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14 changed files with 148 additions and 196 deletions
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@ -164,10 +164,8 @@
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\]
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\end{notation}
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\gist{%
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The following similar to Fubini,
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but for meager sets:
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}{}
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\begin{theorem}[Kuratowski-Ulam]
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\yalabel{Kuratowski-Ulam}{Kuratowski-Ulam}{thm:kuratowskiulam}
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@ -195,7 +193,6 @@ but for meager sets:
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\end{enumerate}
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\end{theorem}
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\begin{refproof}{thm:kuratowskiulam}
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\gist{
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(ii) and (iii) are equivalent by passing to the complement.
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\begin{claim}%[1a]
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@ -289,11 +286,16 @@ but for meager sets:
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$M_x$ is comeager
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as a countable intersection of comeager sets.
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\end{refproof}
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}{}
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% \phantom\qedhere
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% \end{refproof}
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% TODO fix claim numbers
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\gist{%
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\begin{remark}
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Suppose that $A$ has the BP.
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Then there is an open $U$ such that
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$A \symdif U \mathbin{\text{\reflectbox{$\coloneqq$}}} M$ is meager.
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Then $A = U \symdif M$.
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\end{remark}
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}{}
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@ -1,8 +1,8 @@
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\lecture{06}{2023-11-03}{}
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\gist{%
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% \begin{refproof}{thm:kuratowskiulam}
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\begin{enumerate}[(i)]
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\item Let $A$ be a set with the Baire property.
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\item Let $A$ be a set with the Baire Property.
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Write $A = U \symdif M$
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for $U$ open and $M$ meager.
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Then for all $x$,
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@ -51,8 +51,8 @@
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Towards a contradiction suppose that $A$ is not meager.
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Then $U$ is not meager.
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Since $X \times Y$ is second countable,
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we have that $U$ is a countable union of open rectangles.
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At least one of them, say $G \times H \subseteq U$,
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we have that $A$ is a countable union of open rectangles.
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At least one of them, say $G \times H \subseteq A$,
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is not meager.
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By \yaref{thm:kuratowskiulam:c2},
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both $G$ and $H$ are not meager.
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@ -71,59 +71,7 @@
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``$\implies$''
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This is \yaref{thm:kuratowskiulam:c1b}.
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\end{enumerate}
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}{%
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\begin{itemize}
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\item (ii) $\iff$ (iii): pass to complement.
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\item $F \overset{\text{closed}}{\subseteq} X \times Y$ nwd.
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$\implies \{x \in X : F_x \text{ nwd}\} $ comeager:
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\begin{itemize}
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\item $W = F^c$ is open and dense, show that $\{x : W_x \text{ dense}\}$
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is comeager.
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\item $(V_n)$ enumeration of basis. Show that $U_n \coloneqq \{x : V_n \cap W_x \neq \emptyset\}$
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is comeager for all $n$.
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\item $U_n$ is open (projection of open) and dense ($W$ is dense, hence $W \cap ( U \times V_n) \neq \emptyset$ for $U$ open).
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\end{itemize}
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\item $F \subseteq X \times Y$ is nwd $\implies \{x \in X: F_x \text{ nwd}\}$ comeager.
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(consider $\overline{F}$).
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\item (ii) $\implies$:
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$M \subseteq X \times Y$ meager $\implies \{x \in X: M_x \text{ meager}\}$ comeager
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(write $M$ as ctbl. union of nwd.)
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\item (i): If $A$ has the Baire Property,
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then $A = U \symdif M$, $A_x = U_x \symdif M_x$,
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$U_x$ open and $\{x : M_x \text{ meager}\}$ comeager
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$\implies$ (i).
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\item $P \subseteq X$, $Q \subseteq Y$ BP,
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then $P \times Q$ meager $\iff$ $P$ or $Q$ meager.
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\begin{itemize}
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\item $\impliedby$ easy
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\item $\implies$ Suppose $P \times Q$ meager, $P$ not meager.
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$\emptyset\neq P \cap \underbrace{\{x : (P \times Q)_x \text{ meager} \}}_{\text{comeager}} \ni x$.
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$(P \times Q)_x = Q$ is meager.
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\end{itemize}
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\item (ii) $\impliedby$:
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\begin{itemize}
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\item $A$ BP, $\{x : A_x \text{ meager}\}$ comeager.
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\item $A = U \symdif M$.
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\item Suppose $A$ not meager $\leadsto$ $U$ not meager
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$\leadsto \exists G \times H \subseteq U$ not meager.
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\item $G$ and $H$ are not meager.
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\item $\exists x_0 \in G \cap \underbrace{\{x: A_x \text{ meager } \land M_x \text{ meager}\}}_\text{comeager}$.
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\item $H$ meager, as
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\[
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H \subseteq U_{x_0} \subseteq A_{x_0} \cup M_{x_0}.
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\]
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\end{itemize}
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\end{itemize}
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}
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\end{refproof}
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\gist{%
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\begin{remark}
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Suppose that $A$ has the BP.
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Then there is an open $U$ such that
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$A \symdif U \mathbin{\text{\reflectbox{$\coloneqq$}}} M$ is meager.
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Then $A = U \symdif M$.
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\end{remark}
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}{}
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\section{Borel sets} % TODO: fix chapters
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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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\yalabel{Ellis-Numakura Lemma}{Ellis-Numakura}{lem:ellisnumakura}
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Every non-empty compact semigroup
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Every compact semigroup
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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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\end{lemma}
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@ -20,6 +20,9 @@
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\end{remark}
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}{}
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% TODO ANKI-MARKER
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We will be studying projections to the first $d$ coordinates,
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i.e.
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\[
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@ -46,9 +49,6 @@ Let $\pi_n\colon X \to (S^1)^n$ be the projection to the first $n$
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coordinates.
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% TODO ANKI-MARKER
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\begin{lemma}
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\label{lem:lec20:1}
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Let $x,x' \in X$ with $\pi_n(x) = \pi_n(x')$
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@ -146,7 +146,9 @@ For this we define
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% TODO since for $\overline{x}, \overline{y} \in \mathbb{K}^I$,
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% $d(x_\alpha, y_\alpha) = d((f(\overline{x})_\alpha, (f(\overline{y})_\alpha))$.
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\item Minimality:%
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\notexaminable{%
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\gist{%
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\footnote{This is not relevant for the exam.}
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Let $\langle E_n : n < \omega \rangle$
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be an enumeration of a countable basis for $\mathbb{K}^I$.
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@ -163,10 +165,11 @@ For this we define
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is dense in $\overline{x} \mapsto f(\overline{x})$.
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Since the flow is distal, it suffices to show
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that one orbit is dense (cf.~\yaref{thm:distalflowpartition}).
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}
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}{ Not relevant for the exam.}
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\item The order of the flow is $\eta$:%
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\notexaminable{%
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\gist{%
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\footnote{This is not relevant for the exam.}
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Let $\overline{f} = (f_i)_{i \in I} \in \mathbb{K}_I$.
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Consider the flows we get from $(f_i)_{i < j}$
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resp.~$(f_i)_{i \le j}$
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@ -190,6 +193,6 @@ For this we define
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\end{IEEEeqnarray*}
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Beleznay and Foreman show that this is open and dense.%
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% TODO similarities to the lemma used today
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}
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}{ Not relevant for the exam.}
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\end{itemize}
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\end{proof}
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@ -37,9 +37,10 @@ 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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&&& \text{for any $t$, there is $t \oplus 1$, the successor of $t$.}\}
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\end{IEEEeqnarray*}
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$S$ is Borel.\footnote{cf.~\yaref{s12e1}}
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\todo{Exercise sheet 12}
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$S$ is Borel.
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We will
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We will % TODO ?
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construct a reduction
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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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@ -1,10 +1,10 @@
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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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% TODO Define Ultrafilter
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\begin{definition}
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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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@ -44,7 +44,6 @@
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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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\end{notation}
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\gist{%
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\begin{observe}
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Let $\phi(\cdot )$, $\psi(\cdot )$ be formulas.
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@ -54,7 +53,6 @@
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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{observe}
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}{}
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\begin{lemma}
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\label{lem:ultrafilterlimit}
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Let $X $ be a compact Hausdorff space.
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@ -72,10 +70,7 @@
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\begin{notation}
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In this case we write $x = \ulim{\cU}_n x_n$.
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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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for metric spaces.}
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\gist{
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For metric spaces:
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\begin{refproof}{lem:ultrafilterlimit}[sketch]
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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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or $(\cU n) x_n \in Z$.
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@ -90,13 +85,8 @@
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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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$C_1 \supseteq C_2 \supseteq \ldots$, $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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}{}
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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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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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\end{refproof}
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Let $\beta \N$ be the Čech-Stone compactification of $\N$,
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@ -130,14 +120,15 @@ This gives $+ \colon \beta\N \times \beta\N \to \beta\N$.
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This is not commutative,
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but associative and $a \mapsto a + b$ is continuous
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for a fixed $b$,
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i.e.~it is a left compact topological semigroup.
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for a fixed $b$.
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This is called a left compact topological semigroup.
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Let $X$ be a compact Hausdorff space
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and let $T \colon X \to X$ be continuous.%
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\footnote{Note that this may not be a homeomorphism, i.e.~we only get a $\N$-action
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\footnote{Note that this need not be a homeomorphism, i.e.~we only get a $\N$-action
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but not a $\Z$-action.}
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For any $\cU \in \beta\N$, we define $T^{\cU}$ by
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@ -166,6 +157,7 @@ is not necessarily continuous.
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\[
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\forall n.~\exists k < M.~ T^{n+k}(x) \in G.
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\]
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\end{definition}
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\begin{fact}
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Let $\cU, \cV \in \beta\N$
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|
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@ -7,17 +7,15 @@ Let $\beta\N$ denote the set of ultrafilters on $\N$.
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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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\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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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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$V_{A \cup B} = V_A \cup V_B$,
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$V_{A^c} = \beta\N \setminus V_A$.
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\end{itemize}
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\end{fact}
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\gist{%
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\begin{observe}
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\label{ob:bNclopenbasis}
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Note that the basis is clopen. In particular
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@ -27,7 +25,6 @@ Let $\beta\N$ denote the set of ultrafilters on $\N$.
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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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\end{observe}
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}{}
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\begin{fact}
|
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\label{fact:bNhd}
|
||||
|
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@ -57,14 +54,12 @@ Let $\beta\N$ denote the set of ultrafilters on $\N$.
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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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\gist{%
|
||||
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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||||
(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 get $\{F_i = V_{A_i} : i \in I\}$
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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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$\{A_i : i \in I\} \mathbin{\text{\reflectbox{$\coloneqq$}}} \cF_0$
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has the finite intersection property.
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|
@ -83,10 +78,11 @@ Let $\beta\N$ denote the set of ultrafilters on $\N$.
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\item $ \{\hat{n}\} $ is open in $\beta\N$ for all $n \in \N$.
|
||||
\item $\N \subseteq \beta\N$ is dense.
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||||
\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 $\cU \in \beta\N$,
|
||||
|
|
@ -136,11 +132,6 @@ Let $\beta\N$ denote the set of ultrafilters on $\N$.
|
|||
% TODO general fact: continuous functions agreeing on a dense set
|
||||
% agree everywhere (fact section)
|
||||
\end{proof}
|
||||
\begin{trivial}+
|
||||
$\beta$ is a functor from the category of topological
|
||||
spaces to the category of compact Hausdorff spaces.
|
||||
It is left adjoint to the inclusion functor.
|
||||
\end{trivial}
|
||||
|
||||
% RECAP
|
||||
\gist{%
|
||||
|
|
@ -225,12 +216,13 @@ 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
|
||||
\begin{IEEEeqnarray*}{rCl}
|
||||
(\cU n)&& n \in P\\
|
||||
&\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& (\cU_k)\left( \forall {I \subseteq n}.~ (\sum_{i \in I} x_i + n + k) \in P\right).
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||||
\end{IEEEeqnarray*}
|
||||
\[
|
||||
(\cU n)[
|
||||
n \in P
|
||||
\land (\cU_k) n + k \in P
|
||||
\land \forall {I \subseteq n}.~ (\sum_{i \in I} x_i + n \in P)
|
||||
\land (\cU_k)\left( \forall {I \subseteq n}.~ (\sum_{i \in I} x_i + n + k) \in P\right).
|
||||
\]
|
||||
Chose $x_n > x_{n-1}$ that satisfies this.
|
||||
\end{itemize}
|
||||
Set $H \coloneqq \{x_i : i < \omega\}$.
|
||||
|
|
@ -239,3 +231,6 @@ to obtain
|
|||
|
||||
Next time we'll see another proof of this theorem.
|
||||
|
||||
|
||||
|
||||
|
||||
|
|
|
|||
|
|
@ -3,36 +3,12 @@
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|||
% Points: 15 / 16
|
||||
|
||||
\nr 1
|
||||
Let $(X,d)$ be a metric space and $\emptyset \neq A \subseteq X$.
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||||
Let $d(x,A) \coloneqq \inf(d(x,a) : a \in A\}$.
|
||||
|
||||
\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}
|
||||
|
||||
\todo{handwritten solution}
|
||||
|
||||
\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 \vocab{ultrametric},
|
||||
i.e.~$d(f,g) \le \max \{d(f,h), d(g,h)\}$ for all $f,g,h \in X^{\N}$ :
|
||||
\item $d$ is an ultrametric:
|
||||
|
||||
Let $f,g,h \in X^{\N}$.
|
||||
|
||||
|
|
@ -94,15 +70,10 @@ d(f,g) \coloneqq \begin{cases}
|
|||
|
||||
\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 \yaref{s1e2} we know that $\N^{\N}$ is Polish.
|
||||
From (2) we know that $\N^{\N}$ is Polish.
|
||||
Hence it suffices to show that $S_{\infty}$ is $G_{\delta}$
|
||||
with respect to $\N^\N$.
|
||||
|
||||
|
|
@ -140,9 +111,69 @@ S_{\infty} = \{f\colon \N \to \N \text{ bijective}\} \subseteq \N^{\N}.
|
|||
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:
|
||||
|
|
@ -174,7 +205,7 @@ S_{\infty} = \{f\colon \N \to \N \text{ bijective}\} \subseteq \N^{\N}.
|
|||
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 \vocab{uniform metric} $d_u(f,g) \coloneqq \sup_{x \in X} |d(f(x), g(x))|$.
|
||||
Consider the metric $d_u(f,g) \coloneqq \sup_{x \in X} |d(f(x), g(x))|$.
|
||||
Clearly $d_u$ is a metric.
|
||||
|
||||
\begin{claim}
|
||||
|
|
@ -212,7 +243,7 @@ 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 find $n > \frac{3}{\eta}$ and $m$ such that $f \in C_{m,n}$,
|
||||
we finde $n > \frac{3}{\eta}$ 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 $d_u(f,g) \le \eta$,
|
||||
|
|
|
|||
|
|
@ -12,14 +12,6 @@
|
|||
|
||||
\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}
|
||||
|
|
@ -27,7 +19,6 @@ 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.
|
||||
|
|
@ -60,21 +51,17 @@ Since $2^{n} \setminus \left( \prod_{i < n} X_i \right)$
|
|||
is finite, we get that
|
||||
$f^{-1}(2^{\omega} \setminus ((\prod_{i <n} X_{i}) \times 2^{\omega}))$
|
||||
is $G_\delta$ as a finite union of $G_{\delta}$ sets.
|
||||
}{}
|
||||
|
||||
|
||||
|
||||
\nr 2
|
||||
Let $X$ be a Polish space. Then $X$ is homeomorphic to a closed subspace of $\R^{ \omega}$ :
|
||||
\todo{handwritten solution}
|
||||
% \begin{itemize}
|
||||
% \item
|
||||
% Let $f(x^{(i)})$ be a sequence in $f(X)$.
|
||||
% Suppose that $f(x^{(i)}) \to y$.
|
||||
% We have that $f^{-1} = \pi_{\text{odd}}$ is continuous.
|
||||
% 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}
|
||||
(b)
|
||||
Let $f(x^{(i)})$ be a sequence in $f(X)$.
|
||||
Suppose that $f(x^{(i)}) \to y$.
|
||||
We have that $f^{-1} = \pi_{\text{odd}}$ is continuous.
|
||||
Then $\pi_{\text{odd}}(f(x^{(i)}) \to \pi_{\text{odd}}(y)$.
|
||||
Since $\pi_{\text{even}}$ converges, we have $\pi_{\text{odd}}(y) \in X$.
|
||||
|
||||
|
||||
\nr 3
|
||||
|
|
@ -143,13 +130,6 @@ Let $X$ be a Polish space. Then $X$ is homeomorphic to a closed subspace of $\R^
|
|||
\end{proof}
|
||||
|
||||
\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)]
|
||||
\item $f$ is a topological embedding:
|
||||
Consider a basic open set
|
||||
|
|
|
|||
|
|
@ -8,7 +8,7 @@ Let $A \neq \emptyset$ be discrete.
|
|||
For $D \subseteq A^{\omega}$,
|
||||
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)]
|
||||
\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}
|
||||
(i) $\implies$ (ii):
|
||||
Let $(Y,T)$ be a subflow of $(X,T)$.
|
||||
take $y \in Y$. Then $Ty$ is dense in $X$.
|
||||
take $y \in Y$. Then $Ty$ is dense in mKX.
|
||||
But $Ty \subseteq Y$, so $Y$ is dense in $X$.
|
||||
Since $Y$ is closed, we get $Y = X$.
|
||||
|
||||
|
|
|
|||
|
|
@ -84,12 +84,11 @@ with parameter $\alpha \in \R$, $1 \cdot x \coloneqq x + \alpha$.
|
|||
\nr 4
|
||||
|
||||
% Examinable!
|
||||
% TODO THINK!
|
||||
|
||||
\gist{%
|
||||
% RECAP
|
||||
Let $X$ be a metrizable topological space
|
||||
and let $K(X) \coloneqq \{ K \subseteq X : K \text{ compact}\}$.
|
||||
Let $X$ be a metrizable topological space.
|
||||
|
||||
Let $K(X) \coloneqq \{ K \subseteq X : \text{ compact}\}$.
|
||||
|
||||
The Vietoris topology has a basis given by
|
||||
$\{K \subseteq U\}$, $U$ open (type 1)
|
||||
|
|
@ -104,21 +103,19 @@ $\max_{a \in A} d(a,B)$.
|
|||
On previous sheets, we checked that $d_H$ is a metric.
|
||||
If $X$ is separable, then so is $K(X)$.
|
||||
% END RECAP
|
||||
}{}
|
||||
|
||||
\begin{fact}
|
||||
\label{fact:s12e4}
|
||||
Let $(X,d)$ be a complete metric space.
|
||||
Then so is $(K(X), d_H)$.
|
||||
\end{fact}
|
||||
\begin{refproof}{fact:s12e4}
|
||||
\begin{proof}
|
||||
We need to show that $(K(X), d_H)$ is complete.
|
||||
|
||||
Let $(K_n)_{ n< \omega}$ be Cauchy in $(K(X), d_H)$.
|
||||
Wlog.~$K_n \neq \emptyset$ for all $n$.
|
||||
|
||||
Let $K = \{ x \in X : \forall x \in U \overset{\text{open}}{\subseteq} X.~
|
||||
\text{ $U \cap K_n \neq \emptyset$ for infinitely many $n$}\}$.
|
||||
\text{ $X$ intersects $K_n$ for infinitely many $n$}\}$.
|
||||
|
||||
Equivalently,
|
||||
$K = \{x : x \text{ is a cluster point of some subsequence $(x_n)$ with $x_n \in K_n$ for all $K_n$}\}$.
|
||||
|
|
@ -126,12 +123,12 @@ Then so is $(K(X), d_H)$.
|
|||
(A cluster point is a limit of some subsequence).
|
||||
|
||||
\begin{claim}
|
||||
\label{fact:s12e4:c1}
|
||||
$K_n \to K$.
|
||||
\end{claim}
|
||||
\begin{refproof}{fact:s12e4:c1}
|
||||
\begin{subproof}
|
||||
Note that $K$ is closed (the complement is open).
|
||||
|
||||
|
||||
\begin{claim}
|
||||
$K \neq \emptyset$.
|
||||
\end{claim}
|
||||
|
|
@ -162,7 +159,7 @@ Then so is $(K(X), d_H)$.
|
|||
space, it is complete.
|
||||
|
||||
So it suffices to show that $K$ is totally bounded.
|
||||
Let $\epsilon > 0$.
|
||||
Let $\epsilon > 0$
|
||||
Take $N$ such that $d_H(K_i,K_j) < \epsilon$
|
||||
for all $i,j \ge N$.
|
||||
|
||||
|
|
@ -203,8 +200,9 @@ Then so is $(K(X), d_H)$.
|
|||
To do this, construct a sequence of $y_{n_i} \in K_{n_i}$
|
||||
starting with $y$ such that $d(y_{n_i}, y_{n_{i+1}}) < \frac{\epsilon}{2^{i+2}}$.
|
||||
(same trick as before).
|
||||
\end{refproof}
|
||||
\end{refproof}
|
||||
\end{subproof}
|
||||
|
||||
\end{proof}
|
||||
|
||||
\begin{fact}
|
||||
If $X$ is compact metrisable,
|
||||
|
|
@ -225,3 +223,9 @@ Then so is $(K(X), d_H)$.
|
|||
|
||||
|
||||
% TODO complete and totally bounded Sutherland metric and topological spaces
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
|
|
|||
|
|
@ -2,7 +2,7 @@
|
|||
\tutorial{15}{2024-01-31}{Additions}
|
||||
|
||||
The following is not relevant for the exam,
|
||||
but aims to give a more general picture.
|
||||
but gives a more general picture.
|
||||
|
||||
Let $X$ be a topological space
|
||||
and let $\cF$ be a filter on $ X$.
|
||||
|
|
@ -12,7 +12,6 @@ all sets containing an open neighbourhood of $x$,
|
|||
is contained in $\cF$.
|
||||
|
||||
\begin{fact}
|
||||
\label{fact:hdifffilterlimit}
|
||||
$X$ is Hausdorff iff every filter has at most one limit point.
|
||||
\end{fact}
|
||||
\begin{proof}
|
||||
|
|
@ -22,7 +21,6 @@ is contained in $\cF$.
|
|||
\end{proof}
|
||||
|
||||
\begin{fact}
|
||||
\label{fact:compactiffufconv}
|
||||
$X$ is (quasi-) compact
|
||||
iff every ultrafilter converges.
|
||||
\end{fact}
|
||||
|
|
@ -31,7 +29,7 @@ is contained in $\cF$.
|
|||
Let $\cU$ be an ultrafilter.
|
||||
Consider the family $\cV = \{\overline{A} : A \in \cU\}$
|
||||
of closed sets.
|
||||
By the FIP we get that there exist
|
||||
By the FIP we geht that there exist
|
||||
$c \in X$ such that $c \in \overline{A}$ for all $A \in \cU$.
|
||||
Let $N$ be an open neighbourhood of $c$.
|
||||
If $N^c \in \cU$, then $c \in N^c \lightning$
|
||||
|
|
@ -71,19 +69,17 @@ so is $f(\cB)$.
|
|||
\end{fact}
|
||||
\begin{proof}
|
||||
Consider $(f,g)^{-1}(\Delta) \supseteq A$.
|
||||
The RHS is a dense closed set, i.e.~the entire space.
|
||||
\end{proof}
|
||||
|
||||
We can uniquely extend a continuous $f\colon X \to Y$
|
||||
We can uniquely extend $f\colon X \to Y$ continuous
|
||||
to a continuous $\overline{f}\colon \beta X \to Y$
|
||||
by setting $\overline{f}(\cU) \coloneqq \lim_\cU f$.
|
||||
|
||||
% Let $V$ be an open neighbourhood of $y \in \overline{f}\left( U \right)$.
|
||||
% Consider $f^{-1}(V)$.
|
||||
% Then
|
||||
% \[
|
||||
% \{\cF \in \beta\N : \cF \ni f^{-1}(V)\}
|
||||
% \]
|
||||
% is a basic open set.
|
||||
Let $V$ be an open neighbourhood of $Y$ in $\overline{f}\left( U) \right) $.
|
||||
Consider $f^{-1}(V)$.
|
||||
Consider the basic open set
|
||||
\[
|
||||
\{\cF \in \beta\N : \cF \ni f^{-1}(V)\}.
|
||||
\]
|
||||
|
||||
\todo{I missed the last 5 minutes}
|
||||
|
|
|
|||
Loading…
Reference in a new issue