fixed some typesetting problems
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@ -48,8 +48,6 @@ year = {2012},
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TITLE = {Embedding of countable linear orders into $\Bbb Q$ as topological spaces},
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AUTHOR = {Eric Wofsey},
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HOWPUBLISHED = {Mathematics Stack Exchange},
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NOTE = {URL:https://math.stackexchange.com/q/3722713 (version: 2020-06-16)},
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EPRINT = {https://math.stackexchange.com/q/3722713},
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URL = {https://math.stackexchange.com/q/3722713}
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}
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@ -1,6 +1,6 @@
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\subsection{The Lusin Separation Theorem}
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\lecture{10}{2023-11-17}{}
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\begin{theorem}[\vocab{Lusin separation theorem}]
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\yalabel{Lusin Separation Theorem}{Lusin Separation}{thm:lusinseparation}
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Let $X$ be Polish and $A,B \subseteq X$ disjoint analytic.
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@ -143,6 +143,7 @@ with $(f^{-1}(\{1\}), <)$.
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\end{itemize}
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\end{definition}
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\begin{remark}
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\leavevmode
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\begin{itemize}
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\item A prewellordering may not be a linear order since
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it is not necessarily antisymmetric.
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@ -193,7 +193,8 @@ Let $\rho(\prec) \coloneqq \sup \{\rho_{\prec}(x) + 1 : x \in X\}$.
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Clearly $|W| \le \aleph_0$.
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Define $\prec^\ast$ on $W$
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by setting
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\[(s_0,u_1,s_1,\ldots, u_n,s_n) \succ^\ast (s_0',u_1', s_1', \ldots, u_m', s_m') :\iff\]
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\[(s_0,u_1,s_1,\ldots, u_n,s_n) \succ^\ast (s_0',u_1', s_1', \ldots, u_m', s_m')\]
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iff
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\begin{itemize}
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\item $n < m$ and
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\item $\forall i \le n.~s_i \subsetneq s_i' \land u_i \subsetneq u_i'$.
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@ -95,7 +95,8 @@ coordinates.
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\]
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\gist{%
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\begin{claim}
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$F(x_k, x') = \inf_m d(\sigma^m(w_k), 1)$,
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It is
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\[F(x_k, x') = \inf_m d(\sigma^m(w_k), 1),\]
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where $\sigma(\xi_1, \xi_2, \ldots) = (\xi_1, \xi_1\xi_2, \xi_2\xi_3, \ldots)$.
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\end{claim}
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\begin{subproof}
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@ -111,7 +112,6 @@ coordinates.
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By minimality of $(X,T)$ for any $\epsilon >0$,
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there exists $m \in \Z$ such that
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$d(\sigma^m w_k, w^\ast) < \epsilon$.
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% TODO Think about this
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Then
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\begin{IEEEeqnarray*}{rCl}
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\inf_m d(\sigma^m w_k, 1) &\le & \inf_m d(\sigma^m w_k, w^\ast) + d(w^\ast, 1)\\
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@ -14,7 +14,7 @@
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\arrow["h"', dashed, from=2-3, to=2-2]
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\arrow["{\overline{g}, \text{ max. isom.}}", curve={height=-12pt}, from=2-3, to=3-2]
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\end{tikzcd}\]
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We want to show that this tower is normal,
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i.e.~the isometric extensions are maximal isometric extension.
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\gist{%
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@ -25,18 +25,18 @@
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Then there are $x,x' \in X$ with
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$\pi'(x) \neq \pi'(x')$ but $\pi_n(x) = \pi_n(x') =t \in X_n$.
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Then $h^{-1}(t) \ni \pi'(x), \pi'(x')$.
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By a \yaref{lem:lec20:1}
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there is a sequence $(x_k)$ in $X$
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with $\pi_{n-1}(x_k) = \pi_{n-1}(x) = \pi_{n-1}(x')$ for all $k$,
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such that $F(x_k, x) \to 0$ and $F(x_k, x') \to 0$.
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Let $\rho$ be a metric witnessing that $\overline{g}$
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is an isometric extension,
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i.e.~
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$\rho$ is defined on $\bigcup_{x \in X_{n-1}} (\overline{g}^{-1}(x))^2 \overset{\text{closed}}{\subseteq} Y \times Y$,
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continuous and $\rho(Ta, Tb) = \rho(a,b)$ for $\overline{g}(a) = \overline{g}(b)$.
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For $a,b \in X$ such that
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\[
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\overline{g}(\pi'(a)) = \overline{g}(\pi'(b))
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@ -45,14 +45,14 @@
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\[
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R(a,b) \coloneqq \rho(\pi'(a), \pi'(b)).
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\]
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\begin{itemize}
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\item For any two out of $x,x',(x_k)$, $R$ is defined.
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\item $R(x,x_k) = R(\tau^m x, \tau^m x_k)$ for all $m$.
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\item $F(x,x_k) \xrightarrow{k\to \infty} 0$,
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so there is a sequence $(m_k)$
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such that
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$d(\tau^{m_k}x, \tau^{m_k} x_k) \xrightarrow{k \to \infty} 0$.
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\[d(\tau^{m_k}x, \tau^{m_k} x_k) \xrightarrow{k \to \infty} 0.\]
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\end{itemize}
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By continuity of $\rho$,
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we have that $R(x,x_k) = R(\tau^{m_k} x, \tau^{m_k} x_k) \xrightarrow{k \to \infty} 0$,
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@ -95,7 +95,109 @@ We do a second proof of \yaref{thm:hindman}:
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Consider $y(0)$.
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We will prove that this color works and construct a corresponding $H$.
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\begin{itemize}
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% def power(s):
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% if len(s) == 0:
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% return [[]]
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% else:
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% p = power(s[0:-1])
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% return [q + [s[-1]] for q in p] + [q for q in p]
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%
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%
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% def draw(hs):
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% s = "\\begin{tikzpicture}"
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% s += "\n\t\\node at (-0.5,0.5) {$x$};";
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% for (i,h) in enumerate(hs):
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% s += "\n\t\\node at (" + str(hs[i]) + ",0.5) {$h_"+str(i)+"$};";
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%
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% for subset in power(range(0,len(hs))):
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% if len(subset) <= 1:
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% continue
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% c = sum([hs[i] for i in subset])
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% c2 = 0.9 if 0 in subset else 0.7
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% s += "\n\t\\node[black!40!white] at (" + str(c) + ", " + str(c2) + ") {\\tiny{$" + " + ".join(map(lambda x : "h_{" + str(x) + "}", subset)) + "$}};"
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% s += "\n\t\\draw[black!40!white, very thin] (" + str(c) + ", " + str(c2 - 0.1) + ") -- (" + str(c) + ",-0.1);"
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% if subset != list(range(0,len(subset))):
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% s += "\n\t\\node[blue!40!white] at (" + str(c) + ", " + str(c2-2) + "){\\tiny{$y(" + " + ".join(map(lambda x: "h_{" + str(x) + "}", subset[0:-1])) + ")$}};"
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% s += "\n\t\\draw[blue!40!white, very thin] (" + str(c) + ", -0.1) -- (" + str(c) + ", " + str(c2-1.8) + ");"
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% for (i,h) in enumerate(hs):
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% hsum = sum(hs[0:i+1])
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% s += '\n\t\\draw[blue] ('\
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% + str(h) + ', 0) -- ('\
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% + str(h) + ', -0.2) -- ('\
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% + str(hsum) + ', -0.2) -- ('\
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% + str(hsum) + ', 0);'
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% s += "\n\t\\node[blue] at (" + str(h) + ", -0.5) {$y(0)$};"
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% if i > 0:
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% s += "\n\t\\node[blue] at (" + str(hsum)\
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% + ", -0.5) {$y(" + ("\
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% + ".join(list(map(lambda x : "h_{" + str(x) + "}", range(0,i)))))\
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% + ")$};";
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% s += "\n\t\\draw[thick] (0,0) -- ("+ str(sum(hs) + 2) + ",0);"
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% s += "\n\\end{tikzpicture}"
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% return s
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% print(draw(np.cumsum([0.8,1,2.4,4.5])))
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\adjustbox{scale=0.7,center}{%
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\begin{tikzpicture}
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\node at (-0.5,0.5) {$x$};
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\node at (0.8,0.5) {$h_0$};
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\node at (1.8,0.5) {$h_1$};
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\node at (4.2,0.5) {$h_2$};
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\node at (8.7,0.5) {$h_3$};
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\node[black!40!white] at (15.5, 0.9) {\tiny{$h_{0} + h_{1} + h_{2} + h_{3}$}};
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\draw[black!40!white, very thin] (15.5, 0.8) -- (15.5,-0.1);
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\node[black!40!white] at (14.7, 0.7) {\tiny{$h_{1} + h_{2} + h_{3}$}};
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\draw[black!40!white, very thin] (14.7, 0.6) -- (14.7,-0.1);
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\node[blue!40!white] at (14.7, -1.3){\tiny{$y(h_{1} + h_{2})$}};
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\draw[blue!40!white, very thin] (14.7, -0.1) -- (14.7, -1.1);
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\node[black!40!white] at (13.7, 0.9) {\tiny{$h_{0} + h_{2} + h_{3}$}};
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\draw[black!40!white, very thin] (13.7, 0.8) -- (13.7,-0.1);
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\node[blue!40!white] at (13.7, -1.1){\tiny{$y(h_{0} + h_{2})$}};
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\draw[blue!40!white, very thin] (13.7, -0.1) -- (13.7, -0.9);
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\node[black!40!white] at (12.899999999999999, 0.7) {\tiny{$h_{2} + h_{3}$}};
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\draw[black!40!white, very thin] (12.899999999999999, 0.6) -- (12.899999999999999,-0.1);
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\node[blue!40!white] at (12.899999999999999, -1.3){\tiny{$y(h_{2})$}};
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\draw[blue!40!white, very thin] (12.899999999999999, -0.1) -- (12.899999999999999, -1.1);
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\node[black!40!white] at (11.299999999999999, 0.9) {\tiny{$h_{0} + h_{1} + h_{3}$}};
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\draw[black!40!white, very thin] (11.299999999999999, 0.8) -- (11.299999999999999,-0.1);
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\node[blue!40!white] at (11.299999999999999, -1.1){\tiny{$y(h_{0} + h_{1})$}};
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\draw[blue!40!white, very thin] (11.299999999999999, -0.1) -- (11.299999999999999, -0.9);
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\node[black!40!white] at (10.5, 0.7) {\tiny{$h_{1} + h_{3}$}};
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\draw[black!40!white, very thin] (10.5, 0.6) -- (10.5,-0.1);
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\node[blue!40!white] at (10.5, -1.3){\tiny{$y(h_{1})$}};
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\draw[blue!40!white, very thin] (10.5, -0.1) -- (10.5, -1.1);
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\node[black!40!white] at (9.5, 0.9) {\tiny{$h_{0} + h_{3}$}};
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\draw[black!40!white, very thin] (9.5, 0.8) -- (9.5,-0.1);
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\node[blue!40!white] at (9.5, -1.1){\tiny{$y(h_{0})$}};
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\draw[blue!40!white, very thin] (9.5, -0.1) -- (9.5, -0.9);
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\node[black!40!white] at (6.800000000000001, 0.9) {\tiny{$h_{0} + h_{1} + h_{2}$}};
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\draw[black!40!white, very thin] (6.800000000000001, 0.8) -- (6.800000000000001,-0.1);
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\node[black!40!white] at (6.0, 0.7) {\tiny{$h_{1} + h_{2}$}};
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\draw[black!40!white, very thin] (6.0, 0.6) -- (6.0,-0.1);
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\node[blue!40!white] at (6.0, -1.3){\tiny{$y(h_{1})$}};
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\draw[blue!40!white, very thin] (6.0, -0.1) -- (6.0, -1.1);
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\node[black!40!white] at (5.0, 0.9) {\tiny{$h_{0} + h_{2}$}};
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\draw[black!40!white, very thin] (5.0, 0.8) -- (5.0,-0.1);
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\node[blue!40!white] at (5.0, -1.1){\tiny{$y(h_{0})$}};
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\draw[blue!40!white, very thin] (5.0, -0.1) -- (5.0, -0.9);
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\node[black!40!white] at (2.6, 0.9) {\tiny{$h_{0} + h_{1}$}};
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\draw[black!40!white, very thin] (2.6, 0.8) -- (2.6,-0.1);
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\draw[blue] (0.8, 0) -- (0.8, -0.2) -- (0.8, -0.2) -- (0.8, 0);
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\node[blue] at (0.8, -0.5) {$y(0)$};
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\draw[blue] (1.8, 0) -- (1.8, -0.2) -- (2.6, -0.2) -- (2.6, 0);
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\node[blue] at (1.8, -0.5) {$y(0)$};
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\node[blue] at (2.6, -0.5) {$y(h_{0})$};
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\draw[blue] (4.2, 0) -- (4.2, -0.2) -- (6.800000000000001, -0.2) -- (6.800000000000001, 0);
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\node[blue] at (4.2, -0.5) {$y(0)$};
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\node[blue] at (6.800000000000001, -0.5) {$y(h_{0} + h_{1})$};
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\draw[blue] (8.7, 0) -- (8.7, -0.2) -- (15.5, -0.2) -- (15.5, 0);
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\node[blue] at (8.7, -0.5) {$y(0)$};
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\node[blue] at (15.5, -0.5) {$y(h_{0} + h_{1} + h_{2})$};
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\draw[thick] (0,0) -- (17.5,0);
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\end{tikzpicture}
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}
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\begin{itemize}
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\item % Step 1
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Let $G_0 \coloneqq [y(0)]$
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and let $N_0$ be such that
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@ -152,10 +254,10 @@ We do a second proof of \yaref{thm:hindman}:
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\item proximal $\leadsto$ $\forall N$.~$T^n(x)\defon{N} = T^n(y)\defon{N}$
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for infinitely many $n$.
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\item uniform recurrence $\leadsto$
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\[
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\forall n .~\exists N.~\forall r.y\defon{\{r,\ldots,r+N-1\}}
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\text{ contains } y\defon{\{0,\ldots,n\}} \text{ as a subsequence.}
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\]
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\begin{IEEEeqnarray*}{rl}
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\forall n .~\exists N.~\forall r.~&y\defon{\{r,\ldots,r+N-1\}}\\
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&\text{ contains } y\defon{\{0,\ldots,n\}} \text{ as a subsequence.}
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\end{IEEEeqnarray*}
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(consider neighbourhood $G_n = \{z \in X : z\defon{n} = y\defon{n} \}$).
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\end{itemize}
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\item Consider $c \coloneqq y(0)$. This color works:
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@ -101,20 +101,7 @@ let
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\item Then $[T]$ is compact:
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\todo{TODO}
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% Let $\langle s_n, n <\omega \rangle$
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% be a Cauchy sequence in $[T]$.
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% Then for every $m < \omega$
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% there exists an $N < \omega$ such that
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% $s_n\defon{m} = s_{n'}\defon{m}$
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% for all $n, n' > N$.
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% Thus there exists a pointwise limit $s$ of the $s_n$.
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% Since for all $m$ we have $s\defon{m} = s_n\defon{m} \in [T]$
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% for $m$ large enough,
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% we get $s \in [T]$.
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% Hence $[T]$ is sequentially compact.
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% https://alanmath.wordpress.com/2011/06/16/on-trees-compactness-and-finite-splitting/
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\end{enumerate}
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\nr 2
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\subsection{Sheet 10}
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\todo{Copy from Abdelrahman and Shiguma}
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\nr 2
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\todo{Def skew shift flow (on $(\R / \Z)^2$!)}
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