diff --git a/inputs/lecture_17.tex b/inputs/lecture_17.tex index c9b265d..79ced2a 100644 --- a/inputs/lecture_17.tex +++ b/inputs/lecture_17.tex @@ -127,7 +127,7 @@ since $X^X$ has these properties. \begin{lemma}[Ellis–Numakura] \yalabel{Ellis-Numakura Lemma}{Ellis-Numakura}{lem:ellisnumakura} - Every compact semigroup + Every non-empty compact semigroup contains an \vocab{idempotent} element, i.e.~$f$ such that $f^2 = f$. \end{lemma} diff --git a/inputs/lecture_23.tex b/inputs/lecture_23.tex index aedf7f5..6068d5d 100644 --- a/inputs/lecture_23.tex +++ b/inputs/lecture_23.tex @@ -37,10 +37,9 @@ Let $I$ be a linear order S & \coloneqq & \{ x \in \LO(\N) :& x \text{ has a least element},\\ &&& \text{for any $t$, there is $t \oplus 1$, the successor of $t$.}\} \end{IEEEeqnarray*} - \todo{Exercise sheet 12} - $S$ is Borel. + $S$ is Borel.\footnote{cf.~\yaref{s12e1}} - We will % TODO ? + We will construct a reduction \begin{IEEEeqnarray*}{rCl} M \colon S &\longrightarrow & C(\mathbb{K}^\N,\mathbb{K})^\N. %\\ diff --git a/inputs/lecture_24.tex b/inputs/lecture_24.tex index 366e135..958fa5f 100644 --- a/inputs/lecture_24.tex +++ b/inputs/lecture_24.tex @@ -1,10 +1,10 @@ \lecture{24}{2024-01-23}{Combinatorics!} +% ANKI 2 + \subsection{Applications to Combinatorics} % Ramsey Theory} -% TODO Define Ultrafilter - \begin{definition} An \vocab{ultrafilter} on $\N$ (or any other set) is a family $\cU \subseteq \cP(\N)$ @@ -44,6 +44,7 @@ for $\{ n \in \N : \phi(n)\} \in \cU$. We say that $\phi(n)$ holds for \vocab{$\cU$-almost all} $n$. \end{notation} +\gist{% \begin{observe} Let $\phi(\cdot )$, $\psi(\cdot )$ be formulas. @@ -53,6 +54,7 @@ \item $(\cU n) ~\lnot \phi(n) \iff \lnot (\cU n)~ \phi(n)$. \end{enumerate} \end{observe} +}{} \begin{lemma} \label{lem:ultrafilterlimit} Let $X $ be a compact Hausdorff space. @@ -72,6 +74,8 @@ \end{notation} \begin{refproof}{lem:ultrafilterlimit}\footnote{The proof from the lecture only works for metric spaces.} +\gist{ + For metric spaces: Whenever we write $X = Y \cup Z$ we have $(\cU n) x_n \in Y$ or $(\cU n) x_n \in Z$. @@ -86,8 +90,13 @@ $C \in \cP_{n+1} \implies \exists C \subseteq D \in \cP_{n}$ and $C_1 \supseteq C_2 \supseteq \ldots$, $C_i \in \cP_i $ $\implies | \bigcap_{i} C_i| = 1$. - It is clear that we can do this for metric spaces, - but such partition can be found for compact Hausdorff spaces as well. + It is clear that we can do this for metric spaces. + +}{} + See \yaref{thm:uflimit} for the full proof. + See + \yaref{fact:compactiffufconv} and + \yaref{fact:hdifffilterlimit} for a more general statement. \end{refproof} Let $\beta \N$ be the Čech-Stone compactification of $\N$, @@ -157,7 +166,6 @@ is not necessarily continuous. \[ \forall n.~\exists k < M.~ T^{n+k}(x) \in G. \] - \end{definition} \begin{fact} Let $\cU, \cV \in \beta\N$ diff --git a/inputs/lecture_25.tex b/inputs/lecture_25.tex index b2e1ff0..27acec3 100644 --- a/inputs/lecture_25.tex +++ b/inputs/lecture_25.tex @@ -7,15 +7,17 @@ Let $\beta\N$ denote the set of ultrafilters on $\N$. where a basis consist of sets $V_A \coloneqq \{p \in \beta\N : A \in p\}, A \subseteq \N$. + \gist{% (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} - +\gist{% \begin{observe} \label{ob:bNclopenbasis} Note that the basis is clopen. In particular @@ -25,6 +27,7 @@ Let $\beta\N$ denote the set of ultrafilters on $\N$. 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} @@ -54,12 +57,14 @@ Let $\beta\N$ denote the set of ultrafilters on $\N$. $\bigcap_{j=1}^k F_{i_j} \neq \emptyset$. We need to show that $\bigcap_{i \in I} F_i \neq \emptyset$. + \gist{% 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. + }{Wlog.~$F_i = V_{A_i}$.} Hence $\{A_i : i \in I\} \mathbin{\text{\reflectbox{$\coloneqq$}}} \cF_0$ has the finite intersection property. @@ -78,11 +83,10 @@ Let $\beta\N$ denote the set of ultrafilters on $\N$. \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} + \label{thm:uflimit} For every compact Hausdorff space $X$, a sequence $(x_n)$ in $X$, and $\cU \in \beta\N$, @@ -133,7 +137,6 @@ Let $\beta\N$ denote the set of ultrafilters on $\N$. % agree everywhere (fact section) \end{proof} \begin{trivial}+ - More generally, $\beta$ is a functor from the category of topological spaces to the category of compact Hausdorff spaces. It is left adjoint to the inclusion functor. @@ -222,13 +225,12 @@ to obtain 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 - \[ - (\cU n)[ - n \in P - \land (\cU_k) n + k \in P - \land \forall {I \subseteq n}.~ (\sum_{i \in I} x_i + n \in P) - \land (\cU_k)\left( \forall {I \subseteq n}.~ (\sum_{i \in I} x_i + n + k) \in P\right). - \] + \begin{IEEEeqnarray*}{rCl} + (\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). + \end{IEEEeqnarray*} Chose $x_n > x_{n-1}$ that satisfies this. \end{itemize} Set $H \coloneqq \{x_i : i < \omega\}$. @@ -237,6 +239,3 @@ to obtain Next time we'll see another proof of this theorem. - - - diff --git a/inputs/tutorial_02.tex b/inputs/tutorial_02.tex index b039cf5..32d26df 100644 --- a/inputs/tutorial_02.tex +++ b/inputs/tutorial_02.tex @@ -3,12 +3,36 @@ % Points: 15 / 16 \nr 1 -\todo{handwritten solution} +Let $(X,d)$ be a metric space and $\emptyset \neq A \subseteq X$. +Let $d(x,A) \coloneqq \inf(d(x,a) : a \in A\}$. + +\begin{itemize} + \item $d(-,A)$ is uniformly continuous: + + Clearly $|d(x,A) - d(y,A)| \le d(x,y)$. + \todo{Add details} + \item $d(x,A) = 0 \iff x \in \overline{A}$. + + $d(x,A) = 0$ iff there is a sequence in $A$ + converging towards $x$ iff $x \in \overline{A}$. +\end{itemize} + \nr 2 +Let $X$ be a discrete space. +For $f,g \in X^{\N}$ define +\[ +d(f,g) \coloneqq \begin{cases} + (1 + \min \{n: f(n) \neq g(n)\})^{-1} &: f \neq g,\\ + 0 &: f= g. +\end{cases} +\] + + \begin{enumerate}[(a)] - \item $d$ is an ultrametric: + \item $d$ is an \vocab{ultrametric}, + i.e.~$d(f,g) \le \max \{d(f,h), d(g,h)\}$ for all $f,g,h \in X^{\N}$ : Let $f,g,h \in X^{\N}$. @@ -70,10 +94,15 @@ \nr 3 +Consider $\N$ as a discrete space and $\N^{\N}$ with the product topology. +Let +\[ +S_{\infty} = \{f\colon \N \to \N \text{ bijective}\} \subseteq \N^{\N}. +\] \begin{enumerate}[(a)] \item $S_{\infty}$ is a Polish space: - From (2) we know that $\N^{\N}$ is Polish. + From \yaref{s1e2} we know that $\N^{\N}$ is Polish. Hence it suffices to show that $S_{\infty}$ is $G_{\delta}$ with respect to $\N^\N$. @@ -111,69 +140,9 @@ Clearly there cannot exist a finite subcover as $B$ is the disjoint union of the $B_j$. - % TODO Think about this \end{enumerate} \nr 4 - -% (uniform metric) -% -% \begin{enumerate}[(a)] -% \item $d_u$ is a metric on $\cC(X,Y)$: -% -% It is clear that $d_u(f,f) = 0$. -% -% Let $f \neq g$. Then there exists $x \in X$ with -% $f(x) \neq g(x)$, hence $d_u(f,g) \ge d(f(x), g(x)) > 0$. -% -% Since $d$ is symmetric, so is $d_u$. -% -% Let $f,g,h \in \cC(X,Y)$. -% Take some $\epsilon > 0$ -% choose $x_1, x_2 \in X$ -% with $d_u(f,g) \le d(f(x_1), g(x_1)) + \epsilon$, -% $d_u(g,h) \le d(g(x_2), h(x_2)) + \epsilon$. -% -% Then for all $x \in X$ -% \begin{IEEEeqnarray*}{rCl} -% d(f(x), h(x)) &\le & -% d(f(x), g(x)) + d(g(x), h(x))\\ -% &\le & d(f(x_1), g(x_1)) + d(g(x_2), h(x_2))-2\epsilon\\ -% &\le & d_u(f,g) + d_u(g,h) - 2\epsilon. -% \end{IEEEeqnarray*} -% Thus $d_u(f,g) \le d_u(f,g) + d_u(g,h) - 2\epsilon$. -% Taking $\epsilon \to 0$ yields the triangle inequality. -% -% \item $\cC(X,Y)$ is a Polish space: -% \todo{handwritten solution} -% -% \begin{itemize} -% \item $d_u$ is a complete metric: -% -% Let $(f_n)_n$ be a Cauchy series with respect to $d_u$. -% -% Then clearly $(f_n(x))_n$ is a Cauchy sequence with respect -% to $d$ for every $x$. -% Hence there exists a pointwise limit $f$ of the $f_n$. -% We need to show that $f$ is continuous. -% -% %\todo{something something uniform convergence theorem} -% -% \item $(\cC(X,Y), d_u)$ is separable: -% -% Since $Y$ is separable, there exists a countable -% dense subset $S \subseteq Y$. -% -% Consider $\cC(X,S) \subseteq \cC(X,Y)$. -% Take some $f \in \cC(X,Y)$. -% Since $X$ is compact, -% -% -% % TODO -% -% \end{itemize} -% \end{enumerate} - \begin{fact} Let $X $ be a compact Hausdorff space. Then the following are equivalent: @@ -205,7 +174,7 @@ Let $X$ be compact Polish\footnote{compact metrisable $\implies$ compact Polish} and $Y $ Polish. Let $\cC(X,Y)$ be the set of continuous functions $X \to Y$. -Consider the metric $d_u(f,g) \coloneqq \sup_{x \in X} |d(f(x), g(x))|$. +Consider the \vocab{uniform metric} $d_u(f,g) \coloneqq \sup_{x \in X} |d(f(x), g(x))|$. Clearly $d_u$ is a metric. \begin{claim} @@ -243,7 +212,7 @@ Clearly $d_u$ is a metric. for each $y \in X_m$. Then $\bigcup_{m,n} D_{m,n}$ is dense in $\cC(X,Y)$: Indeed if $f \in \cC(X,Y)$ and $\eta > 0$, - we finde $n > \frac{3}{\eta}$ and $m$ such that $f \in C_{m,n}$, + we find $n > \frac{3}{\eta}$ and $m$ such that $f \in C_{m,n}$, since $f$ is uniformly continuous. Let $g \in D_{m,n}$ be such that $\forall y \in X_m.~d(f(y), g(y)) < \frac{1}{n+1}$. We have $d_u(f,g) \le \eta$, diff --git a/inputs/tutorial_03.tex b/inputs/tutorial_03.tex index 5aeb648..a760624 100644 --- a/inputs/tutorial_03.tex +++ b/inputs/tutorial_03.tex @@ -12,6 +12,14 @@ \nr 1 +Let $X$ be a Polish space. +Then there exists an injection $f\colon X \to 2^\omega$ +such that for each $n < \omega$, +the set $f^{-1}(\{(y_n) \in 2^\omega : y_n = 1\})$ +is open. +Moreover if $V \subseteq 2^{ \omega}$ is closed, +then $f^{-1}(V)$ is $G_\delta$. + Let $(U_i)_{i < \omega}$ be a countable base of $X$. Define \begin{IEEEeqnarray*}{rCl} @@ -19,6 +27,7 @@ Define x &\longmapsto & (x_i)_{i < \omega} \end{IEEEeqnarray*} where $x_i = 1$ iff $x \in U_i$ and $x_i = 0$ otherwise. +\gist{ Then $f^{-1}(\{y = (y_n) \in 2^\omega | y_n = 1\}) = U_n$ is open. We have that $f$ is injective since $X$ is T1. @@ -51,17 +60,21 @@ Since $2^{n} \setminus \left( \prod_{i < n} X_i \right)$ is finite, we get that $f^{-1}(2^{\omega} \setminus ((\prod_{i