lecture 12
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(continuous wrt.~to the topology of $X$)
On the other hand
X \hookrightarrow\cN \hookrightarrow[\text{continuous embedding}]\cC
X \hookrightarrow\cN \overset{\text{continuous embedding}}{\hookrightarrow}\cC
\todo{second inclusion was on a homework sheet}
For the first inclusion,
@ -132,7 +132,7 @@
\arrow["\subseteq", hook, from=1-2, to=2-3]
\arrow["\subseteq", hook, from=2-3, to=1-4]
\arrow["\subseteq", hook, from=2-3, to=3-4]
Let $X$ be a Polish space.
@ -174,7 +174,7 @@ Let $\Tr = \{T \in {2^{\N}}^{<\N} : T \text{ is a tree}\} \subseteq {2^{\N}}^{<\
\Tr \subseteq {2^{\N}}^{<\N}
is closed (where we take the topology of the Cantor space).
Indeed, for any $ s \in \N^{<\N}$

inputs/lecture_12.tex Normal file
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A tree $T$ is \vocab{ill-founded}
if it has an infinite branch,
i.e. $[T] \neq \emptyset$
Otherwise it is called \vocab{well-founded}.
\[\IF \coloneqq \{T \in \Tr : T \text{ is ill-founded}\}\]
\[\WF \coloneqq \{T \in \Tr : T \text{ is well-founded}\}\]
$\IF \in \Sigma^1_1(\Tr)$.
We have
T \in \IF &\iff& \exists \beta \in \cN .~\forall n \in \N.~T(\beta\defon{n}) = 1.
Consider $\{(T, \beta) \in \Tr \times \cN : \forall n.~ T(\beta\defon{n}) = 1\}$.
Note that this set is closed in $\Tr \times \cN$,
since it is a countable intersection of clopen sets.
% TODO Why clopen?
Then $\IF = \proj_{\Tr}(D) \in \Sigma^1_1$.
An analytic set $B$ in some Polish space $Y$
is \vocab{complete analytic} (\vocab{$\Sigma^1_1$-complete})
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 $x \in A \iff f(x) \in B$.
Similarly, define \vocab{complete coanalytic} (\vocab{$\Pi^1_1$-complete}).
\item Complements of $\Sigma^1_1$-complete sets are $\Pi^1_1$-complete.
\item $\Sigma^1_1$-complete sets are never Borel:
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.
Suppose that $A \subseteq \cN$ is analytic.
Then there is $f\colon \cN \to \Tr$
such that $x \in A \iff f(x)$ is ill-founded.
For the proof we need some prerequisites:
\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 $S^{\N}$:
T &\longmapsto & [T]\\
\{\alpha\defon{n} : \alpha \in D, n \in \N\} &\longmapsfrom & D\\
\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\}.
\item \leavevmode
Let $A \subseteq \cN$.
The following are equivalent:
\item $A$ is analytic.
\item There is a pruned tree on $\N \times \N$
such that
\[A = \proj_1 ([T]) = \{x \in \cN : \exists y \in \cN.~ (x,y) \in [T]\}.\]
$A$ 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.
Take a tree $T$ on $\N \times \N$
as in \autoref{prop:lec12:2}, i.e.~$A = \proj_1([T])$.
f\colon \cN &\longrightarrow & \Tr \\
x &\longmapsto & T(x).
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
\item $t \in T(x) \iff (s,t) \in T$,
\item $t \in T(y) \iff (s,t) \in T$.
So if $x\defon{n} = y\defon{n}$,
then $t \in T(x) \iff t \in T(y)$ as long as $|t| \le n$..
$\IF$ is $\Sigma^1_1$-complete.
Let $A \subseteq X$ is analytic
and $X$ Polish and uncountable,
% https://q.uiver.app/#q=WzAsNSxbMCwwLCJYIl0sWzEsMCwiXFxjTiJdLFsyLDAsIlxcVHIiXSxbMCwxLCJBIl0sWzEsMSwiYihBKSJdLFsxLDIsImYiXSxbMCwxLCJiIl0sWzMsMCwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbNCwxLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dXQ==
X & \cN & \Tr \\
A & {b(A)}
\arrow["f", from=1-2, to=1-3]
\arrow["b", from=1-1, to=1-2]
\arrow[hook, from=2-1, to=1-1]
\arrow[hook, from=2-2, to=1-2]
If $X$ is Polish and countable and $A \subseteq X$ analytic,
just consider
g \colon X &\longrightarrow & \Tr \\
x &\longmapsto & \begin{cases}
a &: x \in A,\\
b &: x \not\in A,\\
where $a \in \IF$ and $b \not\in \IF$
are chosen arbitrarily.
\subsection{Linear Orders}
Let us consider the space
\[\LO \coloneqq \{x \in 2^{\N\times \N} : x \text{ is a linear order on $\N$}\},\]
where we code a linear order $(\N, <)$
by $x \in 2^{\N \times \N}$ with $x(m,n) = 1 \iff m \le n$.
\WO \coloneqq \{x \in \LO: x \text{ is a well ordering}\}.
Recall that
\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 $(B, <_B)$.
A \vocab{rank} on some set $C$
is a function
\phi\colon C \to \Ord.
Let $C = \WO$
\phi\colon \WO &\longrightarrow & \Ord \\
where $\phi((A,<_A))$ is the unique ordinal
isomorphic to $(A, <_A)$.

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\DeclareSimpleMathOperator{LO} % linear orders
\DeclareSimpleMathOperator{WO} % well orderings

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