_0(0) \ge s_1(0) \ge s_2(0) \ge \ldots$ + stabilizes for $n > n_0$. + Let $a_0 \coloneqq s_{n_0}(0)$. + Now for $n \ge n_0$ we have that $s_n(0)$ is constant, + hence for $n > n_0$ the value $s_{n}(1)$ must be defined. + Thus there is $n_1 \ge n_0$ such that $s_n(1)$ + is constant for all $n \ge n_1$. + Let $a_1 \coloneqq s_{n_1}(1)$ + and so on. + Then $(a_0,a_1,a_2, \ldots) \in [T]$. + }{easy} \end{proof} + +% TODO ANKI-MARKER + \begin{theorem}[Lusin-Sierpinski] The set $\LO \setminus \WO$ (resp.~$2^{\Q} \setminus \WO$) diff --git a/jrpie-gist.sty b/jrpie-gist.sty index ea7caa3..405fecf 100644 --- a/jrpie-gist.sty +++ b/jrpie-gist.sty @@ -4,6 +4,7 @@ % TODO gist info % TODO link to long version (provide link to main document) +% TODO \phantomsection to cross link \newcommand{\gist}[2]{% \ifcsname EnableGist\endcsname% #2%