gist 12
Some checks failed
Build latex and deploy / checkout (push) Failing after 16m52s

This commit is contained in:
Josia Pietsch 2024-01-24 15:35:14 +01:00
parent 3917993d93
commit acb56df1c2
Signed by: josia
GPG key ID: E70B571D66986A2D
3 changed files with 148 additions and 107 deletions

View file

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

View file

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

View file

@ -21,63 +21,92 @@
T \in \IF &\iff& \exists \beta \in \cN .~\forall n \in \N.~T(\beta\defon{n}) = 1. T \in \IF &\iff& \exists \beta \in \cN .~\forall n \in \N.~T(\beta\defon{n}) = 1.
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
Consider $D \coloneqq \{(T, \beta) \in \Tr \times \cN : \forall n.~ T(\beta\defon{n}) = 1\}$. Consider
\[D \coloneqq \{(T, \beta) \in \Tr \times \cN : \forall n.~ T(\beta\defon{n}) = 1\}.\]
\gist{%
Note that this set is closed in $\Tr \times \cN$, Note that this set is closed in $\Tr \times \cN$,
since it is a countable intersection of clopen sets. since it is a countable intersection of clopen sets.
% TODO Why clopen?
Then $\IF = \proj_{\Tr}(D) \in \Sigma^1_1$. Then $\IF = \proj_{\Tr}(D) \in \Sigma^1_1$.
}{$D \overset{\text{closed}}{\subseteq} \Tr \times \cN$ and $\IF = \proj_{\Tr}(D)$.}
\end{proof} \end{proof}
\begin{definition} \begin{definition}
An analytic set $B$ in some Polish space $Y$ An analytic set $B$ in some Polish space $Y$
is \vocab{complete analytic} (\vocab{$\Sigma^1_1$-complete}) is \vocab{complete analytic} (\vocab{$\Sigma^1_1$-complete})
\gist{%
iff for any analytic $A \in \Sigma^1_1(X)$ for some Polish space $X$, iff for any analytic $A \in \Sigma^1_1(X)$ for some Polish space $X$,
there exists a Borel function $f\colon X\to Y$ 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} \end{definition}
\begin{observe} \begin{observe}
\leavevmode \leavevmode
\begin{itemize} \begin{itemize}
\item Complements of $\Sigma^1_1$-complete sets are $\Pi^1_1$-complete. \item Complements of $\Sigma^1_1$-complete sets are $\Pi^1_1$-complete.
\item $\Sigma^1_1$-complete sets are never Borel: \item $\Sigma^1_1$-complete sets are never Borel%
\gist{:
Suppose there is a $\Sigma^1_1$-complete set $B \in \cB(Y)$. Suppose there is a $\Sigma^1_1$-complete set $B \in \cB(Y)$.
Take $A \in \Sigma^1_1(X) \setminus \cB(X)$ Take $A \in \Sigma^1_1(X) \setminus \cB(X)$%
\footnote{e.g.~\yaref{thm:universals11}}
and $f\colon X \to Y$ Borel. and $f\colon X \to Y$ Borel.
But then $f^{-1}(B)$ is Borel. But then we get that $f^{-1}(B)$ is Borel $\lightnig$.
}{.}
\end{itemize} \end{itemize}
\end{observe} \end{observe}
% TODO ANKI-MARKER
\begin{theorem} \begin{theorem}
\label{thm:lec12:1} \label{thm:lec12:1}
Suppose that $A \subseteq \cN$ is analytic. Suppose that $A \subseteq \cN$ is analytic.
Then there is $f\colon \cN \to \Tr$\todo{Borel?} \gist{%
such that $x \in A \iff f(x)$ is ill-founded. 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} \end{theorem}
For the proof we need some prerequisites: For the proof we need some prerequisites:
\begin{enumerate}[1.]
\item Recall that for $S$ countable, \gist{%
the pruned\footnote{no maximal elements, in particular this implies ill-founded if the tree is non empty.} trees Recall that for $S$ countable,
$T \subseteq S^{<\N}$ on $S$ correspond the pruned%
to closed subsets of $S^{\N}$: \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} \begin{IEEEeqnarray*}{rCl}
T &\longmapsto & [T]\\ T &\longmapsto & [T]\\
\{\alpha\defon{n} : \alpha \in D, n \in \N\} &\longmapsfrom & D\\ \{\alpha\defon{n} : \alpha \in D, n \in \N\} &\longmapsfrom & D\\
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
\todo{Copy from exercises} }{%
\item \leavevmode\begin{definition} 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 $\N \times \N$ If $T$ is a tree on $\N \times \N$
and $x \in \cN$, and $x \in \cN$,
then the \vocab{section at $x$} then the \vocab{section at $x$}
%denoted $T(x)$, denoted $T(x)$,
is the following tree on $\N$ : is the following tree on $\N$ :
\[ \[
T(x) = \{s \in \N^{<\N} : (x\defon{|s|}, s) \in T\}. T(x) = \{s \in \N^{<\N} : (x\defon{|s|}, s) \in T\}.
\] \]
\end{definition} \end{definition}
\item \leavevmode \begin{proposition}
\begin{proposition}
\label{prop:lec12:2} \label{prop:lec12:2}
Let $A \subseteq \cN$. Let $A \subseteq \cN$.
The following are equivalent: The following are equivalent:
@ -87,24 +116,28 @@ For the proof we need some prerequisites:
such that such that
\[A = \proj_1 ([T]) = \{x \in \cN : \exists y \in \cN.~ (x,y) \in [T]\}.\] \[A = \proj_1 ([T]) = \{x \in \cN : \exists y \in \cN.~ (x,y) \in [T]\}.\]
\end{itemize} \end{itemize}
\end{proposition} \end{proposition}
\begin{proof} \begin{proof}
\gist{%
$A$ is analytic iff $A$ is analytic iff
there exists $F \overset{\text{closed}}{\subseteq} \N \times \N$ there exists $F \overset{\text{closed}}{\subseteq} (\N \times \N)^{\N}$
such that $A = \proj_1(F)$. such that $A = \proj_1(F)$.
But closed sets of $\N \times \N$ correspond to pruned trees, But closed sets of $\N^\N \times \N^{\N}$ correspond to pruned trees,
by the first point. by the first point.
\end{proof} }{Closed subsets of $\N^\N \times \N^\N$ correspond to pruned trees.}
\end{enumerate} \end{proof}
\begin{refproof}{thm:lec12:1} \begin{refproof}{thm:lec12:1}
\gist{%
Take a tree $T$ on $\N \times \N$ Take a tree $T$ on $\N \times \N$
as in \autoref{prop:lec12:2}, i.e.~$A = \proj_1([T])$. 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 Consider
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
f\colon \cN &\longrightarrow & \Tr \\ f\colon \cN &\longrightarrow & \Tr \\
x &\longmapsto & T(x). x &\longmapsto & T(x).
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
Clearly $x \in A \iff f(x)$ is ill-founded. \gist{%
Clearly $x \in A \iff f(x) \in \IF$.
$f$ is continuous: $f$ is continuous:
Let $x\defon{n} = y\defon{n}$ for some $n \in \N$. Let $x\defon{n} = y\defon{n}$ for some $n \in \N$.
Then for all $m \le n, s,t \in \N^{<\N}$ Then for all $m \le n, s,t \in \N^{<\N}$
@ -116,7 +149,8 @@ For the proof we need some prerequisites:
\end{itemize} \end{itemize}
So if $x\defon{n} = y\defon{n}$, So if $x\defon{n} = y\defon{n}$,
then $t \in T(x) \iff t \in T(y)$ as long as $|t| \le n$.. then $t \in T(x) \iff t \in T(y)$ as long as $|t| \le n$.
}{}
\end{refproof} \end{refproof}
\begin{corollary} \begin{corollary}
@ -125,7 +159,9 @@ For the proof we need some prerequisites:
\end{corollary} \end{corollary}
\begin{proof} \begin{proof}
Let $X$ be Polish. 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 Then
% https://q.uiver.app/#q=WzAsNSxbMCwwLCJYIl0sWzEsMCwiXFxjTiJdLFsyLDAsIlxcVHIiXSxbMCwxLCJBIl0sWzEsMSwiYihBKSJdLFsxLDIsImYiXSxbMCwxLCJiIl0sWzMsMCwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbNCwxLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dXQ== % https://q.uiver.app/#q=WzAsNSxbMCwwLCJYIl0sWzEsMCwiXFxjTiJdLFsyLDAsIlxcVHIiXSxbMCwxLCJBIl0sWzEsMSwiYihBKSJdLFsxLDIsImYiXSxbMCwxLCJiIl0sWzMsMCwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbNCwxLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dXQ==
\[\begin{tikzcd} \[\begin{tikzcd}
@ -138,6 +174,7 @@ For the proof we need some prerequisites:
\end{tikzcd}\] \end{tikzcd}\]
where $f$ is chosen as in \yaref{thm:lec12:1}. where $f$ is chosen as in \yaref{thm:lec12:1}.
\gist{%
If $X$ is Polish and countable and $A \subseteq X$ analytic, If $X$ is Polish and countable and $A \subseteq X$ analytic,
just consider just consider
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
@ -148,6 +185,7 @@ For the proof we need some prerequisites:
\end{cases} \end{cases}
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
where $a \in \IF$ and $b \not\in \IF$ are chosen arbitrarily. where $a \in \IF$ and $b \not\in \IF$ are chosen arbitrarily.
}{}
\end{proof} \end{proof}
\subsection{Linear Orders} \subsection{Linear Orders}
@ -161,9 +199,9 @@ Let
\[ \[
\WO \coloneqq \{x \in \LO: x \text{ is a well ordering}\}. \WO \coloneqq \{x \in \LO: x \text{ is a well ordering}\}.
\] \]
\gist{%
Recall that Recall that
\begin{itemize} \begin{itemize}
\item $(A,<)$ is a well ordering iff there are no infinite descending chains. \item $(A,<)$ is a well ordering iff there are no infinite descending chains.
\item Every well ordering is isomorphic to an ordinal. \item Every well ordering is isomorphic to an ordinal.
\item Any two well orderings are comparable, \item Any two well orderings are comparable,
@ -172,7 +210,8 @@ Recall that
Let $(A, <_A) \prec (B, <_B)$ denote that Let $(A, <_A) \prec (B, <_B)$ denote that
$(A, <_A)$ is isomorphic to a proper initial segment of $(B, <_B)$. $(A, <_A)$ is isomorphic to a proper initial segment of $(B, <_B)$.
\end{itemize} \end{itemize}
}{}
\begin{definition} \begin{definition}
A \vocab{rank} on some set $C$ A \vocab{rank} on some set $C$
@ -181,7 +220,8 @@ Recall that
\phi\colon C \to \Ord. \phi\colon C \to \Ord.
\] \]
\end{definition} \end{definition}
\begin{example} \gist{%
\begin{example}
Let $C = \WO$ Let $C = \WO$
and and
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
@ -189,5 +229,5 @@ Recall that
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
where $\phi((A,<_A))$ is the unique ordinal where $\phi((A,<_A))$ is the unique ordinal
isomorphic to $(A, <_A)$. isomorphic to $(A, <_A)$.
\end{example} \end{example}
}{}