lecture 4

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Josia Pietsch 2023-10-20 11:53:35 +02:00
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Let $X \neq \emptyset$
be a perfect Polish space.
Then there is an embedding
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\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 Normal file
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Some of $F_s$ might be empty.
\[D = \{x \in \cN : \bigcap_{n} F_{x\defon{n}} \neq \emptyset\}.\]
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
$D$ is closed.
Let $x_n$ be a series in $D$
converging to $x$ in $\cN$.
Then $x \in \cN$.
$(f(x_n))$ is Cauchy.
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 $y \in \overline{F_{x\defon{N'}}}$.
\[y \in \bigcap_{n} \overline{F_{x\defon{n}}} = \bigcap_{n} F_{x\defon{n}},\]
i.e.~$y \in D$ and $y = f(x)$.
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
\item $s \subseteq t \implies \phi(s) \subseteq \phi(t)$,
\item $|s| = \phi(|s|)$,
\item if $s \in S$, then $\phi(s) = s$.
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.
\subsection{Meager and Comeager Sets}
Let $X$ be a topological space, $A \subseteq X$.
We say that $A$ is \vocab{nowhere dense} (\vocab{nwd}),
if $\inter(\overline{A}) = \emptyset$.
\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 $U$).
\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
$\Q \subseteq \R$ is meager.
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.
$=^\ast$ is an equivalence relation.
A set $A \subseteq X$
has the \vocab{Baire property} (\vocab{BP})
if $A =^\ast U$ for some $U \overset{\text{open}}{\subseteq} X$.
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}

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\RequirePackage{mkessler-mathfixes} % Load this last since it renews behaviour
\DeclareMathOperator{\inter}{int} % interior
\newcommand{\defon}[1]{|_{#1}} % TODO

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