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In \gls{ZFC}:

\begin{flashcard}[sum-product-defn]
\begin{definition*}[Cardinal sum and product]
  \cloze{Given a set $I$ and \glspl{card} $m_i$, $i \in
  I$, we define
  \[ \sum_{i \in I} m_i = \card \left(
  \bigsqcup_{i \in I} M_i \right) \]
  (here $M_i$ is a set of \gls{cardty} $m_i$, $i
  \in I$, $\bigsqcup_{i \in I} M_i = \bigcup_{i \in
  I} M_i \times \{i\}$). We also define
  \[ \prod_{i \in I} m_i = \card \left( \prod_{i
  \in I} M_i \right) \]
  ($\prod_{i \in I} M_i = \{f : I \to \bigcup_{i
  \in I} M_i \st f(i) \in M_i ~\forall i \in
  I\}$).}
\end{definition*}
\end{flashcard}

\vspace{-1em}
Need \gls{aoc} as we need to be able to choose
$M_i$ for each $i \in I$ and to prove these
operations are well-defined, given $M_i \equivcard
M_i'$, $i \in I$, we need to choose for each $i
\in I$, a bijection $f_i : M_i \to M_i'$, and show
$\bigcup_i M_i \equivcard \bigsqcup_i M_i'$, $\prod_i
M_u \equivcard \prod_i M_i'$.

\begin{example*}
  If $\card I \cle \caleph_\alpha$, $m_i \cle
  \caleph_\aleph$ for all $I \in I$, then $\sum_{i
  \in I} m_i \cle \caleph_\alpha$.
\end{example*}

\begin{note*}
  If $n = \card I$ and $m_i = m ~\forall i \in I$,
  $\prod_{i \in I} m_i = m^n$.
\end{note*}

\vspace{-1em}
If $\alpha \wle \beta$, then
\[ 2^{\caleph_\beta} \cle
\caleph_\alpha^{\caleph_\beta} = 2^{\caleph_\alpha
\caleph_\beta} \cle 2^{\caleph_\beta
\caleph_\beta} = 2^{\caleph_\beta} .\]
So we've reduced to studying $2^{\caleph_\beta}$.
Hard. $\caleph_\alpha \clt 2^{\caleph_\alpha}$, so
$\caleph_0 \clt 2^{\caleph_0} = \card (\RR)$.

\glsnoundefn{CH}{Continuum Hypothesis}{N/A}
Continuum Hypothesis (CH): $2^{\caleph_0} =
\caleph_1$.

Paul Cohen proved: if \gls{ZFC} is \gls{cons},
then so are \gls{ZFC} + \gls{CH} and \gls{ZFC} + $\neg$
\gls{CH}.

\textit{\textbf{THIS IS THE END OF ALL THE EXAMINABLE MATERIAL FOR
THIS COURSE (THE NEXT SECTION IS COMPLETELY NON
EXAMINABLE).}}

\newpage
\section{Classical Descriptive Set Theory
(Non-examinable)}

Study of ``definable sets'' in Polish spaces.
Borel hierarchy, projective hierarchy.

\textbf{Aim:} \gls{CH} holds for analytic sets.

We show that the analogous statement to $P \neq
NP$ holds in this setting.

\begin{definition*}[Polish space]
  A topological space $X$ is a \emph{Polish space} if it
  is separable and complete metrizable.
\end{definition*}

\begin{example*}
  Baire space $\mathcal{N} = \NN^{\NN}$. Basic
  open sets are:
  \[ U_{m_1, \ldots, m_k} = \{\bf{n} = (n_i)_{i =
  1}^\infty \in \mathcal{N} \st n_i = m_i, 1 \le i
  \le k\} .\]
  $d(\bf{m}, \bf{n}) = \sum_{k, m_k \neq n_k}
  2^{-k}$.

  $\{0, 1\}^{\NN} \subset \mathcal{N}$.
\end{example*}

\begin{lemma}
  \label{polish_cts_image}
  Any Polish space is a continuous image of
  $\mathcal{N}$.
\end{lemma}

\begin{proof}
  Let $X$ be a Polish space with complete metric
  $d$. Let $X = \bigcup_{n \in \NN} U_n$, $U_n$
  non-empty and open, $\diam(U_n) < 1$ (since $X$
  separable). Let $U_n = \bigcup_{p \in \NN} U_{n,
  p}$, $U_{n, p}$ non-empty and open, $\diam(U_{n,
  p}) < \half$.

  Continue infinitely, by letting
  \[ U_{n_1, \ldots, n_k} = \bigcup_{n_{k + 1} \in
  \NN} U_{n_1, \ldots, n_{k + 1}} \]
  with $U_{n_1, \ldots, n_{k + 1}}$ always
  non-empty and open, and $\diam < \frac{1}{k +
  1}$.

  Now pick $x_{n_1, \ldots, n_k} \in U_{n_1,
  \ldots, n_k}$. Define
  \begin{align*}
    \phi : \mathcal{N}
    &\to X \\
    \varphi(\bf{n})
    &= \lim_{k \to \infty} x_{n_1, \ldots, n_k}
    \qedhere
  \end{align*}
\end{proof}

\begin{lemma}
  $\mathcal{N}$ is homeomorphic to the set of
  irrationals on $[0, 1]$.
\end{lemma}

\begin{proof}
  Continued fractions (for a definition and some
  properties, see \courseref[Number
  Theory]{NT}).
\end{proof}

\begin{definition*}[Borel hierarchy]
  Let $X$ be a set. A \emph{$\sigma$-field} on $X$ is a
  subset $\mathcal{F} \subset \PP X$ such that
  \begin{enumerate}[(i)]
    \item $\emptyset \in \mathcal{F}$
    \item $A_1, A_2, \ldots \in \mathcal{F}
      \implies \bigcup_{n \in \NN} A_n \in \mathcal{F}$
    \item $A \in \mathcal{F} \implies X \setminus
      A \in \mathcal{F}$
  \end{enumerate}
  If $X$ is a Polish space, then the \emph{Borel
  $\sigma$-field} $\mathcal{B}$ on $X$ is the
  smallest $\sigma$-field on $X$ that contains the
  open sets.
\end{definition*}

\begin{remark*}
  This is a field under the operations of
  symmetric difference and intersection, with
  identity $\emptyset$ (alternatively, it is also
  a field under the operations of symmetric
  difference and union, with identity $X$).
\end{remark*}

\begin{definition*}[$\Sigma_1^0$, $\Pi_1^0$]
  We let $\Sigma_1^0$ be the set of open subsets
  of $X$, and $\pi_1^0$ be the set of closed
  subsets of $X$. We define $\Sigma_\alpha^0$,
  $\Pi_\alpha^1$ for $1 \le \alpha < \omega_1$ b
  recursion:
  \begin{itemize}
    \item $\Sigma_{\alpha + 1}^0$ is the countable
      unions of members of $\Pi_\alpha^0$ (for
      example, $\Sigma_2^0$ are the
      $F_\sigma$-sets).
    \item $\Pi_{\alpha + 1}^0$ are the
      complemenets of membets of $\Sigma_{\alpha +
      1}^0$ (for example, $\Pi_2^0$ are the
      $G_\delta$-sets).
  \end{itemize}
  For $\alpha \neq 0$ \glsref[limord]{limit}:
  \begin{itemize}
    \item $\Sigma_\alpha^0$ consists of sets of
      the form $\bigcup_{n \in \NN} A_n$, where
      $\forall n < \omega$, $\exists \beta <
      \alpha$ with $A_n \in \Pi_\beta^0$.
    \item $\Pi_\alpha^0$ is the complements of
      members of $\Sigma_\alpha^0$.
  \end{itemize}
\end{definition*}

\begin{definition*}[$\Delta_\alpha^0$]
  We define $\Delta_\alpha^0 = \Sigma_\alpha^0
  \cap \Pi_\alpha^0$.
\end{definition*}

\vspace{-1em}
We have:
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/b80675cb506342c6.png}
\end{center}
Prove the $\subset$ property by induction,
starting with $\Sigma_1^0 \subset \Sigma_2^0$,
using the fact that we are a metric space (not
just a topological space).

\begin{proposition}
  $\bigcup_{\alpha \wlt \omegaone} \Sigma_\alpha^0 =
  \bigcup_{\alpha \wlt \omegaone} \Pi_\alpha^0 = \mathcal{B}$
  (the set of Borel sets).
\end{proposition}

\begin{proof}
  First notice:
  \[ \bigcup_{\alpha \wlt \omegaone}
  \Sigma_\alpha^0 =
  \bigcup_{\alpha \wlt \omegaone} \Pi_\alpha^0
  \subset \mathcal{B} .\]
  Need: $\mathcal{F} = \bigcup_{\alpha \wlt
  \omegaone} \Sigma_\alpha^0$ is a $\sigma$-field.
  For example, if $A_n \in \mathcal{F}$, $n \in
  \NN$, then $A_n \in \Pi_{\alpha_n}^0$ for some
  $\alpha_n \wlt \omegaone$. Let $\alpha =
  \sup(\alpha_n + 1)$. Then $\bigcup_n A_n \in
  \Sigma_\alpha^0$ etc.
\end{proof}

\begin{definition*}[Universal subset]
  A subset $A \subset \mathcal{N} \times
  \mathcal{N}$ is a \emph{universal
  $\Sigma_\alpha^0$-set} if:
  \begin{enumerate}[(i)]
    \item $A$ is $\Sigma_\alpha^0$
    \item If $B \subset \mathcal{N}$ is
      $\Sigma_\alpha^0$ then $\exists \bf{m} \in
      \mathcal{N}$, $B = \{\bf{n} \in \mathcal{N}
      \st (\bf{m}, \bf{n}) \in A\}$.
  \end{enumerate}
\end{definition*}

\begin{theorem}
  $\forall \alpha$, $1 \wle \alpha \wlt
  \omegaone$, there exists a universal
  $\Sigma_\alpha^0$ set.
\end{theorem}

\begin{proof}
  \phantom{}
  \begin{enumerate}[$\alpha \wgt 1$:]
    \item[$\alpha = 1$:] Can enumerate the basic
      open set of $\mathcal{N}$ as $U_1, U_2,
      U_3, \ldots$. If $B \subset \mathcal{N}$ is
      open, then $B = \bigcup_{i \in \NN} U_{m_i}$
      for some $\bf{m} = (m_i) \in \mathcal{N}$.
      So $\bf{n} \in B \iff \exists i ~\bf{n} \in
      U_{m_i}$. So define
      \[ A = \{(\bf{m}, \bf{n}) \in \mathcal{N}
      \times \mathcal{N} \st \exists i ~\bf{n} \in
      U_{m_i}\} .\]
      This is open and universal by above.
    \item[$\alpha \wgt 1$:] use induction.
      \qedhere
  \end{enumerate}
\end{proof}

\begin{corollary}
  For every $\alpha$, $1 \wle \alpha \wlt
  \omegaone$, there exiss a set $A \in
  \Sigma_\alpha^0 \setminus \Pi_\alpha^0$.
\end{corollary}

\begin{note*}
  It follows that
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/4b9127c86c3f4359.png}
  \end{center}
\end{note*}

\begin{proof}
  Let $A \subset \mathcal{N} \times \mathcal{N}$
  be a universal $\Sigma_\alpha^0$ set.
  \[ B = \{\bf{n} \in \mathcal{N} \st (\bf{n},
  \bf{n}) \in A\} .\]
  $B$ is $\Sigma_\alpha^0$ ($\bf{n} \mapsto
  (\bf{n}, \bf{n})$ is continuous). If $B$ is
  $\Pi_\alpha^0$ then $\exists \bf{m}$ with $B =
  \{\bf{n} \st (\bf{n}, \bf{n}) \notin A\}$.
  $\bf{m} \in B$? contradiction.
\end{proof}