diff --git a/inputs/lecture_15.tex b/inputs/lecture_15.tex new file mode 100644 index 0000000..13f2138 --- /dev/null +++ b/inputs/lecture_15.tex @@ -0,0 +1,3 @@ +\lecture{15}{2023-12-05}{} + +\todo{Lecture 15 is missing} diff --git a/inputs/tutorial_01.tex b/inputs/tutorial_01.tex index bce909e..a6a4894 100644 --- a/inputs/tutorial_01.tex +++ b/inputs/tutorial_01.tex @@ -1,6 +1,5 @@ -\tutorial{01}{202-10-17}{} +\tutorial{01}{2023-10-17}{} -% TODO MAIL \begin{fact} A countable product of separable spaces $(X_n)_{n \in \N}$ is separable. @@ -66,5 +65,4 @@ In arbitrary topological spaces, Lindelöf is the strongest of these notions. - \end{remark} diff --git a/inputs/tutorial_02.tex b/inputs/tutorial_02.tex index 2560209..3916370 100644 --- a/inputs/tutorial_02.tex +++ b/inputs/tutorial_02.tex @@ -1,11 +1,181 @@ +\subsection{Sheet 1} \tutorial{02}{2023-10-24}{} - % Points: 15 / 16 -\subsubsection{Exercise 4} +\nr 1 +\todo{handwritten solution} + +\nr 2 + +\begin{enumerate}[(a)] + \item $d$ is an ultrametric: + + Let $f,g,h \in X^{\N}$. + + We need to show that $d(f,g) \le \max(d(f,h), d(g,h))$. + + If $f = g$ this is trivial. + Otherwise let $n$ be minimal such that $f(n) \neq g(n)$. + Then $h(n) \neq f(n)$ or $ h(n) \neq g(n)$ + must be the case. + W.l.o.g.~$h(n) \neq f(n)$. + Then $d(f,g) = \frac{1}{1+n} \le d(f,h)$. + + \item $d$ induces the product topology on $X^{\N}$: + + It suffices to show that the $\epsilon$-balls with respect to $d$ + are exactly the basic open set of the product topology, + i.e.~the sets of the form + \[ + \{x_1\} \times \ldots \times \{x_n\} \times X^{\N} + \] + for some $n \in \N$, $x_1,\ldots,x _n \in X$. + + Let $\epsilon > 0$. Let $n$ be minimal such that $\frac{1}{1+n} \ge \epsilon$. + Then $B_{\epsilon}((x_i)_{i \in \N}) = \{x_1\} \times \{x_n\} \times X^{\N}$. + Since $\N \ni n \mapsto \frac{1}{1+n}$ + is injective, every basic open set of the product topology + can be written in this way. + + \item $d$ is complete: + + Let $(f_n)_{n \in \N}$ be a Cauchy sequence with respect to $d$. + For $n \in \N$ take $N_n \in \N$ + such that $d(f_i, f_j) < \frac{1}{1 + n}$. + Clearly $f_i(n) = f_j(n)$ for all $n > N_n$. + + Define $f \in X^\N$ by $f(n) \coloneqq f_{N_n}(n)$. + Then $ (f_n)_{n \in \N}$ + converges to $f$, + since for all $n > N_n$ $f_n$ + + + \item If $X$ is countable, then $X^{\N}$ with the product topology + is a Polish space: + + (We assume that $X$ is non-empty, as otherwise the claim is wrong) + + We need to show that there exists a countable dense subset. + To this end, pick some $x_0 \in X$ and + consider the set $D \coloneqq \bigcup_{n\in \N} (X^n \times \{x_{0}\}^{\N})$. + Since $X$ is countable, so is $D$. + Take some $(a_n)_{n \in \N} \in X^{\N}$ + and consider $B \coloneqq B_{\epsilon}((a_n)_{n \in \N})$. + Let $m$ be such that $\frac{1}{1+m} < \epsilon$. + Then $(b_{n})_{n \in \N} \in B \cap D$, + where $b_n \coloneqq a_n$ for $n \le m$ and $b_n \coloneqq x_0$ + otherwise. + Hence $D$ is dense. +\end{enumerate} + +\nr 3 + +\begin{enumerate}[(a)] + \item $S_{\infty}$ is a Polish space: + + From (2) we know that $\N^{\N}$ is Polish. + Hence it suffices to show that $S_{\infty}$ is $G_{\delta}$ + with respect to $\N^\N$. + + Consider the sets $I \coloneqq \bigcap_{(i,j) \in \N^2, i < j} \{f \in \N^{\N} | f(i) \neq f(j)\}$ + and $S \coloneqq \bigcap_{n \in \N} \{f \in \N^\N | n \in \im f\}$. + + We have that $\{f \in \N^\N | f(i) \neq f(j)\} = \bigcup_{n \in \N} \N^{i-1} \times \{n\} \times \N^{i - j -1 } \times (\N \setminus \{n\} ) \times \N^\N$ + is open. + Hence $I$ is $G_{\delta}$. + + Furthermore $\{f \in \N^{\N} | n \in \im f\} = \bigcup_{k \in \N} \N^k \times \{n\} \times \N^\N$j + is open, + thus $S$ is $G_\delta$ as well. + In particular $S \cap G$ is $G_\delta$. + Since $I$ is the subset of injective functions + and $S$ is the subset of surjective functions, + we have that $S_{\infty} = I \cap S$. + + \item $S_{\infty}$ is not locally compact: + + Consider the point $x = (i)_{i \in \N} \in S_{\infty}$. + Let $x \in B$ be open. We need to show that there + is no closed compact set $C \supseteq B$ + W.l.o.g.~let $B = (\{0\} \times \ldots \times \{n\} \times \N^\N) \cap S_\infty$ + for some $n \in \N$. + Let $C \supseteq B$ be some closed set. + Consider the open covering + \[ + \{S_{\infty} \setminus B\} \cup \{ B_j | j > n\}. + \] + where + \[ + B_j \coloneqq (\{0\} \times \ldots \times \{n\} \times \{j\} \times \N^{\N}) \cap S_\infty. + \] + Clearly there cannot exist a finite subcover + as $B$ is the disjoint union of the $B_j$. + + % TODO Think about this +\end{enumerate} + +\nr 4 + +% (uniform metric) +% +% \begin{enumerate}[(a)] +% \item $d_u$ is a metric on $\cC(X,Y)$: +% +% It is clear that $d_u(f,f) = 0$. +% +% Let $f \neq g$. Then there exists $x \in X$ with +% $f(x) \neq g(x)$, hence $d_u(f,g) \ge d(f(x), g(x)) > 0$. +% +% Since $d$ is symmetric, so is $d_u$. +% +% Let $f,g,h \in \cC(X,Y)$. +% Take some $\epsilon > 0$ +% choose $x_1, x_2 \in X$ +% with $d_u(f,g) \le d(f(x_1), g(x_1)) + \epsilon$, +% $d_u(g,h) \le d(g(x_2), h(x_2)) + \epsilon$. +% +% Then for all $x \in X$ +% \begin{IEEEeqnarray*}{rCl} +% d(f(x), h(x)) &\le & +% d(f(x), g(x)) + d(g(x), h(x))\\ +% &\le & d(f(x_1), g(x_1)) + d(g(x_2), h(x_2))-2\epsilon\\ +% &\le & d_u(f,g) + d_u(g,h) - 2\epsilon. +% \end{IEEEeqnarray*} +% Thus $d_u(f,g) \le d_u(f,g) + d_u(g,h) - 2\epsilon$. +% Taking $\epsilon \to 0$ yields the triangle inequality. +% +% \item $\cC(X,Y)$ is a Polish space: +% \todo{handwritten solution} +% +% \begin{itemize} +% \item $d_u$ is a complete metric: +% +% Let $(f_n)_n$ be a Cauchy series with respect to $d_u$. +% +% Then clearly $(f_n(x))_n$ is a Cauchy sequence with respect +% to $d$ for every $x$. +% Hence there exists a pointwise limit $f$ of the $f_n$. +% We need to show that $f$ is continuous. +% +% %\todo{something something uniform convergence theorem} +% +% \item $(\cC(X,Y), d_u)$ is separable: +% +% Since $Y$ is separable, there exists a countable +% dense subset $S \subseteq Y$. +% +% Consider $\cC(X,S) \subseteq \cC(X,Y)$. +% Take some $f \in \cC(X,Y)$. +% Since $X$ is compact, +% +% +% % TODO +% +% \end{itemize} +% \end{enumerate} \begin{fact} - Let $X $ be a compact Hausdorffspace. + Let $X $ be a compact Hausdorff space. Then the following are equivalent: \begin{enumerate}[(i)] \item $X$ is Polish, @@ -32,7 +202,7 @@ Hausdorff spaces are normal \end{proof} -Let $X$ be compact Polish (compact metrisable $\implies$ compact Polish) +Let $X$ be compact Polish\footnote{compact metrisable $\implies$ compact Polish} and $Y $ Polish. Let $\cC(X,Y)$ be the set of continuous functions $X \to Y$. Consider the metric $d_u(f,g) \coloneqq \sup_{x \in X} |d(f(x), g(x))|$. @@ -59,3 +229,4 @@ Clearly $d_u$ is a metric. \begin{claim} There exists a countable dense subset. \end{claim} +\todo{handwritten solution} diff --git a/inputs/tutorial_03.tex b/inputs/tutorial_03.tex index 93e170b..d5e685f 100644 --- a/inputs/tutorial_03.tex +++ b/inputs/tutorial_03.tex @@ -1,15 +1,61 @@ -\ctutorial{03}{2023-10-31}{} +\subsection{Sheet 2} + +\tutorial{03}{2023-10-31}{} + % 15 / 16 \begin{remark} - $F_\sigma$ - stands for ferm\'e sum denumerable. + $F_\sigma$ stands for \vocab{ferm\'e sum denumerable}. \end{remark} -\subsection{Exercise 2} +\nr 1 +Let $(U_i)_{i < \omega}$ be a countable base of $X$. +Define +\begin{IEEEeqnarray*}{rCl} + f\colon X &\longrightarrow & 2^{\omega} \\ + x &\longmapsto & (x_i)_{i < \omega} +\end{IEEEeqnarray*} +where $x_i = 1$ iff $x \in U_i$ and $x_i = 0$ otherwise. +Then $f^{-1}(\{y = (y_n) \in 2^\omega | y_n = 1\}) = U_n$ +is open. +We have that $f$ is injective since $X$ is T1. + +Let $f\colon X \hookrightarrow 2^\omega$ be such that +$f^{-1}(\{y = (y_n) \in 2^\omega | y_n = 1\})$. + +Let $V \subseteq 2^{\omega}$ be closed. +Then $2^{\omega} \setminus V$ is open, i.e.~has the form +$\bigcup_{i \in I} ((\prod_{j a_k} \{(a_0,a_1,\ldots,a_{k-1}, b)\} \times \omega^{\omega}. + \] + In both cases the preimage is open. + + + \item $C \coloneqq 2^{\omega} \setminus f(\omega^\omega)$ + is countable and dense in $2^{\omega}$. + + We have $C = \{x \in 2^{\omega} | x_i = 0 \text{ for all but finitely many $i$}\} = \bigcup_{i < \omega} (2^{i} \times 1^{\omega})$. + Clearly this is countable. + + For denseness take some $x \in 2^\omega$. + Let $x^{(n)}$ be defined by $x^{(n)}_i = x_i$ for $i < n$ + and $x^{(n)}_i = 0$ for $i \ge n$. + Then $x^{(n)} \in C$ for all $n$, + and $x^{(n)}$ converges to $x$. + + \item $f(\omega^\omega)$ is $G_\delta$: + + We have + \begin{IEEEeqnarray*}{rCl} + f(\omega^\omega) &=& 2^\omega \setminus \left(\bigcup_{i < \omega} (2^{i} \times 1^{\omega})\right)\\ + &=& \bigcap_{i < \omega} \left(2^{\omega} \setminus(2^{i} \times 1^{\omega})\right). + \end{IEEEeqnarray*} + \item $C$ as in (2) is homeomorphic to $\Q$. + + Go to the right in the even digits, go to the left for the odd digits, + i.e.~let $C = (1,-1,1,-1, \ldots)$ + and set $x < y$ iff $C \cdot x <_{\text{lex}} C \cdot y$, + where $<_{\text{lex}}$ denotes the lexicographical ordering. + Note that the order topology of $<$ on $C$ + agrees with the subspace topology from $2^\omega$. + + By Cantor's theorem for countable, unbounded, dense linear + linear orders, + we get an order isomorphism $C \leftrightarrow \Q$. + This is also a homeomorphism, as the topologies + on $C$ and $\Q$ are the respective order topologies. +\end{enumerate} diff --git a/inputs/tutorial_04.tex b/inputs/tutorial_04.tex index cc4a069..33b943b 100644 --- a/inputs/tutorial_04.tex +++ b/inputs/tutorial_04.tex @@ -1,70 +1,205 @@ -\tutorial{04}{2023-11-14}{} +\subsection{Sheet 3} -\subsection{Sheet 4} +\tutorial{04}{}{} -% 14 / 20 - -\subsubsection{Exercise 1} +\nr 1 +Let $A \neq \emptyset$ be discrete. +For $D \subseteq A^{\omega}$, +let +\[ + T_D \coloneqq \{x\defon{n} \in A^{<\omega} | x \in D, n \in \N\}.. +\] \begin{enumerate}[(a)] - \item $\langle X_\alpha : \alpha\rangle$ - is a descending chain of closed sets (transfinite induction). + \item For any $D \subseteq A^\omega$, $T_D$ is a pruned tree: - Since $X$ is second countable, there cannot exist - uncountable strictly decreasing chains of closed sets: + Clearly $T_D$ is a tree. + Let $x \in T_D$. + Then there exists $d \in D$ such that $x = d\defon{n}$. + Hence $x \subseteq d\defon{n+1} \in T_D$. + Thus $x$ is not a leaf, i.e.~$T_D$ is pruned. + \item For any $T \subseteq A^{<\omega}$, $[T]$ is a closed subset + of $A^{\omega}$: - Suppose $\langle X_{\alpha}, \alpha < \omega_1\rangle$ - was such a sequence, - then $X \setminus X_{\alpha}$ is open for every $\alpha$, - Let $\{U_n : n < \omega\}$ be a countable basis. - Then $N(\alpha) = \{n | U_n \cap (X \setminus X_\alpha) \neq \emptyset\}$, - is a strictly ascending chain in $\omega$. + Let $a \in A^{\omega} \setminus [T]$. + Then there exists some $n$ such that $a\defon{n} \not\in T$. + Hence $\{a_0\} \times \ldots \times \{a_{n+1}\} \times A^{\omega}$ + is an open neighbourhood of $a$ disjoint from $[T]$. - \item We need to show that $X_{\alpha_0}$ is perfect and closed. - It is closed since all $X_{\alpha}$ are, - and perfect, as a closed set $F$ is perfect - iff it coincides $F'$. + \item $T \mapsto [T]$ is a bijection between the + pruned trees on $A$ and the closed subsets of $A^{\omega}$. - $X \setminus X_{\alpha_0}$ - is countable: - $X_{\alpha} \setminus X_{\alpha + 1}$ is - countable as for every $x$ there exists a basic open set $U$, - such that $U \cap X_{\alpha} = \{x\}$, - and the space is second countable. - Hence $X \setminus X_{\alpha_0}$ - is countable as a countable union of countable sets. + \begin{claim} + $[T_D] = D$ for any closed subset $D \subseteq A^{\omega}$. + \end{claim} + \begin{subproof} + Clearly $D \subseteq [T_D]$. + Let $x \in [T_D]$. + Then for every $n < \omega$, + there exists some $d_n \in D$ such that + $d_n\defon{n} = x\defon{n}$. + Clearly the $d_n$ converge to $x$. + Since $D$ is closed, we get $x \in D$. + \end{subproof} + + This shows that $T \mapsto [T]$ is surjective. + + Now let $T \neq T'$ be pruned trees. + Then there exists $x \in T \mathop{\triangle} T'$, + wlog.~$x \in T \setminus T'$. + Since $T$ is pruned + by applying the axiom of countable choice + we get an infinite branch $x' \in [T] \setminus [T']$. + Hence the map is injective. + + \item Let $N_s \coloneqq \{x \in A^{\omega} | s \subseteq x\}$. + Show that every open $U \subseteq A^{\omega}$ + can be written as $U = \bigcup_{s \in S} N_s$ + for some set of pairwise incompatible $S \subseteq A^{<\omega}$. + + + Let $U$ be open. + Then $U$ has the form + \[ + U = \bigcup_{i \in I} X_i \times A^{\omega} + \] + for some $X_i \subseteq A^{n_i}$, $n_i < \omega$. + Clearly + $U = \bigcup_{s \in S'} N_s$ + for + $S' \coloneqq \bigcup_{i \in I} X_i$. + Define + \[ + S \coloneqq \{s \in S' | \lnot\exists t \in S'.~t\subseteq s \land |t| < |s|\}. + \] + Then the elements of $S$ are pairwise incompatible + and $U = \bigcup_{s \in S} N_s$. + + \item Let $T \subseteq A^{<\omega}$ be an infinite tree which is finitely + splitting. + Then $[T]$ is nonempty: + + Let us recursively construct a sequence of compatible $s_n \in T$ + with $|s_n| = n$ + such that $\{s_n\} \times A^{<\omega} \cap T$ is infinite. + Let $s_0$ be the empty sequence; + by assumption $T$ is infinite. + Suppose that $s_n$ has been chosen. + Since $T$ is finitely splitting, there are only finitely + many $a \in A$ with $s_n\concat a \in T$. + Since $ \{s_n\} \times A^{<\omega} \cap T$ + is infinite, + there must exist at least on $a \in A$ + such that $\{s_n\concat a\} \times A^{<\omega} \cap T$ + is infinite. + Define $s_{n+1} \coloneqq s_n \concat a$. + + Then the union of the $s_n$ is an infinite branch of $T$, + i.e.~$[T]$ is nonempty. + + + \item Then $[T]$ is compact: + \todo{TODO} + + % Let $\langle s_n, n <\omega \rangle$ + % be a Cauchy sequence in $[T]$. + + % Then for every $m < \omega$ + % there exists an $N < \omega$ such that + % $s_n\defon{m} = s_{n'}\defon{m}$ + % for all $n, n' > N$. + % Thus there exists a pointwise limit $s$ of the $s_n$. + + % Since for all $m$ we have $s\defon{m} = s_n\defon{m} \in [T]$ + % for $m$ large enough, + % we get $s \in [T]$. + + % Hence $[T]$ is sequentially compact. \end{enumerate} +\nr 2 +\todo{handwritten} +\nr 3 +\todo{handwritten} -\subsection{Exercise 3} +\nr 4 -\begin{itemize} - \item Let $Y \subseteq \R$ be $G_\delta$ - such that $Y$ and $\R \setminus Y$ are dense in $\R$. - Then $Y \cong \cN$. +\begin{notation} + For $A \subseteq X$ let $A'$ denote the set of + accumulation points of $A$. +\end{notation} - $Y$ is Polish, since it is $G_\delta$. +\begin{theorem} + Let $X$ be a Polish space. + Then there exists a unique partition $X = P \sqcup U$ + of $X$ into a perfect closed subset $P$ and a countable open subset $U$. +\end{theorem} +\begin{proof} - $Y$ is 0-dimensional, - since the sets $(a,b) \cap Y$ for $a, b \in \R \setminus Y$ - form a clopen basis. + Let $P$ be the set of condensation points of $X$ + and $U \coloneqq X \setminus P$. - Each compact subset of $Y$ has empty interior: - Let $K \subseteq Y$ be compact - and $U \subseteq K$ be open in $Y$. - Then we can find cover of $U$ that has no finite subcover $\lightning$. + \begin{claim} + $U$ is open and countable. + \end{claim} + \begin{subproof} + Let $S$ be a countable dense subset. + For each $x \in U$, + there is an $\epsilon_x > 0$, $s_x \in S$ such that $x \in B_{\epsilon_x}(s_x)$ + is at most countable. + Clearly $B_{\epsilon_x}(s_x) \subseteq U$, + as for every $y \in B_{\epsilon_x}(s_x)$, + $B_{\epsilon_x}(s_x)$ witnesses that $y \not\in P$. + Thus $U = \bigcup_{x \in U} B_{\epsilon_x}(s_x)$ + is open. + Wlog.~$\epsilon_x \in \Q$ for all $x$. + Then the RHS is the union of at most + countably many countable sets, as $S \times \Q$ + is countable. + \end{subproof} - \item Let $Y \subseteq \R$ be $G_\delta$ and dense - such that $\R \setminus Y$ is dense as well. - Define $Z \coloneqq \{x \in \R^2 | |x| \in Y\} \subseteq \R^2$. - Then $Z$ is dense in $\R^2$ - and $\R^2 \setminus \Z$ is dense in $\R^2$. + \begin{claim} + $P$ is perfect. + \end{claim} + \begin{subproof} + Let $x \in P'$ and $x \in U$ an open neighbourhood. + Then there exists $y \in P \cap U$. + In particular, $U$ is an open neighbourhood of $ y$, + hence $U$ is uncountable. + It follows that $x \in P$. + On the other hand let $x \in P$ + and let $U$ be an open neighbourhood. + We need to show that $U \cap P \setminus \{x\}$ + is not empty. + Suppose that for all $y \in U \cap P \setminus \{x\}$, + there is an open neighbourhood $U_y$ + such that $U_y$ is at most countable. + Wlog.~$U_y = B_{\epsilon_y}(s_y)$ for some $s_y \in S$, $\epsilon_y > 0$, + where $S$ is again a countable dense subset. + Wlog.~$\epsilon_y \in \Q$. + But then + \[ + U = \{x\} \cup \bigcup_{y \in U} B_{\epsilon_y}(s_y) + \] + is at most countable as a countable union of countable sets, + contradiction $x \in P$. + \end{subproof} - We have that for every $y \in Y$ - $\partial B_y(0) \subseteq Z$. + \begin{claim} + Let $P,U$ be defined as above + and let $P_2 \subseteq X$, $U_2 \subseteq X$ + be such that $P_2$ is perfect and closed, + $U_2$ is countable and open + and $X = P_2 \sqcup U_2$. + Then $P_2 = P$ and $U_2 = U$. + \end{claim} + \todo{TODO} +\end{proof} - - Other example: - Consider $\R^2 \setminus \Q^2$. -\end{itemize} +\begin{corollary} + Any Polish space is either countable or has cardinality equal to $\fc$. +\end{corollary} +\begin{subproof} + \todo{TODO} +\end{subproof} diff --git a/inputs/tutorial_05.tex b/inputs/tutorial_05.tex index 7c9af84..a0fdcad 100644 --- a/inputs/tutorial_05.tex +++ b/inputs/tutorial_05.tex @@ -1,35 +1,123 @@ -\tutorial{05}{}{} +\subsection{Sheet 4} +\tutorial{05}{2023-11-14}{} -% Sheet 5 - 18.5 / 20 +% 14 / 20 -\subsection{Exercise 1} - -Let $B \subseteq C$ be comeager. -Then $B = B_1 \cup B_2$, -where $B_1$ is dense $G_\delta$ -and $B_2$ is meager. +\nr 1 -\begin{fact} - $X$ is Baire iff every non-empty open set is non-meager. +\begin{enumerate}[(a)] + \item $\langle X_\alpha : \alpha\rangle$ + is a descending chain of closed sets (transfinite induction). - In particular, let $X$ be Baire, - then $U \overset{\text{open}}{\subseteq} X$ - is Baire. -\end{fact} + Since $X$ is second countable, there cannot exist + uncountable strictly decreasing chains of closed sets: -\subsection{Exercise 4} + Suppose $\langle X_{\alpha}, \alpha < \omega_1\rangle$ + was such a sequence, + then $X \setminus X_{\alpha}$ is open for every $\alpha$, + Let $\{U_n : n < \omega\}$ be a countable basis. + Then $N(\alpha) = \{n | U_n \cap (X \setminus X_\alpha) \neq \emptyset\}$, + is a strictly ascending chain in $\omega$. -\begin{enumerate}[(i)] - \item $|B| = \fc$, since $B$ contains a comeager - $G_\delta$ set, $B'$: - $B'$ is Polish, - hence $B' = P \cup C$ - for $P$ perfect and $C$ countable, - and $|P| \in \{\fc, 0\}$. - But $B'$ can't contain isolated point$. + \item We need to show that $X_{\alpha_0}$ is perfect and closed. + It is closed since all $X_{\alpha}$ are, + and perfect, as a closed set $F$ is perfect + iff it coincides $F'$. + + $X \setminus X_{\alpha_0}$ + is countable: + $X_{\alpha} \setminus X_{\alpha + 1}$ is + countable as for every $x$ there exists a basic open set $U$, + such that $U \cap X_{\alpha} = \{x\}$, + and the space is second countable. + Hence $X \setminus X_{\alpha_0}$ + is countable as a countable union of countable sets. +\end{enumerate} + +\nr 2 +\todo{handwritten} + + +\nr 3 +\begin{itemize} + \item Let $Y \subseteq \R$ be $G_\delta$ + such that $Y$ and $\R \setminus Y$ are dense in $\R$. + Then $Y \cong \cN$. + + $Y$ is Polish, since it is $G_\delta$. + + $Y$ is 0-dimensional, + since the sets $(a,b) \cap Y$ for $a, b \in \R \setminus Y$ + form a clopen basis. + + Each compact subset of $Y$ has empty interior: + Let $K \subseteq Y$ be compact + and $U \subseteq K$ be open in $Y$. + Then we can find cover of $U$ that has no finite subcover $\lightning$. + + \item Let $Y \subseteq \R$ be $G_\delta$ and dense + such that $\R \setminus Y$ is dense as well. + Define $Z \coloneqq \{x \in \R^2 | |x| \in Y\} \subseteq \R^2$. + Then $Z$ is dense in $\R^2$ + and $\R^2 \setminus \Z$ is dense in $\R^2$. + + + We have that for every $y \in Y$ + $\partial B_y(0) \subseteq Z$. + + + Other example: + Consider $\R^2 \setminus \Q^2$. +\end{itemize} + +\nr 4 + +\begin{enumerate}[(a)] + \item Let $d$ be a compatible, complete metric on $X$, wlog.~$d \le 1$. + Set $ U_{\emptyset} \coloneqq X$. + Suppose that $U_{s}$ has already been chosen. + Then $D_s \coloneqq X \setminus U_s$ is closed. + Hence $U_s^{(n)} \coloneqq \{x \in X | \dist(x,D_s) > \frac{1}{n}\}$ + is open. + Let $m$ be such that $D_s^{(m)} \neq \emptyset$. + Clearly $\overline{U_s^{(n)}} \subseteq U_s$. + Let $(B_k)_{k < \omega}$ be a countable cover of $X$ + consisting of balls of diameter $2^{-|s|-2}$. + Take some bijection $\phi\colon \omega \to \omega \times (\omega \setminus m)$ + and set $U_{s\concat i} \coloneqq U_s^{(\pi_1\left( \phi(i) \right))} \cap B_{\pi_2(\phi(i))}$, + where there $\pi_i$ denote the projections + (if this is empty, set $U_{s \concat i} \coloneqq U_s^{\pi_1(\phi(j))} \cap B_{\pi_2(\phi(j))}$ + for some arbitarily chosen $j < \omega$ such that it is not empty). + Then $\overline{U_{s \concat i}} \subseteq \overline{U_s^{(n)}} \subseteq U_s$, + \[ + \bigcup_{i < \omega} U_{s \concat i} = \bigcup_{n < \omega} U_{s}^{\left( n \right) } = U_s + \] + and $\diam(U_{s \concat i}) \le \diam(B_{\pi_2(\phi(i))})$. + \item Let $s \in \omega^\omega$. + Then + \[ + \bigcap_{n < \omega} \overline{U_{s\defon{n}}} + \] + contains exactly one point. + Let $f$ be the function that maps an $s \in \omega^{\omega}$ + to the unique point in the intersection of the $\overline{U_{s\defon{n}}}$. + Let $x \in X$ be some point. + Then by induction we can construct a sequence $s \in \omega^{\omega}$ + such that $x \in U_{s\defon{n}}$ for all $n$, + hence $x = f(s)$, i.e.~$f$ is surjective. + + Let $B \overset{\text{open}}{\subseteq} X$. + Then $B = \bigcup_{i \in I} U_i$ + for some $i \subseteq \omega^{<\omega}$, + as every basic open set can be recovered as a union of $U_i$ + and $f^{-1}(B) = \bigcup_{i \in I} \left( \{i_0\} \times \ldots \{i_{|i|-1}\} \right) \times \omega^{\omega}$ is open, + hence $f$ is continuous. + + + On the other hand, consider an open ball $B \coloneqq \{\prod_{i < n} \{x_i\}\} \times \omega^{\omega} \subseteq \omega^{\omega}$. + Then $f(B) = U_{(x_0,\ldots,x_{n-1})}$ is open, + hence $f$ is open. \end{enumerate} - - diff --git a/inputs/tutorial_06.tex b/inputs/tutorial_06.tex index 577ef2c..b39a77b 100644 --- a/inputs/tutorial_06.tex +++ b/inputs/tutorial_06.tex @@ -1,180 +1,91 @@ -\tutorial{06}{2023-11-28}{} +\subsection{Sheet 5} -% 5 / 20 +\tutorial{06}{}{} -\subsection{Exercise 1} +% Sheet 5 - 18.5 / 20 -\begin{warning} - Note that not every set has a density! -\end{warning} - -\begin{enumerate}[(a)] - \item Let $X = \bI^{\omega}$. - Let $C_0 = \{(x_n) : x_n \to 0\}$. - Claim: $C_0 \in \Pi^0_3(X)$ (intersections of $F_\sigma$ sets). - - We have - \[ - x \in C_0 \iff \forall q \in \Q^+.~\exists N.~\forall n \ge N.~x_n \le q, - \] - i.e. - \[ - C_0 = \bigcap_{q \in \Q^+}\bigcup_{N < \omega} \bigcap_{n > N} \{x_n : x_n \le q\}. - \] - Clearly this is a $\Pi^0_3$ set. - - \item Let $Z \coloneqq \{f \in 2^{\omega} : f(\N) \text{ has density 0}\}$. - Claim: $Z \in \Pi^0_3(2^{\N})$. - It is - \[ - Z = \bigcap_{q \in \Q^+} \bigcup_{N < \omega} - \bigcap_{n \ge N}\{f \in 2^{\omega} : \frac{\sum_{i < n} f(i)}{n} \le nq\}. - \] - Clearly this is a $\Pi^0_3$-set. -\end{enumerate} - -\subsection{Exercise 2} +\nr 1 +\todo{handwritten} +Let $B \subseteq C$ be comeager. +Then $B = B_1 \cup B_2$, +where $B_1$ is dense $G_\delta$ +and $B_2$ is meager. \begin{fact} - Let $(X,\tau)$ be a Polish space and - $A \in \cB(X)$. - Then there exists $\tau' \supseteq \tau$ - with the same Borel sets as $\tau$ - such that $A$ is clopen. + $X$ is Baire iff every non-empty open set is non-meager. - (Do it for $A$ closed, - then show that the sets which work - form a $\sigma$-algebra). + In particular, let $X$ be Baire, + then $U \overset{\text{open}}{\subseteq} X$ + is Baire. \end{fact} +\nr 2 + +Let $(U_i)_{i < \omega}$ be a countable base of $Y$. +We want to find a $G_\delta$ set $A \subseteq X$ +such that $f\defon{A}$ is continuous. +It suffices make sure that $f^{-1}\defon{A}(U_i)$ is open for all $i < \omega$. +Take some $i < \omega$. +Then $V_i \setminus M_i \subseteq f^{-1}(U_i) \subseteq V_i \cup M_i$, +where $V_i$ is open and $M_i$ is meager. +Let $M'_i \supseteq M_i$ be a meager $F_\sigma$-set. +Now let $A \coloneqq X \setminus \bigcup_{i <\omega} M_i'$. +We have that $A$ is a countable intersection of open dense sets, +hence it is dense and $G_\delta$. +For any $i < \omega$, +$V_i \cap A \subseteq f\defon{A}^{-1}(U_i) \subseteq (V_i \cup M_i) \cap A = V_i \cap A$, +so $f\defon{A}^{-1}(U_i) = V_i \cap A$ is open. + +\nr 3 +\todo{handwritten} + + +\nr 4 \begin{enumerate}[(a)] - \item Let $(X, \tau)$ be Polish. - We want to expand $\tau$ to a Polish topology - $\tau_0$ maintaining the Borel sets, - such that $(X, \tau')$ is 0d. + \item $|B| = \fc$, since $B$ contains a comeager + $G_\delta$ set, $B'$: + $B'$ is Polish, + hence $B' = P \cup C$ + for $P$ perfect and $C$ countable, + and $|P| \in \{\fc, 0\}$. + 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 $|\{a_i | j < i\}| = |i| < \fc$, + there exists some $x \in B \setminus \{\pi_1(a_j)| j N} \{x_n : x_n \le q\}. + \] + Clearly this is a $\Pi^0_3$ set. + + \item Let $Z \coloneqq \{f \in 2^{\omega} : f(\N) \text{ has density 0}\}$. + Claim: $Z \in \Pi^0_3(2^{\N})$. + It is + \[ + Z = \bigcap_{q \in \Q^+} \bigcup_{N < \omega} + \bigcap_{n \ge N}\{f \in 2^{\omega} : \frac{\sum_{i < n} f(i)}{n} \le nq\}. + \] + Clearly this is a $\Pi^0_3$-set. +\end{enumerate} + +\subsection{Exercise 2} + +\begin{fact} + Let $(X,\tau)$ be a Polish space and + $A \in \cB(X)$. + Then there exists $\tau' \supseteq \tau$ + with the same Borel sets as $\tau$ + such that $A$ is clopen. + + (Do it for $A$ closed, + then show that the sets which work + form a $\sigma$-algebra). +\end{fact} + +\begin{enumerate}[(a)] + \item Let $(X, \tau)$ be Polish. + We want to expand $\tau$ to a Polish topology + $\tau_0$ maintaining the Borel sets, + such that $(X, \tau')$ is 0d. + + Let $(U_n)_{n < \omega}$ be a countable base of $(X,\tau)$. + Each $U_n$ is open, hence Borel, + so by a theorem from the lecture$^{\text{tm}}$ + there exists a Polish topology $\tau_n$ + such that $U_n$ is clopen, preserving Borel sets. + + + Hence we get $\tau_\infty$ + such that all the $V_n$ are clopen in $\tau_\infty$. + Let $\tau^{1} \coloneqq \tau_\infty$. + Do this $\omega$-many times to get $\tau^{\omega}$. + $\tau^{\omega}$ has a base consisting + of finite intersections $A_1 \cap \ldots \cap A_n$, + where $A_i$ is a basis element we chose + to construct $\tau_i$, + hence clopen. + \item Let $(X, \tau_X), Y$ be Polish + and $f\colon X \to Y$ Borel. + Show $\exists \tau' \supseteq \tau$ maintaining the Borel structure + with $f$ continuous. + + Let $(U_n)_n$ be a countable base of $Y$. + Clopenize all the preimages of the $(U_n)_n$. + + \item Let $f\colon X \to Y$ be a Borel isomorphism. + Then there are finer topologies preserving the Borel + structure + such that $f\colon X' \to Y'$ is a homeomorphism. + + Repeatedly apply (c). + Get $\tau_X^1$ to make $f$ continuous. + Then get $\tau_Y^1$ to make $f^{-1}$ continuous + (possibly violating continuity of $f$) + and so on. + + Let $\tau_X^\omega \coloneqq \langle \tau_X^n \rangle$ + and similarly for $\tau_Y^\omega$. +\end{enumerate} +\begin{idea} + If you do something and it didn't work, + try doing it again ($\omega$-many times). +\end{idea} + +\subsection{Exercise 3} + +\begin{enumerate}[(a)] + \item Show that if $\Gamma$ is self-dual (closed under complements) + and closed under continuous preimages, + then for any topological space $X$, + there does not exist an $X$-universal set for $\Gamma(X)$. + + + Suppose there is an $X$-universal set for $\Gamma(X)$, + i.e.~$U \subseteq X \times X$ + such that $U \in \Gamma(X \times X) \land \{U_x : \in X\} = \Gamma(X)$. + + Consider $X \xrightarrow[x\mapsto (x,x)]{d} X \times X$. + + Let $V = U^c$. + Then $V \in \Gamma(X \times X)$ and $d^{-1}(V) \in \Gamma(X)$. + Then $d^{-1}(V) = U_x$ for some $x$. + But then $(x,x) \in U \iff x \in d^{-1}(V) \iff (x,x) \not\in U \lightning$. + + + + \item Let $\xi$ be an ordinal + and let $X$ be a topological space. + Show that neither $\cB(X)$ nor $\Delta^0_\xi(X)$ can have $X$-universal + sets. + + Clearly $\cB(X)$ is self-dual and closed under continuous preimages. + Clearly $\Delta^0_\xi(X)$ is self-dual + and closed under continuous preimages (by a trivial induction). +\end{enumerate} + +\subsection{Exercise 4} + +Recall: +\begin{fact}[Sheet 5, Exercise 1] +Let $\emptyset\neq X$ be a Baire space. +Then $\forall A \subseteq X$, +$A$ is either meager or locally comeager. +\end{fact} + +\begin{theorem}[Kechris 16.1] + Let $X, Y$ be Polish. + + Let + \[\cA \coloneqq \{A \in \cB(X \times Y) : \forall \emptyset \neq U \overset{\text{open}}{\subseteq} Y.~ + A_U \coloneqq \{ x \in X : A_x \text{ is not meager in $U$}\} \text{ is Borel}\}.\] + + Then $\cA$ contains all Borel sets. +\end{theorem} +\begin{proof} + \begin{enumerate}[(i)] + \item Show for $V \in \cB(X), W \overset{\text{open}}{\subseteq} Y$ + that $V \times W \in \cA$. + + Clearly $V \times W$ is Borel + and $\{x \in X: W \cap U \text{ is not meager}\} \in \{\emptyset, V\}$. + \item Let $(A_n)_{n < \omega} \in \cA^{\omega}$. + Then $\bigcap_n A_n \in \cA$. + ($(\bigcup_n A_n)_U = \bigcup_n (A_{n})_U$). + \item Let $A \in \cA$ and $B = A^c$. + Fix $\emptyset\neq U \subseteq Y$. + Then $\{x : A_x \text{is not meager in $U$}\}$ is Borel, + i.e.~$\{x : A_x^c \text{ is not meager in $U$}\}$ is Borel. + + Since $A$ is Borel, $A_x$ is Borel as well. + Hence by the fact: + \begin{IEEEeqnarray*}{rCl} + && \{x : A_x^c \text{ is not meager in $U$}\}\\ + &=& \{x \colon A_x^c \text{ is locally comeager in $U$}\}\\ + &=& \{x \colon \exists \emptyset\neq V \overset{\text{open}}{\subseteq} V.~ A_x \text{ is meager in $V$}\}\\ + &=& \bigcup_{\emptyset \neq V \overset{\text{open}}{\subseteq} U} A_V^c + \end{IEEEeqnarray*} + (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} + + + + + +