Josia Pietsch
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92 lines
3.1 KiB
TeX
92 lines
3.1 KiB
TeX
\subsection{Sheet 5}
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\tutorial{06}{}{}
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% Sheet 5 - 18.5 / 20
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\nr 1
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\todo{handwritten}
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Let $B \subseteq C$ be comeager.
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Then $B = B_1 \cup B_2$,
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where $B_1$ is dense $G_\delta$
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and $B_2$ is meager.
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\begin{fact}
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$X$ is Baire iff every non-empty open set is non-meager.
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In particular, let $X$ be Baire,
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then $U \overset{\text{open}}{\subseteq} X$
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is Baire.
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\end{fact}
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\nr 2
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Let $(U_i)_{i < \omega}$ be a countable base of $Y$.
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We want to find a $G_\delta$ set $A \subseteq X$
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such that $f\defon{A}$ is continuous.
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It suffices make sure that $f^{-1}\defon{A}(U_i)$ is open for all $i < \omega$.
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Take some $i < \omega$.
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Then $V_i \setminus M_i \subseteq f^{-1}(U_i) \subseteq V_i \cup M_i$,
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where $V_i$ is open and $M_i$ is meager.
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Let $M'_i \supseteq M_i$ be a meager $F_\sigma$-set.
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Now let $A \coloneqq X \setminus \bigcup_{i <\omega} M_i'$.
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We have that $A$ is a countable intersection of open dense sets,
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hence it is dense and $G_\delta$.
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For any $i < \omega$,
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$V_i \cap A \subseteq f\defon{A}^{-1}(U_i) \subseteq (V_i \cup M_i) \cap A = V_i \cap A$,
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so $f\defon{A}^{-1}(U_i) = V_i \cap A$ is open.
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\nr 3
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\todo{handwritten}
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\nr 4
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\begin{enumerate}[(a)]
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\item $|B| = \fc$, since $B$ contains a comeager
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$G_\delta$ set, $B'$:
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$B'$ is Polish,
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hence $B' = P \cup C$
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for $P$ perfect and $C$ countable,
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and $|P| \in \{\fc, 0\}$.
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But $B'$ can't contain isolated point.
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\item To ensure that (a) holds, it suffices to chose
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$a_i \not\in F_i$.
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Since $|B| = \fc$ and $|\{a_i | j < i\}| = |i| < \fc$,
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there exists some $x \in B \setminus \{\pi_1(a_j)| j <i\}$,
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where $\pi$ denotes the projection.
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Choose one such $x$.
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We need to find $y \in \R$,
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such that $(x,y) \not\in F_i$
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and $\{a_j | j < i\} \cup \{(x,y)\}$
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does not contain three collinear points.
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Since $(F_i)_x$ is meager, we have that
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$|\{x\} \times \R \setminus F_i| = |\R \setminus (F_i)_x| = \fc$.
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Let $L \coloneqq \{y \in \R | \exists j < k < i. ~a_j, a_k,(x,y) \text{are collinear}\}$.
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Since every pair $a_j \neq a_k, j < k < i$,
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adds at most one point to $L$,
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we get $|L| \le |i|^2 < \fc$.
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Hence $|\R \setminus (F_i)_x \setminus L| = \fc$.
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In particular, the set is non empty, and we find $y$ as desired
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and can set $a_i \coloneqq (x,y)$.
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% We have not chosen too many points so far.
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% So there are not too many lines,
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% we can not choose a point from,
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% but there are many points in $B$.
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\item $A$ is by construction not a subset of any $F_\sigma$ meager set.
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Hence it is not meager, since any meager set is contained
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in an $F_\sigma$ meager set.
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\item For every $x \in \R$ we have that $A_x$ contains at most
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two points, hence it is meager.
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In particular $\{x \in \R | A_x \text{ is meager}\} = \R$
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is comeager.
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However $A$ is not meager.
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Hence $A$ can not be a set with the Baire property
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by the theorem.
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In particular, the assumption of the set having
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the BP is necessary.
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\end{enumerate}
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