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Josia Pietsch 2024-01-25 12:38:52 +01:00
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inputs/lecture_09.tex
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@ -6,6 +6,7 @@
we have that $\Sigma^0_\xi(X) \neq \Pi^0_\xi(X)$. we have that $\Sigma^0_\xi(X) \neq \Pi^0_\xi(X)$.
\end{theorem} \end{theorem}
\begin{proof} \begin{proof}
\gist{%
Fix $\xi < \omega_1$. Fix $\xi < \omega_1$.
Towards a contradiction assume $\Sigma^0_\xi(X) = \Pi^0_\xi(X)$. Towards a contradiction assume $\Sigma^0_\xi(X) = \Pi^0_\xi(X)$.
By \autoref{thm:cantoruniversal}, By \autoref{thm:cantoruniversal},
@ -19,24 +20,30 @@
\[z \in A \iff z \in \cU_z \iff (z,z) \in \cU.\] \[z \in A \iff z \in \cU_z \iff (z,z) \in \cU.\]
But by the definition of $A$, But by the definition of $A$,
we have $z \in A \iff (z,z) \not\in \cU \lightning$. we have $z \in A \iff (z,z) \not\in \cU \lightning$.
}{%
Let $\cU$ be $X$-universal for $\Sigma^0_\xi(X)$.
Consider $\{y \in X : (y,y) \not\in \cU\} \in \Pi^0_\xi(X) \setminus \Sigma^0_\xi(X)$.
}
\end{proof} \end{proof}
\begin{definition} \begin{definition}
Let $X$ be a Polish space. Let $X$ be a Polish space.
A set $A \subseteq X$ A set $A \subseteq X$
is called \vocab{analytic} is called \vocab{analytic}
iff iff
\[ \[
\exists Y \text{ Polish}.~\exists B \in \cB(Y).~ \exists Y \text{ Polish}.~\exists B \in \cB(Y).~
\exists \underbrace{f\colon Y \to X}_{\text{continuous}}.~ \exists \underbrace{f\colon Y \to X}_{\text{continuous}}.~
f(B) = A. f(B) = A.
\] \]
\end{definition} \end{definition}
\gist{%
Trivially, every Borel set is analytic. Trivially, every Borel set is analytic.
We will see that not every analytic set is Borel. We will see that not every analytic set is Borel.
}{}
\begin{remark} \begin{remark}
In the definition we can replace the assertion that In the definition we can replace the assertion that
$f$ 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}
@ -50,7 +57,7 @@ We will see that not every analytic set is Borel.
Then the following are equivalent: Then the following are equivalent:
\begin{enumerate}[(i)] \begin{enumerate}[(i)]
\item $A$ is analytic. \item $A$ is analytic.
\item There exists a Polish space $Y$ \item There exists a Polish space $Y$
and $f\colon Y \to X$ and $f\colon Y \to X$
continuous\footnote{or Borel} continuous\footnote{or Borel}
such that $A = f(Y)$. such that $A = f(Y)$.
@ -64,6 +71,7 @@ We will see that not every analytic set is Borel.
\end{enumerate} \end{enumerate}
\end{theorem} \end{theorem}
\begin{proof} \begin{proof}
\gist{%
To show (i) $\implies$ (ii): To show (i) $\implies$ (ii):
take $B \in \cB(Y')$ take $B \in \cB(Y')$
and $f\colon Y' \to X$ and $f\colon Y' \to X$
@ -73,11 +81,16 @@ We will see that not every analytic set is Borel.
such that $B$ is clopen with respect to the new topology. such that $B$ is clopen with respect to the new topology.
Then let $g = f\defon{B}$ Then let $g = f\defon{B}$
and $Y = (B, \cT\defon{B})$. and $Y = (B, \cT\defon{B})$.
}{(i) $\implies$ (ii):
Clopenize the Borel set, then restrict.
}
(ii) $\implies$ (iii): (ii) $\implies$ (iii):
Any Polish space is the continuous image of $\cN$. Any Polish space is the continuous image of $\cN$.
\gist{%
Let $g_1: \cN \to Y$ Let $g_1: \cN \to Y$
and $h \coloneqq g \circ g_1$. and $h \coloneqq g \circ g_1$.
}{}
(iii) $\implies$ (iv): (iii) $\implies$ (iv):
Let $h\colon \cN \to X$ with $h(\cN) = A$. Let $h\colon \cN \to X$ with $h(\cN) = A$.