From 774686637362465e95450a126b33f850d551bed7 Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Sat, 3 Feb 2024 02:02:09 +0100 Subject: [PATCH] ultrafilter limit --- inputs/lecture_08.tex | 4 ++-- inputs/lecture_24.tex | 16 ++++++++-------- inputs/lecture_25.tex | 6 +++--- inputs/lecture_26.tex | 2 +- inputs/lecture_27.tex | 8 +++----- logic.sty | 1 + 6 files changed, 18 insertions(+), 19 deletions(-) diff --git a/inputs/lecture_08.tex b/inputs/lecture_08.tex index 898ceb0..97d494d 100644 --- a/inputs/lecture_08.tex +++ b/inputs/lecture_08.tex @@ -93,8 +93,8 @@ where $X$ is a metrizable, usually second countable space. \end{proof} \begin{remark} Since $2^{\omega}$ embeds - into any uncountable polish space $Y$ - such that the image is closed, + into any uncountable polish space $Y$, + % such that the image is closed, we can replace $2^{\omega}$ by $Y$ in the statement of the theorem.% \footnote{By definition of the subspace topology diff --git a/inputs/lecture_24.tex b/inputs/lecture_24.tex index 87185cd..09b6a77 100644 --- a/inputs/lecture_24.tex +++ b/inputs/lecture_24.tex @@ -59,14 +59,14 @@ there is a unique $x \in X$, such that \[ - (\cU_n) (x_n \in G) + (\cU n) (x_n \in G) \] for every neighbourhood% \footnote{$G \subseteq X$ is a neighbourhood iff $x \in \inter G$.} $G$ of $x$. \end{lemma} \begin{notation} - In this case we write $x = \cU-\lim_n x_n$. + In this case we write $x = \ulim{\cU}_n x_n$. \end{notation} \begin{refproof}{lem:ultrafilterlimit}[sketch] Whenever we write $X = Y \cup Z$ @@ -101,13 +101,13 @@ This gives an operation on principal ultrafilters (we identify $n \in \N$ with the corresponding principal filter). We want to extend this to all of $\beta\N$. Fix the first argument to get a function $\N \to \N, n \mapsto k+n$. -For $\cU \in \beta\N$ consider $\cU-\lim_n (k+n)$. +For $\cU \in \beta\N$ consider $\ulim{\cU}_n (k+n)$. So for a fixed $k \in \N$ we get $k+ \cdot \colon\beta\N \to \beta\N$, i.e.~$+ \colon \N \times \beta\N \to \beta\N$. Fixing the second coordinate to be $\cV \in \beta\N$, we get a function $+\cV \colon \N \to \beta\N$. For $ \cU \in \beta\N$ -consider $\cU-\lim_n n + \cV$. +consider $\ulim{\cU}_n n + \cV$. This gives $+ \colon \beta\N \times \beta\N \to \beta\N$. % TODO ? @@ -130,7 +130,7 @@ and let $T \colon X \to X$ be continuous.% but not a $\Z$-action.} For any $\cU \in \beta\N$, we define $T^{\cU}$ by -$T^\cU(x) \coloneqq \cU-\lim_n T^n(x)$ for $x \in X$. +$T^\cU(x) \coloneqq \ulim{\cU}_n T^n(x)$ for $x \in X$. For fixed $x$, the map $\cU \mapsto T^{\cU}(x)$ is continuous. @@ -165,9 +165,9 @@ is not necessarily continuous. \end{fact} \begin{proof} \begin{IEEEeqnarray*}{rCl} - T^{\cU + \cV}(x) &=& (\cU + \cV)-\lim_k T^k(x)\\ - &=& \cU-\lim_m \cV-\lim_n T^{m+n}(x)\\ - &\overset{T^m \text{ continuous}}{=}& \cU-\lim_m T^m (\cV-\lim_n T^n(x))\\ + T^{\cU + \cV}(x) &=& \ulim{(\cU + \cV)}_k T^k(x)\\ + &=& \ulim{\cU}_m \ulim{\cV}_n T^{m+n}(x)\\ + &\overset{T^m \text{ continuous}}{=}& \ulim{\cU}_m T^m (\ulim{\cV}_n T^n(x))\\ &=& T^\cU(T^\cV(x)). \end{IEEEeqnarray*} \end{proof} diff --git a/inputs/lecture_25.tex b/inputs/lecture_25.tex index 0703780..e92aff3 100644 --- a/inputs/lecture_25.tex +++ b/inputs/lecture_25.tex @@ -86,7 +86,7 @@ Let $\beta\N$ denote the set of ultrafilters on $\N$. For every compact Hausdorff space $X$, a sequence $(x_n)$ in $X$, and $\cU \in \beta\N$, - we have that $\cU-\lim_n x_n = x$ + we have that $\ulim{\cU}_n x_n = x$ exists and is unique, i.e.~for all $x \in G \overset{\text{open}}{\subseteq} X$ we have $\{n \in \N : x_n \in G\} \in \cU$. @@ -110,7 +110,7 @@ Let $\beta\N$ denote the set of ultrafilters on $\N$. \end{IEEEeqnarray*} since $B_1 \cup \ldots \cup B_m \in \cU \iff \exists i < m.~B_i \in \cU$. - It is clear that $\cU-\lim_n x_n$ + It is clear that $\ulim{\cU}_n x_n$ is unique, since $X$ is Hausdorff. \end{proof} @@ -123,7 +123,7 @@ Let $\beta\N$ denote the set of ultrafilters on $\N$. Let \begin{IEEEeqnarray*}{rCl} \tilde{f}\colon \beta\N &\longrightarrow & X \\ - \cU &\longmapsto & \cU-\lim_n f(n). + \cU &\longmapsto & \ulim{\cU}_n f(n). \end{IEEEeqnarray*} \todo{Exercise: Check that $\tilde{f}$ is continuous.} diff --git a/inputs/lecture_26.tex b/inputs/lecture_26.tex index f4342ac..d075850 100644 --- a/inputs/lecture_26.tex +++ b/inputs/lecture_26.tex @@ -189,7 +189,7 @@ we need the following: (\cV n) T^n(x) \in \underbrace{T^{-k}(\overbrace{G}^{\text{closed}})}_{\text{closed}}. \] - Therefore $\cV-\lim_n T^n(x) \in T^{-k}(G)$. + Therefore $\ulim{\cV}_n T^n(x) \in T^{-k}(G)$. So $T^k(T^\cV(x)) \in G \subseteq G_0$. diff --git a/inputs/lecture_27.tex b/inputs/lecture_27.tex index d7ac9da..e825c2a 100644 --- a/inputs/lecture_27.tex +++ b/inputs/lecture_27.tex @@ -30,7 +30,7 @@ \] Thus \[ - \underbrace{\cV-\lim_n T^n(x)}_{T^\cV(x)} \not\in T^{-k}(G). + \underbrace{\ulim{\cV}_n T^n(x)}_{T^\cV(x)} \not\in T^{-k}(G). \] We get that \[ @@ -48,8 +48,8 @@ $S \colon \beta\N \to \beta\N$, $S(\cU ) = \hat{1}+ \cU$. Then \[ -S^\cV(\cU) = \cV-\lim_n S^n(\cU) = \cV-\lim_n(\hat{n} + \cU) = -\cV-\lim_n \hat{n} + \cU = \cV + \cU. +S^\cV(\cU) = \ulim{\cV}_n S^n(\cU) = \ulim{\cV}_n(\hat{n} + \cU) = +\ulim{\cV}_n \hat{n} + \cU = \cV + \cU. \] % TODO check @@ -75,8 +75,6 @@ S^\cV(\cU) = \cV-\lim_n S^n(\cU) = \cV-\lim_n(\hat{n} + \cU) = \[ \forall \cU \in I.~\forall \cV \in \beta\N.~\cV + \cU \in I. \] - - \end{definition} \begin{theorem} diff --git a/logic.sty b/logic.sty index 556746b..057fb5c 100644 --- a/logic.sty +++ b/logic.sty @@ -151,6 +151,7 @@ \DeclareSimpleMathOperator{proj} \newcommand{\fc}{\mathfrak{c}} \DeclareMathOperator{\acts}{\curvearrowright} +\newcommand{\ulim}[1]{\mathop{#1\text{-lim}}\limits} \newcommand\lecture[3]{\hrule{\color{darkgray}\hfill{\tiny[Lecture #1, #2]}}} \newcommand\tutorial[3]{\hrule{\color{darkgray}\hfill{\tiny[Tutorial #1, #2]}}}