Josia Pietsch
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233 lines
7.6 KiB
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233 lines
7.6 KiB
TeX
\lecture{12}{2023-11-24}{}
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\begin{definition}
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A tree $T$ is \vocab{ill-founded}
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if it has an infinite branch,
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i.e. $[T] \neq \emptyset$
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Otherwise it is called \vocab{well-founded}.
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Let
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\[\IF \coloneqq \{T \in \Tr : T \text{ is ill-founded}\}\]
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and
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\[\WF \coloneqq \{T \in \Tr : T \text{ is well-founded}\}\]
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\end{definition}
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\begin{proposition}
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\label{prop:ifs11}
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$\IF \in \Sigma^1_1(\Tr)$.
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\end{proposition}
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\begin{proof}
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We have
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\begin{IEEEeqnarray*}{rCl}
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T \in \IF &\iff& \exists \beta \in \cN .~\forall n \in \N.~T(\beta\defon{n}) = 1.
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\end{IEEEeqnarray*}
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Consider
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\[D \coloneqq \{(T, \beta) \in \Tr \times \cN : \forall n.~ T(\beta\defon{n}) = 1\}.\]
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\gist{%
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Note that this set is closed in $\Tr \times \cN$,
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since it is a countable intersection of clopen sets.
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Then $\IF = \proj_{\Tr}(D) \in \Sigma^1_1$.
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}{$D \overset{\text{closed}}{\subseteq} \Tr \times \cN$ and $\IF = \proj_{\Tr}(D)$.}
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\end{proof}
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\begin{definition}
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An analytic set $B$ in some Polish space $Y$
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is \vocab{complete analytic} (\vocab{$\Sigma^1_1$-complete})
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\gist{%
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iff for any analytic $A \in \Sigma^1_1(X)$ for some Polish space $X$,
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there exists a Borel function $f\colon X\to Y$
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such that $x \in A \iff f(x) \in B$,
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i.e.~$f^{-1}(B) = A$.
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}{%
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iff for any $A \in \Sigma^1_1(X)$, $X$ Polish
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there exists $f\colon X \to Y$ Borel such that $f^{-1}(B) = A$.}
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\gist{%
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Similarly, a conalytic set $B$ is called
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\vocab{complete coanalytic} (\vocab{$\Pi^1_1$-complete})
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iff for any $A \subseteq \Pi^1_1(X)$
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there exists $f\colon X \to Y$ Borel such that $f^{-1}(B) = A$.
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}{Similarly we define \vocab{complete coanalytic} / \vocab{$\Pi_1^1$-complete}.}
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\end{definition}
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\begin{observe}
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\leavevmode
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\begin{itemize}
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\item Complements of $\Sigma^1_1$-complete sets are $\Pi^1_1$-complete.
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\item $\Sigma^1_1$-complete sets are never Borel%
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\gist{:
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Suppose there is a $\Sigma^1_1$-complete set $B \in \cB(Y)$.
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Take $A \in \Sigma^1_1(X) \setminus \cB(X)$%
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\footnote{e.g.~\yaref{thm:universals11}}
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and $f\colon X \to Y$ Borel.
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But then we get that $f^{-1}(B)$ is Borel $\lightning$.
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}{.}
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\end{itemize}
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\end{observe}
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\begin{theorem}
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\label{thm:lec12:1}
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Suppose that $A \subseteq \cN$ is analytic.
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\gist{%
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Then there is a continuous function $f\colon \cN \to \Tr$
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such that $x \in A \iff f(x)$ is ill-founded,
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i.e.~$A = f^{-1}(\IF)$.
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}{%
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Then there exists $f\colon \cN \to \Tr$ continuous
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such that $A = f^{-1}(\IF)$.
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}
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\end{theorem}
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For the proof we need some prerequisites:
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\gist{%
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Recall that for $S$ countable,
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the pruned%
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\footnote{no maximal elements,
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in particular this implies ill-founded if the tree is non empty.
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} trees $T \subseteq S^{<\N}$ on $S$ correspond
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to closed subsets of $S^{\N}$:%
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\footnote{cf.~\yaref{s3e1} (c)}
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\begin{IEEEeqnarray*}{rCl}
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T &\longmapsto & [T]\\
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\{\alpha\defon{n} : \alpha \in D, n \in \N\} &\longmapsfrom & D\\
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\end{IEEEeqnarray*}
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}{%
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For $S$ countable,
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pruned trees on $S$ correspond to closed subsets of $S^{\N}$
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via $T \mapsto [T]$.
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}
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\begin{definition}
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If $T$ is a tree on $\N \times \N$
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and $x \in \cN$,
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then the \vocab{section at $x$}
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denoted $T(x)$,
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is the following tree on $\N$ :
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\[
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T(x) = \{s \in \N^{<\N} : (x\defon{|s|}, s) \in T\}.
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\]
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\end{definition}
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\begin{proposition}
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\label{prop:lec12:2}
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Let $A \subseteq \cN$.
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The following are equivalent:
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\begin{itemize}
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\item $A$ is analytic.
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\item There is a pruned tree on $\N \times \N$
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such that
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\[A = \proj_1 ([T]) = \{x \in \cN : \exists y \in \cN.~ (x,y) \in [T]\}.\]
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\end{itemize}
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\end{proposition}
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\begin{proof}
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\gist{%
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$A$ is analytic iff
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there exists $F \overset{\text{closed}}{\subseteq} (\N \times \N)^{\N}$
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such that $A = \proj_1(F)$.
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But closed sets of $\N^\N \times \N^{\N}$ correspond to pruned trees,
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by the first point.
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}{Closed subsets of $\N^\N \times \N^\N$ correspond to pruned trees.}
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\end{proof}
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\begin{refproof}{thm:lec12:1}
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\gist{%
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Take a tree $T$ on $\N \times \N$
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as in \autoref{prop:lec12:2}, i.e.~$A = \proj_1([T])$.
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}{Write $A = \proj_1([T])$ for a pruned tree $T$ on $\N \times \N$.}
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Consider
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\begin{IEEEeqnarray*}{rCl}
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f\colon \cN &\longrightarrow & \Tr \\
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x &\longmapsto & T(x).
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\end{IEEEeqnarray*}
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\gist{%
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Clearly $x \in A \iff f(x) \in \IF$.
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$f$ is continuous:
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Let $x\defon{n} = y\defon{n}$ for some $n \in \N$.
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Then for all $m \le n, s,t \in \N^{<\N}$
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such that $s = x\defon{m} = y \defon{m}$ and $|t| = |s|$,
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we have
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\begin{itemize}
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\item $t \in T(x) \iff (s,t) \in T$,
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\item $t \in T(y) \iff (s,t) \in T$.
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\end{itemize}
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So if $x\defon{n} = y\defon{n}$,
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then $t \in T(x) \iff t \in T(y)$ as long as $|t| \le n$.
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}{}
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\end{refproof}
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\begin{corollary}
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\label{cor:ifs11c}
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$\IF$ is $\Sigma^1_1$-complete.
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\end{corollary}
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\begin{proof}
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Let $X$ be Polish.
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Suppose that $A \subseteq X$ is analytic and uncountable%
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\gist{}{ (trivial for countable)}.
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Then
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% https://q.uiver.app/#q=WzAsNSxbMCwwLCJYIl0sWzEsMCwiXFxjTiJdLFsyLDAsIlxcVHIiXSxbMCwxLCJBIl0sWzEsMSwiYihBKSJdLFsxLDIsImYiXSxbMCwxLCJiIl0sWzMsMCwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbNCwxLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dXQ==
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\[\begin{tikzcd}
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X & \cN & \Tr \\
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A & {b(A)}
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\arrow["f", from=1-2, to=1-3]
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\arrow["b", from=1-1, to=1-2]
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\arrow[hook, from=2-1, to=1-1]
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\arrow[hook, from=2-2, to=1-2]
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\end{tikzcd}\]
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where $f$ is chosen as in \yaref{thm:lec12:1}.
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\gist{%
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If $X$ is Polish and countable and $A \subseteq X$ analytic,
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just consider
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\begin{IEEEeqnarray*}{rCl}
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g \colon X &\longrightarrow & \Tr \\
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x &\longmapsto & \begin{cases}
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a &: x \in A,\\
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b &: x \not\in A,\\
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\end{cases}
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\end{IEEEeqnarray*}
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where $a \in \IF$ and $b \not\in \IF$ are chosen arbitrarily.
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}{}
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\end{proof}
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\subsection{Linear Orders}
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Let us consider the space
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\[\LO \coloneqq \{x \in 2^{\N\times \N} : x \text{ is a linear order on $\N$}\},\]
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where we code a linear order $(\N, <)$
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by $x \in 2^{\N \times \N}$ with $x(m,n) = 1 \iff m \le n$.
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Let
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\[
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\WO \coloneqq \{x \in \LO: x \text{ is a well ordering}\}.
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\]
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\gist{%
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Recall that
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\begin{itemize}
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\item $(A,<)$ is a well ordering iff there are no infinite descending chains.
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\item Every well ordering is isomorphic to an ordinal.
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\item Any two well orderings are comparable,
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i.e.~they are isomorphic,
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or one is isomorphic to an initial segment of the other.
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Let $(A, <_A) \prec (B, <_B)$ denote that
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$(A, <_A)$ is isomorphic to a proper initial segment of $(B, <_B)$.
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\end{itemize}
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}{}
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\begin{definition}
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A \vocab{rank} on some set $C$
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is a function
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\[
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\phi\colon C \to \Ord.
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\]
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\end{definition}
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\gist{%
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\begin{example}
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Let $C = \WO$
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and
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\begin{IEEEeqnarray*}{rCl}
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\phi\colon \WO &\longrightarrow & \Ord \\
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\end{IEEEeqnarray*}
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where $\phi((A,<_A))$ is the unique ordinal
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isomorphic to $(A, <_A)$.
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\end{example}
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}{}
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