ultrafilter limit
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Josia Pietsch 2024-02-03 02:02:09 +01:00
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6 changed files with 18 additions and 19 deletions

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@ -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

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@ -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}

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@ -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.}

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@ -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 $T^k(T^\cV(x)) \in G \subseteq G_0$.

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@ -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}

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@ -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]}}}