%! TEX root = LST.tex
% vim: tw=50
% 12/03/2024 09AM

\subsubsection*{Projective Hierarchy}

\begin{definition*}[Analytic st]
  An \emph{analytic set} (in a Polish space) is
  the continuous image of $\mathcal{N}$.
\end{definition*}

\begin{example*}
  Every Polish space (by \cref{polish_cts_image}).
  Every closed subset of Polish space.
\end{example*}

\begin{proposition}
  \label{analytic_tfae}
  Let $A \subset X$, $X$ Polish. Then the
  following are equivalent:
  \begin{enumerate}[(i)]
    \item $A$ is analytic.
    \item $A$ is a continuous image of a Borel
      set.
    \item $A$ is the projection onto $X$ of some
      Borel subset of $Y \times X$, $Y$ Polish.
    \item $A$ is the projection onto $X$ of some
      closed subset of $Y \times X$, $Y$ Polish.
    \item $A$ is the projection onto $X$ of some
      Borel subset of $\mathcal{N} \times X$.
    \item $A$ is the projection onto $X$ of some
      closed subset of $\mathcal{N} \times X$.
  \end{enumerate}
\end{proposition}

\begin{note*}
  $\mathcal{N} = \NN^{\NN}$, $\mathcal{N} \times
  \mathcal{N} = \NN^{\NN \sqcup \NN}$ homeomorphic
  to $\mathcal{N}$, and $\mathcal{N}^{\NN} = \NN^{\NN
  \times \NN}$ is also homeomorphic to $\NN$.
\end{note*}

\begin{proof}
  Enough to show (ii) $\Rightarrow$ (I) $\Rightarrow$
  (vi).
  \begin{enumerate}[(i) $\Rightarrow$ (vi)]
    \item[(i) $\Rightarrow$ (vi)] $A = f(\mathcal{N})$, $f$ closed.
      $A$ is the projection onto $X$ of
      \[ \{(\bf{n}, f(\bf{n})) \st \bf{n} \in
      \mathcal{N}\} \]
      which is closed.
    \item[(ii) $\Rightarrow$ (i)] Need: Borel
      $\implies$ analytic. Enough: every Borel set
      satisfies (vi). $\Pi_1^0$ is a subset of the
      sets satisfying (vi). Need that the set of
      sets satisfying (vi) is closed under
      countable union and intersection. Assume
      $A_n$ is the projection of $F_n \subset
      \mathcal{N} \times X$, $F_n$ closed. So $x
      \in A_n$ $\iff$ $\exists \bf{n} \in
      \mathcal{N}, (\bf{n}, x) \in F_n$. Then
      \[ x \in \bigcup_n A_n \iff \exists n \in
      \NN ~\exists \bf{n} \in \mathcal{N}~~
      (\bf{n}, x) \in F_n .\]
      Let
      \[ F = \{(n, \bf{n}, x) \in \ub{\NN \times
      \mathcal{N}}_{\mathcal{N}} \times X \st (\bf{n}, x \in
      F_n)\} \]
      which is closed and projects onto $\bigcup_n
      A_n$.

      For intersection: $x \in \bigcap_n A_n$ if
      and only if $\forall n ~\exists \bf{n},
      (\bf{n}, x) \in F_n$. Then
      \[ G = \{(\bf{n}_1, \bf{n}_2, \bf{n}_3,
      \ldots, x) \in \mathcal{N}^{\NN} \times X
      \st (n_i, x) \in F_i ~\forall i\} \]
      is closed and projects onto $\bigcap_n A_n$.
      \qedhere
  \end{enumerate}
\end{proof}

\begin{definition*}[$\Sigma_n^1$, $\Pi_n^1$]
  Let $\Sigma_1^1$ be the set of analytic sets.
  Let $\Pi_1^1$ be the set of coanalytic sets,
  i.e. complements of analytic sets. For $1 \le n
  \le \omega$, let $\Sigma_{n + 1}^1$ be the
  continuous image of $\Pi_n^1$ sets. Let $\Pi_{n
  + 1}^1$ be the complements of $\Sigma_{n + 1}^1$
  sets.
\end{definition*}

\vspace{-1em}
As before:
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/db7990cfe8ab4592.png}
\end{center}
projective hierarchy
\[ P = \bigcup_{1 \le n < \omega} \Sigma_n^1 =
\bigcup_{1 \le n < \omega} \Pi_n^1 .\]

\begin{theorem}
  There exists a universal analytic set $A \subset
  \mathcal{N} \times \mathcal{N}$.
\end{theorem}

\begin{proof}
  Let $U$ be a universal open set in $\mathcal{N}
  \times (\mathcal{N} \times \mathcal{N})$. So if
  $V \subset \mathcal{N} \times \mathcal{N}$ is
  open then there exists $\bf{p} \in \mathcal{N}$
  such that
  \[ V = \{(\bf{m}, \bf{n}) \in \mathcal{N}
  \times \mathcal{N} \st (\bf{p}, \bf{m}, \bf{n})
  \in U\} .\]
  Suppose $B \subset \mathcal{N}$ is analytic. So
  there exists closed $F \subset \mathcal{N}
  \times \mathcal{N}$ such that
  \[ B = \{\bf{n} \in \mathcal{N} \st \exists
  \bf{m} \in \mathcal{N}, (\bf{m}, \bf{n}) \in F\}
  .\]
  So $\exists \bf{p} \in \NN$ such that
  \[ B = \{\bf{n} \in \mathcal{N} \st \exists
  \bf{m} \in \mathcal{N}, (\bf{p}, \bf{m}, \bf{n})
  \notin U\} .\]
  Let
  \[ A = \{(\bf{r}, \bf{s}) \in \mathcal{N} \times
  \mathcal{N} \st \exists \bf{m} \in \mathcal{N},
  (\bf{r}, \bf{m}, \bf{s}) \notin U\} \]
  This is a projection of a closed set, so
  analytic.
  \[ B = \{\bf{n} \in \mathcal{N} \st (\bf{p},
  \bf{n}) \in A\} . \qedhere \]
\end{proof}

\begin{corollary}
  There exists an analytic, not coanalytic set in
  $\mathcal{N}$.
\end{corollary}

\begin{proof}
  Let $A \subset \mathcal{N} \times \mathcal{N}$
  be a universal analytic set, and $B = \{\bf{n}
  \in \mathcal{N} \st (\bf{n}, \bf{n}) \in A\}$
  analytic. If $B$ is coanalytic, then
  \[ \exists \bf{m} \in \mathcal{N} \qquad B =
  \{\bf{n} \in \mathcal{N} \st (\bf{m}, \bf{n})
  \notin A\} .\]
  Is $\bf{m} \in B$? No, contradiction.
\end{proof}

\begin{remark*}
  So $B \in \Sigma_1^1 \setminus \Pi_1^1$, so $B$
  is not Borel (``$P \neq NP$'').
\end{remark*}

\vspace{-1em}
\textbf{Aim:} $\Sigma_1^1 \cap \Pi_1^1 =
\mathcal{B}$. ``$\supset$'' is
\cref{analytic_tfae}.

\begin{theorem}[Lusin's Separation Theorem]
  If $A_1, A_2$ are disjoint analytic sets, then
  there exists a orel set $B$, $A_1 \subset B$,
  $A_2 \subset X \setminus B$.
\end{theorem}

\begin{proof}
  First: if $Y = \bigcup_n Y_n$, $Z = \bigcup_n
  Z_n$ and $\forall m, n$ $Y_m, Z_n$ can be
  separated by Borel sets, then so can $Y$, $Z$.
  So for all $m, n$, find $Y_m \subset B_{m, n}
  \subset X \setminus Z_n$, $B_{m, n}$ Borel. Then
  \[ B = \bigcup_m \bigcap_n B_{m, n} \]
  is Borel, and $Y \subset B \subset X \setminus
  Z$. Now suppose $f, g$ are continuous and
  $f(\mathcal{N})$, $g(\mathcal{N})$ are disjoint,
  but cannot be separated. Recall
  \[ U_{m_1, m_2, \ldots, m_k} = \{\bf{n} \in
  \mathcal{N} \st n_i = m_i, 1 \le i \le k\} \]
  is our notation for the basic open sets in
  $\mathcal{N}$. $f(\mathcal{N}) = \bigcup_n
  f(U_n)$, $g(\mathcal{N}) = \bigcup_n g(U_n)$.
  There exists $m_1, n_1$ such that $f(U_{m_1})$,
  $g(U_{n_1})$ cannot be separated. Inductively,
  we get $\bf{m}, \bf{n} \in \mathcal{N}$ such
  that for all $\bf{m}, \bf{n} \in \mathcal{N}$,
  $f(U_{m_1, \ldots, m_k})$, $g(U_{n_1, \ldots,
  n_k})$ cannot be separated. But $\mathcal{N}$ is
  Hausdorff (and in fact we can separate points
  using the basic open sets $U$), which gives a
  contradiction.
\end{proof}

\begin{corollary}
  $\Sigma_1^1 \cap \Pi_1^1 = \mathcal{B}$.
\end{corollary}

\begin{example*}
  Let $\Sigma = \bigcup_{k \in \NN_0} \NN^k$. $s,
  t \in \Sigma$, we write $s \prec t$ if $s =
  (n_1, \ldots, n_j)$, $t = (n_1, \ldots, n_i)$, $0
  \le j \le i$. $s \in \Sigma$, $\bf{n} \in
  \mathcal{N}$, $s \prec \bf{n}$ if $s = (n_1,
  \ldots, n_j)$, $j \in \NN_0$. $\PP \Sigma = \{0,
  1\}^{\Sigma}$ Polish space. $T \subset \Sigma$
  is a \emph{tree} if $s \prec t$, $t \in T
  \implies s \in T$. $T$ is \emph{well-founded} if
  $\not\exists \bf{n} \in \mathcal{N}$ such that
  $\forall i$, $(n_1, \ldots, n_i) \in T$.
  \[ WFT = \{T \subset \Sigma \st \text{$T$ is
  well-founded}\} \]
\end{example*}

\vspace{-1em}
A subset $A$ of a Polish space is \emph{perfect}
if $A$ is closed and contains no isolated points.
($x \in A$ is \emph{isolated} if $\exists r > 0$,
$B_r(x) \cap A = \{x\}$).

\begin{lemma}
  $A \neq \emptyset$ perfect set has cardinality
  $2^{\aleph_0}$.
\end{lemma}

\begin{proof}
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/1df767c5d9c14235.png}
  \end{center}
  Closed balls, disjoint, diameter $<1$, centres
  in $A$.
  $\{0,1\}^{\NN} \hookrightarrow A$, so $\card A
  \ge 2^{\aleph_0}$. $\card A \le \card
  \mathcal{N} = \aleph_0^{\aleph_0} = 2^{\aleph_0}$.
\end{proof}

\begin{theorem}
  An analytic set is either countable or contains
  a non-empty perfect set. So \gls{CH} holds for
  analytic sets.
\end{theorem}

\vspace{-1em}
$f(\mathcal{N})$. $T$ tree
\[ [T] = \{\bf{n} \in \mathcal{N} \st (n_1,
\ldots, n_i) \in T ~\forall i\} .\]
$[\Sigma] = \mathcal{N}$, $s \in \Sigma$,
\[ T(s) = \{t \in \Sigma \st t \prec s \text{ or }
s \prec t\} .\]
$T^{(0)} = \Sigma$,
\[ T^{(\alpha + 1)} =
(T^{(\alpha)})' = \{s \in T^{(\alpha)} \st
f([T^{(\alpha)}(s)]) \text{ is uncountable}\} .\]
\[ T^{(\lambda)} = \bigcap_{\alpha < \lambda}
T^{(\alpha)} .\]
$\exists \alpha < \omegaone$, $T^{(\alpha)} =
T^{(\alpha + 1)}$, ($\Sigma$ countable). Either
$T^{(\alpha)} = \emptyset$ implies
$f(\mathcal{N})$ countable or $T^{(\alpha)} \neq
\emptyset$. Find a copy of $\{0, 1\}^\NN \subset
[T^{(\alpha)}]$. The image of $\{0, 1\}^\NN$ is
perfect.