lecture 4

inputs/lecture_03.tex

@@ -107,6 +107,7 @@
 \end{definition}
 
 \begin{theorem}
+    \label{thm:cantortopolish}
     Let $X \neq \emptyset$ 
     be a perfect Polish space.
     Then there is an embedding
@@ -216,7 +217,7 @@
         \item $F_\emptyset = X$,
         \item $F_s$ is $F_\sigma$ for all  $s$.
         \item The $F_{s \concat i}$ partition $F_s$,
-            i.e.~$F_{s} = \bigsqcup_i F_{s \concat i}$.
+            i.e.~$F_{s} = \bigsqcup_i F_{s \concat i}$. % TODO change notation?
 
             Furthermore we want that
             $\overline{F_{s \concat i}} \subseteq F_s$

inputs/lecture_04.tex

@@ -0,0 +1,156 @@
+\lecture{04}{2023-10-20}{}
+\begin{remark}
+        Some of $F_s$ might be empty.
+\end{remark}
+
+\begin{refproof}{thm:bairetopolish}
+    Take
+    \[D = \{x \in \cN : \bigcap_{n} F_{x\defon{n}} \neq \emptyset\}.\]
+
+    Since $\ldots \supseteq F_{x\defon{n}} \supseteq \overline{F_{x\defon{n+1}}} \supseteq F_{x\defon{n+1}} \supseteq \ldots$
+    we have
+    \[
+    \bigcap_{n} F_{x\defon{n}} = \bigcap_{n} \overline{F_{x\defon{n}}}.
+    \] 
+
+    $f\colon D \to  X$ is determined by
+    \[
+    \{f(x)\} = \bigcap_{n} F_{x\defon{n}}
+    \] 
+
+    $f$ is injective and continuous.
+    The proof of this is exactly the same as in 
+    \yaref{thm:cantortopolish}.
+
+    \begin{claim}
+        \label{thm:bairetopolish:c1}
+        $D$ is closed.
+    \end{claim}
+    \begin{refproof}{thm:bairetopolish:c1}
+        Let $x_n$ be a series in $D$
+        converging to $x$ in $\cN$.
+        Then $x \in \cN$.
+        \begin{claim}
+            $(f(x_n))$ is Cauchy.
+        \end{claim}
+        \begin{subproof}
+            Let $\epsilon >  0$.
+            Take $N$ such that $\diam(F_{x\defon{n}}) < \epsilon$.
+            Take $M$ such that for all $m \ge M$,
+            $x_m\defon{N} = x\defon{N}$.
+            Then for all $m, n \ge M$,
+            we have that $f(x_m), f(x_n) \in F_{x\defon{N}}$.
+            So $d(f(x_m), f(x_n)) < \epsilon$ 
+            we have that $(f(x_n))$ is Cauchy.
+            
+            Since $(X,d)$ is complete,
+            there exists $y = \lim_n f(x_n)$.
+
+            Since for all $m \ge M$, $f(x_m) \in F_{x\defon{N}}$,
+            we get that $y \in  \overline{F_{x\defon{N}}}$.
+            
+            Note that for $N' > N$ by the same argument
+            we get $y \in \overline{F_{x\defon{N'}}}$.
+            Hence
+            \[y \in \bigcap_{n} \overline{F_{x\defon{n}}} = \bigcap_{n} F_{x\defon{n}},\]
+            i.e.~$y \in D$ and $y = f(x)$.
+        \end{subproof}
+    \end{refproof}
+
+    We extend $f$ to $g\colon\cN \to X$ 
+    in the following way:
+
+    Take $S \coloneqq \{s \in \N^{<\N}: \exists x \in D, n \in \N.~x=s\defon{n}\}$.
+    Clearly $S$ is a pruned tree.
+    Moreover, since $D$ is closed, we have that\todo{Proof this (homework?)}
+    \[
+    D = [S] = \{x \in \N^\N : \forall n \in \N.~x\defon{n} \in S\}.
+    \] 
+    We construct a \vocab{retraction} $r\colon\cN \to  D$ 
+    (i.e.~$r = \id$ on $D$ and $r$ is a continuous surjection).
+    Then $g \coloneqq f \circ r$.
+    
+    To construct $r$, we will define by induction
+    $\phi\colon \N^{<\N} \to  S$ by induction on the length
+    such that
+    \begin{itemize}
+        \item $s \subseteq t \implies \phi(s) \subseteq \phi(t)$,
+        \item $|s| = \phi(|s|)$,
+        \item if $s \in S$, then $\phi(s) = s$.
+    \end{itemize}
+    Let $\phi(\emptyset) = \emptyset$.
+    Suppose that $\phi(t)$ is defined.
+    If $t\concat a \in S$, then set
+     $\phi(t\concat a) \coloneqq t\concat a$.
+    Otherwise take some $b$ such that
+    $t\concat b \in S$ and define
+    $\phi(t\concat a) \coloneqq \phi(t)\concat b$.
+    This is possible since $S$ is pruned.
+
+    Let $r\colon \cN = [\N^{<\N}] \to [S] = D$
+    be the function defined by $r(x) = \bigcup_n f(x\defon{n})$.
+
+    $r$ is continuous, since
+    $d_{\cN}(r(x), r(y)) \le d_{\cN}(x,y)$. % Lipschitz
+    It is immediate that $r$ is a retraction.
+\end{refproof}
+
+\subsection{Meager and Comeager Sets}
+
+\begin{definition}
+    Let $X$ be a topological space, $A \subseteq  X$.
+    We say that $A$ is \vocab{nowhere dense} (\vocab{nwd}),
+    if $\inter(\overline{A}) = \emptyset$.
+    Equivalently
+    \begin{itemize}
+        \item $\overline{A}$ is nwd,
+        \item $X \setminus A$ is dense in $X$,
+        \item $\forall  \emptyset \neq  U \overset{\text{open}}{\subseteq} X.~
+            \exists  \emptyset \neq V \overset{\text{open}}{\subseteq} U.~
+            V\cap A = \emptyset$.
+            (If we intersect $A$ with an open $U$,
+            then $A \cap U$ is not dense in $U$).
+    \end{itemize}
+    \todo{Think about this}
+
+    A set $B \subseteq  X$ is \vocab{meager} 
+    (or \vocab{first category}),
+    iff it is a countable union of nwd sets.
+
+    The complement of a meager set is called
+    \vocab{comeager}.
+    
+\end{definition}
+\begin{example}
+    $\Q \subseteq \R$ is meager.
+\end{example}
+\begin{notation}
+    Let $A, B \subseteq  X$.
+    We write $A =^\ast B$ 
+    iff the \vocab{symmetric difference},
+    $A \mathop{\triangle} B \coloneqq  (A\setminus B) \cup (B \setminus A)$,
+    is meager.
+\end{notation}
+\begin{remark}
+    $=^\ast$ is an equivalence relation.
+\end{remark}
+\begin{definition}
+    A set $A \subseteq  X$
+    has the \vocab{Baire property} (\vocab{BP})
+    if $A =^\ast U$ for some $U \overset{\text{open}}{\subseteq} X$.
+\end{definition}
+Note that open sets and meager sets have the Baire property.
+
+
+
+
+% \begin{example}
+%     $\Q \subseteq  \R$ is $F_\sigma$.
+% 
+%     $\R \setminus \Q \subseteq \R$ is $G_\delta$.
+% 
+%     $\Q \subseteq \R$ is not $G_{\delta}$.
+%     (It is dense and meager,
+%     hence it can not be $G_\delta$,
+%     by the Baire category theorem).
+% \end{example}

jrpie-math.sty

@@ -48,4 +48,5 @@
 \RequirePackage{mkessler-mathfig}
 \RequirePackage{mkessler-unicodechar}
 \RequirePackage{mkessler-mathfixes} % Load this last since it renews behaviour
+\DeclareMathOperator{\inter}{int} % interior
 \newcommand{\defon}[1]{|_{#1}} % TODO

logic3.tex

@@ -27,6 +27,7 @@
 \input{inputs/lecture_01}
 \input{inputs/lecture_02}
 \input{inputs/lecture_03}
+\input{inputs/lecture_04}