Josia Pietsch
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365 lines
15 KiB
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365 lines
15 KiB
TeX
\lecture{26}{2024-01-30}{}
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Let $T\colon X \to X$ be a continuous map.
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This gives $\N \acts X$.
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\begin{definition}
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\label{def:unifrec}
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A point $x \in X$
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is called \vocab{uniformly recurrent}
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iff
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for each neighbourhood $G$ of $x$,
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there is $M \in \N_+$,
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such that
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\[
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\forall n \in \N.~\exists k < m.~T^{n+k}(x) \in G.
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\]
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\end{definition}
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\begin{definition}
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A pair $x,y \in X$ is \vocab{proximal}%
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\footnote{see also \yaref{def:flow}, where we defined proximal for metric spaces}
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iff for all neighbourhoods $G$ of
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the diagonal%
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\gist{\footnote{recall that the diagonal is defined to be $\Delta \coloneqq \{(x,x) : x \in X\}$}}{}
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infinitely many $n$ satisfy $(T^n(x), T^n(y)) \in G$.
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\end{definition}
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\begin{theorem}
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\label{thm:unifrprox}
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Let $X$ be a compact Hausdorff space and $T\colon X \to X$
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continuous.
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Consider $(X,T)$.%TODO different notations
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Then for every $x \in X$
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there is a uniformly recurrent $y \in X$
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such that $y $ is proximal to $x$.
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\end{theorem}
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We do a second proof of \yaref{thm:hindman}:
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\begin{refproof}{thm:hindmanfurstenberg}[Furstenberg]
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\gist{%
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A partition of $\N$ into $k$-many pieces can be viewed
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as a function $f\colon \N \to k$.
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Let $X = k^\N$ be the set of all such functions.
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Equip $X$ with the product topology.
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Then $X$ is compact and Hausdorff.
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Let $T\colon X \to X$ be the shift
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given by \gist{%
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\begin{IEEEeqnarray*}{rCl}
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T\colon k^{\N} &\longrightarrow & k^{\N} \\
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(y\colon \N \to k)&\longmapsto &
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\begin{pmatrix*}[l]
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\N &\longrightarrow & k \\
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n &\longmapsto & y(n+1),
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\end{pmatrix*}
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\end{IEEEeqnarray*}
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i.e.~}{}%
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$T(y)(n) = y(n+1)$.
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Let $x $ be the given partition.
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We want to find an infinite set $H$ for $x$ as in the theorem.
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Let $y$ be uniformly recurrent and proximal to $x$.
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\begin{itemize}
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\item % Gist: proximal
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Since $x$ and $y$ are proximal,
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we get that for every $N \in \N$,
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there are infinitely many $n$ such that
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$T^n(x)\defon{N} = T^n(y)\defon{N}$.%
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\footnote{%
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Consider $G_N = \{(a,b) \in X^2 : a\defon N = b\defon N\}$
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This is a neighbourhood of the diagonal.%
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}
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\item % Gist: unif. recurrent
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Consider the neighbourhood
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\[
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G_n \coloneqq \{z \in X: z\defon{n} = y\defon{n}\}
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\]
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of $y$.
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By the uniform recurrence of $y$,
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we get that%
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\footnote{Note that here we might need to choose
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a bigger $N$ than the $M$ in \yaref{def:unifrec},
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but $2M$ suffices.}%
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\begin{IEEEeqnarray*}{rl}
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\forall n.~\exists N.~\forall r.~&(y(r), y(r+1), \ldots, y(r+N - 1)\\
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&\text{ contains } (y(0), y(1), \ldots, y(n)) \text{ as a subsequence.}\\
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\end{IEEEeqnarray*}
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\end{itemize}
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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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% 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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\[
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\forall r.~(y(r), \ldots, y(r + N_0 - 1)) \text{ contains $y(0)$.}
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\]
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By proximality, there exist infinitely many $r$
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such that $(y(r), \ldots, (y(r+N_0-1)) = (x(r), \ldots, x(r+N_0-1))$.
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Fix $h_0 \in \N$ such that $x(h_0) = y(0)$.
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\item%
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% Step 2
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Let $G_{n_0} = [(y(0), \ldots, y(h_0)]$.
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Choose $N_1$.
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For all $r$,
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$(y(r), \ldots, y(r+N-1))$ contains
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$(\underbrace{y(0)}_{= C}, \ldots, \underbrace{y(h_0)}_{= C})$.
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Pick $r > h_0$ such that
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$(x(r), \ldots, x(r+N-1))$ contains
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$(y(0), \ldots y(h_0))$.
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Let $(x(r+s), \ldots, x(r+s+h_0)) = (y(0), \ldots, y(h_0))$.
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Then set $h_1 = r + s$.
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Then $x(h_0) = c$, $x(h_1) = y(0) = c$
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and $x(h_0+h_1) = y(h_0) = c$.
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\item%
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% Step 3
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Let $G_{h_0 + h_1} = [y(0), \ldots, y(h_0+h_1)]$.
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Let $r > h_0 + h_1$.
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Choose $N_2$ large enough
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such that
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$(y(0), \ldots, y(h_0+h_1))$
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is contained in $(x(r), \ldots, x(r+N-1))$.
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Let $(y(0), \ldots, y(h_0+h_1)) = (x(r+s), \ldots, x(r+s+N-1))$.
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\item Repeat this:
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Inductively choose $h_i$ such that $x(s+h_i) = y(s+h_i) = c$
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for all sums $s$ of subsets of $\{h_0,\ldots,h_{i-1}\}$.
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To do this, find $N_i$ such that every $N_i$ consecutive
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terms of $y$ contain $(y(0), \ldots, y(\sum_{j < i} h_j))$.
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Then find $h_i > h_{i-1}$ such that
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$(x(h_i), \ldots, x(\sum_{j < i} h_j)) = (y(0), \ldots, y(\sum_{j < i} h_j))$.
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\end{itemize}
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}{
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\begin{itemize}
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\item View partition as $f\colon \N \to k$. Consider $X \coloneqq k^{\N}$ (product topology, compact and Hausdorff).
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Let $x \in X$ be the given partition.
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\item $T\colon X \to X$ shift: $T(y)(n) \coloneqq y(n+1)$.
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\item Let $y$ proximal to $x$, uniformly recurrent.
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\begin{itemize}
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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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\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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\begin{itemize}
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\item $G_0 \coloneqq y\defon{\{0\}}$,
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take $N_0$ such that $y\defon{\{r, \ldots, r + N_0 - 1\}} $
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contains $y(0)$ for all $r$ (unif.~recurrence).
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$y\defon{\{r,\ldots,r+N_0 - 1\} } = x\defon{\{r,\ldots,r+N_0 -1\} }$
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for infinitely many $r$ (proximality).
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Fix $h_0 \in \N$ such that $x(h_0) = y(0)$.
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\item $G_1 \coloneqq y\defon{\{0,\ldots,h_0\} }$,
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take $N_1$ such that $y\defon{\{r,\ldots,r +N_1-1\}}$
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contains $y\defon{\{0,\ldots,h_0\} }$
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for all $r$ (unif.~recurrence).
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So among ever $N_1$ terms, there are two of distance $h_0$
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where $y$ has value $c$.
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So $\exists h_1 > h_0$ such that $x(h_1) = x(h_1 + h_0) = c$
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(proximality).
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\item Repeat:
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Choose $h_i$ such that
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for all sums $s$ of subsets of $\{h_0,\ldots, h_{i-1}\}$,
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$x(s+h_i) = y(s+h_i) = c$:
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Find $N_i$ such that every $N_i$ consecutive
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terms of $y$ contain a segment that coincides
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with the initial segment of $y$
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up to the largest $s$,
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then find a segment of length $N_i$ beyond $h_{i-1}$
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where $x$ and $y$ coincide.
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\end{itemize}
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\end{itemize}
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}
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\end{refproof}
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In order to prove \yaref{thm:unifrprox},
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we need to rephrase the problem in terms of $\beta\N$:
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\begin{theorem}
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\label{thm:unifrprox:helper}
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Let $X$ be a compact Hausdorff space. % ?
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Let $T\colon X \to X$ be continuous.
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\begin{enumerate}[(1)]
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\item $x \in X$ is recurrent
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iff $T^\cU(x) = x$ for some $\cU \in \beta\N \setminus \N$.
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\item $x \in X$ is uniformly recurrent
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iff for every $\cV \in \beta\N$,
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there is $\cU \in \beta\N$
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with $T^{\cU}(T^{\cV}(x)) = x$.
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\item $x, y \in X$ are proximal
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iff there is $\cU \in \beta\N$
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such that $T^\cU(x) = T^\cU(y)$.
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% TODO compare with the statement for the ellis semigroup.
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\end{enumerate}
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\end{theorem}
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\gist{%
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\begin{refproof}{thm:unifrprox:helper}[sketch]
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We only prove (2) here, as it is the most interesting point.%
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\todo{other parts will be in the official notes}
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\begin{subproof}[(2), $\implies$]
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Suppose that $ x$ is uniformly recurrent.
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Take some $\cV \in \beta\N$.
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Let $G_0$ be a neighbourhood of $x$.
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Then $x \in G \subseteq G_0$,
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where $G$ is a closed neighbourhood,
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i.e.~$X \in \inter G$.
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Let $M$ be such that
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\[
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\forall n .~ \exists k < M.~T^{n+k}(x) \in G.
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\]
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So there is a $k < M$ such that
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\[
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(\cV n) T^{n +k}(x) \in G.
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\]
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Hence
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\[
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(\cV n) T^n(x) \in \underbrace{T^{-k}(\overbrace{G}^{\text{closed}})}_{\text{closed}}.
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\]
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Therefore $\ulim{\cV}_n T^n(x) \in T^{-k}(G)$.
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So $T^k(T^\cV(x)) \in G \subseteq G_0$.
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We have shown that for every open neighbourhood $G$ of $x$,
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the set $Y_G = \{k \in \N : T^k(T^\cV(x)) \in G\} \neq \emptyset$.
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The sets $\{Y_G : G \text{ open neighbourhood of } G\}$
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form a filter basis,\footnote{The sets and their supersets form a filter.}
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since $Y_{G_1} \cap Y_{G_2} = Y_{G_1 \cap G_2}$.
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Let $\cU$ be an ultrafilter containing all the $Y_G$.
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Then
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\[
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(\cU k) R^k(T^\cV)(x)) \in G
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\]
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i.e.~$T^\cU(T^\cV(x)) \in \overline{G}$.
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Since we get this for every neighbourhood,
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it follows that $T^\cU ( T^\cV(x)) = x$.
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\end{subproof}
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\phantom\qedhere
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\end{refproof}
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}{}
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