w23-logic-3/inputs/tutorial_06.tex

\subsection{Sheet 5}

\tutorial{06}{}{}

% 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.

     In particular, let $X$ be Baire,
     then $U \overset{\text{open}}{\subseteq} X$
     is Baire.
\end{fact}

\nr 2

Let $(U_i)_{i < \omega}$ be a countable base of $Y$.
We want to find a $G_\delta$ set $A \subseteq X$
such that $f\defon{A}$ is continuous.
It suffices make sure that $f^{-1}\defon{A}(U_i)$ is open for all  $i < \omega$.
Take some  $i < \omega$.
Then  $V_i \setminus M_i \subseteq f^{-1}(U_i) \subseteq V_i \cup M_i$,
where $V_i$ is open and $M_i$ is meager.
Let $M'_i \supseteq M_i$ be a meager $F_\sigma$-set.
Now let $A \coloneqq X \setminus \bigcup_{i <\omega} M_i'$.
We have that $A$ is a countable intersection of open dense sets,
hence it is dense and $G_\delta$.
For any $i < \omega$,
$V_i \cap A \subseteq  f\defon{A}^{-1}(U_i) \subseteq  (V_i \cup M_i) \cap A = V_i \cap A$,
so $f\defon{A}^{-1}(U_i) = V_i \cap A$ is open.

\nr 3
\todo{handwritten}


\nr 4
\begin{enumerate}[(a)]
    \item $|B| = \fc$, 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
        $a_i \not\in F_i$.
        Since $|B| = \fc$ and $|\{a_i | j < i\}| = |i| < \fc$,
        there exists some $x \in B \setminus \{\pi_1(a_j)| j <i\}$,
        where $\pi$ denotes the projection.
        Choose one such $x$.
        We need to find $y \in \R$,
        such that $(x,y) \not\in F_i$
        and $\{a_j | j < i\} \cup \{(x,y)\}$
        does not contain three collinear points.

        Since $(F_i)_x$ is meager, we have that
        $|\{x\} \times \R \setminus F_i| = |\R \setminus (F_i)_x| = \fc$.
        Let $L \coloneqq \{y \in \R | \exists j < k < i. ~a_j, a_k,(x,y) \text{are collinear}\}$.
        Since every pair $a_j \neq a_k, j < k < i$,
        adds at most one point to $L$,
        we get $|L| \le |i|^2 < \fc$.
        Hence $|\R \setminus (F_i)_x \setminus L| = \fc$.
        In particular, the set is non empty, and we find $y$ as desired
        and can set $a_i \coloneqq (x,y)$.

        % We have not chosen too many points so far.
        % So there are not too many lines,
        % we can not choose a point from,
        % but there are many points in $B$.
    \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
        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.
        In particular, the assumption of the set having
        the BP is necessary.

\end{enumerate}