Merge branch 'main' of https://git.abstractnonsen.se/josia-notes/w23-logic-3
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67a851748e
20 changed files with 624 additions and 381 deletions
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@ -1,4 +1,6 @@
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These are my notes on the lecture Probability Theory,
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\gist{%
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These are my notes on the lecture
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Logic 3: Abstract Topological Dynamics and Descriptive Set Theory
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taught by \textsc{Aleksandra Kwiatkowska}
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in the summer term 2023 at the University Münster.
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|
@ -18,3 +20,15 @@ I could not attend!
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This notes follow the way the material was presented in the lecture rather
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closely. Additions (e.g.~from exercise sheets)
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and slight modifications have been marked with $\dagger$.
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}{
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||||
This document aims to give a very brief summary of
|
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my \href{https://josia-notes.users.abstractnonsen.se/w23-logic-3/logic3.pdf}{notes on the course Logic 3}.
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||||
I try to omit most technical details and only summarize the most important
|
||||
ideas.
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|
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Note that this is currently work in progress.
|
||||
Currently the differences to the original document are only
|
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minor (this is still mostly a technical test),
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but this document will get shorter as I work through
|
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and summarize it.
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}
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|
|
|
@ -76,7 +76,7 @@ However the converse of this does not hold.
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\end{itemize}
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\end{fact}
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\begin{fact}
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Compact\footnote{It is not clear whether compact means compact and Hausdorff in this lecture.} Hausdorff spaces are \vocab{normal} (T4)
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Compact Hausdorff spaces are \vocab{normal} (T4)
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i.e.~two disjoint closed subsets can be separated
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by open sets.
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\end{fact}
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|
@ -114,7 +114,7 @@ However the converse of this does not hold.
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\end{absolutelynopagebreak}
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\subsection{Some facts about polish spaces}
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\gist{%
|
||||
\begin{fact}
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Let $(X, \tau)$ be a topological space.
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Let $d$ be a metric on $X$.
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|
@ -130,9 +130,10 @@ To show that $\tau_d = \tau_{d'}$
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for two metrics $d, d'$,
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suffices to show that open balls in one metric are unions of open balls in the other.
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\end{fact}
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}{}
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|
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\begin{notation}
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We sometimes denote $\min(a,b)$ by $a \wedge b$.
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We sometimes\footnote{only in this subsection?} denote $\min(a,b)$ by $a \wedge b$.
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\end{notation}
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\begin{proposition}
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|
@ -142,6 +143,7 @@ suffices to show that open balls in one metric are unions of open balls in the o
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Then $d' \coloneqq \min(d,1)$ is also a metric compatible with $\tau$.
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\end{proposition}
|
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\gist{%
|
||||
\begin{proof}
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To check the triangle inequality:
|
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\begin{IEEEeqnarray*}{rCl}
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|
@ -154,6 +156,7 @@ suffices to show that open balls in one metric are unions of open balls in the o
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|
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Since $d$ is complete, we have that $d'$ is complete.
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\end{proof}
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}{}
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\begin{proposition}
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Let $A$ be a Polish space.
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Then $A^{\omega}$ Polish.
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|
@ -252,15 +255,15 @@ suffices to show that open balls in one metric are unions of open balls in the o
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\begin{proposition}
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Closed subspaces of Polish spaces are Polish.
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\end{proposition}
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\gist{}{
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\gist{%
|
||||
\begin{proof}
|
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Let $X$ be Polish and $V \subseteq X$ closed.
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Let $d$ be a complete metric on $X$.
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Then $d\defon{V}$ is complete.
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Subspaces of second countable spaces
|
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are second countable.
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\end{proof}
|
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}
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\end{proof}%
|
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}{}
|
||||
|
||||
\begin{definition}
|
||||
Let $X$ be a topological space.
|
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|
|
|
@ -72,41 +72,51 @@
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\[d_1((x_1,y_1), (x_2, y_2)) \coloneqq d(x_1,x_2) + |y_1 - y_2|\]
|
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metric is complete.
|
||||
|
||||
$f_U$ is an embedding of $U$ into $X \times \R$\gist{:
|
||||
\begin{itemize}
|
||||
\item It is injective because of the first coordinate.
|
||||
\item It is continuous since $d(x, U^c)$ is continuous
|
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and only takes strictly positive values. % TODO
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\item The inverse is continuous because projections
|
||||
are continuous.
|
||||
\end{itemize}
|
||||
}{.}
|
||||
$f_U$ is an embedding of $U$ into $X \times \R$%
|
||||
\gist{:
|
||||
\begin{itemize}
|
||||
\item It is injective because of the first coordinate.
|
||||
\item It is continuous since $d(x, U^c)$ is continuous
|
||||
and only takes strictly positive values. % TODO
|
||||
\item The inverse is continuous because projections
|
||||
are continuous.
|
||||
\end{itemize}
|
||||
}{.}
|
||||
|
||||
So we have shown that $U$ and
|
||||
the graph of $\tilde{f_U}\colon x \mapsto \frac{1}{d(x, U^c)}$
|
||||
are homeomorphic.
|
||||
The graph is closed \gist{in $U \times \R$,
|
||||
because $\tilde{f_U}$ is continuous.
|
||||
It is closed}{} in $X \times \R$ \gist{because
|
||||
$\tilde{f_U} \to \infty$ for $d(x, U^c) \to 0$}{}.
|
||||
\todo{Make this precise}
|
||||
\gist{%
|
||||
So we have shown that $U$ and
|
||||
the graph of $\tilde{f_U}\colon x \mapsto \frac{1}{d(x, U^c)}$
|
||||
are homeomorphic.
|
||||
The graph is closed \gist{in $U \times \R$,
|
||||
because $\tilde{f_U}$ is continuous.
|
||||
It is closed}{} in $X \times \R$ \gist{because
|
||||
$\tilde{f_U} \to \infty$ for $d(x, U^c) \to 0$}{}.
|
||||
\todo{Make this precise}
|
||||
|
||||
Therefore we identified $U$ with a closed subspace of
|
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the Polish space $(X \times \R, d_1)$.
|
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}{%
|
||||
So $U \cong \mathop{Graph}(x \mapsto \frac{1}{d(x, U^c)})$
|
||||
and the RHS is a close subspace of the Polish space
|
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$(X \times \R, d_1)$.
|
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}
|
||||
|
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Therefore we identified $U$ with a closed subspace of
|
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the Polish space $(X \times \R, d_1)$.
|
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\end{refproof}
|
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|
||||
Let $Y = \bigcap_{n \in \N} U_n$ be $G_{\delta}$.
|
||||
Take
|
||||
Consider
|
||||
\begin{IEEEeqnarray*}{rCl}
|
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f_Y\colon Y &\longrightarrow & X \times \R^{\N} \\
|
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x &\longmapsto &
|
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\left(x, \left( \frac{1}{\delta(x,U_n^c)} \right)_{n \in \N}\right)
|
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\end{IEEEeqnarray*}
|
||||
|
||||
As for an open $U$, $f_Y$ is an embedding.
|
||||
Since $X \times \R^{\N}$
|
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is completely metrizable,
|
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so is the closed set $f_Y(Y) \subseteq X \times \R^\N$.
|
||||
\gist{
|
||||
As for an open $U$, $f_Y$ is an embedding.
|
||||
Since $X \times \R^{\N}$
|
||||
is completely metrizable,
|
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so is the closed set $f_Y(Y) \subseteq X \times \R^\N$.
|
||||
}{}
|
||||
|
||||
\begin{claim}
|
||||
\label{psubspacegdelta:c2}
|
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|
@ -123,36 +133,35 @@
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|||
\item $\diam_d(U) \le \frac{1}{n}$,
|
||||
\item $\diam_{d_Y}(U \cap Y) \le \frac{1}{n}$.
|
||||
\end{enumerate}
|
||||
\gist{
|
||||
We want to show that $Y = \bigcap_{n \in \N} V_n$.
|
||||
For $x \in Y$, $n \in \N$ we have $x \in V_n$,
|
||||
as we can choose two neighbourhoods
|
||||
$U_1$ (open in $Y)$ and $U_2$ (open in $X$ ) of $x$,
|
||||
such that $\diam_{d_Y}(U) < \frac{1}{n}$
|
||||
and $U_2 \cap Y = U_1$.
|
||||
Additionally choose $x \in U_3$ open in $X$
|
||||
with $\diam_{d}(U_3) < \frac{1}{n}$.
|
||||
Then consider $U_2 \cap U_3 \subseteq V_n$.
|
||||
Hence $Y \subseteq \bigcap_{n \in \N} V_n$.
|
||||
\gist{%
|
||||
We want to show that $Y = \bigcap_{n \in \N} V_n$.
|
||||
For $x \in Y$, $n \in \N$ we have $x \in V_n$,
|
||||
as we can choose two neighbourhoods
|
||||
$U_1$ (open in $Y)$ and $U_2$ (open in $X$ ) of $x$,
|
||||
such that $\diam_{d_Y}(U) < \frac{1}{n}$
|
||||
and $U_2 \cap Y = U_1$.
|
||||
Additionally choose $x \in U_3$ open in $X$
|
||||
with $\diam_{d}(U_3) < \frac{1}{n}$.
|
||||
Then consider $U_2 \cap U_3 \subseteq V_n$.
|
||||
Hence $Y \subseteq \bigcap_{n \in \N} V_n$.
|
||||
|
||||
Now let $x \in \bigcap_{n \in \N} V_n$.
|
||||
For each $n$ pick $x \in U_n \subseteq X$ open
|
||||
satisfying (i), (ii), (iii).
|
||||
From (i) and (ii) it follows that $x \in \overline{Y}$,
|
||||
since we can consider a sequence of points $y_n \in U_n \cap Y$
|
||||
and get $y_n \xrightarrow{d} x$.
|
||||
For all $n$ we have that $U_n' \coloneqq U_1 \cap \ldots \cap U_n$
|
||||
is an open set containing $x$,
|
||||
hence $U_n' \cap Y \neq \emptyset$.
|
||||
Thus we may assume that the $U_i$ form a decreasing sequence.
|
||||
We have that $\diam_{d_Y}(U_n \cap Y) \le \frac{1}{n}$.
|
||||
If follows that the $y_n$ form a Cauchy sequence with respect to $d_Y$,
|
||||
since $\diam(U_n \cap Y) \xrightarrow{d_Y} 0$
|
||||
and thus $\diam(\overline{U_n \cap Y}) \xrightarrow{d_Y} 0$.
|
||||
The sequence $y_n$ converges to the unique point in
|
||||
$\bigcap_{n} \overline{U_n \cap Y}$.
|
||||
Since the topologies agree, this point is $x$.
|
||||
}{Then $Y = \bigcap_n U_n$.}
|
||||
Now let $x \in \bigcap_{n \in \N} V_n$.
|
||||
For each $n$ pick $x \in U_n \subseteq X$ open
|
||||
satisfying (i), (ii), (iii).
|
||||
From (i) and (ii) it follows that $x \in \overline{Y}$,
|
||||
since we can consider a sequence of points $y_n \in U_n \cap Y$
|
||||
and get $y_n \xrightarrow{d} x$.
|
||||
For all $n$ we have that $U_n' \coloneqq U_1 \cap \ldots \cap U_n$
|
||||
is an open set containing $x$,
|
||||
hence $U_n' \cap Y \neq \emptyset$.
|
||||
Thus we may assume that the $U_i$ form a decreasing sequence.
|
||||
We have that $\diam_{d_Y}(U_n \cap Y) \le \frac{1}{n}$.
|
||||
If follows that the $y_n$ form a Cauchy sequence with respect to $d_Y$,
|
||||
since $\diam(U_n \cap Y) \xrightarrow{d_Y} 0$
|
||||
and thus $\diam(\overline{U_n \cap Y}) \xrightarrow{d_Y} 0$.
|
||||
The sequence $y_n$ converges to the unique point in
|
||||
$\bigcap_{n} \overline{U_n \cap Y}$.
|
||||
Since the topologies agree, this point is $x$.
|
||||
}{Then $Y = \bigcap_n U_n$.}
|
||||
\end{refproof}
|
||||
\end{refproof}
|
||||
|
||||
|
|
|
@ -2,7 +2,7 @@
|
|||
|
||||
\subsection{Trees}
|
||||
|
||||
|
||||
\gist{%
|
||||
\begin{notation}
|
||||
Let $A \neq \emptyset$, $n \in \N$.
|
||||
Then
|
||||
|
@ -59,6 +59,7 @@
|
|||
define extension, initial segments
|
||||
and concatenation of a finite sequence with an infinite one.
|
||||
\end{notation}
|
||||
}{}
|
||||
|
||||
\begin{definition}
|
||||
A \vocab{tree}
|
||||
|
@ -127,16 +128,19 @@
|
|||
|
||||
We define $U_s$ inductively on the length of $s$.
|
||||
|
||||
For $U_{\emptyset}$ take any non-empty open set
|
||||
with small enough diameter.
|
||||
\gist{%
|
||||
For $U_{\emptyset}$ take any non-empty open set
|
||||
with small enough diameter.
|
||||
|
||||
Given $U_s$, pick $x \neq y \in U_s$
|
||||
and let $U_{s \concat 0} \ni x$,
|
||||
$U_{s \concat 1} \ni y$
|
||||
be disjoint, open,
|
||||
of diameter $\le \frac{1}{2^{|s| +1}}$
|
||||
and such that $\overline{U_{s\concat 0}}, \overline{U_{S \concat 1}} \subseteq U_s$.
|
||||
Given $U_s$, pick $x \neq y \in U_s$
|
||||
and let $U_{s \concat 0} \ni x$,
|
||||
$U_{s \concat 1} \ni y$
|
||||
be disjoint, open,
|
||||
of diameter $\le \frac{1}{2^{|s| +1}}$
|
||||
and such that $\overline{U_{s\concat 0}}, \overline{U_{S \concat 1}} \subseteq U_s$.
|
||||
}{}
|
||||
|
||||
\gist{%
|
||||
Let $x \in 2^{\N}$.
|
||||
Then let $f(x)$ be the unique point in $X$
|
||||
such that
|
||||
|
@ -147,19 +151,23 @@
|
|||
It is clear that $f$ is injective and continuous.
|
||||
% TODO: more details
|
||||
$2^{\N}$ is compact, hence $f^{-1}$ is also continuous.
|
||||
}{Consider $f\colon 2^{\N} \hookrightarrow X, x \mapsto y$, where $\{y\} = \bigcap_n U_{x\defon n}$.
|
||||
By compactness of $2^{\N}$, we get that $f^{-1}$ is continuous.}
|
||||
\end{proof}
|
||||
|
||||
\begin{corollary}
|
||||
\label{cor:perfectpolishcard}
|
||||
Every nonempty perfect Polish
|
||||
space $X$ has cardinality $\fc = 2^{\aleph_0}$
|
||||
% TODO: eulerscript C ?
|
||||
\end{corollary}
|
||||
\begin{proof}
|
||||
Since the cantor space embeds into $X$,
|
||||
we get the lower bound.
|
||||
Since $X$ is second countable and Hausdorff,
|
||||
we get the upper bound.
|
||||
\gist{%
|
||||
Since the cantor space embeds into $X$,
|
||||
we get the lower bound.
|
||||
Since $X$ is second countable and Hausdorff,
|
||||
we get the upper bound.%
|
||||
}{Lower bound: $2^{\N} \hookrightarrow X$,
|
||||
upper bound: \nth{2} countable and Hausdorff.}
|
||||
\end{proof}
|
||||
|
||||
\begin{theorem}
|
||||
|
@ -203,12 +211,14 @@
|
|||
countable union of closed sets,
|
||||
i.e.~the complement of a $G_\delta$ set.
|
||||
\end{definition}
|
||||
\gist{%
|
||||
\begin{observe}
|
||||
\begin{itemize}
|
||||
\item Any open set is $F {\sigma}$.
|
||||
\item In metric spaces the intersection of an open and closed set is $F_\sigma$.
|
||||
\end{itemize}
|
||||
\end{observe}
|
||||
}{}
|
||||
\begin{refproof}{thm:bairetopolish}
|
||||
Let $d$ be a complete metric on $X$.
|
||||
W.l.o.g.~$\diam(X) \le 1$.
|
||||
|
@ -220,7 +230,7 @@
|
|||
\item $F_\emptyset = X$,
|
||||
\item $F_s$ is $F_\sigma$ for all $s$.
|
||||
\item The $F_{s \concat i}$ partition $F_s$,
|
||||
i.e.~$F_{s} = \bigsqcup_i F_{s \concat i}$. % TODO change notation?
|
||||
i.e.~$F_{s} = \bigsqcup_i F_{s \concat i}$.
|
||||
|
||||
Furthermore we want that
|
||||
$\overline{F_{s \concat i}} \subseteq F_s$
|
||||
|
@ -228,6 +238,7 @@
|
|||
\item $\diam(F_s) \le 2^{-|s|}$.
|
||||
\end{enumerate}
|
||||
|
||||
\gist{%
|
||||
Suppose we already have $F_s \text{\reflectbox{$\coloneqq$}} F$.
|
||||
We need to construct a partition $(F_i)_{i \in \N}$
|
||||
of $F$ with $\overline{F_i} \subseteq F$
|
||||
|
@ -252,6 +263,7 @@
|
|||
The sets $D_i \coloneqq F_i^0 \cap B_i \setminus (B_1 \cup \ldots \cup B_{i-1})$
|
||||
are $F_\sigma$, disjoint
|
||||
and $F_i^0 = \bigcup_{j} D_j$.
|
||||
}{Induction.}
|
||||
|
||||
|
||||
|
||||
|
|
|
@ -5,6 +5,7 @@
|
|||
\end{remark}
|
||||
|
||||
\begin{refproof}{thm:bairetopolish}
|
||||
\gist{%
|
||||
Take
|
||||
\[D = \{x \in \cN : \bigcap_{n} F_{x\defon{n}} \neq \emptyset\}.\]
|
||||
|
||||
|
@ -13,15 +14,18 @@
|
|||
\[
|
||||
\bigcap_{n} F_{x\defon{n}} = \bigcap_{n} \overline{F_{x\defon{n}}}.
|
||||
\]
|
||||
}{}
|
||||
|
||||
$f\colon D \to X$ is determined by
|
||||
\[
|
||||
\{f(x)\} = \bigcap_{n} F_{x\defon{n}}
|
||||
\]
|
||||
|
||||
$f$ is injective and continuous.
|
||||
The proof of this is exactly the same as in
|
||||
\yaref{thm:cantortopolish}.
|
||||
\gist{%
|
||||
$f$ is injective and continuous.
|
||||
The proof of this is exactly the same as in
|
||||
\yaref{thm:cantortopolish}.
|
||||
}{}
|
||||
|
||||
\begin{claim}
|
||||
\label{thm:bairetopolish:c1}
|
||||
|
@ -60,7 +64,7 @@
|
|||
|
||||
Take $S \coloneqq \{s \in \N^{<\N}: \exists x \in D, n \in \N.~x=s\defon{n}\}$.
|
||||
Clearly $S$ is a pruned tree.
|
||||
Moreover, since $D$ is closed, we have that\todo{Proof this (homework?)}
|
||||
Moreover, since $D$ is closed, we have that (cf.~\yaref{s3e1})
|
||||
\[
|
||||
D = [S] = \{x \in \N^\N : \forall n \in \N.~x\defon{n} \in S\}.
|
||||
\]
|
||||
|
@ -76,21 +80,27 @@
|
|||
\item $|s| = \phi(|s|)$,
|
||||
\item if $s \in S$, then $\phi(s) = s$.
|
||||
\end{itemize}
|
||||
Let $\phi(\emptyset) = \emptyset$.
|
||||
Suppose that $\phi(t)$ is defined.
|
||||
If $t\concat a \in S$, then set
|
||||
$\phi(t\concat a) \coloneqq t\concat a$.
|
||||
Otherwise take some $b$ such that
|
||||
$t\concat b \in S$ and define
|
||||
$\phi(t\concat a) \coloneqq \phi(t)\concat b$.
|
||||
\gist{%
|
||||
Let $\phi(\emptyset) = \emptyset$.
|
||||
Suppose that $\phi(t)$ is defined.
|
||||
If $t\concat a \in S$, then set
|
||||
$\phi(t\concat a) \coloneqq t\concat a$.
|
||||
Otherwise take some $b$ such that
|
||||
$t\concat b \in S$ and define
|
||||
$\phi(t\concat a) \coloneqq \phi(t)\concat b$.%
|
||||
}{}%
|
||||
This is possible since $S$ is pruned.
|
||||
|
||||
Let $r\colon \cN = [\N^{<\N}] \to [S] = D$
|
||||
be the function defined by $r(x) = \bigcup_n f(x\defon{n})$.
|
||||
\gist{%
|
||||
Let $r\colon \cN = [\N^{<\N}] \to [S] = D$
|
||||
be the function defined by $r(x) = \bigcup_n f(x\defon{n})$.
|
||||
}{}
|
||||
|
||||
$r$ is continuous, since
|
||||
$d_{\cN}(r(x), r(y)) \le d_{\cN}(x,y)$. % Lipschitz
|
||||
It is immediate that $r$ is a retraction.
|
||||
\gist{%
|
||||
It is immediate that $r$ is a retraction.
|
||||
}{}
|
||||
\end{refproof}
|
||||
|
||||
\section{Meager and Comeager Sets}
|
||||
|
@ -117,9 +127,11 @@
|
|||
The complement of a meager set is called
|
||||
\vocab{comeager}.
|
||||
\end{definition}
|
||||
\gist{%
|
||||
\begin{example}
|
||||
$\Q \subseteq \R$ is meager.
|
||||
\end{example}
|
||||
}{}
|
||||
\begin{notation}
|
||||
Let $A, B \subseteq X$.
|
||||
We write $A =^\ast B$
|
||||
|
@ -127,25 +139,29 @@
|
|||
$A \symdif B \coloneqq (A\setminus B) \cup (B \setminus A)$,
|
||||
is meager.
|
||||
\end{notation}
|
||||
\gist{%
|
||||
\begin{remark}
|
||||
$=^\ast$ is an equivalence relation.
|
||||
\end{remark}
|
||||
}{}
|
||||
\begin{definition}
|
||||
A set $A \subseteq X$
|
||||
has the \vocab{Baire property} (\vocab{BP})
|
||||
if $A =^\ast U$ for some $U \overset{\text{open}}{\subseteq} X$.
|
||||
\end{definition}
|
||||
\gist{%
|
||||
Note that open sets and meager sets have the Baire property.
|
||||
}{}
|
||||
|
||||
|
||||
\gist{%
|
||||
\begin{example}
|
||||
\begin{itemize}
|
||||
\item $\Q \subseteq \R$ is $F_\sigma$.
|
||||
\item $\R \setminus \Q \subseteq \R$ is $G_\delta$.
|
||||
\item $\Q \subseteq \R$ is not $G_{\delta}$.
|
||||
(It is dense and meager,
|
||||
\item $\Q \subseteq \R$ is not $G_{\delta}$:
|
||||
It is dense and meager,
|
||||
hence it can not be $G_\delta$
|
||||
by the Baire category theorem).
|
||||
by the \yaref{thm:bct}.
|
||||
\end{itemize}
|
||||
|
||||
\end{example}
|
||||
}{}
|
||||
|
|
|
@ -6,10 +6,9 @@
|
|||
\item If $F$ is closed then $F$ is nwd iff $X \setminus F$ is open and dense.
|
||||
\item Any meager set $B$ is contained in a meager $F_{\sigma}$-set.
|
||||
\end{itemize}
|
||||
|
||||
|
||||
\end{fact}
|
||||
\begin{proof} % remove?
|
||||
\gist{%
|
||||
\begin{proof}
|
||||
\begin{itemize}
|
||||
\item This follows from the definition as $\overline{\overline{A}} = \overline{A}$.
|
||||
\item Trivial.
|
||||
|
@ -17,7 +16,9 @@
|
|||
Then $B \subseteq \bigcup_{n < \omega} \overline{B_n}$.
|
||||
\end{itemize}
|
||||
\end{proof}
|
||||
}{}
|
||||
|
||||
\gist{%
|
||||
\begin{definition}
|
||||
A \vocab{$\sigma$-algebra} on a set $X$
|
||||
is a collection of subsets of $X$
|
||||
|
@ -32,14 +33,15 @@
|
|||
Since $\bigcap_{i < \omega} A_i = \left( \bigcup_{i < \omega} A_i^c \right)^c$
|
||||
we have that $\sigma$-algebras are closed under countable intersections.
|
||||
\end{fact}
|
||||
}{}
|
||||
|
||||
\begin{theorem}
|
||||
\label{thm:bairesigma}
|
||||
Let $X$ be a topological space.
|
||||
Then the collection of sets with the Baire property
|
||||
is a $\sigma$-algebra on $X$.
|
||||
is \gist{a $\sigma$-algebra on $X$.
|
||||
|
||||
It is the smallest $\sigma$-algebra
|
||||
It is}{} the smallest $\sigma$-algebra
|
||||
containing all meager and open sets.
|
||||
\end{theorem}
|
||||
\begin{refproof}{thm:bairesigma}
|
||||
|
@ -97,6 +99,7 @@
|
|||
%\end{example}
|
||||
|
||||
\begin{theorem}[Baire Category theorem]
|
||||
\yalabel{Baire Category Theorem}{Baire Category Theorem}{thm:bct}
|
||||
Let $X$ be a completely metrizable space.
|
||||
Then every comeager set of $X$ is dense in $X$.
|
||||
\end{theorem}
|
||||
|
@ -111,8 +114,8 @@
|
|||
\item The intersection of countable many
|
||||
open dense sets is dense.
|
||||
\end{enumerate}
|
||||
In this case $X$ is called a \vocab{Baire space}.
|
||||
\footnote{see \yaref{s5e1}}
|
||||
In this case $X$ is called a \vocab{Baire space}.%
|
||||
\footnote{cf.~\yaref{s5e1}}
|
||||
\end{theoremdef}
|
||||
\begin{proof}
|
||||
\todo{Proof (short)}
|
||||
|
@ -129,7 +132,6 @@
|
|||
This proof can be adapted to other open sets $X$.
|
||||
\end{proof}
|
||||
|
||||
|
||||
\begin{notation}
|
||||
Let $X ,Y$ be topological spaces,
|
||||
$A \subseteq X \times Y$
|
||||
|
@ -273,9 +275,11 @@ but for meager sets:
|
|||
% \end{refproof}
|
||||
% TODO fix claim numbers
|
||||
|
||||
\gist{%
|
||||
\begin{remark}
|
||||
Suppose that $A$ has the BP.
|
||||
Then there is an open $U$ such that
|
||||
$A \symdif U \mathbin{\text{\reflectbox{$\coloneqq$}}} M$ is meager.
|
||||
Then $A = U \symdif M$.
|
||||
\end{remark}
|
||||
}{}
|
||||
|
|
|
@ -101,7 +101,6 @@ Then define
|
|||
\Pi^0_\alpha(X) \coloneqq \lnot \Sigma^0_\alpha(X) \coloneqq
|
||||
\{X \setminus A | A \in \Sigma^0_\alpha(X)\},
|
||||
\]
|
||||
% \todo{Define $\lnot$ (element-wise complement)}
|
||||
and for $\alpha > 1$
|
||||
\[
|
||||
\Sigma^0_\alpha \coloneqq \{\bigcup_{n < \omega} A_n : A_n \in \Pi^0_{\alpha_n}(X) \text{ for some $\alpha_n < \alpha$}\}.
|
||||
|
|
|
@ -37,17 +37,18 @@
|
|||
\item \begin{itemize}
|
||||
\item $\Sigma^0_\xi(X)$ is closed under countable unions.
|
||||
\item $\Pi^0_\xi(X)$ is closed under countable intersections.
|
||||
\item $\Delta^0_\xi(X)$ is closed under complements,
|
||||
countable unions and
|
||||
countable intersections.
|
||||
\item $\Delta^0_\xi(X)$ is closed under complements.
|
||||
\end{itemize}
|
||||
\item \begin{itemize}
|
||||
\item $\Sigma^0_\xi(X)$ is closed under \emph{finite} intersections.
|
||||
\item $\Pi^0_\xi(X)$ is closed under \emph{finite} unions.
|
||||
\item $\Delta^0_\xi(X)$ is closed under finite unions and
|
||||
finite intersections.
|
||||
\end{itemize}
|
||||
|
||||
\end{enumerate}
|
||||
\end{proposition}
|
||||
\gist{%
|
||||
\begin{proof}
|
||||
\begin{enumerate}[(a)]
|
||||
\item This follows directly from the definition.
|
||||
|
@ -67,24 +68,27 @@
|
|||
|
||||
\end{enumerate}
|
||||
\end{proof}
|
||||
}{}
|
||||
\begin{example}
|
||||
Consider the cantor space $2^{\omega}$.
|
||||
We have that $\Delta^0_1(2^{\omega})$
|
||||
is not closed under countable unions
|
||||
(countable unions yield all open sets, but there are open
|
||||
sets that are not clopen).
|
||||
is not closed under countable unions%
|
||||
\gist{ (countable unions yield all open sets, but there are open
|
||||
sets that are not clopen)}{}.
|
||||
\end{example}
|
||||
|
||||
\subsection{Turning Borels Sets into Clopens}
|
||||
|
||||
\begin{theorem}%
|
||||
\gist{%
|
||||
\footnote{Whilst strikingly concise the verb ``\vocab[Clopenization™]{to clopenize}''
|
||||
unfortunately seems to be non-standard vocabulary.
|
||||
Our tutor repeatedly advised against using it in the final exam.
|
||||
Contrary to popular belief
|
||||
the very same tutor was \textit{not} the one first to introduce it,
|
||||
as it would certainly be spelled ``to clopenise'' if that were the case.
|
||||
}
|
||||
}%
|
||||
}{}%
|
||||
\label{thm:clopenize}
|
||||
Let $(X, \cT)$ be a Polish space.
|
||||
For any Borel set $A \subseteq X$,
|
||||
|
@ -163,7 +167,7 @@
|
|||
such that $\cT_n \supseteq \cT$
|
||||
and $\cB(\cT_n) = \cB(\cT)$.
|
||||
Then the topology $\cT_\infty$ generated by $\bigcup_{n} \cT_n$
|
||||
is still Polish
|
||||
is Polish
|
||||
and $\cB(\cT_\infty) = \cB(T)$.
|
||||
\end{lemma}
|
||||
\begin{proof}
|
||||
|
@ -183,7 +187,51 @@
|
|||
definition of $\cF$ belong to
|
||||
a countable basis of the respective $\cT_n$).
|
||||
|
||||
\todo{This proof will be finished in the next lecture}
|
||||
% Proof was finished in lecture 8
|
||||
Let $Y = \prod_{n \in \N} (X, \cT_n)$.
|
||||
Then $Y$ is Polish.
|
||||
Let $\delta\colon (X, \cT_\infty) \to Y$
|
||||
defined by $\delta(x) = (x,x,x,\ldots)$.
|
||||
\begin{claim}
|
||||
$\delta$ is a homeomorphism.
|
||||
\end{claim}
|
||||
\begin{subproof}
|
||||
Clearly $\delta$ is a bijection.
|
||||
We need to show that it is continuous and open.
|
||||
|
||||
Let $U \in \cT_i$.
|
||||
Then
|
||||
\[
|
||||
\delta^{-1}(D \cap \left( X \times X \times \ldots\times U \times \ldots) \right)) = U \in \cT_i \subseteq \cT_\infty,
|
||||
\]
|
||||
hence $\delta$ is continuous.
|
||||
Let $U \in \cT_\infty$.
|
||||
Then $U$ is the union of sets of the form
|
||||
\[
|
||||
V = U_{n_1} \cap U_{n_2} \cap \ldots \cap U_{nu}
|
||||
\]
|
||||
for some $n_1 < n_2 < \ldots < n_u$
|
||||
and $U_{n_i} \in \cT_i$.
|
||||
|
||||
Thus is suffices to consider sets of this form.
|
||||
We have that
|
||||
\[
|
||||
\delta(V) = D \cap (X \times X \times \ldots \times U_{n_1} \times \ldots \times U_{n_2} \times \ldots \times U_{n_u} \times X \times \ldots) \overset{\text{open}}{\subseteq} D.
|
||||
\]
|
||||
\end{subproof}
|
||||
|
||||
This will finish the proof since
|
||||
\[
|
||||
D = \{(x,x,\ldots) \in Y : x \in X\} \overset{\text{closed}}{\subseteq} Y
|
||||
\]
|
||||
Why? Let $(x_n) \in Y \setminus D$.
|
||||
Then there are $i < j$ such that $x_i \neq x_j$.
|
||||
Take disjoint open $x_i \in U$, $x_j \in V$.
|
||||
Then
|
||||
\[(x_n) \in X \times X \times \ldots \times U \times \ldots \times X \times \ldots \times V \times X \times \ldots\]
|
||||
is open in $Y\setminus D$.
|
||||
Hence $Y \setminus D$ is open, thus $D$ is closed.
|
||||
It follows that $D$ is Polish.
|
||||
\end{proof}
|
||||
|
||||
We need to show that $A$ is closed under countable unions.
|
||||
|
|
|
@ -1,61 +1,8 @@
|
|||
\lecture{08}{2023-11-10}{}
|
||||
|
||||
\todo{put this lemma in the right place}
|
||||
\begin{lemma}[Lemma 2]
|
||||
Let $(X, \cT)$ be a Polish space.
|
||||
Let $\cT_n \supseteq \cT$ be Polish
|
||||
with $\cB(X, \cT_n) = \cB(X, \cT)$.
|
||||
Let $\cT_\infty$ be the topology generated
|
||||
by $\bigcup_n \cT_n$.
|
||||
Then $(X, \cT_\infty)$ is Polish
|
||||
and $\cB(X, \cT_\infty) = \cB(X, \cT)$.
|
||||
\end{lemma}
|
||||
\begin{proof}
|
||||
Let $Y = \prod_{n \in \N} (X, \cT_n)$.
|
||||
Then $Y$ is Polish.
|
||||
Let $\delta\colon (X, \cT_\infty) \to Y$
|
||||
defined by $\delta(x) = (x,x,x,\ldots)$.
|
||||
\begin{claim}
|
||||
$\delta$ is a homeomorphism.
|
||||
\end{claim}
|
||||
\begin{subproof}
|
||||
Clearly $\delta$ is a bijection.
|
||||
We need to show that it is continuous and open.
|
||||
|
||||
Let $U \in \cT_i$.
|
||||
Then
|
||||
\[
|
||||
\delta^{-1}(D \cap \left( X \times X \times \ldots\times U \times \ldots) \right)) = U \in \cT_i \subseteq \cT_\infty,
|
||||
\]
|
||||
hence $\delta$ is continuous.
|
||||
Let $U \in \cT_\infty$.
|
||||
Then $U$ is the union of sets of the form
|
||||
\[
|
||||
V = U_{n_1} \cap U_{n_2} \cap \ldots \cap U_{nu}
|
||||
\]
|
||||
for some $n_1 < n_2 < \ldots < n_u$
|
||||
and $U_{n_i} \in \cT_i$.
|
||||
|
||||
Thus is suffices to consider sets of this form.
|
||||
We have that
|
||||
\[
|
||||
\delta(V) = D \cap (X \times X \times \ldots \times U_{n_1} \times \ldots \times U_{n_2} \times \ldots \times U_{n_u} \times X \times \ldots) \overset{\text{open}}{\subseteq} D.
|
||||
\]
|
||||
\end{subproof}
|
||||
|
||||
This will finish the proof since
|
||||
\[
|
||||
D = \{(x,x,\ldots) \in Y : x \in X\} \overset{\text{closed}}{\subseteq} Y
|
||||
\]
|
||||
Why? Let $(x_n) \in Y \setminus D$.
|
||||
Then there are $i < j$ such that $x_i \neq x_j$.
|
||||
Take disjoint open $x_i \in U$, $x_j \in V$.
|
||||
Then
|
||||
\[(x_n) \in X \times X \times \ldots \times U \times \ldots \times X \times \ldots \times V \times X \times \ldots\]
|
||||
is open in $Y\setminus D$.
|
||||
Hence $Y \setminus D$ is open, thus $D$ is closed.
|
||||
It follows that $D$ is Polish.
|
||||
\end{proof}
|
||||
\lecture{08}{2023-11-10}{}\footnote{%
|
||||
In the beginning of the lecture, we finished
|
||||
the proof of \yaref{thm:clopenize:l2}.
|
||||
This has been moved to the notes on lecture 7.%
|
||||
}
|
||||
|
||||
\subsection{Parametrizations}
|
||||
%\todo{choose better title}
|
||||
|
|
|
@ -148,6 +148,7 @@ We will see later that $\Sigma^1_1(X) \cap \Pi^1_1(X) = \cB(X)$.
|
|||
|
||||
|
||||
\begin{theorem}
|
||||
\label{thm:universals11}
|
||||
Let $X,Y$ be uncountable Polish spaces.
|
||||
There exists a $Y$-universal $\Sigma^1_1(X)$ set.
|
||||
\end{theorem}
|
||||
|
|
|
@ -153,13 +153,14 @@ We will not proof this in this lecture.
|
|||
|
||||
\subsection{Ill-Founded Trees}
|
||||
|
||||
|
||||
\gist{%
|
||||
Recall that a \vocab{tree} on $\N$ is a subset of
|
||||
$\N^{<\N}$
|
||||
closed under taking initial segments.
|
||||
|
||||
We now identify trees with their characteristic functions,
|
||||
i.e.~we want to associate a tree $T \subseteq \N^{<\N}$
|
||||
i.e.~we want to associate a tree $T \subseteq \N^{<\N}$}%
|
||||
{We identify trees $T \subseteq \N^{<\N}$ with their characteristic functions:}
|
||||
\begin{IEEEeqnarray*}{rCl}
|
||||
\One_T\colon \omega^{<\omega} &\longrightarrow & \{0,1\} \\
|
||||
x &\longmapsto & \begin{cases}
|
||||
|
@ -167,16 +168,15 @@ i.e.~we want to associate a tree $T \subseteq \N^{<\N}$
|
|||
0 &: x \not\in T.
|
||||
\end{cases}
|
||||
\end{IEEEeqnarray*}
|
||||
Note that $\One_T \in {\{0,1\}^\N}^{< \N}$.
|
||||
\gist{Note that $\One_T \in {\{0,1\}^\N}^{< \N}$.}{}
|
||||
|
||||
Let $\Tr = \{T \in {2^{\N}}^{<\N} : T \text{ is a tree}\} \subseteq {2^{\N}}^{<\N}$.
|
||||
Let \vocab{$\Tr$} $ \coloneqq \{T \in {2^{\N}}^{<\N} : T \text{ is a tree}\} \subseteq {2^{\N}}^{<\N}$.
|
||||
|
||||
\begin{observe}
|
||||
\[
|
||||
\Tr \subseteq {2^{\N}}^{<\N}
|
||||
\]
|
||||
is closed (where we take the topology of the Cantor space).
|
||||
$\Tr \subseteq {2^{\N}}^{<\N}$ is closed
|
||||
(where we take the topology of the Cantor space).
|
||||
\end{observe}
|
||||
\gist{%
|
||||
Indeed, for any $ s \in \N^{<\N}$
|
||||
we have that $\{T \in {2^{\N}}^{<\N} : s \in T\}$
|
||||
and $\{T \in {2^{\N}}^{<\N} : s\not\in T\}$ are clopen.
|
||||
|
@ -185,6 +185,4 @@ In particular for $s$ fixed,
|
|||
we have that
|
||||
\[\{A \in {2^{\N}}^{<\N} : s \in A \text{ and } s' \in A \text{ for any initial segment $s' \subseteq s$}\}\]
|
||||
is clopen in ${2^{\N}}^{<\N}$.
|
||||
|
||||
|
||||
|
||||
}{}
|
||||
|
|
|
@ -21,102 +21,135 @@
|
|||
T \in \IF &\iff& \exists \beta \in \cN .~\forall n \in \N.~T(\beta\defon{n}) = 1.
|
||||
\end{IEEEeqnarray*}
|
||||
|
||||
Consider $D \coloneqq \{(T, \beta) \in \Tr \times \cN : \forall n.~ T(\beta\defon{n}) = 1\}$.
|
||||
Note that this set is closed in $\Tr \times \cN$,
|
||||
since it is a countable intersection of clopen sets.
|
||||
% TODO Why clopen?
|
||||
Then $\IF = \proj_{\Tr}(D) \in \Sigma^1_1$.
|
||||
Consider
|
||||
\[D \coloneqq \{(T, \beta) \in \Tr \times \cN : \forall n.~ T(\beta\defon{n}) = 1\}.\]
|
||||
\gist{%
|
||||
Note that this set is closed in $\Tr \times \cN$,
|
||||
since it is a countable intersection of clopen sets.
|
||||
Then $\IF = \proj_{\Tr}(D) \in \Sigma^1_1$.
|
||||
}{$D \overset{\text{closed}}{\subseteq} \Tr \times \cN$ and $\IF = \proj_{\Tr}(D)$.}
|
||||
\end{proof}
|
||||
|
||||
\begin{definition}
|
||||
An analytic set $B$ in some Polish space $Y$
|
||||
is \vocab{complete analytic} (\vocab{$\Sigma^1_1$-complete})
|
||||
\gist{%
|
||||
iff for any analytic $A \in \Sigma^1_1(X)$ for some Polish space $X$,
|
||||
there exists a Borel function $f\colon X\to Y$
|
||||
such that $x \in A \iff f(x) \in B$.
|
||||
such that $x \in A \iff f(x) \in B$,
|
||||
i.e.~$f^{-1}(B) = A$.
|
||||
}{%
|
||||
iff for any $A \in \Sigma^1_1(X)$, $X$ Polish
|
||||
there exists $f\colon X \to Y$ Borel such that $f^{-1}(B) = A$.}
|
||||
|
||||
Similarly, define \vocab{complete coanalytic} (\vocab{$\Pi^1_1$-complete}).
|
||||
\gist{%
|
||||
Similarly, a conalytic set $B$ is called
|
||||
\vocab{complete coanalytic} (\vocab{$\Pi^1_1$-complete})
|
||||
iff for any $A \subseteq \Pi^1_1(X)$
|
||||
there exists $f\colon X \to Y$ Borel such that $f^{-1}(B) = A$.
|
||||
}{Similarly we define \vocab{complete coanalytic} / \vocab{$\Pi_1^1$-complete}.}
|
||||
\end{definition}
|
||||
\begin{observe}
|
||||
\leavevmode
|
||||
\begin{itemize}
|
||||
\item Complements of $\Sigma^1_1$-complete sets are $\Pi^1_1$-complete.
|
||||
\item $\Sigma^1_1$-complete sets are never Borel:
|
||||
Suppose there is a $\Sigma^1_1$-complete set $B \in \cB(Y)$.
|
||||
Take $A \in \Sigma^1_1(X) \setminus \cB(X)$
|
||||
and $f\colon X \to Y$ Borel.
|
||||
But then $f^{-1}(B)$ is Borel.
|
||||
\item $\Sigma^1_1$-complete sets are never Borel%
|
||||
\gist{:
|
||||
Suppose there is a $\Sigma^1_1$-complete set $B \in \cB(Y)$.
|
||||
Take $A \in \Sigma^1_1(X) \setminus \cB(X)$%
|
||||
\footnote{e.g.~\yaref{thm:universals11}}
|
||||
and $f\colon X \to Y$ Borel.
|
||||
But then we get that $f^{-1}(B)$ is Borel $\lightning$.
|
||||
}{.}
|
||||
\end{itemize}
|
||||
\end{observe}
|
||||
|
||||
\begin{theorem}
|
||||
\label{thm:lec12:1}
|
||||
Suppose that $A \subseteq \cN$ is analytic.
|
||||
Then there is $f\colon \cN \to \Tr$\todo{Borel?}
|
||||
such that $x \in A \iff f(x)$ is ill-founded.
|
||||
\gist{%
|
||||
Then there is a continuous function $f\colon \cN \to \Tr$
|
||||
such that $x \in A \iff f(x)$ is ill-founded,
|
||||
i.e.~$A = f^{-1}(\IF)$.
|
||||
}{%
|
||||
Then there exists $f\colon \cN \to \Tr$ continuous
|
||||
such that $A = f^{-1}(\IF)$.
|
||||
}
|
||||
\end{theorem}
|
||||
For the proof we need some prerequisites:
|
||||
\begin{enumerate}[1.]
|
||||
\item Recall that for $S$ countable,
|
||||
the pruned\footnote{no maximal elements, in particular this implies ill-founded if the tree is non empty.} trees
|
||||
$T \subseteq S^{<\N}$ on $S$ correspond
|
||||
to closed subsets of $S^{\N}$:
|
||||
\begin{IEEEeqnarray*}{rCl}
|
||||
T &\longmapsto & [T]\\
|
||||
\{\alpha\defon{n} : \alpha \in D, n \in \N\} &\longmapsfrom & D\\
|
||||
\end{IEEEeqnarray*}
|
||||
\todo{Copy from exercises}
|
||||
\item \leavevmode\begin{definition}
|
||||
If $T$ is a tree on $\N \times \N$
|
||||
and $x \in \cN$,
|
||||
then the \vocab{section at $x$}
|
||||
%denoted $T(x)$,
|
||||
is the following tree on $\N$ :
|
||||
\[
|
||||
T(x) = \{s \in \N^{<\N} : (x\defon{|s|}, s) \in T\}.
|
||||
\]
|
||||
\end{definition}
|
||||
\item \leavevmode
|
||||
\begin{proposition}
|
||||
\label{prop:lec12:2}
|
||||
Let $A \subseteq \cN$.
|
||||
The following are equivalent:
|
||||
\begin{itemize}
|
||||
\item $A$ is analytic.
|
||||
\item There is a pruned tree on $\N \times \N$
|
||||
such that
|
||||
\[A = \proj_1 ([T]) = \{x \in \cN : \exists y \in \cN.~ (x,y) \in [T]\}.\]
|
||||
\end{itemize}
|
||||
\end{proposition}
|
||||
\begin{proof}
|
||||
$A$ is analytic iff
|
||||
there exists $F \overset{\text{closed}}{\subseteq} \N \times \N$
|
||||
such that $A = \proj_1(F)$.
|
||||
But closed sets of $\N \times \N$ correspond to pruned trees,
|
||||
by the first point.
|
||||
\end{proof}
|
||||
\end{enumerate}
|
||||
|
||||
\gist{%
|
||||
Recall that for $S$ countable,
|
||||
the pruned%
|
||||
\footnote{no maximal elements,
|
||||
in particular this implies ill-founded if the tree is non empty.
|
||||
} trees $T \subseteq S^{<\N}$ on $S$ correspond
|
||||
to closed subsets of $S^{\N}$:%
|
||||
\footnote{cf.~\yaref{s3e1} (c)}
|
||||
\begin{IEEEeqnarray*}{rCl}
|
||||
T &\longmapsto & [T]\\
|
||||
\{\alpha\defon{n} : \alpha \in D, n \in \N\} &\longmapsfrom & D\\
|
||||
\end{IEEEeqnarray*}
|
||||
}{%
|
||||
For $S$ countable,
|
||||
pruned trees on $S$ correspond to closed subsets of $S^{\N}$
|
||||
via $T \mapsto [T]$.
|
||||
}
|
||||
\begin{definition}
|
||||
If $T$ is a tree on $\N \times \N$
|
||||
and $x \in \cN$,
|
||||
then the \vocab{section at $x$}
|
||||
denoted $T(x)$,
|
||||
is the following tree on $\N$ :
|
||||
\[
|
||||
T(x) = \{s \in \N^{<\N} : (x\defon{|s|}, s) \in T\}.
|
||||
\]
|
||||
\end{definition}
|
||||
\begin{proposition}
|
||||
\label{prop:lec12:2}
|
||||
Let $A \subseteq \cN$.
|
||||
The following are equivalent:
|
||||
\begin{itemize}
|
||||
\item $A$ is analytic.
|
||||
\item There is a pruned tree on $\N \times \N$
|
||||
such that
|
||||
\[A = \proj_1 ([T]) = \{x \in \cN : \exists y \in \cN.~ (x,y) \in [T]\}.\]
|
||||
\end{itemize}
|
||||
\end{proposition}
|
||||
\begin{proof}
|
||||
\gist{%
|
||||
$A$ is analytic iff
|
||||
there exists $F \overset{\text{closed}}{\subseteq} (\N \times \N)^{\N}$
|
||||
such that $A = \proj_1(F)$.
|
||||
But closed sets of $\N^\N \times \N^{\N}$ correspond to pruned trees,
|
||||
by the first point.
|
||||
}{Closed subsets of $\N^\N \times \N^\N$ correspond to pruned trees.}
|
||||
\end{proof}
|
||||
\begin{refproof}{thm:lec12:1}
|
||||
Take a tree $T$ on $\N \times \N$
|
||||
as in \autoref{prop:lec12:2}, i.e.~$A = \proj_1([T])$.
|
||||
\gist{%
|
||||
Take a tree $T$ on $\N \times \N$
|
||||
as in \autoref{prop:lec12:2}, i.e.~$A = \proj_1([T])$.
|
||||
}{Write $A = \proj_1([T])$ for a pruned tree $T$ on $\N \times \N$.}
|
||||
Consider
|
||||
\begin{IEEEeqnarray*}{rCl}
|
||||
f\colon \cN &\longrightarrow & \Tr \\
|
||||
x &\longmapsto & T(x).
|
||||
\end{IEEEeqnarray*}
|
||||
Clearly $x \in A \iff f(x)$ is ill-founded.
|
||||
$f$ is continuous:
|
||||
Let $x\defon{n} = y\defon{n}$ for some $n \in \N$.
|
||||
Then for all $m \le n, s,t \in \N^{<\N}$
|
||||
such that $s = x\defon{m} = y \defon{m}$ and $|t| = |s|$,
|
||||
we have
|
||||
\begin{itemize}
|
||||
\item $t \in T(x) \iff (s,t) \in T$,
|
||||
\item $t \in T(y) \iff (s,t) \in T$.
|
||||
\end{itemize}
|
||||
\gist{%
|
||||
Clearly $x \in A \iff f(x) \in \IF$.
|
||||
$f$ is continuous:
|
||||
Let $x\defon{n} = y\defon{n}$ for some $n \in \N$.
|
||||
Then for all $m \le n, s,t \in \N^{<\N}$
|
||||
such that $s = x\defon{m} = y \defon{m}$ and $|t| = |s|$,
|
||||
we have
|
||||
\begin{itemize}
|
||||
\item $t \in T(x) \iff (s,t) \in T$,
|
||||
\item $t \in T(y) \iff (s,t) \in T$.
|
||||
\end{itemize}
|
||||
|
||||
So if $x\defon{n} = y\defon{n}$,
|
||||
then $t \in T(x) \iff t \in T(y)$ as long as $|t| \le n$..
|
||||
So if $x\defon{n} = y\defon{n}$,
|
||||
then $t \in T(x) \iff t \in T(y)$ as long as $|t| \le n$.
|
||||
}{}
|
||||
\end{refproof}
|
||||
|
||||
\begin{corollary}
|
||||
|
@ -125,7 +158,9 @@ For the proof we need some prerequisites:
|
|||
\end{corollary}
|
||||
\begin{proof}
|
||||
Let $X$ be Polish.
|
||||
Suppose that $A \subseteq X$ is analytic and uncountable.
|
||||
Suppose that $A \subseteq X$ is analytic and uncountable%
|
||||
\gist{}{ (trivial for countable)}.
|
||||
|
||||
Then
|
||||
% https://q.uiver.app/#q=WzAsNSxbMCwwLCJYIl0sWzEsMCwiXFxjTiJdLFsyLDAsIlxcVHIiXSxbMCwxLCJBIl0sWzEsMSwiYihBKSJdLFsxLDIsImYiXSxbMCwxLCJiIl0sWzMsMCwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbNCwxLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dXQ==
|
||||
\[\begin{tikzcd}
|
||||
|
@ -138,16 +173,18 @@ For the proof we need some prerequisites:
|
|||
\end{tikzcd}\]
|
||||
where $f$ is chosen as in \yaref{thm:lec12:1}.
|
||||
|
||||
If $X$ is Polish and countable and $A \subseteq X$ analytic,
|
||||
just consider
|
||||
\begin{IEEEeqnarray*}{rCl}
|
||||
g \colon X &\longrightarrow & \Tr \\
|
||||
x &\longmapsto & \begin{cases}
|
||||
a &: x \in A,\\
|
||||
b &: x \not\in A,\\
|
||||
\end{cases}
|
||||
\end{IEEEeqnarray*}
|
||||
where $a \in \IF$ and $b \not\in \IF$ are chosen arbitrarily.
|
||||
\gist{%
|
||||
If $X$ is Polish and countable and $A \subseteq X$ analytic,
|
||||
just consider
|
||||
\begin{IEEEeqnarray*}{rCl}
|
||||
g \colon X &\longrightarrow & \Tr \\
|
||||
x &\longmapsto & \begin{cases}
|
||||
a &: x \in A,\\
|
||||
b &: x \not\in A,\\
|
||||
\end{cases}
|
||||
\end{IEEEeqnarray*}
|
||||
where $a \in \IF$ and $b \not\in \IF$ are chosen arbitrarily.
|
||||
}{}
|
||||
\end{proof}
|
||||
|
||||
\subsection{Linear Orders}
|
||||
|
@ -161,18 +198,19 @@ Let
|
|||
\[
|
||||
\WO \coloneqq \{x \in \LO: x \text{ is a well ordering}\}.
|
||||
\]
|
||||
\gist{%
|
||||
Recall that
|
||||
\begin{itemize}
|
||||
\item $(A,<)$ is a well ordering iff there are no infinite descending chains.
|
||||
\item Every well ordering is isomorphic to an ordinal.
|
||||
\item Any two well orderings are comparable,
|
||||
i.e.~they are isomorphic,
|
||||
or one is isomorphic to an initial segment of the other.
|
||||
|
||||
Recall that
|
||||
\begin{itemize}
|
||||
\item $(A,<)$ is a well ordering iff there are no infinite descending chains.
|
||||
\item Every well ordering is isomorphic to an ordinal.
|
||||
\item Any two well orderings are comparable,
|
||||
i.e.~they are isomorphic,
|
||||
or one is isomorphic to an initial segment of the other.
|
||||
|
||||
Let $(A, <_A) \prec (B, <_B)$ denote that
|
||||
$(A, <_A)$ is isomorphic to a proper initial segment of $(B, <_B)$.
|
||||
\end{itemize}
|
||||
Let $(A, <_A) \prec (B, <_B)$ denote that
|
||||
$(A, <_A)$ is isomorphic to a proper initial segment of $(B, <_B)$.
|
||||
\end{itemize}
|
||||
}{}
|
||||
|
||||
\begin{definition}
|
||||
A \vocab{rank} on some set $C$
|
||||
|
@ -181,13 +219,14 @@ Recall that
|
|||
\phi\colon C \to \Ord.
|
||||
\]
|
||||
\end{definition}
|
||||
\begin{example}
|
||||
Let $C = \WO$
|
||||
and
|
||||
\begin{IEEEeqnarray*}{rCl}
|
||||
\phi\colon \WO &\longrightarrow & \Ord \\
|
||||
\end{IEEEeqnarray*}
|
||||
where $\phi((A,<_A))$ is the unique ordinal
|
||||
isomorphic to $(A, <_A)$.
|
||||
\end{example}
|
||||
|
||||
\gist{%
|
||||
\begin{example}
|
||||
Let $C = \WO$
|
||||
and
|
||||
\begin{IEEEeqnarray*}{rCl}
|
||||
\phi\colon \WO &\longrightarrow & \Ord \\
|
||||
\end{IEEEeqnarray*}
|
||||
where $\phi((A,<_A))$ is the unique ordinal
|
||||
isomorphic to $(A, <_A)$.
|
||||
\end{example}
|
||||
}{}
|
||||
|
|
|
@ -1,11 +1,12 @@
|
|||
\lecture{13}{2023-11-08}{}
|
||||
|
||||
\gist{%
|
||||
% Recap
|
||||
$\LO = \{x \in 2^{\N\times \N} : x \text{ is a linear order}\} $.
|
||||
$\LO \subseteq 2^{\N \times \N}$ is closed
|
||||
and $\WO = \{x \in \LO: x \text{ is a wellordering}\} $
|
||||
is coanalytic in $\LO$.
|
||||
% End Recap
|
||||
}{}
|
||||
|
||||
Another way to code linear orders:
|
||||
|
||||
|
@ -32,6 +33,7 @@ with $(f^{-1}(\{1\}), <)$.
|
|||
and $[\alpha_i, \alpha_{i+1})$ to $(i,i+1)$.
|
||||
\end{proof}
|
||||
|
||||
|
||||
\begin{definition}[\vocab{Kleene-Brouwer ordering}]
|
||||
Let $(A,<)$ be a linear order and $A$ countable.
|
||||
We define the linear order $<_{KB}$ on $A^{<\N}$
|
||||
|
@ -55,25 +57,30 @@ with $(f^{-1}(\{1\}), <)$.
|
|||
$(T, <_{KB}\defon{T})$ is well ordered.
|
||||
\end{proposition}
|
||||
\begin{proof}
|
||||
If $T$ is ill-founded and $x \in [T]$,
|
||||
then for all $n$, we have $x\defon{n+1} <_{KB} x\defon{n}$.
|
||||
Thus $(T, <_{KB}\defon{T})$ is not well ordered.
|
||||
\gist{%
|
||||
If $T$ is ill-founded and $x \in [T]$,
|
||||
then for all $n$, we have $x\defon{n+1} <_{KB} x\defon{n}$.
|
||||
Thus $(T, <_{KB}\defon{T})$ is not well ordered.
|
||||
|
||||
Conversely, let $<\defon{KB}$ be not a well-ordering
|
||||
on $T$.
|
||||
Let $s_0 >_{KB} s_1 >_{KB} s_2 >_{KB} \ldots$
|
||||
be an infinite descending chain.
|
||||
We have that $s_0(0) \ge s_1(0) \ge s_2(0) \ge \ldots$
|
||||
stabilizes for $n > n_0$.
|
||||
Let $a_0 \coloneqq s_{n_0}(0)$.
|
||||
Now for $n \ge n_0$ we have that $s_n(0)$ is constant,
|
||||
hence for $n > n_0$ the value $s_{n}(1)$ must be defined.
|
||||
Thus there is $n_1 \ge n_0$ such that $s_n(1)$
|
||||
is constant for all $n \ge n_1$.
|
||||
Let $a_1 \coloneqq s_{n_1}(1)$
|
||||
and so on.
|
||||
Then $(a_0,a_1,a_2, \ldots) \in [T]$.
|
||||
Conversely, let $<\defon{KB}$ be not a well-ordering
|
||||
on $T$.
|
||||
Let $s_0 >_{KB} s_1 >_{KB} s_2 >_{KB} \ldots$
|
||||
be an infinite descending chain.
|
||||
We have that $s_0(0) \ge s_1(0) \ge s_2(0) \ge \ldots$
|
||||
stabilizes for $n > n_0$.
|
||||
Let $a_0 \coloneqq s_{n_0}(0)$.
|
||||
Now for $n \ge n_0$ we have that $s_n(0)$ is constant,
|
||||
hence for $n > n_0$ the value $s_{n}(1)$ must be defined.
|
||||
Thus there is $n_1 \ge n_0$ such that $s_n(1)$
|
||||
is constant for all $n \ge n_1$.
|
||||
Let $a_1 \coloneqq s_{n_1}(1)$
|
||||
and so on.
|
||||
Then $(a_0,a_1,a_2, \ldots) \in [T]$.
|
||||
}{easy}
|
||||
\end{proof}
|
||||
|
||||
% TODO ANKI-MARKER
|
||||
|
||||
\begin{theorem}[Lusin-Sierpinski]
|
||||
The set $\LO \setminus \WO$
|
||||
(resp.~$2^{\Q} \setminus \WO$)
|
||||
|
|
|
@ -85,9 +85,8 @@
|
|||
\[
|
||||
\forall x \in X.~(\exists n.~(x,n) \in R \iff \exists! n.~(x,n)\in R^\ast).
|
||||
\]
|
||||
We say that $R^\ast$ \vocab[uniformization]{uniformizes} $R$.
|
||||
\todo{missing picture
|
||||
\url{https://upload.wikimedia.org/wikipedia/commons/4/4c/Uniformization_ill.png}}
|
||||
We say that $R^\ast$ \vocab[uniformization]{uniformizes} $R$.%
|
||||
\footnote{Wikimedia has a \href{https://upload.wikimedia.org/wikipedia/commons/4/4c/Uniformization_ill.png}{nice picture.}}
|
||||
|
||||
\end{theorem}
|
||||
\begin{proof}
|
||||
|
|
|
@ -175,9 +175,10 @@ By Zorn's lemma, this will follow from
|
|||
\end{IEEEeqnarray*}
|
||||
|
||||
The topology induced by the metric
|
||||
is given by basic open subsets\footnote{see Exercise Sheet 9% TODO REF
|
||||
}
|
||||
$[U_0; U_1,\ldots, U_n]$, $U_0,\ldots, U_n \overset{\text{open}}{\subseteq} X$,
|
||||
is given by basic open subsets\footnote{cf.~\yaref{s9e2}}
|
||||
of the form
|
||||
$[U_0; U_1,\ldots, U_n]$,
|
||||
for $U_0,\ldots, U_n \overset{\text{open}}{\subseteq} X$,
|
||||
where
|
||||
\[
|
||||
[U_0; U_1,\ldots,U_n] \coloneqq
|
||||
|
@ -188,8 +189,8 @@ By Zorn's lemma, this will follow from
|
|||
We want to view flows as a metric space.
|
||||
For a fixed compact metric space $X$,
|
||||
we can view the flows $(X,\Z)$ as a subset of $\cC(X,X)$.
|
||||
Note that $\cC(X,X)$ is Polish.
|
||||
Then the minimal flows on $X$ are a Borel subset of $\cC(X,X)$.
|
||||
Note that $\cC(X,X)$ is Polish.\footnote{cf.~\yaref{s1e4}}
|
||||
Then the minimal flows on $X$ are a Borel subset of $\cC(X,X)$.\footnote{Exercise} % TODO
|
||||
|
||||
However we do not want to consider only flows on a fixed space $X$,
|
||||
but we want to look all flows at the same time.
|
||||
|
|
|
@ -27,8 +27,10 @@
|
|||
|
||||
\begin{definition}
|
||||
A topological space is \vocab{Lindelöf}
|
||||
if every open cover has a countable subcover.
|
||||
iff every open cover has a countable subcover.
|
||||
\end{definition}
|
||||
|
||||
|
||||
\begin{fact}
|
||||
Let $X$ be a metric space.
|
||||
If $X$ is Lindelöf,
|
||||
|
@ -64,5 +66,12 @@
|
|||
and Lindelöf coincide.
|
||||
|
||||
In arbitrary topological spaces,
|
||||
Lindelöf is the strongest of these notions.
|
||||
Lindelöf is the weakest of these notions.
|
||||
\end{remark}
|
||||
|
||||
\begin{definition}+
|
||||
A metric space $X$ is \vocab{totally bounded}
|
||||
iff for every $\epsilon > 0$ there exists
|
||||
a finite set of points $x_1,\ldots,x_n$
|
||||
such that $X = \bigcup_{i=1}^n B_{\epsilon}(x_i)$.
|
||||
\end{definition}
|
||||
|
|
|
@ -140,16 +140,7 @@ amounts to a finite number of conditions on the preimage.
|
|||
\end{pmatrix*}&\longmapsfrom & \beta \in \Homeo(X).
|
||||
\end{IEEEeqnarray*}
|
||||
Clearly this has the desired properties.
|
||||
\item We have
|
||||
\begin{IEEEeqnarray*}{Cl}
|
||||
& \Z \circlearrowright X \text{ has a dense orbit}\\
|
||||
\iff& \exists x \in X.~ \overline{\Z\cdot x} = X\\
|
||||
\iff& \exists x \in X.~\forall U\overset{\text{open}}{\subseteq} X.~\exists z \in \Z.~
|
||||
z \cdot x \in U\\
|
||||
\iff&\exists x \in X.~\forall U \overset{\text{open}}{\subseteq} X.~
|
||||
\exists z \in \Z.~f^z(x) \in U.
|
||||
\end{IEEEeqnarray*}
|
||||
\item Let $X$ be a compact Polish space.
|
||||
\item Let $X$ be a compact Polish space.
|
||||
What is the Borel complexity of $\Homeo(X)$ inside $\cC(X,X)$?
|
||||
|
||||
Recall that $\cC(X,X)$ is a Polish space with the uniform topology.
|
||||
|
@ -160,14 +151,9 @@ amounts to a finite number of conditions on the preimage.
|
|||
\end{IEEEeqnarray*}
|
||||
by the general fact
|
||||
\begin{fact}
|
||||
Let $X$ be comapct and $Y$ Hausdorff,
|
||||
Let $X$ be compact and $Y$ Hausdorff,
|
||||
$f\colon X \to Y$ a continuous bijection.
|
||||
Then $f$ is a homeomorphism.
|
||||
\end{fact}
|
||||
\item It suffices to check the condition from part (b)
|
||||
for open sets $U$ of a countable basis
|
||||
and points $x \in X$ belonging to a countable dense subset.
|
||||
Replacing quantifiers by unions resp.~intersections
|
||||
gives that $D$ is Borel.
|
||||
\end{itemize}
|
||||
|
||||
|
|
146
inputs/tutorial_13.tex
Normal file
146
inputs/tutorial_13.tex
Normal file
|
@ -0,0 +1,146 @@
|
|||
\tutorial{13}{2024-01-23}{}
|
||||
|
||||
|
||||
Continuation of sheet 8, exercise 4.
|
||||
% Whiteboard https://wbo.ophir.dev/boards/Dphc7eylJJcIA0WbQsn7jzec1domqyx51gXb5qe6rzw-#263,0,0.5
|
||||
|
||||
\begin{definition}
|
||||
Let $X$ be a compact metric space.
|
||||
For $K \subseteq X$ compact and $U \overset{\text{open}}{\subseteq} X$
|
||||
let
|
||||
\[
|
||||
S_{K,U} \coloneqq \{f \in \cC(X,X): f(K) \subseteq U\}.
|
||||
\]
|
||||
The \vocab{compact open topology} on $\cC(X,X)$
|
||||
is the topology that has $S_{K,U}$ as a subbase.
|
||||
\end{definition}
|
||||
\begin{fact}
|
||||
If $X$ is compact,
|
||||
then the compact open topology
|
||||
is the topology induced by the uniform metric $d_\infty$.
|
||||
\end{fact}
|
||||
\begin{proof}
|
||||
Take some $S_{K,U}$. We need to show that this can be written
|
||||
as a union of open $d_{\infty}$-balls.
|
||||
Let $f_0 \in S_{K,U}$.
|
||||
Consider the continuous function $d(-, U^c)$.
|
||||
Since $f_0(K)$ is compact,
|
||||
there exists $\epsilon \coloneqq \min d(f_0(K), U^c)$
|
||||
and $B_{\epsilon}(f_0) \subseteq S_{K,U}$.
|
||||
|
||||
|
||||
On the other hand, consider $B_{\epsilon}(f_0)$ for some $\epsilon > 0$
|
||||
and $f_0 \in \cC(X,X)$.
|
||||
|
||||
As $f_0$ is uniformly continuous,
|
||||
there exists $\delta > 0$ such that $d(x,x') < \delta \implies d(f_0(x), f_0(x')) < \frac{\epsilon}{3}$.
|
||||
Cover $X$ with finitely many $\delta$-balls $B_\delta(a_1), \ldots, B_{\delta}(a_k)$.
|
||||
Then
|
||||
\[f_0(\overline{B_{\delta}(a_i)}) \subseteq \overline{f_0(B_{\delta}(a_i)} \subseteq \overline{B_{\frac{\epsilon}{3}}(f_0(a_i))} \subseteq B_{\frac{\epsilon}{2}}(f_0(a_i)).\]
|
||||
|
||||
For $i \le k$, let $S_i \coloneqq S_{\overline{B_{\delta}(a_i)}, B_{\frac{\epsilon}{2}}(f_0(a_i))}$.
|
||||
Take $\bigcap_{i \le k} S_i$. This is open
|
||||
in the compact open topology and
|
||||
$B_{\epsilon}(f_0) \subseteq \bigcap_{i \le k} S_i$.
|
||||
\end{proof}
|
||||
|
||||
\begin{claim}
|
||||
$f \in \cC(X,X)$ is surjective
|
||||
iff for all basic open $\emptyset\neq U \subseteq X$
|
||||
there exists a basic open $\emptyset \neq V \subseteq X$
|
||||
with $f(\overline{V}) \subseteq U$.
|
||||
|
||||
Note that we can write this as a $G_\delta$-condition.
|
||||
|
||||
\end{claim}
|
||||
\begin{subproof}
|
||||
Take $B_\epsilon(f(x_0))\subseteq U$.
|
||||
Then there exists $\delta > 0$
|
||||
such that $f(B_{\delta}(x_0)) \subseteq B_{\frac{\epsilon}{2}}(f(x_0))$
|
||||
hence $f(\overline{B_{\delta}(x_0)}) \subseteq B_\epsilon(f(x_0))$.
|
||||
|
||||
|
||||
For the other direction take $y \in X$.
|
||||
We want to find a preimage.
|
||||
For every $B_{\frac{1}{n}}(y)$,
|
||||
there exists a basic open set $V_n$ with $f(\overline{V}) \subseteq B_{\frac{1}{n}}(y)$.
|
||||
Take $x_n \in V_n$.
|
||||
Since $X$ is compact, it is sequentially compact,
|
||||
so there exists a converging subsequence.
|
||||
Wlog.~$x_n \to x$,
|
||||
so $f(x_n) \to f(x) = y$.
|
||||
\end{subproof}
|
||||
|
||||
\begin{claim}
|
||||
$f \in \cC(X,X)$ is injective iff
|
||||
for all basic open $U$,$V$
|
||||
with $\overline{U} \cap \overline{V} = \emptyset$
|
||||
we have $f(\overline{U}) \cap f(\overline{V}) = \emptyset$.
|
||||
|
||||
This is a $G_\delta$-condition,
|
||||
since we can write it as
|
||||
\[
|
||||
\bigcap_{U,V} S_{\overline{U}, f(\overline{V})^c}.
|
||||
\]
|
||||
\end{claim}
|
||||
\begin{subproof}
|
||||
$\implies$ is trivial.
|
||||
|
||||
$\impliedby$ follows since for all pairs $x,y \in X$,
|
||||
we can find $x \in U$, $y \in V$ such that $\overline{U} \cap \overline{V} = \emptyset$.
|
||||
\end{subproof}
|
||||
|
||||
Hence $\Homeo(X,X)$ is $G_\delta$.
|
||||
In particular it is a Polish space.
|
||||
|
||||
|
||||
|
||||
Let $D$ be the set of $\Z$-flows with dense orbit.
|
||||
\begin{claim}
|
||||
$f \in D$ $\iff$
|
||||
for all basic open $U,V \subseteq X$,
|
||||
there exists $n \in \Z$
|
||||
such that $f^n(U) \cap V \neq \emptyset$.
|
||||
|
||||
\end{claim}
|
||||
\begin{subproof}
|
||||
Suppose that the orbit of $x_0 \in X$ is dense.
|
||||
Then there exist $k,l \in \Z$
|
||||
such that $f^k(x_0)\in U$ and $f^l(x_0) \in V$,
|
||||
so $f^{l-k} U \cap V \neq \emptyset$.
|
||||
|
||||
|
||||
For basic open sets $V$
|
||||
let
|
||||
\[
|
||||
A_V \coloneqq \{ x \in X: \exists n.~ f^n(x) \in V\}.
|
||||
\]
|
||||
By assumption, all the $A_V$ are dense.
|
||||
Hence $\bigcap_{V}A_V$ is dense by the \yaref{thm:bct}.
|
||||
|
||||
$A_V = \bigcup_{n \in \Z} f^n(V)$ is open.
|
||||
\end{subproof}
|
||||
|
||||
\begin{claim}
|
||||
The condition can be written as a $G_\delta$ set.
|
||||
\end{claim}
|
||||
\begin{subproof}
|
||||
|
||||
For $f_0(U) \cap V \neq \emptyset$
|
||||
take $u \in U$ such that $f_0(u) \in V$.
|
||||
Then there exists $\epsilon > 0$ such that $B_{\epsilon}(f_0(u)) \subseteq U$,
|
||||
hence $B_{\epsilon}(f_0)$ is an open neighbourhood contained
|
||||
in $\{f : f(U) \cap V \neq \emptyset \} $.
|
||||
|
||||
For $n = 2$ note that
|
||||
$d(f^2(u), f^2_0(u) \le d(f(f(u)), f_0(f(u))) + d(f_0(f(u)), f_0(f_0(u)))$.
|
||||
The first part can be bounded by $d(f,f_0)$.
|
||||
For the second part,
|
||||
note that there exists $\delta$ such that
|
||||
\[d(a,b) < \delta \implies d(f_0(a), f_0(b)) < \frac{\epsilon}{2}.\]
|
||||
Let $\eta \coloneqq \min \{\delta, \frac{\epsilon}{2}\}$
|
||||
and consider $d_\infty(f,f_0) < \epsilon$.
|
||||
|
||||
For other $n$ it is some more work, which is left as an exercise.
|
||||
\end{subproof}
|
||||
|
|
@ -1,6 +1,10 @@
|
|||
\NeedsTeXFormat{LaTeX2e}
|
||||
\ProvidesPackage{jrpie-gist}[2023/01/22 - gist version for lecture notes]
|
||||
|
||||
% TODO gist info
|
||||
% TODO link to long version (provide link to main document)
|
||||
|
||||
% TODO \phantomsection to cross link
|
||||
\newcommand{\gist}[2]{%
|
||||
\ifcsname EnableGist\endcsname%
|
||||
#2%
|
||||
|
|
|
@ -66,6 +66,7 @@
|
|||
\input{inputs/tutorial_07}
|
||||
\input{inputs/tutorial_08}
|
||||
\input{inputs/tutorial_09}
|
||||
\input{inputs/tutorial_13} % sic!
|
||||
\input{inputs/tutorial_10}
|
||||
\input{inputs/tutorial_11}
|
||||
\input{inputs/tutorial_12b}
|
||||
|
|
Loading…
Reference in a new issue