94 lines
3.2 KiB
TeX
94 lines
3.2 KiB
TeX
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\tutorial{07}{2023-12-05}{}
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% 17 / 20
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\subsection{Exercise 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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$\bigcap_i A_i$ 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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\subsection{Exercise 3}
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\begin{itemize}
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\item Make $X$ zero dimensional preserving the Borel structure.
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\item \todo{Find a countable clopen base}
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\item
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\end{itemize}
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\subsection{Exercise 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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% https://q.uiver.app/#q=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\[\begin{tikzcd}
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{X \setminus X_\omega} & {=} & {(X_0 \setminus X_1)} & \cup & {(X_0 \setminus X_1)} & \cup & {(X_0 \setminus X_1)} & \cdots & {} \\
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{Y\setminus Y_\omega} & {=} & {(Y_0 \setminus Y_1)} & \cup & {(Y_0 \setminus Y_1)} & \cup & {(Y_0 \setminus Y_1)} & \cdots & {}
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\arrow["f"'{pos=0.7}, from=1-3, to=2-5]
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\arrow["g"{pos=0.1}, from=2-3, to=1-5]
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\arrow["f"{pos=0.8}, from=1-7, to=2-9]
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\arrow["g"{pos=0.1}, from=2-7, to=1-9]
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\end{tikzcd}\]
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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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