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\begin{flashcard}[upward-lowenheim-skolem-thm]
\begin{corollary}[Upward L\"owenheim-Skolem Theorem]
  \label{upward_ls}
  \cloze{Let $S$ be a first-order \gls{theory}. If $S$
  has an infinite \gls{model}, then $S$ has an
  uncountable \gls{model}.}
\end{corollary}

\begin{proof}
  \cloze{We introduce an uncountable set of new constants
  $\{c_i \st i \in I\}$ to the language. We let
  \[ S' = S \cup \{\pnot c_i = c_j \st i, j \in I,
  i \neq j\} .\]
  Let $A$ be an infinite \gls{model} of $S$. Then $A$ is
  a \gls{model} of any finite subset of $S'$. By
  \nameref{fo_compactness}, $S'$ has a
  \gls{model}.

  A \gls{model} of $S'$ is a \gls{model} $B$ of
  $S$ together with an injection $I \to B$. So $B$
  is uncountable.}
\end{proof}
\end{flashcard}

\begin{remark*}
  For any set $X$, can take $I = \hartgamma(X)$
  (from \nameref{hartogs}). The proof above
  shows that $S$ has a \gls{model} $B$ with an
  injection $I \to B$. So then there will be no
  injection $B \to X$.
\end{remark*}

\begin{flashcard}[downward-lowenheim-skolem-thm]
\begin{corollary}[Downward L\"owenheim-Skolem
  Theorem]
  \label{downward_ls}
  \cloze{Let $S$ be a \gls{cons} first-order \gls{theory}
  in a countable \gls{folang} ($\pOmega, \pPi$ are
  countable). Then if $S$ has a \gls{model}, then $S$
  has a countable \gls{model}.}
\end{corollary}

\begin{proof}
  \cloze{Since $S$ is \gls{cons} (by \nameref{fo_soundness}),
  the proof of \cref{fo_model_existence} builds a
  countable \gls{model} (since the \gls{folang} is
  countable).}
\end{proof}
\end{flashcard}

\subsection{Peano Arithmetic}

\glssymboldefn{pas}{$s$}{$s$}
\glsnoundefn{PA}{PA}{N/A}
We want to axiomatise $\NN$ as a first-order
\gls{theory}. \glsref[folang]{Language}:
\[ \Omega = \{0, s, +, \times\}, \qquad \pPi =
\emptyset \]
with \glspl{arity} $0, 1, 2, 2$. $s$ means
``successor'', and the others are clear.

Axioms of Peano Arithmetic (PA):
\begin{align*}
  &(\pforall x)(\pnot \pas x = 0) \\
  &(\pforall x)(\pforall y)(\pas x = \pas y \pimpl x = y)
  \\
  &(\pforall x)(x \times 0 = 0) \\
  &(\pforall x)(\pforall y)(x \times (\pas y) = (x
  \times y) + x) \\
  &(\pforall t_1) \cdots (\pforall t_n)[(\subst
  p[0 / x] \pand (\pforall x)(p \pimpl \subst
  p[\pas x / x])) \pimpl (\pforall x) p]
\end{align*}
\glsnoundefn{paind}{induction}{N/A}
where the last \gls{sent} is for every \gls{form}
$p$ with $\FV(p) = \{x, t_1, \ldots, t_n\}$. This
is the axiom-scheme for induction.

\begin{remark*}
  Let $p$ be the \gls{form} $x + (y + z) = (x + y)
  + z$. Then you can prove in \gls{PA} that
  $(\pforall x)(\pforall y)(\pforall z) p$ by
  \gls{paind} on $z$ with $x, y$ parameters. You
  \glsref[fproof]{prove}:
  \[ (\pforall x)(\pforall y)(\subst p[0 / z]
  \pand (\pforall z)(p \pimpl \subst p[\pas z /
  z])) \]
\end{remark*}

\begin{note*}
  $\NN_0 = \{0\} \cup \NN$ is a \gls{model} of
  \gls{PA}. We can also interpret $\NN$ as a
  \gls{model} of \gls{PA} by taking a bijection
  with $\NN_0$ (but this would be rather unnatural
  to do).

  By \nameref{upward_ls}, there are uncountable
  \glspl{model} of \gls{PA}. Didn't we lean
  $\NN_0$ is uniquely determined by its
  properties? Yes, but \emph{true} induction says:
  \[ (\forall A \subset \NN_0) ((0 \in A \wedge
  (\forall x)(x \in A \implies \pas x \in A))
  \implies A = \NN_0) \]
  In first-order \gls{theory}, we cannot quantify
  over subsets of \glspl{struc}. The axiom scheme
  for induction captures only countably many
  subsets of $\NN_0$.
\end{note*}

\begin{flashcard}[definable-set-defn]
\begin{definition*}[Definable set]
  \cloze{A subset $A$ of $\NN_0$ is \emph{definable} if
  there's a \gls{form} $p$ in \gls{folang} of
  \gls{PA} with \gls{freev} \gls{var} $x$ such
  that $p_{\NN_0} = A$, i.e.
  \[ \{a \in \NN_0 \st \text{$a$ satisfies $p$}\}
  = A .\]}
\end{definition*}
\end{flashcard}

\begin{example*}
  Set of primes: use
  \[ p = (\pforall y)((\pexists z)(y \cdot z = x)
  \pimpl (y = \ub{1}_{=\pas 0} \por y = x) \]
  Powers of $2$: use
  \[ p = (\pforall y)(((y \mid x) \pand (\text{$y$
  is a prime})) \pimpl y = \ub{2}_{= \pas \pas 0}) \]
\end{example*}

\vspace{-1em}
A consequence of G\"odel's Incompleteness Theorem:
there exists a \gls{sent} $p$ such that $p$ holds
in $\NN_0$, but $\text{\gls{PA}} \not \ssynentails
p$.

\newpage
\section{Set Theory}
\label{sec5}

We will describe set theory as just another
example of first-order \gls{theory}. We want to
understand what the ``universe of sets'' looks
like.

\subsubsection*{Zermelo-Frankel Set Theory (ZF)}

\glsnoundefn{ZF}{ZF}{N/A}
\glssymboldefn{zin}{$\in$}{$\in$}
\glsref[folang]{Language}: $\pOmega = \emptyset$,
$\pPi = \{\in\}$, $\in$ has \gls{arity} $2$.

A \gls{struc} is a set $V$ together with $[\in]_V
\subset V \times V$.

An element of $V$ is called a ``set''. If $a, b
\in V$ and $(a, b) \in [\zin]_V$, we say ``$a$
belongs to $b$'' or ``$a$ is an element of $b$''.
$V$ will be the ``universe of sets'' (when $V$ is
a \gls{model} of \gls{ZF}).
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/741ae34fd8af4c70.png}
\end{center}

There will be $2 + 4 + 3$ axioms of \gls{ZF}.

\begin{enumerate}[(1)]
  \item \begin{flashcard}[extensionality-ax]
      \glsnoundefn{Ext}{(Ext)}{N/A}
      \prompt{\cloze{(1) }}\textbf{\cloze{Axiom of
      Extensionality} (Ext):} \cloze{``If two sets have the same
      members, then they are equal''.
      \[ (\pforall x)(\pforall y)((\pforall z)(z
      \zin x \iff z \zin y) \pimpl x = y) \]}
    \end{flashcard}
  \item
    \begin{flashcard}[separation-ax]
      \glsnoundefn{Sep}{(Sep)}{N/A}
      \prompt{\cloze{(2) }}\textbf{\cloze{Axiom of
      Separation} (Sep):} \cloze{``We can form subsets of
      a set.''
      \[ (\pforall t_1) \cdots (\pforall t_n)
      [(\pforall x)(\pexists y)(\pforall z)(z \zin
      y \iff (z \zin x \pand p))] ,\]
      where $p$ is any \gls{form} with $\FV(p) =
      \{z, t_1, \ldots, t_n\}$. By \gls{Ext}, the
      set $y$ whose existence is asserted is
      unique. We denote it by $\{z \zin x \st
      p\}$. (Formally, we introduce an $(n +
      1)$-ary operation symbol to the
      \gls{folang}; informally, this is an
      abbreviation).}
    \end{flashcard}
    \begin{example*}
      \vspace{1em}
      Given $t, x$, we can form $\{z \zin x \st t
      \zin z\}$.
    \end{example*}
\end{enumerate}
