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Josia Pietsch 2024-01-11 23:57:43 +01:00
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8 changed files with 93 additions and 26 deletions

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@ -109,7 +109,7 @@
(If we intersect $A$ with an open $U$,
then $A \cap U$ is not dense in $U$).
\end{itemize}
\todo{Think about this}
%\todo{Think about this}
A set $B \subseteq X$ is \vocab{meager}
(or \vocab{first category}),
@ -139,8 +139,6 @@
Note that open sets and meager sets have the Baire property.
\begin{example}
\begin{itemize}
\item $\Q \subseteq \R$ is $F_\sigma$.

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@ -91,7 +91,7 @@
Then every comeager set of $X$ is dense in $X$.
\end{theorem}
\todo{Proof (copy from some other lecture)}
\begin{proposition}
\begin{theoremdef}
Let $X$ be a topological space.
The following are equivalent:
\begin{enumerate}[(i)]
@ -101,7 +101,9 @@
\item The intersection of countable many
open dense sets is dense.
\end{enumerate}
\end{proposition}
In this case $X$ is called a \vocab{Baire space}.
\footnote{see \yaref{s5e1}}
\end{theoremdef}
\begin{proof}
\todo{Proof (short)}
@ -118,7 +120,6 @@
\end{proof}
% TODO Fubini
\begin{notation}
Let $X ,Y$ be topological spaces,
$A \subseteq X \times Y$
@ -139,10 +140,11 @@ The following similar to Fubini,
but for meager sets:
\begin{theorem}[Kuratowski-Ulam]
\label{thm:kuratowskiulam}
\yalabel{Kuratowski-Ulam}{Kuratowski-Ulam}{thm:kuratowskiulam}
Let $X,Y$ be second countable topological spaces.
Let $A \subseteq X \times Y$
be a set with the Baire property.
be a set with the Baire property.%
\footnote{It is important that $A$ has the Baire property (cf. \yaref{s5e4}).}
Then
\begin{enumerate}[(i)]

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@ -203,6 +203,3 @@ i.e.~$\Delta^0_1$ is the set of clopen sets.
Consider the cofinite topology on $\omega_1$.
Then the non-empty open sets of this are not $F_\sigma$.
\end{example}

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@ -119,7 +119,7 @@ We will see that not every analytic set is Borel.
Take $f^+\colon X \times Z \to Y \times Z, f^+ = f \times \id$.
Then
\[f^{-1}(B) = \underbrace{\proj_X(\overbrace{(\underbrace{f^+}_{\mathclap{\text{Borel}}})^{-1}(\underbrace{B_0}_{\mathclap{\text{Borel}}})}^{\text{Borel}})}_{\text{analytic}}.\]
\item See \yaref{ex:7.2}.
\item See \yaref{s7e2}.
\end{enumerate}
\end{proof}
@ -167,5 +167,6 @@ We will see later that $\Sigma^1_1(X) \cap \Pi^1_1(X) = \cB(X)$.
\begin{remark}+
Showing that there exist sets that don't have the Baire property
requires the axiom of choice.\todo{Copy exercise from sheet 5}
requires the axiom of choice.
An example of such a set is constructed in \yaref{s5e4}.
\end{remark}

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@ -74,7 +74,7 @@
Then $A$ and $B$ are Borel isomorphic.
\end{theorem}
\begin{proof}
\todo{Homework}
Cf.~\yaref{s7e4}.
\end{proof}
\begin{theorem}[\vocab{Isomorphism Theorem}]

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@ -5,11 +5,6 @@
% Sheet 5 - 18.5 / 20
\nr 1
\todo{handwritten}
Let $B \subseteq C$ be comeager.
Then $B = B_1 \cup B_2$,
where $B_1$ is dense $G_\delta$
and $B_2$ is meager.
\begin{fact}
$X$ is Baire iff every non-empty open set is non-meager.
@ -19,6 +14,60 @@ and $B_2$ is meager.
is Baire.
\end{fact}
\begin{enumerate}[(a)]
\item Let $X$ be a non-empty Baire space
and let $A \subseteq X$.
Show that $A$ cannot be both meager and comeager.
Suppose that $A \subseteq X$ is meager
and comeager.
Then $A = \bigcup_{n < \omega} U_n$
and $X \setminus A = \bigcup_{n < \omega} V_n$
for some nwd sets $U_n, V_n$.
Then $X = A \cup (X \setminus A)$ is meager.
Let $X = \bigcup_{n < \omega} W_n$ be a union of nwd
sets.
Wlog.~the $W_n$ are closed (otherwise replace them $\overline{W_n}$)
Then $\emptyset = \bigcap_{n < \omega} (X \setminus W_n)$
is a countable intersection of open, dense sets,
hence dense $\lightning$
\item Let $X$ be a topological space. The relation $=^\ast$ is transitive:
Suppose $A =^\ast B$ and $B =^\ast C$.
Then $A \symdif C \subseteq (A \symdif B \cup B \symdif C)$
is contained in a meager set.
Since a subset of a nwd set
is nwd, a subset of a meager set is meager.
Hence $A \symdif C$ is meager, thus $A =^\ast C$.
\item Let $X$ be a topological space.
Let $A \subseteq X$ be a set with the Baire property,
then at least one of the following hold:
\begin{enumerate}[(i)]
\item $A $ is meager,
\item there exists $\emptyset = U \overset{\text{open}}{\subseteq} X$
such that $A \cap U$ is comeager in $U$.
\end{enumerate}
Suppose there was $A \subseteq X$ such that (i)
does not hold.
Then there exists $U \overset{\text{open}}{\subseteq} X$
such that $A =^\ast U$.
In particular, $A \symdif U$ is meager,
hence $U \cap (A \symdif U) = U \setminus A$ is meager.
Thus $A \cap U$ is comeager in $U$.
Now suppose that $X$ is a Baire space.
Suppose that for $A$ (i) and (ii) hold.
Let $\emptyset \neq U \overset{\text{open}}{\subseteq} X$
be such that $A \cap U$ is comeager in $U$.
Since $U$ is a Baire space,
this contradicts (a).
\end{enumerate}
\nr 2
Let $(U_i)_{i < \omega}$ be a countable base of $Y$.
@ -41,15 +90,34 @@ so $f\defon{A}^{-1}(U_i) = V_i \cap A$ is open.
\nr 4
\begin{lemma}
There exists a non-meager subset $A \subseteq \R^2$
such that no three points of $A$ are collinear.
\end{lemma}
This requires the use of the axiom of choice.
\begin{proof}
Enumerate the continuum-many $F_\sigma$ subsets of $\R^2$
as $(F_i)_{i < \fc}$.
We will inductively construct a sequence $(a_i)_{i < \fc}$
of points of $\R^2$ such that for each $i < \fc$:
\begin{enumerate}[(i)]
\item $\{a_j | j \le i\} $ is not a subset of $F_i$ and
\item $\{a_j | j \le i\}$ does not contain any three collinear points.
\end{enumerate}
\begin{enumerate}[(a)]
\item $|B| = \fc$, since $B$ contains a comeager
\item Let $B = \{x \in \R | (F_i)_x \text{ is meager}\}$.
Then $B$ is comeager in $\R$ and $|B| = \fc$.
We have $|B| = \fc$, since $B$ contains a comeager
$G_\delta$ set, $B'$:
$B'$ is Polish,
hence $B' = P \cup C$
for $P$ perfect and $C$ countable,
and $|P| \in \{\fc, 0\}$.
But $B'$ can't contain isolated point.
\item To ensure that (a) holds, it suffices to chose
\item We use $B$ to find a suitable point $a_i$:
To ensure that (i) holds, it suffices to chose
$a_i \not\in F_i$.
Since $|B| = \fc$ and $|\{a_i | j < i\}| = |i| < \fc$,
there exists some $x \in B \setminus \{\pi_1(a_j)| j <i\}$,
@ -77,13 +145,16 @@ so $f\defon{A}^{-1}(U_i) = V_i \cap A$ is open.
\item $A$ is by construction not a subset of any $F_\sigma$ meager set.
Hence it is not meager, since any meager set is contained
in an $F_\sigma$ meager set.
\item For every $x \in \R$ we have that $A_x$ contains at most
\end{enumerate}
\end{proof}
\begin{enumerate}
\item[(d)] For every $x \in \R$ we have that $A_x$ contains at most
two points, hence it is meager.
In particular $\{x \in \R | A_x \text{ is meager}\} = \R$
is comeager.
However $A$ is not meager.
Hence $A$ can not be a set with the Baire property
by the theorem.
by \yaref{thm:kuratowskiulam}.
In particular, the assumption of the set having
the BP is necessary.

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@ -1,5 +1,4 @@
\subsection{Sheet 7}
\tutorial{08}{2023-12-05}{}
% 17 / 20
\nr 1
@ -45,7 +44,6 @@
\end{itemize}
\nr 2
\yalabel{Exercise}{}{ex:7.2}
Recall \autoref{thm:analytic}.

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@ -153,6 +153,6 @@
\newcommand\lecture[3]{\hrule{\color{darkgray}\hfill{\tiny[Lecture #1, #2]}}}
\newcommand\tutorial[3]{\hrule{\color{darkgray}\hfill{\tiny[Tutorial #1, #2]}}}
\newcommand\nr[1]{\subsubsection{Exercise #1}}
\newcommand\nr[1]{\subsubsection{Exercise #1}\yalabel{Sheet \arabic{subsection}, Exercise #1}{E \arabic{subsection}.#1}{s\arabic{subsection}e#1}}
\usepackage[bibfile=bibliography/references.bib, imagefile=bibliography/images.bib]{mkessler-bibliography}