Lecture 3

inputs/lecture_03.tex

@@ -0,0 +1,257 @@
+\lecture{03}{2023-10-17}{Embedding of the cantor space into polish spaces}
+
+% ? \subsection{Trees}  TODO
+
+
+\begin{notation}
+    Let $A \neq  \emptyset$, $n \in \N$.
+    Then
+    \[
+        A^n \coloneqq  \{s\colon \{0,1,\ldots, n-1\} \to A \}
+    \]
+    is the set of $n$-element \vocab[Sequence]{sequences}.
+    We often write
+    $(s_0,s_1,\ldots,s_{n-1})$.
+
+    If $s = (s_0,\ldots,s_{n-1})$,
+    then $n$ is the \vocab{length} of $s$,
+    denoted by $|s|$.
+
+    If $n = 0$ there exists only the empty sequence,
+    i.e.~$A^0 = \{\emptyset\}$ and $|\emptyset| = 0$.
+
+    We set
+    \[
+    A^{<\N} \coloneqq  \bigcup_{n=0}^{\infty} A^n
+    \]
+    and
+    \[
+    A^{\N} \coloneqq \{x \colon \N \to  A\}.
+    \]
+
+    If $s \in A^n$ and $m \le n$,
+    we let
+    $s\defon{m} \coloneqq (s_0,\ldots,s_{m-1})$.
+
+    Let $s,t \in A^{<\N}$.
+    We say that $s$ is an \vocab{initial segment} 
+    of $t$ (or $t$ is an \vocab{extension} of $s$)
+    if there exists an $n$ such that $s = t\defon{|s|}$.
+    We write this as $s \subseteq t$.
+
+    We say that $s$ and $t$ are \vocab{compatible}
+    if $s \subseteq t$ or $t \subseteq s$.
+    Otherwise the are \vocab{incompatible},
+    we denote that as $s \perp t$.
+
+    The \vocab{concatenation} 
+    of $s = (s_0,\ldots, s_{n-1})$ 
+    and $t = (t_0,\ldots, t_{m-1})$
+    is the sequence
+    $s\concat t \coloneqq (s_0,\ldots,s_{n-1}, t_0,\ldots, t_{n-1})$
+
+    In the case of $t = (a)$
+    we also write $s\concat a$ for $s\concat (a)$.
+
+    Similarly, if $x \in A^{\N}$
+    we can write $x = (x_0,x_1,\ldots)$.
+    If $n \in \N$, $x\defon{n} \coloneqq (x_0,\ldots,x_{n-1})$,
+    define extension, initial segments
+    and concatenation of a finite sequence with an infinite one.
+\end{notation}
+
+\begin{definition}
+    A \vocab{tree} 
+    on a set $A$ is a subset $T \subseteq  A^{<\N}$ 
+    closed under initial segments,
+    i.e.~if $t \in T, s \subseteq t \implies s \in T$.
+    Elements of trees are called \vocab{nodes}.
+
+    A \vocab{leave} is an element of $T$,
+    that has no extension in $t$.
+
+    An \vocab{infinite branch} of a tree $T$ 
+    is $x \in A^{\N}$ 
+    such that $\forall n.~x\defon{n} \in T$.
+
+    The \vocab{body}  of $T$ is the set of all
+    infinite branches:
+    \[
+    [T] \coloneqq\{x \in A^{\N}: \forall n. x\defon{n} \in T\}.
+    \] 
+
+    We say that $T$ is \vocab{pruned},
+    iff
+    \[
+    \forall  t\in T.\exists  s \supsetneq t.~s \in T.
+    \] 
+\end{definition}
+
+\begin{definition}
+    A \vocab{Cantor scheme}
+    on a set $X$ is a family 
+    $(A_s)_{s \in 2^{< \N}}$ 
+    of subsets of $X$ such that
+    \begin{enumerate}[i)]
+        \item $\forall s \in 2^{<\N}.~A_{s \concat 0} \cap A_{s \concat 1} = \emptyset$.
+        \item $\forall s \in 2^{<\N}, i \in 2.~A_{s \concat i} \subseteq A_s$.
+    \end{enumerate}
+\end{definition}
+
+\begin{definition}
+    A topological space 
+    is \vocab{perfect}
+    if it has no isolated points,
+    i.e.~for any $U \neq \emptyset$ open,
+    there $x \neq y$ such that $x, y \in U$.
+\end{definition}
+
+\begin{theorem}
+    Let $X \neq \emptyset$ 
+    be a perfect Polish space.
+    Then there is an embedding
+    of the Cantor space $2^{\N}$ 
+    into $X$.
+\end{theorem}
+\begin{proof}
+    We will define a Cantor scheme
+    $(U_s)_{s \in 2^{<\N}}$ 
+    such that $\forall s \in  2^{< \N}$.
+    \begin{enumerate}[(i)]
+        \item $U_s \neq  \emptyset$ and open,
+        \item $\diam(U_s) \le  2^{-|s|}$,
+        \item $\overline{U_{s  \concat i}} \subseteq  U_s$
+            for $i \in 2$.
+    \end{enumerate}
+
+    We define $U_s$ inductively on the length of $s$.
+
+    For $U_{\emptyset}$ take any non-empty open set
+    with small enough diameter.
+
+    Given $U_s$, pick $x \neq y \in U_s$
+    and let $U_{s \concat 0} \ni x$,
+    $U_{s \concat 1} \ni y$ 
+    be disjoint, open,
+    of diameter $\le \frac{1}{2^{|s| +1}}$
+    and such that $\overline{U_{s\concat 0}}, \overline{U_{S \concat 1}} \subseteq  U_s$.
+
+    Let $x \in  2^{\N}$.
+    Then let $f(x)$ be the unique point in $X$ 
+    such that
+    \[
+    \{f(x)\}  = \bigcap_{n}  U_{x \defon n} = \bigcap_{n} \overline{U_{x \defon n}.
+    \] 
+    (This is nonempty as $X$ is a completely metrizable space.)
+    It is clear that $f$ is injective and continuous.
+    % TODO: more details
+    $2^{\N}$ is compact, hence $f^{-1}$ is also continuous.
+\end{proof}
+
+\begin{corollary}
+    Every nonempty perfect Polish
+    space $X$ has cardinality $C = 2^{\aleph_0}$ 
+    % TODO: eulerscript C ?
+\end{corollary}
+\begin{proof}
+    Since the cantor space embeds into $X$,
+    we get the lower bound.
+    Since $X$ is second countable and Hausdorff,
+    we get the upper bound.
+\end{proof}
+
+\begin{theorem}
+    Any Polish space is countable
+    or it has cardinality $C$. % TODO C
+\end{theorem}
+\todo{Homework 3}
+
+
+\begin{definition}
+    A \vocab{Lusin scheme} on a set $X$
+    is a family  $(A_s)_{s \in \N^{<\N}}$ 
+    of subsets of $X $
+    such that
+    \begin{enumerate}[(i)]
+        \item $A_{s \concat i} \cap A_{s \concat j} = \emptyset$
+            for all $j \neq i \in \N$, $s \in \N^{<\N}$.
+        \item $A_{s \concat i} \subseteq  A_s$ 
+            for all $i \in \N, s \in \N^{<\N}$.
+    \end{enumerate}
+\end{definition}
+
+\begin{theorem}
+    \label{thm:bairetopolish}
+    Let $X \neq \emptyset$ be a Polish space.
+    Then there is a closed subset
+    \[
+    D \subseteq \N^\N \text{\reflectbox{$\coloneqq$}} \cN
+    % TODO correct N for the Baire space?
+    \] 
+    and a continuous bijection from
+    $D$ onto $X$ (the inverse does not need to be continuous).
+
+    Moreover there is a continuous surjection $g: \cN \to  X$ 
+    extending $f$.
+\end{theorem}
+\begin{definition}
+    An $F_\sigma$ set is the
+    countable union of closed sets,
+    i.e.~the complement of a $G_\delta$ set.
+\end{definition}
+\begin{observe}
+    \begin{itemize}
+        \item Any open set is $F {\sigma}$.
+        \item In metric spaces the intersection of an open and closed set is $F_\sigma$.
+    \end{itemize}
+\end{observe}
+\begin{refproof}{thm:bairetopolish}
+    Let $d$ be a complete metric on $X$.
+    W.l.o.g.~$\diam(X) \le 1$.
+    We construct a Lusin scheme
+    $(F_s)_{s \in \N^{<\N}}$
+    such that $F_s \subseteq  X$
+    and
+    \begin{enumerate}[(i)]
+        \item $F_\emptyset = X$,
+        \item $F_s$ is $F_\sigma$ for all  $s$.
+        \item The $F_{s \concat i}$ partition $F_s$,
+            i.e.~$F_{s} = \bigsqcup_i F_{s \concat i}$.
+
+            Furthermore we want that
+            $\overline{F_{s \concat i}} \subseteq F_s$
+            for all $i$.
+        \item $\diam(F_s) \le 2^{-|s|}$.
+    \end{enumerate}
+
+    Suppose we already have $F_s \text{\reflectbox{$\coloneqq$}} F$.
+    We need to construct a partition $(F_i)_{i \in \N}$
+    of $F$ with $\overline{F_i} \subseteq F$
+    and $\diam(F_i) < \epsilon$
+    for $\epsilon = 2^{-|s| - 1}$,
+    such that the $F_i$ are $F_\sigma$.
+
+    \paragraph{Step 1}
+    Write $F \coloneqq \bigcup_{i \in \N} C_i$
+    for some closed sets $C_i$.
+    W.l.o.g.~$C_i \subseteq C_{i+1}$.
+
+    Let $F_i^0 \coloneqq C_{i+1} \setminus C_i$.
+    These $F_i^0$ are $F_\sigma$,
+    and form a partition of $F$.
+    Furthermore $\overline{F_i^0} \subseteq F$.
+
+    However the diameter might be too large.
+    Fix $i \in \N$ and consider $F_i^0$.
+    Cover it with countably many open balls  $B_1, B_2,\ldots$
+    of diameter smaller than  $\epsilon$.
+    The sets $D_i \coloneqq F_i^0 \cap B_i \setminus (B_1 \cup \ldots \cup B_{i-1})$
+    are $F_\sigma$, disjoint
+    and $F_i^0 = \bigcup_{j} D_j$.
+
+
+
+
+\end{refproof}
+
+

logic.sty

@@ -124,7 +124,12 @@
 \DeclareSimpleMathOperator{Ord}
 \DeclareSimpleMathOperator{trcl}
 \DeclareSimpleMathOperator{tcl}
-\newcommand{\concat}{{}^\frown}
+
+\newcommand{\concat}{\mathop{{}^{\scalebox{.7}{$\smallfrown$}}}}
+
+%https://tex.stackexchange.com/questions/73437/how-do-i-typeset-the-concatenation-of-strings-properly
+%\mathbin{\raisebox{1ex}{\scalebox{.7}{$\frown$}}}%
+    
 \DeclareMathOperator{\hght}{height}
 
 \newcommand\lecture[3]{\hrule{\color{darkgray}\hfill{\tiny[Lecture #1, #2]}}}

logic3.tex

@@ -26,6 +26,7 @@
 
 \input{inputs/lecture_01}
 \input{inputs/lecture_02}
+\input{inputs/lecture_03}