tutorial 06

inputs/tutorial_06.tex

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+\tutorial{06}{2023-11-28}{}
+
+% 5 / 20
+
+\subsection{Exercise 1}
+
+\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}
+
+\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}
+
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