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