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1
inputs/lecture_09.tex
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@ -135,6 +135,7 @@ We will see later that $\Sigma^1_1(X) \cap \Pi^1_1(X) = \cB(X)$.
\begin{theorem}
\label{thm:universals11}
Let $X,Y$ be uncountable Polish spaces.
There exists a $Y$-universal $\Sigma^1_1(X)$ set.
\end{theorem}

14
inputs/lecture_11.tex
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@ -153,13 +153,14 @@ We will not proof this in this lecture.
\subsection{Ill-Founded Trees}
\gist{%
Recall that a \vocab{tree} on $\N$ is a subset of
$\N^{<\N}$
closed under taking initial segments.
We now identify trees with their characteristic functions,
i.e.~we want to associate a tree $T \subseteq \N^{<\N}$
i.e.~we want to associate a tree $T \subseteq \N^{<\N}$}%
{We identify trees $T \subseteq \N^{<\N}$ with their characteristic functions:}
\begin{IEEEeqnarray*}{rCl}
\One_T\colon \omega^{<\omega} &\longrightarrow & \{0,1\} \\
x &\longmapsto & \begin{cases}
@ -167,9 +168,9 @@ i.e.~we want to associate a tree $T \subseteq \N^{<\N}$
0 &: x \not\in T.
\end{cases}
\end{IEEEeqnarray*}
Note that $\One_T \in {\{0,1\}^\N}^{< \N}$.
\gist{Note that $\One_T \in {\{0,1\}^\N}^{< \N}$.}{}
Let $\Tr = \{T \in {2^{\N}}^{<\N} : T \text{ is a tree}\} \subseteq {2^{\N}}^{<\N}$.
Let \vocab{$\Tr$} $ \coloneqq \{T \in {2^{\N}}^{<\N} : T \text{ is a tree}\} \subseteq {2^{\N}}^{<\N}$.
\begin{observe}
\[
@ -177,6 +178,7 @@ Let $\Tr = \{T \in {2^{\N}}^{<\N} : T \text{ is a tree}\} \subseteq {2^{\N}}^{<\
\]
is closed (where we take the topology of the Cantor space).
\end{observe}
\gist{%
Indeed, for any $ s \in \N^{<\N}$
we have that $\{T \in {2^{\N}}^{<\N} : s \in T\}$
and $\{T \in {2^{\N}}^{<\N} : s\not\in T\}$ are clopen.
@ -185,6 +187,4 @@ In particular for $s$ fixed,
we have that
\[\{A \in {2^{\N}}^{<\N} : s \in A \text{ and } s' \in A \text{ for any initial segment $s' \subseteq s$}\}\]
is clopen in ${2^{\N}}^{<\N}$.
}{}

240
inputs/lecture_12.tex
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@ -21,102 +21,136 @@
T \in \IF &\iff& \exists \beta \in \cN .~\forall n \in \N.~T(\beta\defon{n}) = 1.
\end{IEEEeqnarray*}
Consider $D \coloneqq \{(T, \beta) \in \Tr \times \cN : \forall n.~ T(\beta\defon{n}) = 1\}$.
Note that this set is closed in $\Tr \times \cN$,
since it is a countable intersection of clopen sets.
% TODO Why clopen?
Then $\IF = \proj_{\Tr}(D) \in \Sigma^1_1$.
Consider
\[D \coloneqq \{(T, \beta) \in \Tr \times \cN : \forall n.~ T(\beta\defon{n}) = 1\}.\]
\gist{%
Note that this set is closed in $\Tr \times \cN$,
since it is a countable intersection of clopen sets.
Then $\IF = \proj_{\Tr}(D) \in \Sigma^1_1$.
}{$D \overset{\text{closed}}{\subseteq} \Tr \times \cN$ and $\IF = \proj_{\Tr}(D)$.}
\end{proof}
\begin{definition}
An analytic set $B$ in some Polish space $Y$
is \vocab{complete analytic} (\vocab{$\Sigma^1_1$-complete})
\gist{%
iff for any analytic $A \in \Sigma^1_1(X)$ for some Polish space $X$,
there exists a Borel function $f\colon X\to Y$
such that $x \in A \iff f(x) \in B$.
such that $x \in A \iff f(x) \in B$,
i.e.~$f^{-1}(B) = A$.
}{%
iff for any $A \in \Sigma^1_1(X)$, $X$ Polish
there exists $f\colon X \to Y$ Borel such that $f^{-1}(B) = A$.}
Similarly, define \vocab{complete coanalytic} (\vocab{$\Pi^1_1$-complete}).
\gist{%
Similarly, a conalytic set $B$ is called
\vocab{complete coanalytic} (\vocab{$\Pi^1_1$-complete})
iff for any $A \subseteq \Pi^1_1(X)$
there exists $f\colon X \to Y$ Borel such that $f^{-1}(B) = A$.
}{Similarly we define \vocab{complete coanalytic} / \vocab{$\Pi_1^1$-complete}.}
\end{definition}
\begin{observe}
\leavevmode
\begin{itemize}
\item Complements of $\Sigma^1_1$-complete sets are $\Pi^1_1$-complete.
\item $\Sigma^1_1$-complete sets are never Borel:
Suppose there is a $\Sigma^1_1$-complete set $B \in \cB(Y)$.
Take $A \in \Sigma^1_1(X) \setminus \cB(X)$
and $f\colon X \to Y$ Borel.
But then $f^{-1}(B)$ is Borel.
\item $\Sigma^1_1$-complete sets are never Borel%
\gist{:
Suppose there is a $\Sigma^1_1$-complete set $B \in \cB(Y)$.
Take $A \in \Sigma^1_1(X) \setminus \cB(X)$%
\footnote{e.g.~\yaref{thm:universals11}}
and $f\colon X \to Y$ Borel.
But then we get that $f^{-1}(B)$ is Borel $\lightnig$.
}{.}
\end{itemize}
\end{observe}
% TODO ANKI-MARKER
\begin{theorem}
\label{thm:lec12:1}
Suppose that $A \subseteq \cN$ is analytic.
Then there is $f\colon \cN \to \Tr$\todo{Borel?}
such that $x \in A \iff f(x)$ is ill-founded.
\gist{%
Then there is a continuous function $f\colon \cN \to \Tr$
such that $x \in A \iff f(x)$ is ill-founded,
i.e.~$A = f^{-1}(\IF)$.
}{%
Then there exists $f\colon \cN \to \Tr$ continuous
such that $A = f^{-1}(\IF)$.
}
\end{theorem}
For the proof we need some prerequisites:
\begin{enumerate}[1.]
\item Recall that for $S$ countable,
the pruned\footnote{no maximal elements, in particular this implies ill-founded if the tree is non empty.} trees
$T \subseteq S^{<\N}$ on $S$ correspond
to closed subsets of $S^{\N}$:
\begin{IEEEeqnarray*}{rCl}
T &\longmapsto & [T]\\
\{\alpha\defon{n} : \alpha \in D, n \in \N\} &\longmapsfrom & D\\
\end{IEEEeqnarray*}
\todo{Copy from exercises}
\item \leavevmode\begin{definition}
If $T$ is a tree on $\N \times \N$
and $x \in \cN$,
then the \vocab{section at $x$}
%denoted $T(x)$,
is the following tree on $\N$ :
\[
T(x) = \{s \in \N^{<\N} : (x\defon{|s|}, s) \in T\}.
\]
\end{definition}
\item \leavevmode
\begin{proposition}
\label{prop:lec12:2}
Let $A \subseteq \cN$.
The following are equivalent:
\begin{itemize}
\item $A$ is analytic.
\item There is a pruned tree on $\N \times \N$
such that
\[A = \proj_1 ([T]) = \{x \in \cN : \exists y \in \cN.~ (x,y) \in [T]\}.\]
\end{itemize}
\end{proposition}
\begin{proof}
$A$ is analytic iff
there exists $F \overset{\text{closed}}{\subseteq} \N \times \N$
such that $A = \proj_1(F)$.
But closed sets of $\N \times \N$ correspond to pruned trees,
by the first point.
\end{proof}
\end{enumerate}
\gist{%
Recall that for $S$ countable,
the pruned%
\footnote{no maximal elements,
in particular this implies ill-founded if the tree is non empty.
} trees $T \subseteq S^{<\N}$ on $S$ correspond
to closed subsets of $S^{\N}$:%
\footnote{cf.~\yaref{s3e1} (c)}
\begin{IEEEeqnarray*}{rCl}
T &\longmapsto & [T]\\
\{\alpha\defon{n} : \alpha \in D, n \in \N\} &\longmapsfrom & D\\
\end{IEEEeqnarray*}
}{%
For $S$ countable,
pruned trees on $S$ correspond to closed subsets of $S^{\N}$
via $T \mapsto [T]$.
}
\begin{definition}
If $T$ is a tree on $\N \times \N$
and $x \in \cN$,
then the \vocab{section at $x$}
denoted $T(x)$,
is the following tree on $\N$ :
\[
T(x) = \{s \in \N^{<\N} : (x\defon{|s|}, s) \in T\}.
\]
\end{definition}
\begin{proposition}
\label{prop:lec12:2}
Let $A \subseteq \cN$.
The following are equivalent:
\begin{itemize}
\item $A$ is analytic.
\item There is a pruned tree on $\N \times \N$
such that
\[A = \proj_1 ([T]) = \{x \in \cN : \exists y \in \cN.~ (x,y) \in [T]\}.\]
\end{itemize}
\end{proposition}
\begin{proof}
\gist{%
$A$ is analytic iff
there exists $F \overset{\text{closed}}{\subseteq} (\N \times \N)^{\N}$
such that $A = \proj_1(F)$.
But closed sets of $\N^\N \times \N^{\N}$ correspond to pruned trees,
by the first point.
}{Closed subsets of $\N^\N \times \N^\N$ correspond to pruned trees.}
\end{proof}
\begin{refproof}{thm:lec12:1}
Take a tree $T$ on $\N \times \N$
as in \autoref{prop:lec12:2}, i.e.~$A = \proj_1([T])$.
\gist{%
Take a tree $T$ on $\N \times \N$
as in \autoref{prop:lec12:2}, i.e.~$A = \proj_1([T])$.
}{Write $A = \proj_1([T])$ for a pruned tree $T$ on $\N \times \N$.}
Consider
\begin{IEEEeqnarray*}{rCl}
f\colon \cN &\longrightarrow & \Tr \\
x &\longmapsto & T(x).
\end{IEEEeqnarray*}
Clearly $x \in A \iff f(x)$ is ill-founded.
$f$ is continuous:
Let $x\defon{n} = y\defon{n}$ for some $n \in \N$.
Then for all $m \le n, s,t \in \N^{<\N}$
such that $s = x\defon{m} = y \defon{m}$ and $|t| = |s|$,
we have
\begin{itemize}
\item $t \in T(x) \iff (s,t) \in T$,
\item $t \in T(y) \iff (s,t) \in T$.
\end{itemize}
\gist{%
Clearly $x \in A \iff f(x) \in \IF$.
$f$ is continuous:
Let $x\defon{n} = y\defon{n}$ for some $n \in \N$.
Then for all $m \le n, s,t \in \N^{<\N}$
such that $s = x\defon{m} = y \defon{m}$ and $|t| = |s|$,
we have
\begin{itemize}
\item $t \in T(x) \iff (s,t) \in T$,
\item $t \in T(y) \iff (s,t) \in T$.
\end{itemize}
So if $x\defon{n} = y\defon{n}$,
then $t \in T(x) \iff t \in T(y)$ as long as $|t| \le n$..
So if $x\defon{n} = y\defon{n}$,
then $t \in T(x) \iff t \in T(y)$ as long as $|t| \le n$.
}{}
\end{refproof}
\begin{corollary}
@ -125,7 +159,9 @@ For the proof we need some prerequisites:
\end{corollary}
\begin{proof}
Let $X$ be Polish.
Suppose that $A \subseteq X$ is analytic and uncountable.
Suppose that $A \subseteq X$ is analytic and uncountable%
\gist{}{ (trivial for countable)}.
Then
% https://q.uiver.app/#q=WzAsNSxbMCwwLCJYIl0sWzEsMCwiXFxjTiJdLFsyLDAsIlxcVHIiXSxbMCwxLCJBIl0sWzEsMSwiYihBKSJdLFsxLDIsImYiXSxbMCwxLCJiIl0sWzMsMCwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbNCwxLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dXQ==
\[\begin{tikzcd}
@ -138,16 +174,18 @@ For the proof we need some prerequisites:
\end{tikzcd}\]
where $f$ is chosen as in \yaref{thm:lec12:1}.
If $X$ is Polish and countable and $A \subseteq X$ analytic,
just consider
\begin{IEEEeqnarray*}{rCl}
g \colon X &\longrightarrow & \Tr \\
x &\longmapsto & \begin{cases}
a &: x \in A,\\
b &: x \not\in A,\\
\end{cases}
\end{IEEEeqnarray*}
where $a \in \IF$ and $b \not\in \IF$ are chosen arbitrarily.
\gist{%
If $X$ is Polish and countable and $A \subseteq X$ analytic,
just consider
\begin{IEEEeqnarray*}{rCl}
g \colon X &\longrightarrow & \Tr \\
x &\longmapsto & \begin{cases}
a &: x \in A,\\
b &: x \not\in A,\\
\end{cases}
\end{IEEEeqnarray*}
where $a \in \IF$ and $b \not\in \IF$ are chosen arbitrarily.
}{}
\end{proof}
\subsection{Linear Orders}
@ -161,18 +199,19 @@ Let
\[
\WO \coloneqq \{x \in \LO: x \text{ is a well ordering}\}.
\]
\gist{%
Recall that
\begin{itemize}
\item $(A,<)$ is a well ordering iff there are no infinite descending chains.
\item Every well ordering is isomorphic to an ordinal.
\item Any two well orderings are comparable,
i.e.~they are isomorphic,
or one is isomorphic to an initial segment of the other.
Recall that
\begin{itemize}
\item $(A,<)$ is a well ordering iff there are no infinite descending chains.
\item Every well ordering is isomorphic to an ordinal.
\item Any two well orderings are comparable,
i.e.~they are isomorphic,
or one is isomorphic to an initial segment of the other.
Let $(A, <_A) \prec (B, <_B)$ denote that
$(A, <_A)$ is isomorphic to a proper initial segment of $(B, <_B)$.
\end{itemize}
Let $(A, <_A) \prec (B, <_B)$ denote that
$(A, <_A)$ is isomorphic to a proper initial segment of $(B, <_B)$.
\end{itemize}
}{}
\begin{definition}
A \vocab{rank} on some set $C$
@ -181,13 +220,14 @@ Recall that
\phi\colon C \to \Ord.
\]
\end{definition}
\begin{example}
Let $C = \WO$
and
\begin{IEEEeqnarray*}{rCl}
\phi\colon \WO &\longrightarrow & \Ord \\
\end{IEEEeqnarray*}
where $\phi((A,<_A))$ is the unique ordinal
isomorphic to $(A, <_A)$.
\end{example}
\gist{%
\begin{example}
Let $C = \WO$
and
\begin{IEEEeqnarray*}{rCl}
\phi\colon \WO &\longrightarrow & \Ord \\
\end{IEEEeqnarray*}
where $\phi((A,<_A))$ is the unique ordinal
isomorphic to $(A, <_A)$.
\end{example}
}{}