Josia Pietsch
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\subsection{Sheet 5}




\tutorial{06}{}{}




% Sheet 5  18.5 / 20




\nr 1




\begin{fact}


$X$ is Baire iff every nonempty open set is nonmeager.




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 nonempty 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 nonmeager 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 continuummany $F_\sigma$ 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 $B = \fc$:


$B$ contains a comeager $G_\delta$ set, say $B'$.


$B'$ is Polish,


hence $B' = P \cup C$


for $P$ perfect and $C$ countable,


and $P \in \{\fc, 0\}$.


But $B'$ can't contain an 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}


