diff --git a/inputs/lecture_04.tex b/inputs/lecture_04.tex index 51a8ec6..bcd97de 100644 --- a/inputs/lecture_04.tex +++ b/inputs/lecture_04.tex @@ -109,7 +109,7 @@ (If we intersect $A$ with an open $U$, then $A \cap U$ is not dense in $U$). \end{itemize} - \todo{Think about this} + %\todo{Think about this} A set $B \subseteq X$ is \vocab{meager} (or \vocab{first category}), @@ -139,8 +139,6 @@ Note that open sets and meager sets have the Baire property. - - \begin{example} \begin{itemize} \item $\Q \subseteq \R$ is $F_\sigma$. diff --git a/inputs/lecture_05.tex b/inputs/lecture_05.tex index 7169331..f2b55ce 100644 --- a/inputs/lecture_05.tex +++ b/inputs/lecture_05.tex @@ -91,7 +91,7 @@ Then every comeager set of $X$ is dense in $X$. \end{theorem} \todo{Proof (copy from some other lecture)} -\begin{proposition} +\begin{theoremdef} Let $X$ be a topological space. The following are equivalent: \begin{enumerate}[(i)] @@ -101,7 +101,9 @@ \item The intersection of countable many open dense sets is dense. \end{enumerate} -\end{proposition} + In this case $X$ is called a \vocab{Baire space}. + \footnote{see \yaref{s5e1}} +\end{theoremdef} \begin{proof} \todo{Proof (short)} @@ -118,7 +120,6 @@ \end{proof} -% TODO Fubini \begin{notation} Let $X ,Y$ be topological spaces, $A \subseteq X \times Y$ @@ -139,10 +140,11 @@ The following similar to Fubini, but for meager sets: \begin{theorem}[Kuratowski-Ulam] - \label{thm:kuratowskiulam} + \yalabel{Kuratowski-Ulam}{Kuratowski-Ulam}{thm:kuratowskiulam} Let $X,Y$ be second countable topological spaces. Let $A \subseteq X \times Y$ - be a set with the Baire property. + be a set with the Baire property.% + \footnote{It is important that $A$ has the Baire property (cf. \yaref{s5e4}).} Then \begin{enumerate}[(i)] diff --git a/inputs/lecture_06.tex b/inputs/lecture_06.tex index 02c5b30..cc218da 100644 --- a/inputs/lecture_06.tex +++ b/inputs/lecture_06.tex @@ -203,6 +203,3 @@ i.e.~$\Delta^0_1$ is the set of clopen sets. Consider the cofinite topology on $\omega_1$. Then the non-empty open sets of this are not $F_\sigma$. \end{example} - - - diff --git a/inputs/lecture_09.tex b/inputs/lecture_09.tex index 7b90b55..c3a61ce 100644 --- a/inputs/lecture_09.tex +++ b/inputs/lecture_09.tex @@ -119,7 +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 See \yaref{ex:7.2}. + \item See \yaref{s7e2}. \end{enumerate} \end{proof} @@ -167,5 +167,6 @@ We will see later that $\Sigma^1_1(X) \cap \Pi^1_1(X) = \cB(X)$. \begin{remark}+ Showing that there exist sets that don't have the Baire property - requires the axiom of choice.\todo{Copy exercise from sheet 5} + requires the axiom of choice. + An example of such a set is constructed in \yaref{s5e4}. \end{remark} diff --git a/inputs/lecture_11.tex b/inputs/lecture_11.tex index 91bf1a5..60a9e6f 100644 --- a/inputs/lecture_11.tex +++ b/inputs/lecture_11.tex @@ -74,7 +74,7 @@ Then $A$ and $B$ are Borel isomorphic. \end{theorem} \begin{proof} - \todo{Homework} + Cf.~\yaref{s7e4}. \end{proof} \begin{theorem}[\vocab{Isomorphism Theorem}] diff --git a/inputs/tutorial_06.tex b/inputs/tutorial_06.tex index 68c2481..7128041 100644 --- a/inputs/tutorial_06.tex +++ b/inputs/tutorial_06.tex @@ -5,11 +5,6 @@ % Sheet 5 - 18.5 / 20 \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} $X$ is Baire iff every non-empty open set is non-meager. @@ -19,6 +14,60 @@ and $B_2$ is meager. is Baire. \end{fact} + +\begin{enumerate}[(a)] + \item Let $X$ be a non-empty Baire space + and let $A \subseteq X$. + Show that $A$ cannot be both meager and comeager. + + + Suppose that $A \subseteq X$ is meager + and comeager. + Then $A = \bigcup_{n < \omega} U_n$ + and $X \setminus A = \bigcup_{n < \omega} V_n$ + for some nwd sets $U_n, V_n$. + Then $X = A \cup (X \setminus A)$ is meager. + Let $X = \bigcup_{n < \omega} W_n$ be a union of nwd + sets. + Wlog.~the $W_n$ are closed (otherwise replace them $\overline{W_n}$) + Then $\emptyset = \bigcap_{n < \omega} (X \setminus W_n)$ + is a countable intersection of open, dense sets, + hence dense $\lightning$ + + \item Let $X$ be a topological space. The relation $=^\ast$ is transitive: + + Suppose $A =^\ast B$ and $B =^\ast C$. + Then $A \symdif C \subseteq (A \symdif B \cup B \symdif C)$ + is contained in a meager set. + Since a subset of a nwd set + is nwd, a subset of a meager set is meager. + Hence $A \symdif C$ is meager, thus $A =^\ast C$. + + \item Let $X$ be a topological space. + Let $A \subseteq X$ be a set with the Baire property, + then at least one of the following hold: + \begin{enumerate}[(i)] + \item $A $ is meager, + \item there exists $\emptyset = U \overset{\text{open}}{\subseteq} X$ + such that $A \cap U$ is comeager in $U$. + \end{enumerate} + + Suppose there was $A \subseteq X$ such that (i) + does not hold. + Then there exists $U \overset{\text{open}}{\subseteq} X$ + such that $A =^\ast U$. + In particular, $A \symdif U$ is meager, + hence $U \cap (A \symdif U) = U \setminus A$ is meager. + Thus $A \cap U$ is comeager in $U$. + + Now suppose that $X$ is a Baire space. + Suppose that for $A$ (i) and (ii) hold. + Let $\emptyset \neq U \overset{\text{open}}{\subseteq} X$ + be such that $A \cap U$ is comeager in $U$. + Since $U$ is a Baire space, + this contradicts (a). +\end{enumerate} + \nr 2 Let $(U_i)_{i < \omega}$ be a countable base of $Y$. @@ -41,15 +90,34 @@ so $f\defon{A}^{-1}(U_i) = V_i \cap A$ is open. \nr 4 +\begin{lemma} + There exists a non-meager subset $A \subseteq \R^2$ + such that no three points of $A$ are collinear. +\end{lemma} +This requires the use of the axiom of choice. +\begin{proof} + Enumerate the continuum-many $F_\sigma$ subsets of $\R^2$ + as $(F_i)_{i < \fc}$. + We will inductively construct a sequence $(a_i)_{i < \fc}$ + of points of $\R^2$ such that for each $i < \fc$: + \begin{enumerate}[(i)] + \item $\{a_j | j \le i\} $ is not a subset of $F_i$ and + \item $\{a_j | j \le i\}$ does not contain any three collinear points. + \end{enumerate} \begin{enumerate}[(a)] - \item $|B| = \fc$, since $B$ contains a comeager + \item Let $B = \{x \in \R | (F_i)_x \text{ is meager}\}$. + Then $B$ is comeager in $\R$ and $|B| = \fc$. + + We have $|B| = \fc$, 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 + \item We use $B$ to find a suitable point $a_i$: + + To ensure that (i) 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