fixed some typesetting problems

bibliography/references.bib

@@ -48,8 +48,6 @@ year = {2012},
     TITLE = {Embedding of countable linear orders into $\Bbb Q$ as topological spaces},
     AUTHOR = {Eric Wofsey},
     HOWPUBLISHED = {Mathematics Stack Exchange},
-    NOTE = {URL:https://math.stackexchange.com/q/3722713 (version: 2020-06-16)},
-    EPRINT = {https://math.stackexchange.com/q/3722713},
     URL = {https://math.stackexchange.com/q/3722713}
 }
 

inputs/lecture_10.tex

@@ -1,6 +1,6 @@
 \subsection{The Lusin Separation Theorem}
-
 \lecture{10}{2023-11-17}{}
+
 \begin{theorem}[\vocab{Lusin separation theorem}]
     \yalabel{Lusin Separation Theorem}{Lusin Separation}{thm:lusinseparation}
     Let $X$ be Polish and $A,B \subseteq  X$ disjoint analytic.

inputs/lecture_13.tex

@@ -143,6 +143,7 @@ with $(f^{-1}(\{1\}), <)$.
     \end{itemize}
 \end{definition}
 \begin{remark}
+    \leavevmode
     \begin{itemize}
         \item A prewellordering may not be a linear order since
             it is not necessarily antisymmetric.

inputs/lecture_14.tex

@@ -193,7 +193,8 @@ Let $\rho(\prec) \coloneqq  \sup \{\rho_{\prec}(x) + 1 : x \in X\}$.
     Clearly $|W| \le \aleph_0$.
     Define $\prec^\ast$ on $W$
     by setting
-    \[(s_0,u_1,s_1,\ldots, u_n,s_n) \succ^\ast (s_0',u_1', s_1', \ldots, u_m', s_m') :\iff\]
+    \[(s_0,u_1,s_1,\ldots, u_n,s_n) \succ^\ast (s_0',u_1', s_1', \ldots, u_m', s_m')\]
+    iff
     \begin{itemize}
         \item $n < m$ and
         \item $\forall i \le n.~s_i \subsetneq s_i' \land u_i \subsetneq u_i'$.

inputs/lecture_20.tex

@@ -95,7 +95,8 @@ coordinates.
     \]
 \gist{%
     \begin{claim}
-        $F(x_k, x') = \inf_m d(\sigma^m(w_k), 1)$,
+        It is
+        \[F(x_k, x') = \inf_m d(\sigma^m(w_k), 1),\]
         where  $\sigma(\xi_1, \xi_2, \ldots) = (\xi_1, \xi_1\xi_2, \xi_2\xi_3, \ldots)$.
     \end{claim}
     \begin{subproof}
@@ -111,7 +112,6 @@ coordinates.
     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)\\

inputs/lecture_22.tex

@@ -14,7 +14,7 @@
     	\arrow["h"', dashed, from=2-3, to=2-2]
     	\arrow["{\overline{g}, \text{ max. isom.}}", curve={height=-12pt}, from=2-3, to=3-2]
     \end{tikzcd}\]
-    
+
     We want to show that this tower is normal,
     i.e.~the isometric extensions are maximal isometric extension.
 \gist{%
@@ -25,18 +25,18 @@
     Then there are $x,x' \in X$ with
     $\pi'(x) \neq  \pi'(x')$ but $\pi_n(x) = \pi_n(x') =t \in X_n$.
     Then $h^{-1}(t) \ni \pi'(x), \pi'(x')$.
-    
+
     By a \yaref{lem:lec20:1}
     there is a sequence $(x_k)$ in $X$
     with $\pi_{n-1}(x_k) = \pi_{n-1}(x) = \pi_{n-1}(x')$ for all $k$,
     such that $F(x_k, x) \to 0$ and $F(x_k, x') \to 0$.
-    
+
     Let $\rho$ be a metric witnessing that $\overline{g}$
     is an isometric extension,
     i.e.~
     $\rho$ is defined on $\bigcup_{x \in X_{n-1}} (\overline{g}^{-1}(x))^2 \overset{\text{closed}}{\subseteq} Y \times Y$,
     continuous and $\rho(Ta, Tb) = \rho(a,b)$ for $\overline{g}(a) = \overline{g}(b)$.
-    
+
     For $a,b \in X$ such that
     \[
     \overline{g}(\pi'(a)) = \overline{g}(\pi'(b))
@@ -45,14 +45,14 @@
     \[
     R(a,b) \coloneqq  \rho(\pi'(a), \pi'(b)).
     \]
-    
+
     \begin{itemize}
         \item For any two out of $x,x',(x_k)$, $R$ is defined.
         \item $R(x,x_k) = R(\tau^m x, \tau^m x_k)$ for all  $m$.
         \item $F(x,x_k) \xrightarrow{k\to \infty} 0$,
             so there is a sequence $(m_k)$
             such that
-            $d(\tau^{m_k}x, \tau^{m_k} x_k) \xrightarrow{k \to \infty} 0$.
+            \[d(\tau^{m_k}x, \tau^{m_k} x_k) \xrightarrow{k \to \infty} 0.\]
     \end{itemize}
     By continuity of $\rho$,
     we have that  $R(x,x_k) = R(\tau^{m_k} x, \tau^{m_k} x_k) \xrightarrow{k \to  \infty} 0$,

inputs/lecture_26.tex

@@ -95,7 +95,109 @@ We do a second proof of \yaref{thm:hindman}:
     Consider $y(0)$.
     We will prove that this color works and construct a corresponding $H$.
 
-    \begin{itemize}
+% def power(s):
+%     if len(s) == 0:
+%         return [[]]
+%     else:
+%         p = power(s[0:-1])
+%         return [q + [s[-1]] for q in p] + [q for q in p]
+%
+%
+% def draw(hs):
+%     s = "\\begin{tikzpicture}"
+%     s += "\n\t\\node at (-0.5,0.5) {$x$};";
+%     for (i,h) in enumerate(hs):
+%         s += "\n\t\\node at (" + str(hs[i]) + ",0.5) {$h_"+str(i)+"$};";
+%
+%     for subset in power(range(0,len(hs))):
+%         if len(subset) <= 1:
+%             continue
+%         c = sum([hs[i] for i in subset])
+%         c2 = 0.9 if 0 in subset else 0.7
+%         s += "\n\t\\node[black!40!white] at (" + str(c) + ", " + str(c2) + ") {\\tiny{$" + " + ".join(map(lambda x : "h_{" + str(x) + "}", subset)) + "$}};"
+%         s += "\n\t\\draw[black!40!white, very thin] (" + str(c) + ", " + str(c2 - 0.1) + ") -- (" + str(c) + ",-0.1);"
+%         if subset != list(range(0,len(subset))):
+%             s += "\n\t\\node[blue!40!white] at (" + str(c) + ", " + str(c2-2) + "){\\tiny{$y(" + " + ".join(map(lambda x: "h_{" + str(x) + "}", subset[0:-1])) + ")$}};"
+%             s += "\n\t\\draw[blue!40!white, very thin] (" + str(c) + ", -0.1) -- (" + str(c) + ", " + str(c2-1.8) + ");"
+%     for (i,h) in enumerate(hs):
+%         hsum = sum(hs[0:i+1])
+%         s += '\n\t\\draw[blue] ('\
+%             + str(h) + ', 0) -- ('\
+%             + str(h) + ', -0.2) -- ('\
+%             + str(hsum) + ', -0.2) -- ('\
+%             + str(hsum) + ', 0);'
+%         s += "\n\t\\node[blue] at (" + str(h) + ", -0.5) {$y(0)$};"
+%         if i > 0:
+%             s += "\n\t\\node[blue] at (" + str(hsum)\
+%                 + ", -0.5) {$y(" + ("\
+%                 + ".join(list(map(lambda x : "h_{" + str(x) + "}", range(0,i)))))\
+%                 + ")$};";
+%     s += "\n\t\\draw[thick] (0,0) -- ("+ str(sum(hs) + 2) + ",0);"
+%     s += "\n\\end{tikzpicture}"
+%     return s
+% print(draw(np.cumsum([0.8,1,2.4,4.5])))
+
+\adjustbox{scale=0.7,center}{%
+    \begin{tikzpicture}
+        \node at (-0.5,0.5) {$x$};
+        \node at (0.8,0.5) {$h_0$};
+        \node at (1.8,0.5) {$h_1$};
+        \node at (4.2,0.5) {$h_2$};
+        \node at (8.7,0.5) {$h_3$};
+        \node[black!40!white] at (15.5, 0.9) {\tiny{$h_{0} + h_{1} + h_{2} + h_{3}$}};
+        \draw[black!40!white, very thin] (15.5, 0.8) -- (15.5,-0.1);
+        \node[black!40!white] at (14.7, 0.7) {\tiny{$h_{1} + h_{2} + h_{3}$}};
+        \draw[black!40!white, very thin] (14.7, 0.6) -- (14.7,-0.1);
+        \node[blue!40!white] at (14.7, -1.3){\tiny{$y(h_{1} + h_{2})$}};
+        \draw[blue!40!white, very thin] (14.7, -0.1) -- (14.7, -1.1);
+        \node[black!40!white] at (13.7, 0.9) {\tiny{$h_{0} + h_{2} + h_{3}$}};
+        \draw[black!40!white, very thin] (13.7, 0.8) -- (13.7,-0.1);
+        \node[blue!40!white] at (13.7, -1.1){\tiny{$y(h_{0} + h_{2})$}};
+        \draw[blue!40!white, very thin] (13.7, -0.1) -- (13.7, -0.9);
+        \node[black!40!white] at (12.899999999999999, 0.7) {\tiny{$h_{2} + h_{3}$}};
+        \draw[black!40!white, very thin] (12.899999999999999, 0.6) -- (12.899999999999999,-0.1);
+        \node[blue!40!white] at (12.899999999999999, -1.3){\tiny{$y(h_{2})$}};
+        \draw[blue!40!white, very thin] (12.899999999999999, -0.1) -- (12.899999999999999, -1.1);
+        \node[black!40!white] at (11.299999999999999, 0.9) {\tiny{$h_{0} + h_{1} + h_{3}$}};
+        \draw[black!40!white, very thin] (11.299999999999999, 0.8) -- (11.299999999999999,-0.1);
+        \node[blue!40!white] at (11.299999999999999, -1.1){\tiny{$y(h_{0} + h_{1})$}};
+        \draw[blue!40!white, very thin] (11.299999999999999, -0.1) -- (11.299999999999999, -0.9);
+        \node[black!40!white] at (10.5, 0.7) {\tiny{$h_{1} + h_{3}$}};
+        \draw[black!40!white, very thin] (10.5, 0.6) -- (10.5,-0.1);
+        \node[blue!40!white] at (10.5, -1.3){\tiny{$y(h_{1})$}};
+        \draw[blue!40!white, very thin] (10.5, -0.1) -- (10.5, -1.1);
+        \node[black!40!white] at (9.5, 0.9) {\tiny{$h_{0} + h_{3}$}};
+        \draw[black!40!white, very thin] (9.5, 0.8) -- (9.5,-0.1);
+        \node[blue!40!white] at (9.5, -1.1){\tiny{$y(h_{0})$}};
+        \draw[blue!40!white, very thin] (9.5, -0.1) -- (9.5, -0.9);
+        \node[black!40!white] at (6.800000000000001, 0.9) {\tiny{$h_{0} + h_{1} + h_{2}$}};
+        \draw[black!40!white, very thin] (6.800000000000001, 0.8) -- (6.800000000000001,-0.1);
+        \node[black!40!white] at (6.0, 0.7) {\tiny{$h_{1} + h_{2}$}};
+        \draw[black!40!white, very thin] (6.0, 0.6) -- (6.0,-0.1);
+        \node[blue!40!white] at (6.0, -1.3){\tiny{$y(h_{1})$}};
+        \draw[blue!40!white, very thin] (6.0, -0.1) -- (6.0, -1.1);
+        \node[black!40!white] at (5.0, 0.9) {\tiny{$h_{0} + h_{2}$}};
+        \draw[black!40!white, very thin] (5.0, 0.8) -- (5.0,-0.1);
+        \node[blue!40!white] at (5.0, -1.1){\tiny{$y(h_{0})$}};
+        \draw[blue!40!white, very thin] (5.0, -0.1) -- (5.0, -0.9);
+        \node[black!40!white] at (2.6, 0.9) {\tiny{$h_{0} + h_{1}$}};
+        \draw[black!40!white, very thin] (2.6, 0.8) -- (2.6,-0.1);
+        \draw[blue] (0.8, 0) -- (0.8, -0.2) -- (0.8, -0.2) -- (0.8, 0);
+        \node[blue] at (0.8, -0.5) {$y(0)$};
+        \draw[blue] (1.8, 0) -- (1.8, -0.2) -- (2.6, -0.2) -- (2.6, 0);
+        \node[blue] at (1.8, -0.5) {$y(0)$};
+        \node[blue] at (2.6, -0.5) {$y(h_{0})$};
+        \draw[blue] (4.2, 0) -- (4.2, -0.2) -- (6.800000000000001, -0.2) -- (6.800000000000001, 0);
+        \node[blue] at (4.2, -0.5) {$y(0)$};
+        \node[blue] at (6.800000000000001, -0.5) {$y(h_{0} + h_{1})$};
+        \draw[blue] (8.7, 0) -- (8.7, -0.2) -- (15.5, -0.2) -- (15.5, 0);
+        \node[blue] at (8.7, -0.5) {$y(0)$};
+        \node[blue] at (15.5, -0.5) {$y(h_{0} + h_{1} + h_{2})$};
+        \draw[thick] (0,0) -- (17.5,0);
+    \end{tikzpicture}
+}
+
+        \begin{itemize}
         \item % Step 1
             Let $G_0 \coloneqq  [y(0)]$
             and let $N_0$ be such that
@@ -152,10 +254,10 @@ We do a second proof of \yaref{thm:hindman}:
             \item proximal $\leadsto$ $\forall N$.~$T^n(x)\defon{N} = T^n(y)\defon{N}$
                 for infinitely many $n$.
             \item uniform recurrence $\leadsto$
-                \[
+                \begin{IEEEeqnarray*}{rl}
-                    \forall n .~\exists N.~\forall r.y\defon{\{r,\ldots,r+N-1\}}
+                    \forall n .~\exists N.~\forall r.~&y\defon{\{r,\ldots,r+N-1\}}\\
-                    \text{ contains } y\defon{\{0,\ldots,n\}} \text{ as a subsequence.}
+                                                     &\text{ contains } y\defon{\{0,\ldots,n\}} \text{ as a subsequence.}
-                \]
+                \end{IEEEeqnarray*}
                 (consider neighbourhood $G_n = \{z \in X : z\defon{n}  = y\defon{n} \}$).
         \end{itemize}
         \item Consider $c \coloneqq y(0)$. This color works:

inputs/tutorial_04.tex

@@ -101,20 +101,7 @@ let
     \item Then $[T]$ is compact:
         \todo{TODO}
 
-        % Let $\langle s_n, n <\omega \rangle$
+        % https://alanmath.wordpress.com/2011/06/16/on-trees-compactness-and-finite-splitting/
-        % be a Cauchy sequence in $[T]$.
-
-        % Then for every $m < \omega$
-        % there exists an $N < \omega$ such that
-        % $s_n\defon{m} = s_{n'}\defon{m}$
-        % for all  $n, n' > N$.
-        % Thus there exists a pointwise limit $s$ of the $s_n$.
-
-        % Since for all $m$ we have $s\defon{m} = s_n\defon{m} \in [T]$
-        % for $m$ large enough,
-        % we get $s \in [T]$.
-
-        % Hence $[T]$ is sequentially compact.
 \end{enumerate}
 
 \nr 2

inputs/tutorial_12b.tex

@@ -2,6 +2,8 @@
 
 \subsection{Sheet 10}
 
+\todo{Copy from Abdelrahman and Shiguma}
+
 \nr 2
 
 \todo{Def skew shift flow (on $(\R / \Z)^2$!)}