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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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\begin{itemize}
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\item For $\xi = 1$ this holds by the definition of the
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subspace topology.
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We now use transfinite induction, to show that
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the statement holds for all $\xi$.
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Suppose that $\Sigma^0_{\zeta}(Y)$ and $\Pi^0_{\zeta}(Y)$
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are as claimed for all $\zeta < \xi$.
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Then
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\begin{IEEEeqnarray*}{rCl}
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\Sigma^0_\xi(Y) &=& \{\bigcup_{n < \omega} A_n : A_n \in \Pi^0_{\alpha_n}(Y), \alpha_n < \xi\}\\
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&=& \{\bigcup_{n < \omega} (A_n \cap Y) : A_n \in \Pi^0_{\alpha_n}(X), \alpha_n < \xi\}\\
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&=& \{Y \cap \bigcup_{n < \omega} A_n : A_n \in \Pi^0_{\alpha_n}(X), \alpha_n < \xi\}\\
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&=& \{Y \cap A : A \in \Sigma^0_{\xi}(X)\}.
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\end{IEEEeqnarray*}
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and
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\begin{IEEEeqnarray*}{rCl}
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\Pi^0_\xi(Y) &=& \lnot \Sigma^0_\xi(Y)\\
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&=& \{Y \setminus A : A \in \Sigma^0_\xi(Y)\}\\
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&=& \{Y \setminus (A \cap Y) : A \in \Sigma^0_\xi(X)\}\\
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&=& \{Y \cap (X \setminus A) : A \in \Sigma^0_\xi(X)\}\\
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&=& \{Y \cap A : A \in \Pi^0_\xi(X)\}.
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\end{IEEEeqnarray*}
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\item Let $V \in \cB(Y)$.
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We show that $f^{-1}(V) \in \cB(Y)$,
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by induction on the minimal $\xi$ such that $V \in \Sigma_\xi^0$.
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For $\xi = 0$ this is clear.
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Suppose that we have already shown $f^{-1}(V') \in \cB(Y)$
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for all $V' \in \Sigma^0_\zeta$, $\zeta < \xi$.
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Then $f^{-1}(Y \setminus V') = X \setminus f^{-1}(V') \in \cB(V)$,
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since complements of Borel sets are Borel.
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In particular, this also holds for $\Pi^0_\zeta$ sets
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and $\zeta < \xi$.
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Let $V \in \Sigma^0_\xi$.
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Then $V = \bigcap_{n} V_n$ for some $V_n \in \Pi^{0}_{\alpha_n}$,
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$\alpha_n < \xi$.
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In particular $f^{-1}(V) = \bigcup_n f^{-1}(V_n) \in \cB(X)$.
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\end{itemize}
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\nr 2
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Recall \autoref{thm:analytic}.
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Let $(A_i)_{i<\omega}$ be analytic subsets of a Polish space $X$.
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Then there exists Polish spaces $Y_i$ and $f_i\colon Y_i \to X$
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continuous such that $f_i(B_i) = A_i$
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for some $B_i \in \cB(Y_i)$.
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\begin{itemize}
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\item $\bigcup_i A_i$ is analytic:
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Consider the Polish space $Y \coloneqq \coprod_{i < \omega} Y_i$
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and $f \coloneqq \coprod_i f_i$, i.e.~
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$Y_i \ni y \mapsto f_i(y)$.
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$f$ is continuous,
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$\coprod_{i < \omega} B_i \in \cB(Y)$
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and
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\[f(\coprod_{i < \omega} B_i) = \bigcup_i A_i.\]
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\item $\bigcap_i A_i$ is $\Sigma^1_1$:
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% Let $Y_i$ be Polish such that $f_i(Y_i) = A_i$.
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% Let $Y \coloneqq \coprod Y_i$, $f = \coprod f_i$ and $Z = \prod Y_i$.
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% Note that $Y$ and $Z$ are Polish.
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% We can embed $Z$ into $Y^{\N}$.
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%
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% Define a tree $T$ on $Y$ as follows:
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% $(y_0, \ldots, y_n) \in T$ iff
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% \begin{itemize}
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% \item $\forall 0 \le i \le n.~ y_i \in Y_i$ and
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% \item $\forall i,j .~ f(y_i) = f(y_j)$.
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% \end{itemize}
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%
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% Then $[T]$ consists of sequences $y = (y_n)$
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% such that $\forall j \in \N.~f(y) \in \im (f_j)$,
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% so $\forall y \in [T].~f(y) \in \bigcap_{i \in \N} \im(f_i) = \bigcap_{i \in \N} A_i$.
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% $[T] \subseteq i(Z) \subseteq Y^{\N}$,
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% and $[T]$ is closed.
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%
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%
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% Other solution:
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Let $Z = \prod Y_i$
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and let $D \subseteq Z$
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be defined by
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\[
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D \coloneqq \{(y_n) : f_i(y_i) = f_j(y_j) ~ \forall i,j\}.
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\]
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$D$ is closed,
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at it is the preimage of the diagonal
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under $Z \xrightarrow{(f_0,f_1,\ldots)} X^{\N}$.
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Then $\bigcap A_i$ is the image of $D$
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under $Z \xrightarrow{(y_n) \mapsto f_0(y_0)} X$.
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\paragraph{Other solution}
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Let $F_n \subseteq X \times \cN$ be closed,
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and $C \subseteq X \times \cN^{\N}$ defined by
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\[
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C \coloneqq \{(x,(y^{(n)}) ) : \forall n.~(x, y^{(n)}) \in F_n\}.
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\]
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$C$ is closed
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and $\bigcap A_i = \proj_X(C)$.
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\end{itemize}
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\nr 3
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\todo{Wait for mail}
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\todo{Find a countable clopen base}
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\begin{itemize}
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\item We use the same construction as in exercise 2 (a)
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on sheet 6.
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Let $A \subseteq X$ be analytic,
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i.e.~there exists a Polish space $Y$ and $f\colon Y \to X$ Borel
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with $f(Y) = X$.
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Then $f$ is still Borel with respect to the
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new topology, since Borel sets are preserved
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and by exercise 1 (b).
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% Let $(B_i)_{i < \omega}$ be a countable basis of $(X,\tau)$.
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% By a theorem from the lecture, there exists Polish
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% topologies $\cT_i$ such that $B_i$ is clopen wrt.~$\cT_i$
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% and $\cB(\cT_i) = \cB(\tau)$.
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% By a lemma from the lecture,
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% $\tau' \coloneqq \bigcup_i \cT_i$
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% is Polish as well and $\cB(\tau') = \cB(\tau)$.
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% \todo{TODO: Basis}
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\item Suppose that there exist no disjoint clopen sets $U_0,U_1$,
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such that $W \cap U_0$ and $W \cap U_1$ are uncountable.
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Let $W_0 \coloneqq W$
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Then there exist disjoint clopen sets $C_i^{(0)}$
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such that $W_0 \subseteq \bigcup_{i < \omega} C_i^{(0)}$
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and $\diam(C_i) < 1$,
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since $X$ is zero-dimensional.
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By assumption, exactly one of the $C_i^{(0)}$ has
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uncountable intersection with $W_0$.
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Let $i_0$ be such that $W_0 \cap C_{i_0}^{(0)}$ is uncountable
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and set $W_1 \coloneqq W_0 \cap C_{i_0}^{(0)}$.
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Note that $W_0 \setminus W_1 = \bigcup_{i \neq i_0} C_i^{(0)}$ is countable.
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Let us recursively continue this construction:
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Suppose that $W_n$ uncountable has been chosen.
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Then choose $C_{i}^{(n)}$ clopen,
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disjoint with diameter $\le \frac{1}{n}$
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such that $W_n \subseteq \bigcup_{i} C_i^{(n)}$
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and let $i_n$ be the unique index
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such that $W_n \cap C_{i_n}^{(n)}$ is uncountable.
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Since $\diam(C_{i_n}^{(n)}) \xrightarrow{n \to \infty} 0$
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and the $C_{i_n}^{(n)}$ are closed,
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we get that $\bigcap_n C_{i_n}^{(n)}$
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contains exactly one point. Let that point be $x$.
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However then
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\[
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W = \left(\bigcup_{n < \omega} \bigcup_{i \neq i_n} (C_{i}^{(n)} \cap W)\right)
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\cup \bigcap_{n} (W \cap C_{i_n}^{(n)})
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= \left(\bigcup_{n < \omega} \bigcup_{i \neq i_n} (C_{i}^{(n)} \cap W)\right) \cup \{x\}
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\]
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is countable as a countable union of countable sets $\lightning$.
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\item Consider a finer topology $\tau'$ on $X$ such that $(X, \tau')$ is zero dimensional
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as in the first part.
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Clearly $f$ is also continuous with respect to the new topology,
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so we may assume that $X$ is zero dimensional.
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Let $W \subseteq X$ be such that $f\defon{W}$ is injective
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and $f(W) = f(X)$ (this exists by the axiom of choice).
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Since $f(X)$ is uncountable, so is $W$.
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By the second point, there exist disjoint clopen sets
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$U_0, U_1$, such that $W \cap U_0$ and $W\cap U_1$
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are uncountable.
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Inductively construct $U_s$ for $s \in 2^{<\omega}$
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as follows:
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Suppose that $U_{s}$ has already been chosen.
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Then let $U_{s\concat 0}, U_{s\concat 1} \subseteq U_s$
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be disjoint clopen such that $U_{s\concat 1} \cap W$
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and $U_{s\concat 0} \cap W$ are uncountable.
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Such sets exist, since $ U_s \cap W$ is uncountable
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and $U_s$ is a zero dimensional space with the subspace topology.
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And since $U_s$ is clopen, we have that a subset of $U_s$ is clopen
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in $U_s$ iff it is clopen in $X$.
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Clearly this defines a Cantor scheme.
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\item \todo{TODO}
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\end{itemize}
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\nr 4
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Proof of Schröder-Bernstein:
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Let $X_0 \coloneqq X$, $Y_0 \coloneqq Y$
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and define $X_{i+1} \coloneqq g(Y_i)$, $Y_{i+1 } \coloneqq g(X_i)$.
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We have $X_{i+1} \subseteq X_i$ and similarly for $Y_i$.
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$f$ and $g$ are bijections between
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$X_\omega \coloneqq \bigcap X_i$ and $Y_\omega \coloneqq \bigcap Y_i$.
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%\resizebox{\textwidth}{!}{%
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% https://q.uiver.app/#q=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\[\begin{tikzcd}
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{X \setminus X_\omega =} & {(X_0 \setminus X_1)} & \cup & {(X_0 \setminus X_1)} & \cup & {(X_0 \setminus X_1)} & \cdots & {} \\
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{Y\setminus Y_\omega =} & {(Y_0 \setminus Y_1)} & \cup & {(Y_0 \setminus Y_1)} & \cup & {(Y_0 \setminus Y_1)} & \cdots & {}
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\arrow["f"'{pos=0.7}, from=1-2, to=2-4]
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\arrow["g"{pos=0.1}, from=2-2, to=1-4]
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\arrow["f"{pos=0.8}, from=1-6, to=2-8]
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\arrow["g"{pos=0.1}, from=2-6, to=1-8]
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\end{tikzcd}\]
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%}
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By \autoref{thm:lusinsouslin}
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the injective image via a Borel set of a Borel set is Borel.
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\autoref{thm:lusinsouslin} also gives that the inverse
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of a bijective Borel map is Borel.
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So we can just do the same proof and every set will be Borel.
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@ -150,8 +150,9 @@
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\end{proof}
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\begin{corollary}
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\label{cor:perfectpolishcard}
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Every nonempty perfect Polish
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space $X$ has cardinality $C = 2^{\aleph_0}$
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space $X$ has cardinality $\fc = 2^{\aleph_0}$
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% TODO: eulerscript C ?
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\end{corollary}
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\begin{proof}
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|
@ -163,9 +164,11 @@
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\begin{theorem}
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Any Polish space is countable
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or it has cardinality $C$. % TODO C
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or it has cardinality $\fc$. % TODO C
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\end{theorem}
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\todo{Homework 3}
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\begin{proof}
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See \autoref{cor:polishcard}.
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\end{proof}
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\begin{definition}
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|
|
|
@ -1,3 +1,5 @@
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\subsection{Sheet 6}
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\lecture{07}{2023-11-07}{}
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\begin{proposition}
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|
@ -188,5 +190,3 @@
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$\cB(\cT_\infty') = \cB(\cT)$
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and $A $ is clopen in $\cT_{\infty}'$.
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\end{refproof}
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|
|
|
@ -43,6 +43,7 @@ We will see that not every analytic set is Borel.
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\end{remark}
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\begin{theorem}
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\label{thm:borel}
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Let $X$ be Polish,
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$\emptyset \neq A \subseteq X$.
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Then the following are equivalent:
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|
@ -118,7 +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 \todo{Exercise}
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\item See \yaref{ex:7.2}.
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\end{enumerate}
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\end{proof}
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|
|
|
@ -51,7 +51,7 @@
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\begin{theorem}
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\label{thm:lec12:1}
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Suppose that $A \subseteq \cN$ is analytic.
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Then there is $f\colon \cN \to \Tr$
|
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Then there is $f\colon \cN \to \Tr$\todo{Borel?}
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such that $x \in A \iff f(x)$ is ill-founded.
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\end{theorem}
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For the proof we need some prerequisites:
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|
|
|
@ -197,9 +197,12 @@ let
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\todo{TODO}
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\end{proof}
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\begin{corollary}
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\begin{corollary}\label{cor:polishcard}
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Any Polish space is either countable or has cardinality equal to $\fc$.
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\end{corollary}
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\begin{subproof}
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\todo{TODO}
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Let $X = P \sqcup U$
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where $P$ is perfect and $U$ is countable.
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If $P \neq \emptyset$, we have $|P| = \fc$
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by \yaref{cor:perfectpolishcard}.
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\end{subproof}
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||||
|
|
|
@ -48,7 +48,7 @@ so $f\defon{A}^{-1}(U_i) = V_i \cap A$ is open.
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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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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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||||
|
|
|
@ -1,9 +1,9 @@
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|||
\subsection{Sheet 6}
|
||||
\tutorial{07}{2023-11-28}{}
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% 5 / 20
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\subsection{Exercise 1}
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\nr 1
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\begin{warning}
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Note that not every set has a density!
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\end{warning}
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|
@ -33,8 +33,7 @@
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Clearly this is a $\Pi^0_3$-set.
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\end{enumerate}
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\subsection{Exercise 2}
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\nr 2
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\begin{fact}
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Let $(X,\tau)$ be a Polish space and
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$A \in \cB(X)$.
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|
@ -96,8 +95,7 @@
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try doing it again ($\omega$-many times).
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\end{idea}
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\subsection{Exercise 3}
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\nr 3
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||||
\begin{enumerate}[(a)]
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||||
\item Show that if $\Gamma$ is self-dual (closed under complements)
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and closed under continuous preimages,
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|
@ -128,8 +126,7 @@
|
|||
and closed under continuous preimages (by a trivial induction).
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||||
\end{enumerate}
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\subsection{Exercise 4}
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\nr 4
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||||
Recall:
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||||
\begin{fact}[Sheet 5, Exercise 1]
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||||
Let $\emptyset\neq X$ be a Baire space.
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||||
|
@ -137,12 +134,15 @@ Then $\forall A \subseteq X$,
|
|||
$A$ is either meager or locally comeager.
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\end{fact}
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\begin{theorem}[Kechris 16.1]
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\begin{theorem}\footnote{See Kechris 16.1}
|
||||
Let $X, Y$ be Polish.
|
||||
|
||||
Let
|
||||
\[\cA \coloneqq \{A \in \cB(X \times Y) : \forall \emptyset \neq U \overset{\text{open}}{\subseteq} Y.~
|
||||
A_U \coloneqq \{ x \in X : A_x \text{ is not meager in $U$}\} \text{ is Borel}\}.\]
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||||
For $\emptyset \neq U \overset{\text{open}}{\subseteq} Y$
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||||
let
|
||||
\[A_U \coloneqq \{ x \in X : A_x \text{ is not meager in $U$}\}.\]
|
||||
|
||||
Define
|
||||
\[\cA \coloneqq \{A \in \cB(X \times Y) : \forall \emptyset \neq U \overset{\text{open}}{\subseteq} Y.~ A_U \text{ is Borel}\}.\]
|
||||
|
||||
Then $\cA$ contains all Borel sets.
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||||
\end{theorem}
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||||
|
@ -172,9 +172,3 @@ $A$ is either meager or locally comeager.
|
|||
(a countable union suffices, since we only need to check this for $V$ of the basis; if $A \subseteq V$ is nwd, then $A \cap U \subseteq U$ is nwd for all $U \overset{\text{open}}{\subseteq} V$).
|
||||
\end{enumerate}
|
||||
\end{proof}
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
|
223
inputs/tutorial_08.tex
Normal file
223
inputs/tutorial_08.tex
Normal file
|
@ -0,0 +1,223 @@
|
|||
\subsection{Sheet 7}
|
||||
|
||||
\tutorial{08}{2023-12-05}{}
|
||||
% 17 / 20
|
||||
\nr 1
|
||||
\begin{itemize}
|
||||
\item For $\xi = 1$ this holds by the definition of the
|
||||
subspace topology.
|
||||
|
||||
We now use transfinite induction, to show that
|
||||
the statement holds for all $\xi$.
|
||||
Suppose that $\Sigma^0_{\zeta}(Y)$ and $\Pi^0_{\zeta}(Y)$
|
||||
are as claimed for all $\zeta < \xi$.
|
||||
|
||||
Then
|
||||
\begin{IEEEeqnarray*}{rCl}
|
||||
\Sigma^0_\xi(Y) &=& \{\bigcup_{n < \omega} A_n : A_n \in \Pi^0_{\alpha_n}(Y), \alpha_n < \xi\}\\
|
||||
&=& \{\bigcup_{n < \omega} (A_n \cap Y) : A_n \in \Pi^0_{\alpha_n}(X), \alpha_n < \xi\}\\
|
||||
&=& \{Y \cap \bigcup_{n < \omega} A_n : A_n \in \Pi^0_{\alpha_n}(X), \alpha_n < \xi\}\\
|
||||
&=& \{Y \cap A : A \in \Sigma^0_{\xi}(X)\}.
|
||||
\end{IEEEeqnarray*}
|
||||
and
|
||||
\begin{IEEEeqnarray*}{rCl}
|
||||
\Pi^0_\xi(Y) &=& \lnot \Sigma^0_\xi(Y)\\
|
||||
&=& \{Y \setminus A : A \in \Sigma^0_\xi(Y)\}\\
|
||||
&=& \{Y \setminus (A \cap Y) : A \in \Sigma^0_\xi(X)\}\\
|
||||
&=& \{Y \cap (X \setminus A) : A \in \Sigma^0_\xi(X)\}\\
|
||||
&=& \{Y \cap A : A \in \Pi^0_\xi(X)\}.
|
||||
\end{IEEEeqnarray*}
|
||||
\item Let $V \in \cB(Y)$.
|
||||
|
||||
We show that $f^{-1}(V) \in \cB(Y)$,
|
||||
by induction on the minimal $\xi$ such that $V \in \Sigma_\xi^0$.
|
||||
For $\xi = 0$ this is clear.
|
||||
Suppose that we have already shown $f^{-1}(V') \in \cB(Y)$
|
||||
for all $V' \in \Sigma^0_\zeta$, $\zeta < \xi$.
|
||||
Then $f^{-1}(Y \setminus V') = X \setminus f^{-1}(V') \in \cB(V)$,
|
||||
since complements of Borel sets are Borel.
|
||||
In particular, this also holds for $\Pi^0_\zeta$ sets
|
||||
and $\zeta < \xi$.
|
||||
Let $V \in \Sigma^0_\xi$.
|
||||
Then $V = \bigcap_{n} V_n$ for some $V_n \in \Pi^{0}_{\alpha_n}$,
|
||||
$\alpha_n < \xi$.
|
||||
In particular $f^{-1}(V) = \bigcup_n f^{-1}(V_n) \in \cB(X)$.
|
||||
\end{itemize}
|
||||
|
||||
\nr 2
|
||||
\yalabel{Exercise}{}{ex:7.2}
|
||||
|
||||
|
||||
Recall \autoref{thm:analytic}.
|
||||
|
||||
Let $(A_i)_{i<\omega}$ be analytic subsets of a Polish space $X$.
|
||||
Then there exists Polish spaces $Y_i$ and $f_i\colon Y_i \to X$
|
||||
continuous such that $f_i(B_i) = A_i$
|
||||
for some $B_i \in \cB(Y_i)$.
|
||||
|
||||
\begin{itemize}
|
||||
\item $\bigcup_i A_i$ is analytic:
|
||||
Consider the Polish space $Y \coloneqq \coprod_{i < \omega} Y_i$
|
||||
and $f \coloneqq \coprod_i f_i$, i.e.~
|
||||
$Y_i \ni y \mapsto f_i(y)$.
|
||||
$f$ is continuous,
|
||||
$\coprod_{i < \omega} B_i \in \cB(Y)$
|
||||
and
|
||||
\[f(\coprod_{i < \omega} B_i) = \bigcup_i A_i.\]
|
||||
\item $\bigcap_i A_i$ is analytic:
|
||||
% Let $Y_i$ be Polish such that $f_i(Y_i) = A_i$.
|
||||
% Let $Y \coloneqq \coprod Y_i$, $f = \coprod f_i$ and $Z = \prod Y_i$.
|
||||
% Note that $Y$ and $Z$ are Polish.
|
||||
% We can embed $Z$ into $Y^{\N}$.
|
||||
%
|
||||
% Define a tree $T$ on $Y$ as follows:
|
||||
% $(y_0, \ldots, y_n) \in T$ iff
|
||||
% \begin{itemize}
|
||||
% \item $\forall 0 \le i \le n.~ y_i \in Y_i$ and
|
||||
% \item $\forall i,j .~ f(y_i) = f(y_j)$.
|
||||
% \end{itemize}
|
||||
%
|
||||
% Then $[T]$ consists of sequences $y = (y_n)$
|
||||
% such that $\forall j \in \N.~f(y) \in \im (f_j)$,
|
||||
% so $\forall y \in [T].~f(y) \in \bigcap_{i \in \N} \im(f_i) = \bigcap_{i \in \N} A_i$.
|
||||
% $[T] \subseteq i(Z) \subseteq Y^{\N}$,
|
||||
% and $[T]$ is closed.
|
||||
%
|
||||
%
|
||||
% Other solution:
|
||||
Let $Z = \prod Y_i$
|
||||
and let $D \subseteq Z$
|
||||
be defined by
|
||||
\[
|
||||
D \coloneqq \{(y_n) : f_i(y_i) = f_j(y_j) ~ \forall i,j\}.
|
||||
\]
|
||||
$D$ is closed,
|
||||
at it is the preimage of the diagonal
|
||||
under $Z \xrightarrow{(f_0,f_1,\ldots)} X^{\N}$.
|
||||
Then $\bigcap A_i$ is the image of $D$
|
||||
under $Z \xrightarrow{(y_n) \mapsto f_0(y_0)} X$.
|
||||
|
||||
\paragraph{Other solution}
|
||||
|
||||
Let $F_n \subseteq X \times \cN$ be closed,
|
||||
and $C \subseteq X \times \cN^{\N}$ defined by
|
||||
\[
|
||||
C \coloneqq \{(x,(y^{(n)}) ) : \forall n.~(x, y^{(n)}) \in F_n\}.
|
||||
\]
|
||||
$C$ is closed
|
||||
and $\bigcap A_i = \proj_X(C)$.
|
||||
\end{itemize}
|
||||
|
||||
\nr 3
|
||||
\todo{Wait for mail}
|
||||
\todo{Find a countable clopen base}
|
||||
|
||||
\begin{itemize}
|
||||
\item We use the same construction as in exercise 2 (a)
|
||||
on sheet 6.
|
||||
Let $A \subseteq X$ be analytic,
|
||||
i.e.~there exists a Polish space $Y$ and $f\colon Y \to X$ Borel
|
||||
with $f(Y) = X$.
|
||||
Then $f$ is still Borel with respect to the
|
||||
new topology, since Borel sets are preserved
|
||||
and by exercise 1 (b).
|
||||
% Let $(B_i)_{i < \omega}$ be a countable basis of $(X,\tau)$.
|
||||
% By a theorem from the lecture, there exists Polish
|
||||
% topologies $\cT_i$ such that $B_i$ is clopen wrt.~$\cT_i$
|
||||
% and $\cB(\cT_i) = \cB(\tau)$.
|
||||
% By a lemma from the lecture,
|
||||
% $\tau' \coloneqq \bigcup_i \cT_i$
|
||||
% is Polish as well and $\cB(\tau') = \cB(\tau)$.
|
||||
% \todo{TODO: Basis}
|
||||
\item Suppose that there exist no disjoint clopen sets $U_0,U_1$,
|
||||
such that $W \cap U_0$ and $W \cap U_1$ are uncountable.
|
||||
|
||||
Let $W_0 \coloneqq W$
|
||||
Then there exist disjoint clopen sets $C_i^{(0)}$
|
||||
such that $W_0 \subseteq \bigcup_{i < \omega} C_i^{(0)}$
|
||||
and $\diam(C_i) < 1$,
|
||||
since $X$ is zero-dimensional.
|
||||
|
||||
By assumption, exactly one of the $C_i^{(0)}$ has
|
||||
uncountable intersection with $W_0$.
|
||||
Let $i_0$ be such that $W_0 \cap C_{i_0}^{(0)}$ is uncountable
|
||||
and set $W_1 \coloneqq W_0 \cap C_{i_0}^{(0)}$.
|
||||
Note that $W_0 \setminus W_1 = \bigcup_{i \neq i_0} C_i^{(0)}$ is countable.
|
||||
|
||||
Let us recursively continue this construction:
|
||||
Suppose that $W_n$ uncountable has been chosen.
|
||||
Then choose $C_{i}^{(n)}$ clopen,
|
||||
disjoint with diameter $\le \frac{1}{n}$
|
||||
such that $W_n \subseteq \bigcup_{i} C_i^{(n)}$
|
||||
and let $i_n$ be the unique index
|
||||
such that $W_n \cap C_{i_n}^{(n)}$ is uncountable.
|
||||
|
||||
Since $\diam(C_{i_n}^{(n)}) \xrightarrow{n \to \infty} 0$
|
||||
and the $C_{i_n}^{(n)}$ are closed,
|
||||
we get that $\bigcap_n C_{i_n}^{(n)}$
|
||||
contains exactly one point. Let that point be $x$.
|
||||
|
||||
However then
|
||||
\[
|
||||
W = \left(\bigcup_{n < \omega} \bigcup_{i \neq i_n} (C_{i}^{(n)} \cap W)\right)
|
||||
\cup \bigcap_{n} (W \cap C_{i_n}^{(n)})
|
||||
= \left(\bigcup_{n < \omega} \bigcup_{i \neq i_n} (C_{i}^{(n)} \cap W)\right) \cup \{x\}
|
||||
\]
|
||||
is countable as a countable union of countable sets $\lightning$.
|
||||
|
||||
\item Consider a finer topology $\tau'$ on $X$ such that $(X, \tau')$ is zero dimensional
|
||||
as in the first part.
|
||||
Clearly $f$ is also continuous with respect to the new topology,
|
||||
so we may assume that $X$ is zero dimensional.
|
||||
|
||||
Let $W \subseteq X$ be such that $f\defon{W}$ is injective
|
||||
and $f(W) = f(X)$ (this exists by the axiom of choice).
|
||||
Since $f(X)$ is uncountable, so is $W$.
|
||||
By the second point, there exist disjoint clopen sets
|
||||
$U_0, U_1$, such that $W \cap U_0$ and $W\cap U_1$
|
||||
are uncountable.
|
||||
Inductively construct $U_s$ for $s \in 2^{<\omega}$
|
||||
as follows:
|
||||
Suppose that $U_{s}$ has already been chosen.
|
||||
Then let $U_{s\concat 0}, U_{s\concat 1} \subseteq U_s$
|
||||
be disjoint clopen such that $U_{s\concat 1} \cap W$
|
||||
and $U_{s\concat 0} \cap W$ are uncountable.
|
||||
Such sets exist, since $ U_s \cap W$ is uncountable
|
||||
and $U_s$ is a zero dimensional space with the subspace topology.
|
||||
And since $U_s$ is clopen, we have that a subset of $U_s$ is clopen
|
||||
in $U_s$ iff it is clopen in $X$.
|
||||
|
||||
Clearly this defines a Cantor scheme.
|
||||
\item \todo{TODO}
|
||||
|
||||
\end{itemize}
|
||||
|
||||
\nr 4
|
||||
|
||||
Proof of Schröder-Bernstein:
|
||||
|
||||
Let $X_0 \coloneqq X$, $Y_0 \coloneqq Y$
|
||||
and define $X_{i+1} \coloneqq g(Y_i)$, $Y_{i+1 } \coloneqq g(X_i)$.
|
||||
We have $X_{i+1} \subseteq X_i$ and similarly for $Y_i$.
|
||||
$f$ and $g$ are bijections between
|
||||
$X_\omega \coloneqq \bigcap X_i$ and $Y_\omega \coloneqq \bigcap Y_i$.
|
||||
|
||||
|
||||
% https://q.uiver.app/#q=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
|
||||
\adjustbox{scale=0.7,center}{%
|
||||
\begin{tikzcd}
|
||||
{X \setminus X_\omega =} & {(X_0 \setminus X_1)} & \cup & {(X_0 \setminus X_1)} & \cup & {(X_0 \setminus X_1)} & \cdots & {} \\
|
||||
{Y\setminus Y_\omega =} & {(Y_0 \setminus Y_1)} & \cup & {(Y_0 \setminus Y_1)} & \cup & {(Y_0 \setminus Y_1)} & \cdots & {}
|
||||
\arrow["f"'{pos=0.7}, from=1-2, to=2-4]
|
||||
\arrow["g"{pos=0.1}, from=2-2, to=1-4]
|
||||
\arrow["f"{pos=0.8}, from=1-6, to=2-8]
|
||||
\arrow["g"{pos=0.1}, from=2-6, to=1-8]
|
||||
\end{tikzcd}
|
||||
}
|
||||
|
||||
By \autoref{thm:lusinsouslin}
|
||||
the injective image via a Borel set of a Borel set is Borel.
|
||||
|
||||
\autoref{thm:lusinsouslin} also gives that the inverse
|
||||
of a bijective Borel map is Borel.
|
||||
So we can just do the same proof and every set will be Borel.
|
|
@ -22,6 +22,7 @@
|
|||
\usepackage{listings}
|
||||
\usepackage{multirow}
|
||||
\usepackage{float}
|
||||
\usepackage{adjustbox}
|
||||
\usepackage{quiver}
|
||||
%\usepackage{algorithmicx}
|
||||
|
||||
|
@ -132,6 +133,9 @@
|
|||
\DeclareSimpleMathOperator{LO} % linear orders
|
||||
\DeclareSimpleMathOperator{WO} % well orderings
|
||||
|
||||
|
||||
\DeclareSimpleMathOperator{osc} % oscillation
|
||||
|
||||
\newcommand{\concat}{\mathop{{}^{\scalebox{.7}{$\smallfrown$}}}}
|
||||
|
||||
%https://tex.stackexchange.com/questions/73437/how-do-i-typeset-the-concatenation-of-strings-properly
|
||||
|
@ -143,3 +147,5 @@
|
|||
\newcommand{\fc}{\mathfrak{c}}
|
||||
|
||||
\newcommand\lecture[3]{\hrule{\color{darkgray}\hfill{\tiny[Lecture #1, #2]}}}
|
||||
\newcommand\tutorial[3]{\hrule{\color{darkgray}\hfill{\tiny[Tutorial #1, #2]}}}
|
||||
\newcommand\nr[1]{\subsubsection{Exercise #1}}
|
||||
|
|
15
logic3.tex
15
logic3.tex
|
@ -15,6 +15,7 @@
|
|||
|
||||
\cleardoublepage
|
||||
|
||||
\setcounter{tocdepth}{2}
|
||||
\tableofcontents
|
||||
\cleardoublepage
|
||||
|
||||
|
@ -38,6 +39,9 @@
|
|||
\input{inputs/lecture_12}
|
||||
\input{inputs/lecture_13}
|
||||
\input{inputs/lecture_14}
|
||||
\input{inputs/lecture_15}
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
@ -45,6 +49,17 @@
|
|||
|
||||
\appendix
|
||||
|
||||
\section{Tutorial and Exercises}
|
||||
\input{inputs/tutorial_01}
|
||||
\input{inputs/tutorial_02}
|
||||
\input{inputs/tutorial_03}
|
||||
\input{inputs/tutorial_04}
|
||||
\input{inputs/tutorial_05}
|
||||
\input{inputs/tutorial_06}
|
||||
\input{inputs/tutorial_07}
|
||||
\input{inputs/tutorial_08}
|
||||
|
||||
|
||||
\PrintVocabIndex
|
||||
|
||||
|
||||
|
|
Loading…
Reference in a new issue