thank you, Mirko!

inputs/intro.tex

@@ -12,6 +12,9 @@ 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

@@ -61,7 +61,9 @@ 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]
-    \todo{Counterexamples in Topology}
+    Consider $\R$ with the topology given by the basis $\{[a,b) : a,b \in \R\}$.
+    This is T3, but not second countable
+    and not metrizable.
 \end{example}
 
 \begin{fact}
@@ -98,6 +100,7 @@ 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:
@@ -107,6 +110,7 @@ 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}
 
@@ -175,6 +179,7 @@ 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)
@@ -185,7 +190,7 @@ 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.
+    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 $f: X \hookrightarrow [0,1]^{\omega}$ 
@@ -227,7 +232,9 @@ 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}
-        \todo{Exercise!}
+        Consider $B_{\epsilon}(x_n) \subseteq X$ for some $n \in \N$, $\epsilon > 0$.
+        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

@@ -13,7 +13,6 @@
  \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 $\forall n .~ \exists x_n\in C.~d(x,x_n) < \frac{1}{n}$.
@@ -54,7 +53,7 @@
          then there exists a complete metric on $Y$.
      \end{claim}
      \begin{refproof}{psubspacegdelta:c1}
-         Let $Y = U$be open in $X$.
+         Let $Y = U$ be open in $X$.
          Consider the map
          \begin{IEEEeqnarray*}{rCl}
              f_U\colon U &\longrightarrow &

inputs/lecture_03.tex

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

inputs/lecture_04.tex

@@ -1,6 +1,7 @@
 \lecture{04}{2023-10-20}{}
+
 \begin{remark}
-        Some of $F_s$ might be empty.
+        Some of the $F_s$ might be empty.
 \end{remark}
 
 \begin{refproof}{thm:bairetopolish}
@@ -29,7 +30,6 @@
     \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}
@@ -40,21 +40,19 @@
             $x_m\defon{N} = x\defon{N}$.
             Then for all $m, n \ge M$,
             we have that $f(x_m), f(x_n) \in F_{x\defon{N}}$.
-            So $d(f(x_m), f(x_n)) < \epsilon$ 
-            we have that $(f(x_n))$ is Cauchy.
-            
-            Since $(X,d)$ is complete,
-            there exists $y = \lim_n f(x_n)$.
+            So $d(f(x_m), f(x_n)) < \epsilon$, i.e.~$(f(x_n))$ is Cauchy.
 
-            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
-            \[y \in \bigcap_{n} \overline{F_{x\defon{n}}} = \bigcap_{n} F_{x\defon{n}},\]
-            i.e.~$y \in D$ and $y = f(x)$.
         \end{subproof}
+        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 $y \in  \overline{F_{x\defon{N}}}$.
+        
+        Note that for $N' > N$ by the same argument
+        we get $y \in \overline{F_{x\defon{N'}}}$.
+        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{refproof}
 
     We extend $f$ to $g\colon\cN \to X$ 
@@ -70,7 +68,7 @@
     (i.e.~$r = \id$ on $D$ and $r$ is a continuous surjection).
     Then $g \coloneqq f \circ r$.
     
-    To construct $r$, we will define by induction
+    To construct $r$, we will define 
     $\phi\colon \N^{<\N} \to  S$ by induction on the length
     such that
     \begin{itemize}

inputs/lecture_18.tex

@@ -15,6 +15,7 @@ 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

@@ -274,6 +274,7 @@ 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

@@ -0,0 +1,196 @@
+\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 $X$ 2nd countable $\implies$ X separable.
+            Then $X$ \nth{2} countable $\implies$ X separable.
         \item If $X$ is a metric space and separable,
-            then $X$ is 2nd countable.
+            then $X$ is \nth{2} 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 2nd countable.
+    then it is \nth{2} 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 2nd countable,
+    If $X$ is \nth{2} countable,
     then it is Lindelöff.
 \end{fact}
 \begin{proof}
@@ -60,7 +60,7 @@
 \end{proof}
 \begin{remark}
     For metric spaces the notions
-    of being 2nd countable, separable
+    of being \nth{2} countable, separable
     and Lindelöf coincide.
 
     In arbitrary topological spaces,

inputs/tutorial_11.tex

@@ -0,0 +1,113 @@
+\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

@@ -50,3 +50,13 @@
 \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

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