From f0bab3e1ac28930021d472ba68430b140e658b7d Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Fri, 8 Dec 2023 01:55:13 +0100 Subject: [PATCH] removed junk --- + | 223 -------------------------------------------------------------- 1 file changed, 223 deletions(-) delete mode 100644 + diff --git a/+ b/+ deleted file mode 100644 index ad680a2..0000000 --- a/+ +++ /dev/null @@ -1,223 +0,0 @@ -\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 - - -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 $\Sigma^1_1$: - % 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 $Y_\omega \coloneqq \bigcap Y_i$. - - -%\resizebox{\textwidth}{!}{% -% https://q.uiver.app/#q=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 -\[\begin{tikzcd} - {X \setminus X_\omega =} & {(X_0 \setminus X_1)} & \cup & {(X_0 \setminus X_1)} & \cup & {(X_0 \setminus X_1)} & \cdots & {} \\ - {Y\setminus Y_\omega =} & {(Y_0 \setminus Y_1)} & \cup & {(Y_0 \setminus Y_1)} & \cup & {(Y_0 \setminus Y_1)} & \cdots & {} - \arrow["f"'{pos=0.7}, from=1-2, to=2-4] - \arrow["g"{pos=0.1}, from=2-2, to=1-4] - \arrow["f"{pos=0.8}, from=1-6, to=2-8] - \arrow["g"{pos=0.1}, from=2-6, to=1-8] -\end{tikzcd}\] -%} - -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.