inputs/intro.tex

@@ -12,9 +12,6 @@ If you find errors or want to improve something,
 please send me a message:\\
 \texttt{lecturenotes@jrpie.de}.
 
-Many thanks to \textsc{Mirko Bartsch} for providing notes on lectures
-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 $\dagger$.

inputs/lecture_01.tex

@@ -61,9 +61,7 @@ However the converse of this does not hold.
     Then $\{x_0\}$ is dense in $X$, but $X$ is not second countable.
 \end{example}
 \begin{example}[Sorgenfrey line]
-    Consider $\R$ with the topology given by the basis $\{[a,b) : a,b \in \R\}$.
+    \todo{Counterexamples in Topology}
-    This is T3, but not second countable
-    and not metrizable.
 \end{example}
 
 \begin{fact}
@@ -100,7 +98,6 @@ However the converse of this does not hold.
     then $X$ is metrisable.
 \end{theorem}
 
-\begin{absolutelynopagebreak}
 \begin{fact}
     If $X$ is a compact Hausdorff space,
     the following are equivalent:
@@ -110,7 +107,6 @@ However the converse of this does not hold.
         \item $X$ is second countable.
     \end{itemize}
 \end{fact}
-\end{absolutelynopagebreak}
 
 \subsection{Some facts about polish spaces}
 
@@ -179,7 +175,6 @@ suffices to show that open balls in one metric are unions of open balls in the o
 \end{proof}
 
 \begin{definition}[Our favourite Polish spaces]
-    \leavevmode
     \begin{itemize}
         \item $2^{\omega}$ is called the \vocab{Cantor set}.
             (Consider $2$ with the discrete topology)
@@ -190,7 +185,7 @@ suffices to show that open balls in one metric are unions of open balls in the o
     \end{itemize}
 \end{definition}
 \begin{proposition}
-    Let $X$ be a separable, metrisable topological space\footnote{e.g.~Polish, but we don't need completeness.}.
+    Let $X$ be a separable, metrisable topological space.
     Then $X$ topologically embeds into the
     \vocab{Hilbert cube},
     i.e. there is an injective $f: X \hookrightarrow [0,1]^{\omega}$ 
@@ -232,9 +227,7 @@ suffices to show that open balls in one metric are unions of open balls in the o
         $f^{-1}$ is continuous.
     \end{claim}
     \begin{subproof}
-        Consider $B_{\epsilon}(x_n) \subseteq X$ for some $n \in \N$, $\epsilon > 0$.
+        \todo{Exercise!}
-        Then $f(U) = f(X) \cap [0,1]^{n} \times [0,\epsilon) \times [0,1]^{\omega}$
-        is open\footnote{as a subset of $f(X)$!}.
     \end{subproof}
 \end{proof}
 \begin{proposition}

inputs/lecture_02.tex

@@ -13,6 +13,7 @@
  \begin{proof}
      Let $C \subseteq X$ be closed.
      Let $U_{\frac{1}{n}} \coloneqq \{x | d(x, C) < \frac{1}{n}\}$.
+     \todo{Exercise}
      Clearly $C \subseteq \bigcap U_{\frac{1}{n}}$.
      Let $x \in \bigcap U_{\frac{1}{n}}$.
      Then $\forall n .~ \exists x_n\in C.~d(x,x_n) < \frac{1}{n}$.

inputs/lecture_03.tex

@@ -1,6 +1,6 @@
 \lecture{03}{2023-10-17}{Embedding of the cantor space into polish spaces}
 
-\subsection{Trees}
+% ? \subsection{Trees}  TODO
 
 
 \begin{notation}
@@ -255,7 +255,9 @@
 
 
 
+    [To be continued]
     \phantom\qedhere
+
 \end{refproof}
 
 

inputs/lecture_04.tex

@@ -1,7 +1,6 @@
 \lecture{04}{2023-10-20}{}
-
 \begin{remark}
-        Some of the $F_s$ might be empty.
+        Some of $F_s$ might be empty.
 \end{remark}
 
 \begin{refproof}{thm:bairetopolish}
@@ -30,6 +29,7 @@
     \begin{refproof}{thm:bairetopolish:c1}
         Let $x_n$ be a series in $D$
         converging to $x$ in $\cN$.
+        Then $x \in \cN$.
         \begin{claim}
             $(f(x_n))$ is Cauchy.
         \end{claim}
@@ -40,11 +40,12 @@
             $x_m\defon{N} = x\defon{N}$.
             Then for all $m, n \ge M$,
             we have that $f(x_m), f(x_n) \in F_{x\defon{N}}$.
-            So $d(f(x_m), f(x_n)) < \epsilon$, i.e.~$(f(x_n))$ is Cauchy.
+            So $d(f(x_m), f(x_n)) < \epsilon$ 
+            we have that $(f(x_n))$ is Cauchy.
             
-        \end{subproof}
             Since $(X,d)$ is complete,
             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 $y \in  \overline{F_{x\defon{N}}}$.
             
@@ -53,6 +54,7 @@
             Hence
             \[y \in \bigcap_{n} \overline{F_{x\defon{n}}} = \bigcap_{n} F_{x\defon{n}},\]
             i.e.~$y \in D$ and $y = f(x)$.
+        \end{subproof}
     \end{refproof}
 
     We extend $f$ to $g\colon\cN \to X$ 
@@ -68,7 +70,7 @@
     (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 by induction
     $\phi\colon \N^{<\N} \to  S$ by induction on the length
     such that
     \begin{itemize}

inputs/lecture_18.tex

@@ -15,7 +15,6 @@ sketch of \yaref{thm:l16:3} in the case of $|Z| = 1$.
 Let $(X,T)$ be a distal flow.
 Then $G \coloneqq  E(X,T)$ is a group.
 \begin{definition}
-    \label{def:F}
 For $x, x' \in X$ define
 \[
 F(x,x') \coloneqq \inf \{d(gx, gx') : g \in G\}.

inputs/lecture_19.tex

@@ -274,7 +274,6 @@ More generally we can show:
     $(X'_{\xi+1}, T)$ is a factor of $(X_{\xi+1}, T)$.
 \end{proof}
 \begin{example}[{\cite[p. 513]{Furstenberg}}]
-    \label{ex:19:inftorus}
     Let $X$ be the infinite torus
     \[
     X \coloneqq \{(\xi_1, \xi_2, \ldots) : \xi_i \in \C, |\xi_i| = 1\}.

inputs/lecture_20.tex

@@ -1,196 +0,0 @@
-\lecture{20}{2024-01-09}{The infinite Torus}
-
-\begin{example}
-    \footnote{This is the same as \yaref{ex:19:inftorus},
-        but with new notation.}
-    Let $X = (S^1)^{\N}$\footnote{We identify $S^1$ and $\faktor{\R}{\Z}$.}
-    and consider $\left(X, \Z \right)$
-    where the action is generated by
-    \[
-        \tau\colon (x_1,x_2,x_3,\ldots) \mapsto(x_1 + \alpha, x_1 + x_2, x_2 + x_3, \ldots)
-    \]
-    for some irrational $\alpha$.
-\end{example}
-\begin{remark}+
-    Note that we can identify $S^1$ with a subset of  $\C$ (and use multiplication)
-    or with  $\faktor{\R}{\Z}$ (and use addition).
-    In the lecture both notations were used.% to make things extra confusing.
-    Here I'll try to only use multiplicative notation.
-\end{remark}
-We will be studying projections to the first $d$ coordinates,
-i.e.
-\[
-    \tau_d \colon (x_1,\ldots,x_d) \mapsto (e^{\i \alpha} x_1, x_1x_2, \ldots, x_{d-1}x_d).
-\]
-$\tau_d$ is called the \vocab{$d$-skew shift}.
-For $d = 1$ we get the circle rotation $x \mapsto e^{\i \alpha} x$.
-\begin{fact}
-    \label{fact:tau1minimal}
-    The circle rotation $x \mapsto e^{\i \alpha} x$ is minimal.
-    In fact, every subgroup of $S^1$ is either dense in $S^1$
-    or it is of the form
-    \[
-    H_m \coloneqq  \{x \in S^1 : x^m = 0\}
-    \]
-    for some $m \in \Z$.
-\end{fact}
-\todo{Homework!}
-We will show that $\tau_d$ is minimal for all $d$,
-i.e.~every orbit is dense.
-From this it will follow that $\tau$ is minimal.
-
-Let $\pi_n\colon X \to  (S^1)^n$ be the projection to the first $n$
-coordinates.
-
-
-
-
-
-\begin{lemma}
-    Let $x,x' \in X$ with $\pi_n(x) = \pi_n(x')$
-    for some $n$.
-    Then there is a sequence of points $x_k$ with
-    \[\pi_{n-1}(x_k) = \pi_{n-1}(x) = \pi_{n-1}(x')\]
-    for all $k$
-    and
-    \[
-    F(x_k, x) \xrightarrow{k \to \infty} 0,
-    F(x_k, x') \xrightarrow{k \to  \infty} 0,
-    \]
-    where $F$ is as in \yaref{def:F},
-    i.e.~$F(a,b) = \inf_{n \in \Z} d(\tau^n a, \tau^n b)$,
-    where $d$ is the metric on $X$,
-    $d((x_i), (y_i)) = \max_n \frac{1}{2^n} | x_n - y_n|$.% TODO use multiplicative notation
-\end{lemma}
-\begin{proof}
-    Let
-    \begin{IEEEeqnarray*}{rCl}
-        x &=& (\alpha^0_1, \alpha^0_2, \ldots, \alpha^0_{n-1}, \alpha_n, \alpha_{n+1}, \alpha_{n+2},\ldots)\\
-        x' &=& (\alpha^0_1, \alpha^0_2, \ldots, \alpha^0_{n-1}, \alpha_n, \alpha'_{n+1}, \alpha'_{n+2},\ldots).\\
-    \end{IEEEeqnarray*}
-    We will choose $x_k$ of the form
-    \[
-        (\alpha^0_1, \alpha^0_2, \ldots, \alpha^0_{n-1} \alpha_n e^{\i \beta_k}, \alpha_{n+1}, \alpha_{n+2}, \ldots),
-    \]
-    where $\beta_k$ is such that $\frac{\beta_k}{\pi}$ is irrational
-    and $|\beta_k| < 2^{-k}$.
-    Fix a sequence of such $\beta_k$.
-    Then
-    \[d(x_k,x) = 2^{-n} |e^{\i \beta_k} - 1| < 2^{-n-k} \xrightarrow{k\to \infty} 0.\]
-    In particular $F(x_k, x) \to 0$.
-
-    We want to show that $F(x_k, x') < 2^{-n-k}$.
-    For $u, u' \in X$,
-    $u = (\xi_n)_{n \in \N}$,
-    $u' = (\xi'_n)_{n \in \N}$,
-    let $\frac{u}{u'} = (\frac{\xi_n}{\xi'_n})_{n \in \N}$
-    ($X$ is a group).
-    
-    We are interested in $F(x_k, x') = \inf_m d(\tau^m x_k, \tau^m x')$,
-    but it is easier to consider the distance between
-    their quotient and $1$.
-    Consider
-    \[
-        w_k \coloneqq \frac{x_k}{x'} = (\underbrace{1,\ldots,1}_{n-1}, e^{\i \beta_k}, \overbrace{\frac{\alpha_{n+1}}{\alpha'_{n+1}}, \frac{\alpha_{n+2}}{\alpha'_{n+2}}, \ldots}^{\mathclap{\text{not interesting}}}).
-    \]
-    \begin{claim}
-        $F(x_k, x') = \inf_m d(\sigma^m(w_k), 1)$,
-        where  $\sigma(\xi_1, \xi_2, \ldots) = (\xi_1, \xi_1\xi_2, \xi_2\xi_3, \ldots)$.
-    \end{claim}
-    \begin{subproof}
-        We have
-        \begin{IEEEeqnarray*}{rCl}
-            F(u,u') &=& \inf_m d(\tau^m u, \tau^m u')\\
-                    &=& \inf_m d(\frac{\tau^m u}{\tau^m u'}, 1)\\
-                    &=& \inf_m d(\sigma^m\left( \frac{u}{u'} \right), 1).
-        \end{IEEEeqnarray*}
-    \end{subproof}
-
-    Fix $k$. Let $w^\ast = (1,\ldots,1, e^{\i \beta_k}, 1, \ldots)$.
-    By minimality of $(X,T)$ for any $\epsilon >0$,
-    there exists $m \in \Z$ such that 
-    $d(\sigma^m w_k, w^\ast) < \epsilon$.
-    % TODO Think about this
-    Then
-    \begin{IEEEeqnarray*}{rCl}
-        \inf_m d(\sigma^m w_k, 1) &\le & \inf_m d(\sigma^m w_k, w^\ast) + d(w^\ast, 1)\\
-                                  &\le & 2^{-n} | e^{\i \beta_k}- 1|\\
-                                  &<& 2^{-n-k}.
-    \end{IEEEeqnarray*}
-\end{proof}
-
-
-\begin{definition}
-    For every continuous  $f\colon S^1 \to  S^1$, the
-    \vocab{winding number} $[f] \in \Z$
-    is the unique integer such that $f$ is homotopic%
-    \footnote{$f\colon Y \to  Z$ and $g\colon  Y \to  Z$ are homotopic
-        iff there is $H\colon Y \times [0,1] \to  \Z$
-        continuous such that $H(\cdot ,0) = f$ and $H(\cdot ,1) = g$.}
-    to the map
-    $x \mapsto x^{n}$.
-\end{definition}
-
-\begin{remark}
-    Note that for
-    \begin{IEEEeqnarray*}{rCl}
-        \sigma\colon (S^1)^d &\longrightarrow & S^1 \\
-        (x_1,\ldots,x_d) &\longmapsto & x_d
-    \end{IEEEeqnarray*}
-    we have that $T = \tau_{d+1}$,
-    where
-    \begin{IEEEeqnarray*}{rCl}
-        T\colon (S^1)^d \times S^1 &\longrightarrow & (S^1)^d \times S^1 \\
-        (y, x_{d+1}) &\longmapsto & (\tau_d(y), \sigma(y) x_{d+1}).
-    \end{IEEEeqnarray*}
-\end{remark}
-\begin{theorem}
-    \label{thm:taudminimal:help}
-    For every $d$ if $\tau_d$\footnote{more formally $((S^1)^d, \langle \tau_d \rangle)$}
-    is minimal, then $\tau_{d+1}$ is minimal.
-\end{theorem}
-\begin{corollary}
-    $\tau_d$ is minimal for all $d$.
-\end{corollary}
-\begin{proof}
-    $\tau_1$ is minimal (\yaref{fact:tau1minimal}).
-    Apply \yaref{thm:taudminimal:help}.
-\end{proof}
-\begin{corollary}
-    Since all the $\tau_d$ are minimal,
-    $\tau$ is minimal.
-\end{corollary}
-\begin{proof}
-    This follows from the definition of the product topology,
-    since for a basic open set $U = U_1 \times \ldots \times U_d \times (S^1)^{\infty}$
-    it suffices to analyze the first $d$ coordinates.
-\end{proof}
-
-\begin{refproof}{thm:taudminimal:help}
-    Let $s \coloneqq  \tau_d$ and $Y \coloneqq (S^1)^d$.
-    Consider
-    \begin{IEEEeqnarray*}{rCl}
-        \gamma\colon S^1 &\longrightarrow & Y \\
-        x &\longmapsto & (x,x,\ldots,x)
-    \end{IEEEeqnarray*}
-    \begin{enumerate}[(a)]
-        \item $\gamma$ and $s \circ \gamma$ are homotopic
-            via
-            \begin{IEEEeqnarray*}{rCl}
-                H\colon S^1 \times [0,1] &\longrightarrow & (S^1)^d \\
-            (x, t)&\longmapsto & (x e^{\i t  \alpha}, x^{t+1}, x^{t+1}, x^{t+1},\ldots, x^{t+1})
-            \end{IEEEeqnarray*}
-        \item For all $m \in  \Z \setminus \{0\}$, we have
-            $[x \mapsto \left(\sigma(\gamma(x))\right)^m] = m \neq 0$,
-            since $\sigma(\gamma(x)) = \sigma((x,\ldots,x)) = x$.
-    \end{enumerate}
-
-    [to be continued]
-    \phantom\qedhere
-    
-\end{refproof}
-
-
-
-
-

inputs/tutorial_01.tex

@@ -13,9 +13,9 @@
 \begin{fact}
     \begin{itemize}
         \item Let $X$ be a topological space.
-            Then $X$ \nth{2} countable $\implies$ X separable.
+            Then $X$ 2nd countable $\implies$ X separable.
         \item If $X$ is a metric space and separable,
-            then $X$ is \nth{2} countable.
+            then $X$ is 2nd countable.
     \end{itemize}
 \end{fact}
 \begin{proof}
@@ -32,18 +32,18 @@
 \begin{fact}
     Let $X$ be a metric space.
     If $X$ is Lindelöf,
-    then it is \nth{2} countable.
+    then it is 2nd countable.
 \end{fact}
 \begin{proof}
     For all $q \in \Q$
-    consider the cover $B_q(x), x \in X$
+    Consider the cover $B_q(x), x \in X$
     and choose a countable subcover.
     The union of these subcovers is
     a countable base.
 \end{proof}
 \begin{fact}
     Let $X$ be a topological space.
-    If $X$ is \nth{2} countable,
+    If $X$ is 2nd countable,
     then it is Lindelöff.
 \end{fact}
 \begin{proof}
@@ -60,7 +60,7 @@
 \end{proof}
 \begin{remark}
     For metric spaces the notions
-    of being \nth{2} countable, separable
+    of being 2nd countable, separable
     and Lindelöf coincide.
 
     In arbitrary topological spaces,

inputs/tutorial_11.tex

@@ -1,113 +0,0 @@
-\tutorial{11}{2024-01-09}{}
-
-An equivalent definition of subflows can be given as   follows:
-\begin{definition}
-    Let $(X,T)$ be a flow with action $\alpha_x$.
-    Let $Y \subseteq X$ be a compact subspace of $X$.
-    If $Y$ is invariant under $\alpha_x$, we say that
-    $(Y,T)$ (with action  $\alpha_x\defon{T \times Y}$
-    is a subflow of $(X,T)$.
-\end{definition}
-
-\begin{example}[Flows with a non-closed orbit]
-    \begin{enumerate}[1.]
-        \item Consider $(S^1, \Z)$
-            with action given by $1 \cdot x = x + c$ for
-            a fixed $c \in \R\setminus\Q$.\footnote{We identify $S^1$ and $\faktor{\R}{\Z}$.}
-            Then the orbit of $0$, $\{nc : n \in \Z\}$ is dense but consists only of irrationals
-            (except $0$),
-            so it is not closed.
-        \item Consider $(S^1, \Q)$ with action $qx \coloneqq x + q$.
-            The orbit of $0$, $\faktor{\Q}{\Z} \subseteq S^1$,
-            is dense but not closed.
-
-            $(S^1,\Q)$ is minimal.
-    \end{enumerate}
-\end{example}
-\begin{example}[\vocab{Left Bernoulli shift}]
-    Consider $(\{0,1\}^{\Z},  T)$,
-    where $T = \Z$ and the action is given by
-    \begin{IEEEeqnarray*}{rCl}
-       \Z \times \{0,1\}^{\Z}&\longrightarrow & \Z \\
-       (m, (x_n)_{n \in \Z})&\longmapsto & (x_{n+m})_{n \in \Z}.
-    \end{IEEEeqnarray*}
-
-    The orbit of $z \coloneqq (0)_{n \in \Z}$ consist of only on point.
-    In particular it is closed.
-
-    Let $x \coloneqq  ( [n = 0])_{n \in \Z}$.
-    Then $Tx = \{([n = m])_{n \in \Z} | m \in \Z\}$.
-    Clearly $z \not\in Tx$.
-    \begin{claim}
-        $z \in \overline{Tx}$
-    \end{claim}
-    \begin{proof}
-        Consider a basic open $z \in U_I = \{y : y_i = 0, i \in I\}$
-        where $I \subseteq \Z$ is finite.
-        Then $U_I \cap Tx \neq \emptyset$
-        as we can shift the $1$ out of $I$,
-        i.e.~$(\max I + 1) x \in U_I$.
-    \end{proof}
-\end{example}
-
-Flows are always on non-empty spaces $X$.
-\begin{fact}
-    Consider a flow $(X,T)$.
-    The following are equivalent:
-    \begin{enumerate}[(i)]
-        \item Every $T$-orbit is dense.
-        \item There is no proper subflow,
-    \end{enumerate}
-    If these conditions hold, the flow is called \vocab{minimal}.
-\end{fact}
-\begin{proof}
-    (i) $\implies$ (ii):
-    Let $(Y,T)$ be a subflow of $(X,T)$.
-    take  $y \in Y$. Then $Ty$ is dense in mKX.
-    But $Ty \subseteq Y$, so $Y$ is dense in $X$.
-    Since $Y$ is closed, we get $Y = X$.
-
-    (ii) $\implies$ (i):
-    Take $x \in X$. Consider $Tx$.
-    It suffices to show that $\overline{Tx}$ is a subflow.
-    Clearly $\overline{Tx}$ is closed,
-    so it suffices to show that it is  $T$-invariant.
-    Let $y \in \overline{Tx}$ and $t \in T$.
-    Take $ty \in U \overset{\text{open}}{\subseteq} X$.
-    Since $t^{-1}$ acts as a homeomorphism
-    we have $y \in t^{-1} U \overset{\text{open}}{\subseteq} X$.
-    We find some $t'x \in t^{-1}U$ since $y \in \overline{Tx}$.
-    So $tt'x \in Tx \cap U$.
-\end{proof}
-
-\begin{fact}
-    Every flow $(X,T)$ contains a minimal subflow.
-\end{fact}
-\begin{proof}
-    We use Zorn's lemma:
-    Let $S$ be the set of all subflows of $(X,T)$
-    ordered by  $Y \le Y' :\iff Y \supseteq Y'$.
-    We need to show that for a chain $\langle Y_i : i \in I \rangle$,
-    there exists a lower bound.
-    Consider $\bigcap_{i \in I} Y_i$. This a subflow:
-    \begin{itemize}
-        \item It is closed as it is an intersection of closed sets.
-        \item It is $T$-invariant, since each of the $Y_i$ is.
-        \item It is non-empty by \yaref{tut10fact}.
-    \end{itemize}
-\end{proof}
-\begin{fact}
-    \label{tut10fact}
-    Let $X$ be a topological space.
-    Then $X$ is compact iff every family of closed sets with
-    FIP\footnote{finite intersection property, i.e.~the intersection of every finite sub-family is non-empty}
-    has non-empty intersection.
-\end{fact}
-\begin{proof}
-    Note that families of
-    closed sets correspond to families of open sets by taking complements.
-    A family of open sets is a cover iff the corresponding family
-    has empty intersection,
-    and is admits a finite subcover iff the corresponding family
-    has the FIP.
-\end{proof}

jrpie-math.sty

@@ -50,13 +50,3 @@
 \RequirePackage{mkessler-mathfixes} % Load this last since it renews behaviour
 \DeclareMathOperator{\inter}{int} % interior
 \newcommand{\defon}[1]{|_{#1}} % TODO
-
-\RequirePackage[super]{nth}
-
-% TODO MOVE
-% https://tex.stackexchange.com/a/94702
-\newenvironment{absolutelynopagebreak}
-  {\par\nobreak\vfil\penalty0\vfilneg
-   \vtop\bgroup}
-  {\par\xdef\tpd{\the\prevdepth}\egroup
-   \prevdepth=\tpd}

logic3.tex

@@ -44,7 +44,10 @@
 \input{inputs/lecture_17}
 \input{inputs/lecture_18}
 \input{inputs/lecture_19}
-\input{inputs/lecture_20}
+
+
+
+
 
 \cleardoublepage
 
@@ -61,7 +64,6 @@
 \input{inputs/tutorial_08}
 \input{inputs/tutorial_09}
 \input{inputs/tutorial_10}
-\input{inputs/tutorial_11}
 
 \section{Facts}
 \input{inputs/facts}