diff --git a/inputs/lecture_24.tex b/inputs/lecture_24.tex index 7a0c2a8..87185cd 100644 --- a/inputs/lecture_24.tex +++ b/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.} diff --git a/inputs/lecture_25.tex b/inputs/lecture_25.tex new file mode 100644 index 0000000..2b8ea9a --- /dev/null +++ b/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. + + + + diff --git a/logic3.tex b/logic3.tex index b8548c2..99489d9 100644 --- a/logic3.tex +++ b/logic3.tex @@ -51,6 +51,7 @@ \input{inputs/lecture_22} \input{inputs/lecture_23} \input{inputs/lecture_24} +\input{inputs/lecture_25} \cleardoublepage