From c986475c775646dfdddcca81f63cf6a42ea3523f Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Tue, 23 Jan 2024 21:52:45 +0100 Subject: [PATCH] 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}