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Josia Pietsch 2024-01-24 23:13:56 +01:00
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3 changed files with 25 additions and 22 deletions

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@ -173,10 +173,8 @@ i.e.~we want to associate a tree $T \subseteq \N^{<\N}$}%
Let \vocab{$\Tr$} $ \coloneqq \{T \in {2^{\N}}^{<\N} : T \text{ is a tree}\} \subseteq {2^{\N}}^{<\N}$.
\begin{observe}
\[
\Tr \subseteq {2^{\N}}^{<\N}
\]
is closed (where we take the topology of the Cantor space).
$\Tr \subseteq {2^{\N}}^{<\N}$ is closed
(where we take the topology of the Cantor space).
\end{observe}
\gist{%
Indeed, for any $ s \in \N^{<\N}$

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@ -33,7 +33,6 @@ with $(f^{-1}(\{1\}), <)$.
and $[\alpha_i, \alpha_{i+1})$ to $(i,i+1)$.
\end{proof}
% TODO ANKI-MARKER
\begin{definition}[\vocab{Kleene-Brouwer ordering}]
Let $(A,<)$ be a linear order and $A$ countable.
@ -58,25 +57,30 @@ with $(f^{-1}(\{1\}), <)$.
$(T, <_{KB}\defon{T})$ is well ordered.
\end{proposition}
\begin{proof}
If $T$ is ill-founded and $x \in [T]$,
then for all $n$, we have $x\defon{n+1} <_{KB} x\defon{n}$.
Thus $(T, <_{KB}\defon{T})$ is not well ordered.
\gist{%
If $T$ is ill-founded and $x \in [T]$,
then for all $n$, we have $x\defon{n+1} <_{KB} x\defon{n}$.
Thus $(T, <_{KB}\defon{T})$ is not well ordered.
Conversely, let $<\defon{KB}$ be not a well-ordering
on $T$.
Let $s_0 >_{KB} s_1 >_{KB} s_2 >_{KB} \ldots$
be an infinite descending chain.
We have that $s_0(0) \ge s_1(0) \ge s_2(0) \ge \ldots$
stabilizes for $n > n_0$.
Let $a_0 \coloneqq s_{n_0}(0)$.
Now for $n \ge n_0$ we have that $s_n(0)$ is constant,
hence for $n > n_0$ the value $s_{n}(1)$ must be defined.
Thus there is $n_1 \ge n_0$ such that $s_n(1)$
is constant for all $n \ge n_1$.
Let $a_1 \coloneqq s_{n_1}(1)$
and so on.
Then $(a_0,a_1,a_2, \ldots) \in [T]$.
Conversely, let $<\defon{KB}$ be not a well-ordering
on $T$.
Let $s_0 >_{KB} s_1 >_{KB} s_2 >_{KB} \ldots$
be an infinite descending chain.
We have that $s_0(0) \ge s_1(0) \ge s_2(0) \ge \ldots$
stabilizes for $n > n_0$.
Let $a_0 \coloneqq s_{n_0}(0)$.
Now for $n \ge n_0$ we have that $s_n(0)$ is constant,
hence for $n > n_0$ the value $s_{n}(1)$ must be defined.
Thus there is $n_1 \ge n_0$ such that $s_n(1)$
is constant for all $n \ge n_1$.
Let $a_1 \coloneqq s_{n_1}(1)$
and so on.
Then $(a_0,a_1,a_2, \ldots) \in [T]$.
}{easy}
\end{proof}
% TODO ANKI-MARKER
\begin{theorem}[Lusin-Sierpinski]
The set $\LO \setminus \WO$
(resp.~$2^{\Q} \setminus \WO$)

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@ -4,6 +4,7 @@
% TODO gist info
% TODO link to long version (provide link to main document)
% TODO \phantomsection to cross link
\newcommand{\gist}[2]{%
\ifcsname EnableGist\endcsname%
#2%