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206b61941e
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3 changed files with 9 additions and 7 deletions
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@ -40,6 +40,7 @@ We will see that not every analytic set is Borel.
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$f$ is continuous
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$f$ is continuous
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by the weaker assertion of $f$ being Borel.
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by the weaker assertion of $f$ being Borel.
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\todo{Copy exercise from sheet 5}
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\todo{Copy exercise from sheet 5}
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% TODO WHY?
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\end{remark}
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\end{remark}
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\begin{theorem}
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\begin{theorem}
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@ -1,5 +1,6 @@
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\subsection{The Lusin Separation Theorem}
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\lecture{10}{2023-11-17}{}
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\lecture{10}{2023-11-17}{}
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\todo{Start a new subsection here?}
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\begin{theorem}[\vocab{Lusin separation theorem}]
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\begin{theorem}[\vocab{Lusin separation theorem}]
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\yalabel{Lusin Separation Theorem}{Lusin Separation}{thm:lusinseparation}
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\yalabel{Lusin Separation Theorem}{Lusin Separation}{thm:lusinseparation}
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Let $X$ be Polish and $A,B \subseteq X$ disjoint analytic.
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Let $X$ be Polish and $A,B \subseteq X$ disjoint analytic.
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@ -55,7 +56,7 @@ we need the following definition:
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such that $A \cap B = \emptyset$
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such that $A \cap B = \emptyset$
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Then there are continuous surjections
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Then there are continuous surjections
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$f\colon \cN \twoheadrightarrow A \subseteq X$
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$f\colon \cN \twoheadrightarrow A \subseteq X$
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and $g\colon \twoheadrightarrow B \subseteq X$.
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and $g\colon \cN \twoheadrightarrow B \subseteq X$.
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Write $A_s \coloneqq f(\cN_s)$ and $B_s \coloneqq g(\cN_s)$.
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Write $A_s \coloneqq f(\cN_s)$ and $B_s \coloneqq g(\cN_s)$.
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Note that $A_s = \bigcup_m A_{s\concat m}$
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Note that $A_s = \bigcup_m A_{s\concat m}$
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@ -105,8 +106,8 @@ we need the following definition:
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Then $f(A)$ is Borel.
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Then $f(A)$ is Borel.
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\end{theorem}
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\end{theorem}
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\begin{refproof}{thm:lusinsouslin}
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\begin{refproof}{thm:lusinsouslin}
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W.l.o.g.~suppose that $f$ is continuous
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W.l.o.g.~suppose that $f$ is continuous,
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$A$ is closed,\footnote{We might even assume that $A$ is clopen, but we only need closed.}
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$A$ is closed\footnote{We might even assume that $A$ is clopen, but we only need closed.}
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and $X = \cN$ by \yaref{thm:bairetopolish}:
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and $X = \cN$ by \yaref{thm:bairetopolish}:
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@ -139,7 +140,7 @@ we need the following definition:
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\end{itemize}
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\end{itemize}
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The second point follows from injectivity of $f$
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The second point follows from injectivity of $f$
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and the fact that $\cN_{s \concat n} \cap \cN_{s\concat n'} = \emptyset$.
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and the fact that $\cN_{s \concat n} \cap \cN_{s\concat n'} = \emptyset$.
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In particular, the $(B_s)$ for a Lusin scheme.
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In particular, the $(B_s)$ form a Lusin scheme.
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Note that $f(A) = \bigcup_{s \in \omega^k} B_s$
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Note that $f(A) = \bigcup_{s \in \omega^k} B_s$
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for every $k <\omega$,
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for every $k <\omega$,
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@ -118,7 +118,7 @@
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Hence \yaref{thm:bsb} can be applied.
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Hence \yaref{thm:bsb} can be applied.
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\end{proof}
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\end{proof}
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\subsection{Projective Hierarchy}
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\subsection{The Projective Hierarchy}
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% https://q.uiver.app/#q=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@ -151,7 +151,7 @@
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We will not proof this in this lecture.
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We will not proof this in this lecture.
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\subsection{Trees} % TODO section?
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\subsection{Ill-Founded Trees}
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Recall that a \vocab{tree} on $\N$ is a subset of
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Recall that a \vocab{tree} on $\N$ is a subset of
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