diff --git a/+ b/+ new file mode 100644 index 0000000..ad680a2 --- /dev/null +++ b/+ @@ -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 + + +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. diff --git a/inputs/lecture_03.tex b/inputs/lecture_03.tex index 9fa61e1..1f02625 100644 --- a/inputs/lecture_03.tex +++ b/inputs/lecture_03.tex @@ -150,8 +150,9 @@ \end{proof} \begin{corollary} + \label{cor:perfectpolishcard} Every nonempty perfect Polish - space $X$ has cardinality $C = 2^{\aleph_0}$ + space $X$ has cardinality $\fc = 2^{\aleph_0}$ % TODO: eulerscript C ? \end{corollary} \begin{proof} @@ -163,9 +164,11 @@ \begin{theorem} Any Polish space is countable - or it has cardinality $C$. % TODO C + or it has cardinality $\fc$. % TODO C \end{theorem} -\todo{Homework 3} +\begin{proof} + See \autoref{cor:polishcard}. +\end{proof} \begin{definition} diff --git a/inputs/lecture_07.tex b/inputs/lecture_07.tex index 09fc8be..82929d9 100644 --- a/inputs/lecture_07.tex +++ b/inputs/lecture_07.tex @@ -1,3 +1,5 @@ +\subsection{Sheet 6} + \lecture{07}{2023-11-07}{} \begin{proposition} @@ -188,5 +190,3 @@ $\cB(\cT_\infty') = \cB(\cT)$ and $A $ is clopen in $\cT_{\infty}'$. \end{refproof} - - diff --git a/inputs/lecture_09.tex b/inputs/lecture_09.tex index f79c2aa..7b90b55 100644 --- a/inputs/lecture_09.tex +++ b/inputs/lecture_09.tex @@ -43,6 +43,7 @@ We will see that not every analytic set is Borel. \end{remark} \begin{theorem} + \label{thm:borel} Let $X$ be Polish, $\emptyset \neq A \subseteq X$. Then the following are equivalent: @@ -118,7 +119,7 @@ We will see that not every analytic set is Borel. Take $f^+\colon X \times Z \to Y \times Z, f^+ = f \times \id$. Then \[f^{-1}(B) = \underbrace{\proj_X(\overbrace{(\underbrace{f^+}_{\mathclap{\text{Borel}}})^{-1}(\underbrace{B_0}_{\mathclap{\text{Borel}}})}^{\text{Borel}})}_{\text{analytic}}.\] - \item \todo{Exercise} + \item See \yaref{ex:7.2}. \end{enumerate} \end{proof} diff --git a/inputs/lecture_12.tex b/inputs/lecture_12.tex index 2eaf19e..fd37caf 100644 --- a/inputs/lecture_12.tex +++ b/inputs/lecture_12.tex @@ -51,7 +51,7 @@ \begin{theorem} \label{thm:lec12:1} Suppose that $A \subseteq \cN$ is analytic. - Then there is $f\colon \cN \to \Tr$ + Then there is $f\colon \cN \to \Tr$\todo{Borel?} such that $x \in A \iff f(x)$ is ill-founded. \end{theorem} For the proof we need some prerequisites: diff --git a/inputs/tutorial_04.tex b/inputs/tutorial_04.tex index 33b943b..0a34c14 100644 --- a/inputs/tutorial_04.tex +++ b/inputs/tutorial_04.tex @@ -197,9 +197,12 @@ let \todo{TODO} \end{proof} -\begin{corollary} +\begin{corollary}\label{cor:polishcard} Any Polish space is either countable or has cardinality equal to $\fc$. \end{corollary} \begin{subproof} - \todo{TODO} + Let $X = P \sqcup U$ + where $P$ is perfect and $U$ is countable. + If $P \neq \emptyset$, we have $|P| = \fc$ + by \yaref{cor:perfectpolishcard}. \end{subproof} diff --git a/inputs/tutorial_06.tex b/inputs/tutorial_06.tex index b39a77b..68c2481 100644 --- a/inputs/tutorial_06.tex +++ b/inputs/tutorial_06.tex @@ -48,7 +48,7 @@ so $f\defon{A}^{-1}(U_i) = V_i \cap A$ is open. hence $B' = P \cup C$ for $P$ perfect and $C$ countable, and $|P| \in \{\fc, 0\}$. - But $B'$ can't contain isolated point$. + But $B'$ can't contain isolated point. \item To ensure that (a) holds, it suffices to chose $a_i \not\in F_i$. Since $|B| = \fc$ and $|\{a_i | j < i\}| = |i| < \fc$, diff --git a/inputs/tutorial_07.tex b/inputs/tutorial_07.tex index 7e77ec1..d5a064b 100644 --- a/inputs/tutorial_07.tex +++ b/inputs/tutorial_07.tex @@ -1,9 +1,9 @@ +\subsection{Sheet 6} \tutorial{07}{2023-11-28}{} % 5 / 20 -\subsection{Exercise 1} - +\nr 1 \begin{warning} Note that not every set has a density! \end{warning} @@ -33,8 +33,7 @@ Clearly this is a $\Pi^0_3$-set. \end{enumerate} -\subsection{Exercise 2} - +\nr 2 \begin{fact} Let $(X,\tau)$ be a Polish space and $A \in \cB(X)$. @@ -96,8 +95,7 @@ try doing it again ($\omega$-many times). \end{idea} -\subsection{Exercise 3} - +\nr 3 \begin{enumerate}[(a)] \item Show that if $\Gamma$ is self-dual (closed under complements) and closed under continuous preimages, @@ -128,8 +126,7 @@ and closed under continuous preimages (by a trivial induction). \end{enumerate} -\subsection{Exercise 4} - +\nr 4 Recall: \begin{fact}[Sheet 5, Exercise 1] Let $\emptyset\neq X$ be a Baire space. @@ -137,12 +134,15 @@ Then $\forall A \subseteq X$, $A$ is either meager or locally comeager. \end{fact} -\begin{theorem}[Kechris 16.1] +\begin{theorem}\footnote{See Kechris 16.1} Let $X, Y$ be Polish. - Let - \[\cA \coloneqq \{A \in \cB(X \times Y) : \forall \emptyset \neq U \overset{\text{open}}{\subseteq} Y.~ - A_U \coloneqq \{ x \in X : A_x \text{ is not meager in $U$}\} \text{ is Borel}\}.\] + For $\emptyset \neq U \overset{\text{open}}{\subseteq} Y$ + 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. \end{theorem} @@ -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} - - - - - - diff --git a/inputs/tutorial_08.tex b/inputs/tutorial_08.tex new file mode 100644 index 0000000..e5f13dc --- /dev/null +++ b/inputs/tutorial_08.tex @@ -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. diff --git a/logic.sty b/logic.sty index 30a9a8e..81f2429 100644 --- a/logic.sty +++ b/logic.sty @@ -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}} diff --git a/logic3.tex b/logic3.tex index 6e23238..b4f6e4b 100644 --- a/logic3.tex +++ b/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