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fixed some typesetting problems
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Josia Pietsch 2024-02-09 20:23:05 +01:00
parent dcf4851177
commit 458dd9ab1f
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GPG key ID: E70B571D66986A2D
9 changed files with 122 additions and 31 deletions
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2
bibliography/references.bib
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@ -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}
}

2
inputs/lecture_10.tex
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@ -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.

1
inputs/lecture_13.tex
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@ -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.

3
inputs/lecture_14.tex
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@ -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 $\forall i \le n.~s_i \subsetneq s_i' \land u_i \subsetneq u_i'$.

4
inputs/lecture_20.tex
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@ -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)\\

2
inputs/lecture_22.tex
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@ -52,7 +52,7 @@
\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 $R(x,x_k) = R(\tau^{m_k} x, \tau^{m_k} x_k) \xrightarrow{k \to \infty} 0$,

110
inputs/lecture_26.tex
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@ -95,6 +95,108 @@ We do a second proof of \yaref{thm:hindman}:
Consider $y(0)$.
We will prove that this color works and construct a corresponding $H$.
% 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 $G_0 \coloneqq [y(0)]$
@ -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 $\leadsto$
\[
\forall n .~\exists N.~\forall r.y\defon{\{r,\ldots,r+N-1\}}
\text{ contains } y\defon{\{0,\ldots,n\}} \text{ as a subsequence.}
\]
\begin{IEEEeqnarray*}{rl}
\forall n .~\exists N.~\forall r.~&y\defon{\{r,\ldots,r+N-1\}}\\
&\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:

15
inputs/tutorial_04.tex
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@ -101,20 +101,7 @@ let
\item Then $[T]$ is compact:
\todo{TODO}
% Let $\langle s_n, n <\omega \rangle$
% 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.
% https://alanmath.wordpress.com/2011/06/16/on-trees-compactness-and-finite-splitting/
\end{enumerate}
\nr 2

2
inputs/tutorial_12b.tex
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@ -2,6 +2,8 @@
\subsection{Sheet 10}
\todo{Copy from Abdelrahman and Shiguma}
\nr 2
\todo{Def skew shift flow (on $(\R / \Z)^2$!)}