\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 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 $Z \xrightarrow{(y_n) \mapsto f_0(y_0)} X$. \emph{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 \begin{lemma} Let $X$ be a second-countable topological space. Then every base of $X$ contains a countable subset which is also a base of $X$. \end{lemma} \begin{proof} Let $\cC = \{C_n : n < \omega\}$ be a countable base of $X$ and let $\cB = \{B_i: i \in I\}$ be a base of $X$ with (possibly uncountable) index set $I$. Fix $n < \omega$. It suffices to show that $C_n$ is a union of countably many elements of $\cB$. As $\cB$ is a base, $C_n = \bigcup_{j \in J} B_j$ for some $J \subseteq I$. Since $\cC$ is a base, there exists $M_j \subseteq \N$ such that $B_j = \bigcup_{m \in M_j} C_m$ for all $j \in J$. Let $M = \bigcup_{j \in J} M_j \subseteq \N$. For each $m \in M$, there exists $f(m) \in J$ such that $m \in M_{f(m)}$. Then $\bigcup_{m \in M} B_{f(m)} = C_n$. \end{proof} \begin{remark} We don't actually need this. \end{remark} \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 $\lightning$. Other proof (without using the existence of a countable clopen basis): We can cover $X$ by countably many clopen sets of diameter $< \frac{1}{n}$: Cover $X$ with open balls of diameter $< \frac{1}{n}$. Write each open ball as a union of clopen sets. That gives us a cover by clopen sets of diameter $< \frac{1}{n}$. As $X$ is Lindelöf, there exists a countable subcover. Then continue as in the first proof. \item Note that this step does not help us to prove the statement. It was an error on the exercise sheet. % 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 Let $Y$ be a Polish space and $A \subseteq Y$ analytic and uncountable. Expand the topology on $Y$ so that $Y$ is zero dimensional and $A$ is still analytic. Then there exists a Polish space $X$ and a continuous function $f\colon X \to Y$ such that $f(X) = A$. $A$ is uncountable, so by (2) there exists non-empty disjoint clopen $V_0$, $V_1$ such that $V_0 \cap A$ and $V_1 \cap A$ are uncountable. Let $W_0 = f^{-1}(V_0 \cap A)$ and $W_1 = f^{-1}(V_1 \cap A)$. $ W_0$ and $W_1$ are clopen and disjoint. We can cover $W_0$ with countably many open sets of diameter $\le \frac{1}{n}$ and similarly for $W_1$. Then pick open sets such that there image is uncountable. Repeating this construction we get a Cantor scheme on $X$. So we get $2^{\N} \overset{s}{\hookrightarrow} X$ and by construction of the cantor scheme, we get that $f \circ s$ is injective and continuous. \end{itemize} \nr 4 Proof of Schröder-Bernstein: Let $X_0 \coloneqq X$, $Y_0 \coloneqq Y$ and define $X_{i+1} \coloneqq g(Y_i)$, $Y_{i+1 } \coloneqq f(X_i)$. 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_1 \setminus X_2)} & \cup & {(X_2 \setminus X_3)} & \cdots & {} \\ {Y\setminus Y_\omega =} & {(Y_0 \setminus Y_1)} & \cup & {(Y_1 \setminus Y_2)} & \cup & {(Y_2 \setminus Y_3)} & \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 function 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.