w23-logic-3/inputs/tutorial_06.tex

\subsection{Sheet 5}

\tutorial{06}{}{}

% Sheet 5 - 18.5 / 20

\nr 1

\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}


\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 $Y$.
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{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 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 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 $x \in B \setminus \{\pi_1(a_j)| j <i\}$,
        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.
\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 \yaref{thm:kuratowskiulam}.
        In particular, the assumption of the set having
        the BP is necessary.

\end{enumerate}