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Josia Pietsch 2024-01-18 17:17:27 +01:00
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3 changed files with 9 additions and 7 deletions
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1
inputs/lecture_09.tex
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@ -40,6 +40,7 @@ We will see that not every analytic set is Borel.
$f$ is continuous $f$ is continuous
by the weaker assertion of $f$ being Borel. by the weaker assertion of $f$ being Borel.
\todo{Copy exercise from sheet 5} \todo{Copy exercise from sheet 5}
% TODO WHY?
\end{remark} \end{remark}
\begin{theorem} \begin{theorem}

11
inputs/lecture_10.tex
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@ -1,5 +1,6 @@
\subsection{The Lusin Separation Theorem}
\lecture{10}{2023-11-17}{} \lecture{10}{2023-11-17}{}
\todo{Start a new subsection here?}
\begin{theorem}[\vocab{Lusin separation theorem}] \begin{theorem}[\vocab{Lusin separation theorem}]
\yalabel{Lusin Separation Theorem}{Lusin Separation}{thm:lusinseparation} \yalabel{Lusin Separation Theorem}{Lusin Separation}{thm:lusinseparation}
Let $X$ be Polish and $A,B \subseteq X$ disjoint analytic. Let $X$ be Polish and $A,B \subseteq X$ disjoint analytic.
@ -55,7 +56,7 @@ we need the following definition:
such that $A \cap B = \emptyset$ such that $A \cap B = \emptyset$
Then there are continuous surjections Then there are continuous surjections
$f\colon \cN \twoheadrightarrow A \subseteq X$ $f\colon \cN \twoheadrightarrow A \subseteq X$
and $g\colon \twoheadrightarrow B \subseteq X$. and $g\colon \cN \twoheadrightarrow B \subseteq X$.
Write $A_s \coloneqq f(\cN_s)$ and $B_s \coloneqq g(\cN_s)$. Write $A_s \coloneqq f(\cN_s)$ and $B_s \coloneqq g(\cN_s)$.
Note that $A_s = \bigcup_m A_{s\concat m}$ Note that $A_s = \bigcup_m A_{s\concat m}$
@ -105,8 +106,8 @@ we need the following definition:
Then $f(A)$ is Borel. Then $f(A)$ is Borel.
\end{theorem} \end{theorem}
\begin{refproof}{thm:lusinsouslin} \begin{refproof}{thm:lusinsouslin}
W.l.o.g.~suppose that $f$ is continuous W.l.o.g.~suppose that $f$ is continuous,
$A$ is closed,\footnote{We might even assume that $A$ is clopen, but we only need closed.} $A$ is closed\footnote{We might even assume that $A$ is clopen, but we only need closed.}
and $X = \cN$ by \yaref{thm:bairetopolish}: and $X = \cN$ by \yaref{thm:bairetopolish}:
@ -139,7 +140,7 @@ we need the following definition:
\end{itemize} \end{itemize}
The second point follows from injectivity of $f$ The second point follows from injectivity of $f$
and the fact that $\cN_{s \concat n} \cap \cN_{s\concat n'} = \emptyset$. and the fact that $\cN_{s \concat n} \cap \cN_{s\concat n'} = \emptyset$.
In particular, the $(B_s)$ for a Lusin scheme. In particular, the $(B_s)$ form a Lusin scheme.
Note that $f(A) = \bigcup_{s \in \omega^k} B_s$ Note that $f(A) = \bigcup_{s \in \omega^k} B_s$
for every $k <\omega$, for every $k <\omega$,

4
inputs/lecture_11.tex
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@ -118,7 +118,7 @@
Hence \yaref{thm:bsb} can be applied. Hence \yaref{thm:bsb} can be applied.
\end{proof} \end{proof}
\subsection{Projective Hierarchy} \subsection{The Projective Hierarchy}
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@ -151,7 +151,7 @@
We will not proof this in this lecture. We will not proof this in this lecture.
\subsection{Trees} % TODO section? \subsection{Ill-Founded Trees}
Recall that a \vocab{tree} on $\N$ is a subset of Recall that a \vocab{tree} on $\N$ is a subset of