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@ -127,7 +127,7 @@ since $X^X$ has these properties.
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\begin{lemma}[Ellis–Numakura]
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\begin{lemma}[Ellis–Numakura]
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\yalabel{Ellis-Numakura Lemma}{Ellis-Numakura}{lem:ellisnumakura}
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\yalabel{Ellis-Numakura Lemma}{Ellis-Numakura}{lem:ellisnumakura}
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Every compact semigroup
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Every non-empty compact semigroup
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contains an \vocab{idempotent} element,
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contains an \vocab{idempotent} element,
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i.e.~$f$ such that $f^2 = f$.
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i.e.~$f$ such that $f^2 = f$.
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\end{lemma}
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\end{lemma}
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@ -37,10 +37,9 @@ Let $I$ be a linear order
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S & \coloneqq & \{ x \in \LO(\N) :& x \text{ has a least element},\\
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S & \coloneqq & \{ x \in \LO(\N) :& x \text{ has a least element},\\
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&&& \text{for any $t$, there is $t \oplus 1$, the successor of $t$.}\}
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&&& \text{for any $t$, there is $t \oplus 1$, the successor of $t$.}\}
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\end{IEEEeqnarray*}
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\end{IEEEeqnarray*}
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\todo{Exercise sheet 12}
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$S$ is Borel.\footnote{cf.~\yaref{s12e1}}
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$S$ is Borel.
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We will % TODO ?
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We will
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construct a reduction
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construct a reduction
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\begin{IEEEeqnarray*}{rCl}
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\begin{IEEEeqnarray*}{rCl}
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M \colon S &\longrightarrow & C(\mathbb{K}^\N,\mathbb{K})^\N. %\\
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M \colon S &\longrightarrow & C(\mathbb{K}^\N,\mathbb{K})^\N. %\\
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@ -1,10 +1,10 @@
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\lecture{24}{2024-01-23}{Combinatorics!}
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\lecture{24}{2024-01-23}{Combinatorics!}
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% ANKI 2
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\subsection{Applications to Combinatorics} % Ramsey Theory}
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\subsection{Applications to Combinatorics} % Ramsey Theory}
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% TODO Define Ultrafilter
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\begin{definition}
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\begin{definition}
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An \vocab{ultrafilter} on $\N$ (or any other set)
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An \vocab{ultrafilter} on $\N$ (or any other set)
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is a family $\cU \subseteq \cP(\N)$
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is a family $\cU \subseteq \cP(\N)$
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@ -44,6 +44,7 @@
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for $\{ n \in \N : \phi(n)\} \in \cU$.
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for $\{ n \in \N : \phi(n)\} \in \cU$.
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We say that $\phi(n)$ holds for \vocab{$\cU$-almost all} $n$.
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We say that $\phi(n)$ holds for \vocab{$\cU$-almost all} $n$.
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\end{notation}
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\end{notation}
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\gist{%
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\begin{observe}
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\begin{observe}
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Let $\phi(\cdot )$, $\psi(\cdot )$ be formulas.
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Let $\phi(\cdot )$, $\psi(\cdot )$ be formulas.
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@ -53,6 +54,7 @@
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\item $(\cU n) ~\lnot \phi(n) \iff \lnot (\cU n)~ \phi(n)$.
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\item $(\cU n) ~\lnot \phi(n) \iff \lnot (\cU n)~ \phi(n)$.
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\end{enumerate}
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\end{enumerate}
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\end{observe}
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\end{observe}
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}{}
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\begin{lemma}
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\begin{lemma}
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\label{lem:ultrafilterlimit}
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\label{lem:ultrafilterlimit}
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Let $X $ be a compact Hausdorff space.
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Let $X $ be a compact Hausdorff space.
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@ -72,6 +74,8 @@
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\end{notation}
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\end{notation}
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\begin{refproof}{lem:ultrafilterlimit}\footnote{The proof from the lecture only works
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\begin{refproof}{lem:ultrafilterlimit}\footnote{The proof from the lecture only works
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for metric spaces.}
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for metric spaces.}
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\gist{
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For metric spaces:
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Whenever we write $X = Y \cup Z$
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Whenever we write $X = Y \cup Z$
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we have $(\cU n) x_n \in Y$
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we have $(\cU n) x_n \in Y$
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or $(\cU n) x_n \in Z$.
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or $(\cU n) x_n \in Z$.
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@ -86,8 +90,13 @@
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$C \in \cP_{n+1} \implies \exists C \subseteq D \in \cP_{n}$
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$C \in \cP_{n+1} \implies \exists C \subseteq D \in \cP_{n}$
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and
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and
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$C_1 \supseteq C_2 \supseteq \ldots$, $C_i \in \cP_i $ $\implies | \bigcap_{i} C_i| = 1$.
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$C_1 \supseteq C_2 \supseteq \ldots$, $C_i \in \cP_i $ $\implies | \bigcap_{i} C_i| = 1$.
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It is clear that we can do this for metric spaces,
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It is clear that we can do this for metric spaces.
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but such partition can be found for compact Hausdorff spaces as well.
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}{}
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See \yaref{thm:uflimit} for the full proof.
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See
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\yaref{fact:compactiffufconv} and
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\yaref{fact:hdifffilterlimit} for a more general statement.
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\end{refproof}
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\end{refproof}
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Let $\beta \N$ be the Čech-Stone compactification of $\N$,
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Let $\beta \N$ be the Čech-Stone compactification of $\N$,
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@ -157,7 +166,6 @@ is not necessarily continuous.
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\[
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\[
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\forall n.~\exists k < M.~ T^{n+k}(x) \in G.
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\forall n.~\exists k < M.~ T^{n+k}(x) \in G.
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\]
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\]
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\end{definition}
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\end{definition}
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\begin{fact}
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\begin{fact}
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Let $\cU, \cV \in \beta\N$
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Let $\cU, \cV \in \beta\N$
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@ -7,15 +7,17 @@ Let $\beta\N$ denote the set of ultrafilters on $\N$.
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where a basis consist of sets
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where a basis consist of sets
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$V_A \coloneqq \{p \in \beta\N : A \in p\}, A \subseteq \N$.
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$V_A \coloneqq \{p \in \beta\N : A \in p\}, A \subseteq \N$.
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\gist{%
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(For $A, B \subseteq \N$ we have $V_{A \cap B} = V_{A} \cap V_B$
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(For $A, B \subseteq \N$ we have $V_{A \cap B} = V_{A} \cap V_B$
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and $\beta\N = V_\N$.)
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and $\beta\N = V_\N$.)
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}{}
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\item Note also that for $A, B \subseteq \N$,
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\item Note also that for $A, B \subseteq \N$,
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$V_{A \cup B} = V_A \cup V_B$,
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$V_{A \cup B} = V_A \cup V_B$,
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$V_{A^c} = \beta\N \setminus V_A$.
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$V_{A^c} = \beta\N \setminus V_A$.
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\end{itemize}
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\end{itemize}
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\end{fact}
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\end{fact}
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\gist{%
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\begin{observe}
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\begin{observe}
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\label{ob:bNclopenbasis}
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\label{ob:bNclopenbasis}
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Note that the basis is clopen. In particular
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Note that the basis is clopen. In particular
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If $F$ is closed, then $U = \beta\N \setminus F = \bigcup_{i\in I} V_{A_i}$,
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If $F$ is closed, then $U = \beta\N \setminus F = \bigcup_{i\in I} V_{A_i}$,
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so $F = \bigcap_{i \in I} V_{\N \setminus A_i}$.
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so $F = \bigcap_{i \in I} V_{\N \setminus A_i}$.
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\end{observe}
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\end{observe}
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}{}
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\begin{fact}
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\begin{fact}
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\label{fact:bNhd}
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\label{fact:bNhd}
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$\bigcap_{j=1}^k F_{i_j} \neq \emptyset$.
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$\bigcap_{j=1}^k F_{i_j} \neq \emptyset$.
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We need to show that $\bigcap_{i \in I} F_i \neq \emptyset$.
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We need to show that $\bigcap_{i \in I} F_i \neq \emptyset$.
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\gist{%
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Replacing each $F_i$ by $V_{A_j^i}$ such
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Replacing each $F_i$ by $V_{A_j^i}$ such
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that $F_i = \bigcap_{j \in J_i} V_{A_j^i}$
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that $F_i = \bigcap_{j \in J_i} V_{A_j^i}$
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(cf.~\yaref{ob:bNclopenbasis})
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(cf.~\yaref{ob:bNclopenbasis})
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we may assume that $F_i$ is of the form $V_{A_i}$.
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we may assume that $F_i$ is of the form $V_{A_i}$.
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We get $\{F_i = V_{A_i} : i \in I\}$
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We get $\{F_i = V_{A_i} : i \in I\}$
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with the finite intersection property.
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with the finite intersection property.
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}{Wlog.~$F_i = V_{A_i}$.}
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Hence
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Hence
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$\{A_i : i \in I\} \mathbin{\text{\reflectbox{$\coloneqq$}}} \cF_0$
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$\{A_i : i \in I\} \mathbin{\text{\reflectbox{$\coloneqq$}}} \cF_0$
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has the finite intersection property.
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has the finite intersection property.
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\item $ \{\hat{n}\} $ is open in $\beta\N$ for all $n \in \N$.
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\item $ \{\hat{n}\} $ is open in $\beta\N$ for all $n \in \N$.
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\item $\N \subseteq \beta\N$ is dense.
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\item $\N \subseteq \beta\N$ is dense.
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\end{itemize}
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\end{itemize}
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\todo{Easy exercise}
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% TODO write down (exercise)
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\end{fact}
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\end{fact}
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\begin{theorem}
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\begin{theorem}
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\label{thm:uflimit}
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For every compact Hausdorff space $X$,
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For every compact Hausdorff space $X$,
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a sequence $(x_n)$ in $X$,
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a sequence $(x_n)$ in $X$,
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and $\cU \in \beta\N$,
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and $\cU \in \beta\N$,
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% agree everywhere (fact section)
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% agree everywhere (fact section)
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\end{proof}
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\end{proof}
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\begin{trivial}+
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\begin{trivial}+
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More generally,
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$\beta$ is a functor from the category of topological
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$\beta$ is a functor from the category of topological
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spaces to the category of compact Hausdorff spaces.
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spaces to the category of compact Hausdorff spaces.
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It is left adjoint to the inclusion functor.
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It is left adjoint to the inclusion functor.
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Take $x_2 > x_1$ that satisfies this.
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Take $x_2 > x_1$ that satisfies this.
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\item Suppose we have chosen $\langle x_i : i < n \rangle$.
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\item Suppose we have chosen $\langle x_i : i < n \rangle$.
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Since $\cU$ is idempotent, we have
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Since $\cU$ is idempotent, we have
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\[
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\begin{IEEEeqnarray*}{rCl}
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(\cU n)[
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(\cU n)&& n \in P\\
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n \in P
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&\land& (\cU_k) n + k \in P\\
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\land (\cU_k) n + k \in P
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&\land& \forall {I \subseteq n}.~ (\sum_{i \in I} x_i + n \in P)\\
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\land \forall {I \subseteq n}.~ (\sum_{i \in I} x_i + n \in P)
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&\land& (\cU_k)\left( \forall {I \subseteq n}.~ (\sum_{i \in I} x_i + n + k) \in P\right).
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\land (\cU_k)\left( \forall {I \subseteq n}.~ (\sum_{i \in I} x_i + n + k) \in P\right).
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\end{IEEEeqnarray*}
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\]
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Chose $x_n > x_{n-1}$ that satisfies this.
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Chose $x_n > x_{n-1}$ that satisfies this.
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\end{itemize}
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\end{itemize}
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Set $H \coloneqq \{x_i : i < \omega\}$.
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Set $H \coloneqq \{x_i : i < \omega\}$.
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@ -237,6 +239,3 @@ to obtain
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Next time we'll see another proof of this theorem.
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Next time we'll see another proof of this theorem.
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% Points: 15 / 16
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% Points: 15 / 16
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\nr 1
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\nr 1
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\todo{handwritten solution}
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Let $(X,d)$ be a metric space and $\emptyset \neq A \subseteq X$.
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Let $d(x,A) \coloneqq \inf(d(x,a) : a \in A\}$.
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\begin{itemize}
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\item $d(-,A)$ is uniformly continuous:
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Clearly $|d(x,A) - d(y,A)| \le d(x,y)$.
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\todo{Add details}
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\item $d(x,A) = 0 \iff x \in \overline{A}$.
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$d(x,A) = 0$ iff there is a sequence in $A$
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converging towards $x$ iff $x \in \overline{A}$.
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\end{itemize}
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\nr 2
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\nr 2
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Let $X$ be a discrete space.
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For $f,g \in X^{\N}$ define
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\[
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d(f,g) \coloneqq \begin{cases}
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(1 + \min \{n: f(n) \neq g(n)\})^{-1} &: f \neq g,\\
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0 &: f= g.
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\end{cases}
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\]
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\begin{enumerate}[(a)]
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\begin{enumerate}[(a)]
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\item $d$ is an ultrametric:
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\item $d$ is an \vocab{ultrametric},
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i.e.~$d(f,g) \le \max \{d(f,h), d(g,h)\}$ for all $f,g,h \in X^{\N}$ :
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Let $f,g,h \in X^{\N}$.
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Let $f,g,h \in X^{\N}$.
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\nr 3
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\nr 3
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Consider $\N$ as a discrete space and $\N^{\N}$ with the product topology.
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Let
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\[
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S_{\infty} = \{f\colon \N \to \N \text{ bijective}\} \subseteq \N^{\N}.
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\]
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\begin{enumerate}[(a)]
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\begin{enumerate}[(a)]
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\item $S_{\infty}$ is a Polish space:
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\item $S_{\infty}$ is a Polish space:
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From (2) we know that $\N^{\N}$ is Polish.
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From \yaref{s1e2} we know that $\N^{\N}$ is Polish.
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Hence it suffices to show that $S_{\infty}$ is $G_{\delta}$
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Hence it suffices to show that $S_{\infty}$ is $G_{\delta}$
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with respect to $\N^\N$.
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with respect to $\N^\N$.
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Clearly there cannot exist a finite subcover
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Clearly there cannot exist a finite subcover
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as $B$ is the disjoint union of the $B_j$.
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as $B$ is the disjoint union of the $B_j$.
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% TODO Think about this
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\end{enumerate}
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\end{enumerate}
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\nr 4
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\nr 4
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% (uniform metric)
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%
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% \begin{enumerate}[(a)]
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% \item $d_u$ is a metric on $\cC(X,Y)$:
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%
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% It is clear that $d_u(f,f) = 0$.
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%
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% Let $f \neq g$. Then there exists $x \in X$ with
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% $f(x) \neq g(x)$, hence $d_u(f,g) \ge d(f(x), g(x)) > 0$.
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%
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% Since $d$ is symmetric, so is $d_u$.
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%
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% Let $f,g,h \in \cC(X,Y)$.
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% Take some $\epsilon > 0$
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% choose $x_1, x_2 \in X$
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% with $d_u(f,g) \le d(f(x_1), g(x_1)) + \epsilon$,
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% $d_u(g,h) \le d(g(x_2), h(x_2)) + \epsilon$.
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%
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% Then for all $x \in X$
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% \begin{IEEEeqnarray*}{rCl}
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% d(f(x), h(x)) &\le &
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% d(f(x), g(x)) + d(g(x), h(x))\\
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% &\le & d(f(x_1), g(x_1)) + d(g(x_2), h(x_2))-2\epsilon\\
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% &\le & d_u(f,g) + d_u(g,h) - 2\epsilon.
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% \end{IEEEeqnarray*}
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% Thus $d_u(f,g) \le d_u(f,g) + d_u(g,h) - 2\epsilon$.
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% Taking $\epsilon \to 0$ yields the triangle inequality.
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%
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% \item $\cC(X,Y)$ is a Polish space:
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% \todo{handwritten solution}
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%
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% \begin{itemize}
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% \item $d_u$ is a complete metric:
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%
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% Let $(f_n)_n$ be a Cauchy series with respect to $d_u$.
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%
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% Then clearly $(f_n(x))_n$ is a Cauchy sequence with respect
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% to $d$ for every $x$.
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% Hence there exists a pointwise limit $f$ of the $f_n$.
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% We need to show that $f$ is continuous.
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%
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% %\todo{something something uniform convergence theorem}
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%
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% \item $(\cC(X,Y), d_u)$ is separable:
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%
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% 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}
|
\begin{fact}
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||||||
Let $X $ be a compact Hausdorff space.
|
Let $X $ be a compact Hausdorff space.
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Then the following are equivalent:
|
Then the following are equivalent:
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|
@ -205,7 +174,7 @@
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Let $X$ be compact Polish\footnote{compact metrisable $\implies$ compact Polish}
|
Let $X$ be compact Polish\footnote{compact metrisable $\implies$ compact Polish}
|
||||||
and $Y $ Polish.
|
and $Y $ Polish.
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||||||
Let $\cC(X,Y)$ be the set of continuous functions $X \to Y$.
|
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.
|
Clearly $d_u$ is a metric.
|
||||||
|
|
||||||
\begin{claim}
|
\begin{claim}
|
||||||
|
@ -243,7 +212,7 @@ Clearly $d_u$ is a metric.
|
||||||
for each $y \in X_m$.
|
for each $y \in X_m$.
|
||||||
Then $\bigcup_{m,n} D_{m,n}$ is dense in $\cC(X,Y)$:
|
Then $\bigcup_{m,n} D_{m,n}$ is dense in $\cC(X,Y)$:
|
||||||
Indeed if $f \in \cC(X,Y)$ and $\eta > 0$,
|
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.
|
since $f$ is uniformly continuous.
|
||||||
Let $g \in D_{m,n}$ be such that $\forall y \in X_m.~d(f(y), g(y)) < \frac{1}{n+1}$.
|
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 $d_u(f,g) \le \eta$,
|
We have $d_u(f,g) \le \eta$,
|
||||||
|
|
|
@ -12,6 +12,14 @@
|
||||||
|
|
||||||
\nr 1
|
\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 $X$.
|
Let $(U_i)_{i < \omega}$ be a countable base of $X$.
|
||||||
Define
|
Define
|
||||||
\begin{IEEEeqnarray*}{rCl}
|
\begin{IEEEeqnarray*}{rCl}
|
||||||
|
@ -19,6 +27,7 @@ Define
|
||||||
x &\longmapsto & (x_i)_{i < \omega}
|
x &\longmapsto & (x_i)_{i < \omega}
|
||||||
\end{IEEEeqnarray*}
|
\end{IEEEeqnarray*}
|
||||||
where $x_i = 1$ iff $x \in U_i$ and $x_i = 0$ otherwise.
|
where $x_i = 1$ iff $x \in U_i$ and $x_i = 0$ otherwise.
|
||||||
|
\gist{
|
||||||
Then $f^{-1}(\{y = (y_n) \in 2^\omega | y_n = 1\}) = U_n$
|
Then $f^{-1}(\{y = (y_n) \in 2^\omega | y_n = 1\}) = U_n$
|
||||||
is open.
|
is open.
|
||||||
We have that $f$ is injective since $X$ is T1.
|
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
|
is finite, we get that
|
||||||
$f^{-1}(2^{\omega} \setminus ((\prod_{i <n} X_{i}) \times 2^{\omega}))$
|
$f^{-1}(2^{\omega} \setminus ((\prod_{i <n} X_{i}) \times 2^{\omega}))$
|
||||||
is $G_\delta$ as a finite union of $G_{\delta}$ sets.
|
is $G_\delta$ as a finite union of $G_{\delta}$ sets.
|
||||||
|
}{}
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
\nr 2
|
\nr 2
|
||||||
|
Let $X$ be a Polish space. Then $X$ is homeomorphic to a closed subspace of $\R^{ \omega}$ :
|
||||||
\todo{handwritten solution}
|
\todo{handwritten solution}
|
||||||
(b)
|
% \begin{itemize}
|
||||||
Let $f(x^{(i)})$ be a sequence in $f(X)$.
|
% \item
|
||||||
Suppose that $f(x^{(i)}) \to y$.
|
% Let $f(x^{(i)})$ be a sequence in $f(X)$.
|
||||||
We have that $f^{-1} = \pi_{\text{odd}}$ is continuous.
|
% Suppose that $f(x^{(i)}) \to y$.
|
||||||
Then $\pi_{\text{odd}}(f(x^{(i)}) \to \pi_{\text{odd}}(y)$.
|
% We have that $f^{-1} = \pi_{\text{odd}}$ is continuous.
|
||||||
Since $\pi_{\text{even}}$ converges, we have $\pi_{\text{odd}}(y) \in X$.
|
% Then $\pi_{\text{odd}}(f(x^{(i)}) \to \pi_{\text{odd}}(y)$.
|
||||||
|
% Since $\pi_{\text{even}}$ converges, we have $\pi_{\text{odd}}(y) \in X$.
|
||||||
|
% \end{itemize}
|
||||||
|
|
||||||
|
|
||||||
\nr 3
|
\nr 3
|
||||||
|
@ -130,6 +143,13 @@ Since $\pi_{\text{even}}$ converges, we have $\pi_{\text{odd}}(y) \in X$.
|
||||||
\end{proof}
|
\end{proof}
|
||||||
|
|
||||||
\nr 4
|
\nr 4
|
||||||
|
|
||||||
|
Define
|
||||||
|
\begin{IEEEeqnarray*}{rCl}
|
||||||
|
f\colon \omega^{\omega} &\longrightarrow & 2^\omega \\
|
||||||
|
(x_n)&\longmapsto & 0^{x_0} 1 0^{x_1} 1 \ldots.
|
||||||
|
\end{IEEEeqnarray*}
|
||||||
|
|
||||||
\begin{enumerate}[(1)]
|
\begin{enumerate}[(1)]
|
||||||
\item $f$ is a topological embedding:
|
\item $f$ is a topological embedding:
|
||||||
Consider a basic open set
|
Consider a basic open set
|
||||||
|
|
|
@ -8,7 +8,7 @@ Let $A \neq \emptyset$ be discrete.
|
||||||
For $D \subseteq A^{\omega}$,
|
For $D \subseteq A^{\omega}$,
|
||||||
let
|
let
|
||||||
\[
|
\[
|
||||||
T_D \coloneqq \{x\defon{n} \in A^{<\omega} | x \in D, n \in \N\}..
|
T_D \coloneqq \{x\defon{n} \in A^{<\omega} | x \in D, n \in \N\}.
|
||||||
\]
|
\]
|
||||||
\begin{enumerate}[(a)]
|
\begin{enumerate}[(a)]
|
||||||
\item For any $D \subseteq A^\omega$, $T_D$ is a pruned tree:
|
\item For any $D \subseteq A^\omega$, $T_D$ is a pruned tree:
|
||||||
|
|
|
@ -63,7 +63,7 @@ Flows are always on non-empty spaces $X$.
|
||||||
\begin{proof}
|
\begin{proof}
|
||||||
(i) $\implies$ (ii):
|
(i) $\implies$ (ii):
|
||||||
Let $(Y,T)$ be a subflow of $(X,T)$.
|
Let $(Y,T)$ be a subflow of $(X,T)$.
|
||||||
take $y \in Y$. Then $Ty$ is dense in mKX.
|
take $y \in Y$. Then $Ty$ is dense in $X$.
|
||||||
But $Ty \subseteq Y$, so $Y$ is dense in $X$.
|
But $Ty \subseteq Y$, so $Y$ is dense in $X$.
|
||||||
Since $Y$ is closed, we get $Y = X$.
|
Since $Y$ is closed, we get $Y = X$.
|
||||||
|
|
||||||
|
|
|
@ -12,6 +12,7 @@ all sets containing an open neighbourhood of $x$,
|
||||||
is contained in $\cF$.
|
is contained in $\cF$.
|
||||||
|
|
||||||
\begin{fact}
|
\begin{fact}
|
||||||
|
\label{fact:hdifffilterlimit}
|
||||||
$X$ is Hausdorff iff every filter has at most one limit point.
|
$X$ is Hausdorff iff every filter has at most one limit point.
|
||||||
\end{fact}
|
\end{fact}
|
||||||
\begin{proof}
|
\begin{proof}
|
||||||
|
|
Loading…
Reference in a new issue