w23-logic-3/inputs/lecture_04.tex

\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}