lecture 25

inputs/lecture_24.tex

@@ -92,7 +92,9 @@ i.e.~the set of all ultrafilters on $\N$
 with the topology given by open sets $V_{A} = \{ p \in \beta\N : A \in P\} $
 for $A \subseteq  \N$.
 
-This is a compact Hausdorff space.\todo{Homework}
+This is a compact Hausdorff space.%
+\footnote{cf.~\yaref{fact:bNhd}, \yaref{fact:bNcompact}}%
+\todo{move facts}
 We can turn it into a compact semigroup:
 Consider $+ \colon \N \times  \N \to  \N$.
 This gives an operation on principal ultrafilters
@@ -169,8 +171,6 @@ is not necessarily continuous.
                          &=& T^\cU(T^\cV(x)).
     \end{IEEEeqnarray*}
 \end{proof}
-
-
 \todo{Homework: Check the details that were omitted during the lecture.}
 
 

inputs/lecture_25.tex

@@ -0,0 +1,237 @@
+\lecture{25}{2024-01-26}{}
+
+Let $\beta\N$ denote the set of ultrafilters on $\N$.
+\begin{fact}
+    \begin{itemize}
+        \item This is a topological space,
+            where a basis consist of sets
+            $V_A \coloneqq \{p \in \beta\N : A \in p\}, A \subseteq \N$.
+    
+            (For $A, B \subseteq \N$ we have $V_{A \cap B} = V_{A} \cap V_B$
+            and $\beta\N = V_\N$.)
+    
+        \item Note also that for $A, B \subseteq \N$,
+            $V_{A \cup B} = V_A \cup V_B$,
+            $V_{A^c} = \beta\N \setminus V_A$.
+    \end{itemize}
+\end{fact}
+
+\begin{observe}
+    \label{ob:bNclopenbasis}
+    Note that the basis is clopen. In particular
+    any closed set can be written as an intersection of sets
+    of the form  $V_A$:
+
+    If $F$ is closed, then $U = \beta\N \setminus F = \bigcup_{i\in I} V_{A_i}$,
+    so $F = \bigcap_{i \in I} V_{\N \setminus A_i}$.
+\end{observe}
+
+\begin{fact}
+    \label{fact:bNhd}
+    $\beta\N$ is Hausdorff.
+\end{fact}
+\begin{proof}
+    Let $\cU \neq  \cV \in \beta\N$.
+    Then there is some $A \in \cU \setminus \cV$,
+    so $A^c \in \cV$,
+    so $\cU \in V_A$ and $\cV \in V_A^c$.
+\end{proof}
+%\begin{remark}+
+%    This even shows that  $\beta\N$ is totally separated.
+%    In fact, $\beta\N$ is a profinite space,
+%    as the next fact shows.
+%\end{remark}
+\begin{fact}
+    \label{fact:bNcompact}
+    $\beta\N$ is compact.
+\end{fact}
+\begin{proof}
+    Let $\{F_i\}_{i \in I}$ be non-empty
+    and closed
+    such that
+    for any $i_1,\ldots., i_k \in I$,
+    $k \in \N$,
+    $\bigcap_{j=1}^k F_{i_j} \neq \emptyset$.
+
+    We need to show that $\bigcap_{i \in I} F_i \neq \emptyset$.
+    Replacing each $F_i$ by $V_{A_j^i}$ such
+    that $F_i = \bigcap_{j \in J_i} V_{A_j^i}$
+    (cf.~\yaref{ob:bNclopenbasis})
+    we may assume that $F_i$ is of the form $V_{A_i}$.
+    We get $\{F_i = V_{A_i} : i \in I\}$
+    with the finite intersection property.
+    Hence
+    $\{A_i : i \in I\} \mathbin{\text{\reflectbox{$\coloneqq$}}} \cF_0$
+    has the finite intersection property.
+
+    Then $\cF = \{A \subseteq  \N : A \supseteq A_{i_1} \cap \ldots \cap A_{i_k}, k \in \N, i_1, \ldots, i_k \in I\}$
+    is a filter.
+
+    Let $\cU$ be an ultrafilter extending $\cF$.
+    Then  $\cU \in \bigcap_{i \in I} V_{A_i} = \bigcap_{i \in I} F_i$.
+\end{proof}
+\begin{fact}
+    Consider $\N$ as a subspace of $\beta\N$
+    via $\N \hookrightarrow \beta\N, n \mapsto \hat{n} \coloneqq \{A \subseteq \N : n \in A\}$.
+    Then
+    \begin{itemize}
+        \item $ \{\hat{n}\} $ is open in $\beta\N$ for all $n \in \N$.
+        \item $\N \subseteq  \beta\N$ is dense.
+    \end{itemize}
+    \todo{Easy exercise}
+    % TODO write down (exercise)
+\end{fact}
+
+\begin{theorem}
+    For every compact Hausdorff space $X$,
+    a sequence $(x_n)$ in $X$,
+    and $\cU \in \beta\N$,
+    we have that $\cU-\lim_n x_n = x$
+    exists and is unique,
+    i.e.~for all  $x \in G \overset{\text{open}}{\subseteq} X$
+    we have $\{n \in \N : x_n \in G\}  \in \cU$.
+\end{theorem}
+\begin{proof}
+    Towards a contradiction
+    assume that there is no such $x$.
+
+    For every $x$ take $x \in G_x \overset{\text{open}}{\subseteq} X$
+    such that
+    $\{ n \in \N : x_n \in G_x\} \not\in  \cU$.
+    So $\{G_x\}_{x \in X}$ is an open cover of $X$.
+    Since $X$ is compact,
+    there exists a finite subcover
+    $G_{x_1}, \ldots, G_{x_m}$.
+    
+    But then
+    \begin{IEEEeqnarray*}{rCl}
+        \N &=& \{ n \in \N : x_n \in \bigcup_{i=1}^m G_{x_i}\}\\
+           &=& \underbrace{\bigcup_{i=1}^m \overbrace{\{n \in \N : x_n \in G_{x_i}\}}^{\not\in \cU}}_{\not\in \cU},
+    \end{IEEEeqnarray*}
+    since $B_1 \cup \ldots \cup B_m \in \cU \iff \exists  i < m.~B_i \in \cU$.
+
+    It is clear that  $\cU-\lim_n x_n$
+    is unique, since  $X$ is Hausdorff.
+\end{proof}
+
+\begin{theorem}
+    Let $X$ be a compact Hausdorff space.
+    For any $f\colon  \N \to X$
+    there is a unique continuous extension $\tilde{f}\colon \beta\N \to X$.
+\end{theorem}
+\begin{proof}
+    Let
+    \begin{IEEEeqnarray*}{rCl}
+        \tilde{f}\colon \beta\N &\longrightarrow & X \\
+        \cU &\longmapsto & \cU-\lim_n f(n).
+    \end{IEEEeqnarray*}
+    \todo{Exercise: Check that $\tilde{f}$ is continuous.}
+
+    $\tilde{f}$ is uniquely determined,
+    since $\N \subseteq \beta\N$ is dense.
+    % TODO general fact: continuous functions agreeing on a dense set
+    % agree everywhere (fact section)
+\end{proof}
+
+% RECAP
+\gist{%
+$\beta\N$ is equipped with $+$ which extends $+\colon  \N \times \N \to \N$,
+\[
+ \cU + \cV = \{A \subseteq  \N : (\cU m)\left( (\cU n) \{m+n \in A\}  \right)\}.
+\]
+This is associative, but not commutative.
+}{}
+% END RECAP
+\begin{fact}
+    $+\colon \beta\N\times \beta\N \to \beta\N$
+    is left continuous,
+    i.e.~for $\cV$ fixed,
+    $\cU \mapsto \cU + \cV$ is continuous.
+\end{fact}
+\begin{proof}
+    Fix $A$ and consider $V_A$.
+    We need to show that the inverse image of  $V_A$ is open.
+
+    We have
+    \begin{IEEEeqnarray*}{rCl}
+        \cU + \cV \in V_A &\iff& A \in  \cU + \cV\\
+                          &\iff& (\cU_m)(\cV_n) \{m+n \in A\}\\
+                          &\iff& \{m \in \N : (\cV n) m+n \in A\} \in \cU\\
+                          &\iff& \cU \in V_{\{m \in \N: (\cV n) m+n \in A\}}.
+    \end{IEEEeqnarray*}
+\end{proof}
+\begin{corollary}
+    $(\beta\N,+)$ 
+    is a %(left)
+    \vocab{compact semigroup},
+    i.e.~it is comapct, Hausdorff, associative and left-continuous.%
+    %\footnote{There is no convention on left and right.}
+\end{corollary}
+So we can apply the \yaref{lem:ellisnumakura}
+to obtain
+\begin{corollary}
+    There is $\cU \in \beta\N$
+    such that $\cU + \cU = \cU$.
+\end{corollary}
+\begin{observe}
+    Principal ultrafilters $\neq \hat{0}$ are not idempotent.
+    We can restrict to $\beta\N \setminus \N$
+    to get an idempotent element that is not principal.
+    % TODO THINK ABOUT THIS
+\end{observe}
+
+\begin{theorem}[Hindman]
+    If $\N$ is partitioned into finitely many
+    sets,
+    then there is is an infinite subset $H \subseteq \N$
+    such that all finite sums of distinct
+    elements of $H$
+    belong to the same set of the partition.
+\end{theorem}
+\begin{proof}[Galvin,Glazer]
+    Let $\cU \in \beta\N \setminus \N$
+    be such that $\cU + \cU = \cU$.
+    Let $P$ be the piece of the partition
+    that is in $\cU$.
+    So $(\cU n ) n \in P$.
+    Let us define a sequence $x_1,x_2,\ldots$
+    \begin{itemize}
+        \item $\cU$ is idempotent,
+            so $(\cU n)(\cU k) n+k \in P$.
+            We get
+            \[(\cU n) \left( n \in P \land (\cU_k) n+k \in P \right)\].
+            Pick $x_1$ that satisfies this,
+            i.e.~$x_1 \in P$ and $(\cU_k) x_1+k \in P$.
+        \item $\cU$ is idempotent,
+            so
+            \[
+                (\cU n)[
+                    n \in P
+                    \land (\cU_k) n + k \in P
+                    \land x_1 + n \in P
+                    \land (\cU_k) x_1 + n + k \in P
+                ]
+            \]
+            Take $x_2 > x_1$ that satisfies this.
+        \item Suppose we have chosen $\langle x_i : i < n \rangle$.
+            Since $\cU$ is idempotent, we have
+            \[
+                (\cU n)[
+                    n \in P
+                    \land (\cU_k) n + k \in  P
+                    \land \forall {I \subseteq n}.~ (\sum_{i \in I} x_i + n \in P)
+                    \land (\cU_k)\left( \forall {I \subseteq n}.~ (\sum_{i \in I} x_i + n + k) \in P\right).
+            \]
+            Chose $x_n > x_{n-1}$ that satisfies this.
+    \end{itemize}
+    Set $H \coloneqq \{x_i : i < \omega\}$.
+
+    
+\end{proof}
+
+
+Next time we'll see another proof of this theorem.
+
+
+
+

logic3.tex

@@ -51,6 +51,7 @@
 \input{inputs/lecture_22}
 \input{inputs/lecture_23}
 \input{inputs/lecture_24}
+\input{inputs/lecture_25}
 
 \cleardoublepage