w23-logic-3/inputs/lecture_09.tex

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\lecture{09}{2023-11-14}{}
\begin{theorem}
Let $X$ be an uncountable Polish space.
Then for all $\xi < \omega_1$,
we have that $\Sigma^0_\xi(X) \neq \Pi^0_\xi(X)$.
\end{theorem}
\begin{proof}
Fix $\xi < \omega_1$.
Towards a contradiction assume $\Sigma^0_\xi(X) = \Pi^0_\xi(X)$.
By \autoref{thm:cantoruniversal},
there is a $X$-universal set $\cU$ for $\Sigma^0_\xi(X)$.
Take $A \coloneqq \{y \in X : (y,y) \not\in \cU\}$.
Then $A \in \Pi^0_\xi(X)$.\todo{Needs a small proof}
By assumption $A \in \Sigma^0_\xi(X)$,
i.e.~there exists some $z \in X$ such that $A = \cU_z$.
We have
\[z \in A \iff z \in \cU_z \iff (z,z) \in \cU.\]
But by the definition of $A$,
we have $z \in A \iff (z,z) \not\in \cU \lightning$.
\end{proof}
\begin{definition}
Let $X$ be a Polish space.
A set $A \subseteq X$
is called \vocab{analytic}
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iff
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\[
\exists Y \text{ Polish}.~\exists B \in \cB(Y).~
\exists \underbrace{f\colon Y \to X}_{\text{continuous}}.~
f(B) = A.
\]
\end{definition}
Trivially, every Borel set is analytic.
We will see that not every analytic set is Borel.
\begin{remark}
In the definition we can replace the assertion that
$f$ is continuous
by the weaker assertion of $f$ being Borel.
\todo{Copy exercise from sheet 5}
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% TODO WHY?
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\end{remark}
\begin{theorem}
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\label{thm:borel}
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Let $X$ be Polish,
$\emptyset \neq A \subseteq X$.
Then the following are equivalent:
\begin{enumerate}[(i)]
\item $A$ is analytic.
\item There exists a Polish space $Y$
and $f\colon Y \to X$
continuous\footnote{or Borel}
such that $A = f(Y)$.
\item There exists $h\colon \cN \to X$
continuous with $h(\cN) = A$.
\item There is $F \overset{\text{closed}}{\subseteq} X \times \cN$
such that $A = \proj_X(F)$.
\item There is a Borel set $B \subseteq X \times Y$
for some Polish space $Y$,
such that $A = \proj_X(B)$.
\end{enumerate}
\end{theorem}
\begin{proof}
To show (i) $\implies$ (ii):
take $B \in \cB(Y')$
and $f\colon Y' \to X$
continuous with $f(B) = A$.
Take a finer Polish topology $\cT$ on $Y'$
adding no Borel sets,
such that $B$ is clopen with respect to the new topology.
Then let $g = f\defon{B}$
and $Y = (B, \cT\defon{B})$.
(ii) $\implies$ (iii):
Any Polish space is the continuous image of $\cN$.
Let $g_1: \cN \to Y$
and $h \coloneqq g \circ g_1$.
(iii) $\implies$ (iv):
Let $h\colon \cN \to X$ with $h(\cN) = A$.
Let $G(h) \coloneqq \{(a,b) : h(a) = b\} \overset{\text{closed}}{\subseteq} \cN \times X$
be the \vocab{graph} of $h$.
Take $F \coloneqq G(h)^{-1} \coloneqq \{(c,d) | (d,c) \in G(h)\}$
Clearly (iv) $\implies$ (v).
(v) $\implies$ (i):
Take $f \coloneqq \proj_X$.
\end{proof}
\begin{theorem}
Let $X,Y$ be Polish spaces.
Let $f\colon X \to Y$ be \vocab{Borel}
(i.e.~preimages of open sets are Borel).
\begin{enumerate}[(a)]
\item The image of an analytic set is analytic.
\item The preimage of an analytic set is analytic.
\item Analytic sets are closed under countable unions
and countable intersections.
\end{enumerate}
\end{theorem}
\begin{proof}
\begin{enumerate}[(a)]
\item Let $A \subseteq X$ analytic.
Then there exists $Z$ Polish and $g\colon Z \to X$
continuous
with $g(Z) = A$.
We have that $f(A) = (f \circ g)(Z)$
and $f \circ g$ is Borel.
\item Let $f\colon X \to Y$ be Borel
and $B \subseteq Y$ analytic.
Take $Z$ Polish
and $B_0 \subseteq Y \times Z$
such that $\proj_Y(B_0) = B$.
Take $f^+\colon X \times Z \to Y \times Z, f^+ = f \times \id$.
Then
\[f^{-1}(B) = \underbrace{\proj_X(\overbrace{(\underbrace{f^+}_{\mathclap{\text{Borel}}})^{-1}(\underbrace{B_0}_{\mathclap{\text{Borel}}})}^{\text{Borel}})}_{\text{analytic}}.\]
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\item See \yaref{s7e2}.
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\end{enumerate}
\end{proof}
\begin{notation}
Let $X$ be Polish.
Let $\Sigma^1_1(X)$ denote the set of all analytic
subsets of $X$.
$\Pi^1_1(X) \coloneqq \{B \subseteq X : X \setminus B \in \Sigma^1_1(X)\}$
is the set of \vocab{coanalytic} sets.
\end{notation}
We will see later that $\Sigma^1_1(X) \cap \Pi^1_1(X) = \cB(X)$.
\begin{theorem}
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\label{thm:universals11}
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Let $X,Y$ be uncountable Polish spaces.
There exists a $Y$-universal $\Sigma^1_1(X)$ set.
\end{theorem}
\begin{proof}
Take $\cU \subseteq Y \times X \times \cN$
which is $Y$-universal for $\Pi^0_1(X \times \cN)$.
Let $\cV \coloneqq \proj_{Y \times Y}(\cU)$.
Then $\cV$ is $Y$-universal for $\Sigma^1_1(X)$:
\begin{itemize}
\item $\cV \in \Sigma^1_1(Y \times X)$
since $\cV$ is a projection of a closed set.
\item All sections of $\cV$ are analytic.
Let $A \in \Sigma^1_1(X)$.
Let $C \subseteq X \times \cN$
be closed such that $\proj_X(C) = A$.
There is $y \in Y$
such that $\cU_y = C$,
hence $\cV_y = A$.
\end{itemize}
\end{proof}
\begin{remark}
In the same way that we proved
$\Sigma^0_\xi(X) \neq \Pi^0_\xi(X)$ for $\xi < \omega_1$,
we obtain that $\Sigma^1_1(X) \neq \Pi^1_1(X)$.
In fact if $\cU$ is universal for $\Sigma^1_1(X)$,
then $\{y : (y,y) \in \cU\} \in \Sigma^1_1(X) \setminus \Pi^1_1(X)$.
In particular, this set is not Borel.
\end{remark}
\begin{remark}+
Showing that there exist sets that don't have the Baire property
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requires the axiom of choice.
An example of such a set is constructed in \yaref{s5e4}.
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\end{remark}