Josia Pietsch
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286 lines
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\subsection{Sheet 7}


\tutorial{08}{20231205}{}


% 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 secondcountable 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 zerodimensional.




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 nonempty 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 SchröderBernstein:




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 $Y_\omega \coloneqq \bigcap Y_i$.






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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=12, to=24]


\arrow["g"{pos=0.1}, from=22, to=14]


\arrow["f"{pos=0.8}, from=16, to=28]


\arrow["g"{pos=0.1}, from=26, to=18]


\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.
