\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} 
    if
    \[
    \exists Y \text{ Polish}.~\exists  B \in \cB(Y).~
    \exists \underbrace{f\colon Y \to X}_{\text{continuous}}.~
    f(B) = A.
    \] 
\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}
\end{remark}

\begin{theorem}
    \label{thm:borel}
    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
            \[f^{-1}(B) = \underbrace{\proj_X(\overbrace{(\underbrace{f^+}_{\mathclap{\text{Borel}}})^{-1}(\underbrace{B_0}_{\mathclap{\text{Borel}}})}^{\text{Borel}})}_{\text{analytic}}.\]
        \item \todo{Exercise}
    \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}
    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
    requires the axiom of choice.\todo{Copy exercise from sheet 5}
\end{remark}