Josia Pietsch 6c2a76d838
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improved lecture 17
2024-02-05 20:30:13 +01:00

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\subsection{Sheet 5}
% Sheet 5 - 18.5 / 20
\nr 1
$X$ 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.
\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
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:
\item $A $ is meager,
\item there exists $\emptyset = U \overset{\text{open}}{\subseteq} X$
such that $A \cap U$ is comeager in $U$.
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).
\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
\nr 4
There exists a non-meager subset $A \subseteq \R^2$
such that no three points of $A$ are collinear.
This requires the use of the axiom of choice.
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 $i < \fc$:
\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.
\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.
\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.