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small fix
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Josia Pietsch 2023-12-05 01:54:56 +01:00
parent 0612713145
commit 6a9c2956b7
Signed by: josia
GPG key ID: E70B571D66986A2D
1 changed files with 5 additions and 4 deletions
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inputs/lecture_12.tex
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@ -20,7 +20,7 @@
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 $\{(T, \beta) \in \Tr \times \cN : \forall n.~ T(\beta\defon{n}) = 1\}$. Consider $D \coloneqq \{(T, \beta) \in \Tr \times \cN : \forall n.~ T(\beta\defon{n}) = 1\}$.
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? % TODO Why clopen?
@ -123,9 +123,9 @@ For the proof we need some prerequisites:
$\IF$ is $\Sigma^1_1$-complete. $\IF$ is $\Sigma^1_1$-complete.
\end{corollary} \end{corollary}
\begin{proof} \begin{proof}
Let $A \subseteq X$ is analytic Let $X$ be Polish.
and $X$ Polish and uncountable, Suppose that $A \subseteq X$ is analytic and uncountable.
then Then
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\[\begin{tikzcd} \[\begin{tikzcd}
X & \cN & \Tr \\ X & \cN & \Tr \\
@ -135,6 +135,7 @@ For the proof we need some prerequisites:
\arrow[hook, from=2-1, to=1-1] \arrow[hook, from=2-1, to=1-1]
\arrow[hook, from=2-2, to=1-2] \arrow[hook, from=2-2, to=1-2]
\end{tikzcd}\] \end{tikzcd}\]
where $f$ is chosen as in \yaref{thm:lec12:1}.
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