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8 changed files with 93 additions and 26 deletions
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@ -109,7 +109,7 @@
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(If we intersect $A$ with an open $U$,
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then $A \cap U$ is not dense in $U$).
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\end{itemize}
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\todo{Think about this}
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%\todo{Think about this}
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A set $B \subseteq X$ is \vocab{meager}
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(or \vocab{first category}),
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@ -139,8 +139,6 @@
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Note that open sets and meager sets have the Baire property.
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\begin{example}
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\begin{itemize}
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\item $\Q \subseteq \R$ is $F_\sigma$.
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@ -91,7 +91,7 @@
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Then every comeager set of $X$ is dense in $X$.
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\end{theorem}
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\todo{Proof (copy from some other lecture)}
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\begin{proposition}
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\begin{theoremdef}
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Let $X$ be a topological space.
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The following are equivalent:
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\begin{enumerate}[(i)]
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@ -101,7 +101,9 @@
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\item The intersection of countable many
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open dense sets is dense.
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\end{enumerate}
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\end{proposition}
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In this case $X$ is called a \vocab{Baire space}.
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\footnote{see \yaref{s5e1}}
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\end{theoremdef}
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\begin{proof}
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\todo{Proof (short)}
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@ -118,7 +120,6 @@
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\end{proof}
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% TODO Fubini
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\begin{notation}
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Let $X ,Y$ be topological spaces,
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$A \subseteq X \times Y$
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@ -139,10 +140,11 @@ The following similar to Fubini,
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but for meager sets:
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\begin{theorem}[Kuratowski-Ulam]
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\label{thm:kuratowskiulam}
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\yalabel{Kuratowski-Ulam}{Kuratowski-Ulam}{thm:kuratowskiulam}
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Let $X,Y$ be second countable topological spaces.
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Let $A \subseteq X \times Y$
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be a set with the Baire property.
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be a set with the Baire property.%
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\footnote{It is important that $A$ has the Baire property (cf. \yaref{s5e4}).}
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Then
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\begin{enumerate}[(i)]
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@ -203,6 +203,3 @@ i.e.~$\Delta^0_1$ is the set of clopen sets.
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Consider the cofinite topology on $\omega_1$.
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Then the non-empty open sets of this are not $F_\sigma$.
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\end{example}
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@ -119,7 +119,7 @@ We will see that not every analytic set is Borel.
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Take $f^+\colon X \times Z \to Y \times Z, f^+ = f \times \id$.
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Then
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\[f^{-1}(B) = \underbrace{\proj_X(\overbrace{(\underbrace{f^+}_{\mathclap{\text{Borel}}})^{-1}(\underbrace{B_0}_{\mathclap{\text{Borel}}})}^{\text{Borel}})}_{\text{analytic}}.\]
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\item See \yaref{ex:7.2}.
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\item See \yaref{s7e2}.
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\end{enumerate}
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\end{proof}
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@ -167,5 +167,6 @@ We will see later that $\Sigma^1_1(X) \cap \Pi^1_1(X) = \cB(X)$.
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\begin{remark}+
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Showing that there exist sets that don't have the Baire property
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requires the axiom of choice.\todo{Copy exercise from sheet 5}
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requires the axiom of choice.
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An example of such a set is constructed in \yaref{s5e4}.
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\end{remark}
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@ -74,7 +74,7 @@
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Then $A$ and $B$ are Borel isomorphic.
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\end{theorem}
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\begin{proof}
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\todo{Homework}
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Cf.~\yaref{s7e4}.
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\end{proof}
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\begin{theorem}[\vocab{Isomorphism Theorem}]
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@ -5,11 +5,6 @@
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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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@ -19,6 +14,60 @@ and $B_2$ is meager.
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is Baire.
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\end{fact}
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\begin{enumerate}[(a)]
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\item Let $X$ be a non-empty Baire space
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and let $A \subseteq X$.
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Show that $A$ cannot be both meager and comeager.
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Suppose that $A \subseteq X$ is meager
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and comeager.
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Then $A = \bigcup_{n < \omega} U_n$
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and $X \setminus A = \bigcup_{n < \omega} V_n$
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for some nwd sets $U_n, V_n$.
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Then $X = A \cup (X \setminus A)$ is meager.
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Let $X = \bigcup_{n < \omega} W_n$ be a union of nwd
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sets.
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Wlog.~the $W_n$ are closed (otherwise replace them $\overline{W_n}$)
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Then $\emptyset = \bigcap_{n < \omega} (X \setminus W_n)$
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is a countable intersection of open, dense sets,
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hence dense $\lightning$
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\item Let $X$ be a topological space. The relation $=^\ast$ is transitive:
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Suppose $A =^\ast B$ and $B =^\ast C$.
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Then $A \symdif C \subseteq (A \symdif B \cup B \symdif C)$
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is contained in a meager set.
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Since a subset of a nwd set
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is nwd, a subset of a meager set is meager.
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Hence $A \symdif C$ is meager, thus $A =^\ast C$.
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\item Let $X$ be a topological space.
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Let $A \subseteq X$ be a set with the Baire property,
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then at least one of the following hold:
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\begin{enumerate}[(i)]
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\item $A $ is meager,
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\item there exists $\emptyset = U \overset{\text{open}}{\subseteq} X$
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such that $A \cap U$ is comeager in $U$.
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\end{enumerate}
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Suppose there was $A \subseteq X$ such that (i)
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does not hold.
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Then there exists $U \overset{\text{open}}{\subseteq} X$
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such that $A =^\ast U$.
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In particular, $A \symdif U$ is meager,
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hence $U \cap (A \symdif U) = U \setminus A$ is meager.
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Thus $A \cap U$ is comeager in $U$.
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Now suppose that $X$ is a Baire space.
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Suppose that for $A$ (i) and (ii) hold.
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Let $\emptyset \neq U \overset{\text{open}}{\subseteq} X$
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be such that $A \cap U$ is comeager in $U$.
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Since $U$ is a Baire space,
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this contradicts (a).
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\end{enumerate}
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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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\nr 4
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\begin{lemma}
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There exists a non-meager subset $A \subseteq \R^2$
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such that no three points of $A$ are collinear.
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\end{lemma}
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This requires the use of the axiom of choice.
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\begin{proof}
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Enumerate the continuum-many $F_\sigma$ subsets of $\R^2$
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as $(F_i)_{i < \fc}$.
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We will inductively construct a sequence $(a_i)_{i < \fc}$
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of points of $\R^2$ such that for each $i < \fc$:
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\begin{enumerate}[(i)]
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\item $\{a_j | j \le i\} $ is not a subset of $F_i$ and
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\item $\{a_j | j \le i\}$ does not contain any three collinear points.
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\end{enumerate}
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\begin{enumerate}[(a)]
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\item $|B| = \fc$, since $B$ contains a comeager
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\item Let $B = \{x \in \R | (F_i)_x \text{ is meager}\}$.
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Then $B$ is comeager in $\R$ and $|B| = \fc$.
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We have $|B| = \fc$, 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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\item We use $B$ to find a suitable point $a_i$:
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To ensure that (i) 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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\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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\end{enumerate}
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\end{proof}
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\begin{enumerate}
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\item[(d)] 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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by \yaref{thm:kuratowskiulam}.
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In particular, the assumption of the set having
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the BP is necessary.
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@ -1,5 +1,4 @@
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\subsection{Sheet 7}
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\tutorial{08}{2023-12-05}{}
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% 17 / 20
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\nr 1
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\end{itemize}
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\nr 2
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\yalabel{Exercise}{}{ex:7.2}
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Recall \autoref{thm:analytic}.
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@ -153,6 +153,6 @@
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\newcommand\lecture[3]{\hrule{\color{darkgray}\hfill{\tiny[Lecture #1, #2]}}}
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\newcommand\tutorial[3]{\hrule{\color{darkgray}\hfill{\tiny[Tutorial #1, #2]}}}
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\newcommand\nr[1]{\subsubsection{Exercise #1}}
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\newcommand\nr[1]{\subsubsection{Exercise #1}\yalabel{Sheet \arabic{subsection}, Exercise #1}{E \arabic{subsection}.#1}{s\arabic{subsection}e#1}}
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\usepackage[bibfile=bibliography/references.bib, imagefile=bibliography/images.bib]{mkessler-bibliography}
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