%! TEX root = FC.tex
% vim: tw=50
% 25/01/2025 12PM

\newpage
\section{The Continuum Hypothesis}

1398: G\"odel showed $\Con(\ZFC) \implies
\Con(\ZFC + \CH)$.

1962: Cohen showed $\Con(\ZFC)
\implies \Con(\ZFC + \neg\CH)$.

\begin{theorem*}[Hartogs's Theorem]
  For every set $X$, there is a (least) cardinal
  $\alpha$ such that there is no injection from
  $\alpha$ to $X$.
\end{theorem*}

We denote this by Hartogs's aleph of $X$,
$\aleph(X)$.

\begin{theorem*}
  For every set $X$, there is no injection from
  $\mathcal{P}(X)$ into $X$. We denote this by
  $2^{|X|}$.
\end{theorem*}

Using Axiom of Choice, well-order $\mathcal{P}(X)$
and get ordinal $2^{|X|}$.

We have
\[
  \aleph(X) \le 2^{|X|}
.\]
\begin{notation*}
  Define:
  \begin{align*}
    \aleph_0
    &\defeq \Nbb \\
    \aleph_{\alpha + 1}
    &\defeq \aleph(\aleph_\alpha) \\
    \aleph_\lambda
    &\defeq \bigcup_{\alpha < \lambda}
    \aleph_\alpha \\
    \beth_0
    &= \Nbb \\
    \beth_{\alpha + 1}
    &= 2^{\beth_\alpha} \\
    \beth_\lambda
    &= \bigcup_{\alpha < \lambda} \beth_\alpha
  \end{align*}
  Clearly, $\aleph_\alpha \le \beth_\alpha$.
\end{notation*}

Continuum Hypothesis (CH): $\aleph_1 = \beth_1$.

Generalised Continuum Hypothesis (GCH): $\forall \alpha
\aleph_\alpha = \beth_\alpha$.

Why ``continuum''?

\begin{lemma*}
  \begin{iffc}
    \lhs
    CH
    \rhs
    $\forall X \subseteq \Rbb$, $X$ is uncountable
    $\implies$ $X \sim \Rbb$ ($\sim$ means ``there
    is a bijection'').
  \end{iffc}
\end{lemma*}

\begin{iffproof}
  \rightimpl
  Obvious since $\beth_1 = |\Rbb|$.
  \leftimpl
  Suppose $|\Rbb| > \aleph_1$. Well-order the
  reals in order type $\kappa > \aleph_1$:
  \[
    \{r_\alpha : \alpha < \kappa\}
  .\]
  Consider $X \defeq \{r_\alpha : \alpha < w_1\}
  \subseteq \Rbb$. This is a subset of the reals
  of cardinality $\aleph_1$, hence is an
  uncountable subset which is not in bijection
  with it.
\end{iffproof}

Reminder:

G\"odel showed $\Con(\ZFC) \implies
\Con(\ZFC + \CH)$.

Cohen showed $\Con(\ZFC)
\implies \Con(\ZFC + \neg\CH)$.

Relative consistency proofs.

By Completeness Theorem, this means:

If there is $M \models \ZFC$, then I can
construct $N \models \ZFC +
(\neg) \CH$.

Analogy from algebra:
\[
  \mathcal{L}_{\text{fields}}
  = \{0, 1, +, \cdot, -, {}^{-1}\}
.\]
Axioms of fields: Fields.

Let $\varphi_{\sqrt{2}} \defeq \exists x (x \cdot
x = 1 + 1)$. Note $\Qbb \models \neg
\varphi_{\sqrt{2}}$.

\textbf{Idea:} Start with $\Qbb$ and extend $\Qbb$
to get $F \models \text{Fields} +
\varphi_{\sqrt{2}}$.

Construct by \emph{algebraic closure} (not in the
usual sense -- here we just mean adding in
$\sqrt{2}$ and then adding everything else that
this requires us to add).

Obtain $\Qbb(\sqrt{2}) \models \text{Fields} +
\varphi_{\sqrt{2}}$.

This is easy because everything that matters
(Fields and $\varphi_{\sqrt{2}}$) is determined by
equations; all formulas we need to check are
atomic.

\begin{fcdefn}[absolute]
  \glsadjdefn{abs}{absolute}{formula}%
  \glsadjdefn{uabs}{upwards absolute}{formula}%
  \glsadjdefn{dabs}{downwards absolute}{formula}%
  If $M \subseteq N$ and $M, N$ are
  $\mathcal{L}$-structures and $\varphi$ an
  $\mathcal{L}$-formula, then we say
  \emph{$\varphi$ is absolute between $M$ and $N$}
  if for all $x_1, \ldots, x_m \in M$,
  \[
    M \models \varphi(x_1, \ldots, x_n)
    \iff N \models \varphi(x_1, \ldots, x_n)
  .\]
  If the $\Rightarrow$ direction holds, then we
  say ``upwards absolute'', and if the
  $\Leftarrow$ direction holds, then we say
  ``downwards absolute''.
\end{fcdefn}

\begin{theorem*}[Substructure Lemma]
  All atomic formulas are \gls{abs} between
  substructures.
\end{theorem*}

WHat if we have models of ZFC? Have
\[
  \mathcal{L}_\in = \{\in\}
.\]
No function symbols nor constant symbols. So:
almost nothing is atomic.

$M \subseteq N$ if and only if $M$ is an
$\mathcal{L}_\in$-substructure of $N$.

And: the formulas that we care about are
definitely not atomic, but instead very complex.

Try to imagine a proof of:

If $M \models \ZFC$ then there is $N
\supseteq M$ such that $N \models \ZFC +
\CH$.

Without loss of generality $M \models \neg
\CH$ (else we are done). For simplicity,
let's consider the case $\beth_1 = \aleph_2$.
\begin{center}
  \includegraphics[width=0.1\linewidth]{images/1c3d0e1625ef4d79.png}
\end{center}
What can we do to ``get rid of $\aleph_1$''?

Maybe a surjection $f : \Nbb \to \aleph_1$. Maybe
we can form $M[f] \supseteq M$ to get a smallest
model $M[f] \models \ZFC$.

Clearly, in $M[f]$, $\aleph_1^M$ is not a cardinal
anymore.

Does that show CH?

All sorts of things can happen

Assuming it is actually possible to form this
smallest model $M[f]$, there are lots of ways that
this might not end up being enough to deduce CH.
For example:
\begin{itemize}
  \item Maybe $\Rbb^{M[f]} \neq \Rbb^M$
  \item Maybe $\aleph_2^M$ is not a cardinal
    either
\end{itemize}

\subsubsection*{A fundamental problem of
non-absoluteness}

Consider $\varphi_\emptyset(x) \defeq \forall z(z \notin
x)$, which means ``$x$ is empty''.

Consider $M \models \ZFC$. Therefore there
are $e, e'$ such that $M \models
\varphi_\emptyset(e)$ and $M \models \forall z(z
\in e' \iff z = e)$.

Consider $N \defeq M \setminus \{e\}$.

$N$ is an $\mathcal{L}_\in$-substructure of $M$.
But $N \models \varphi_\emptyset(e')$, even though
$M \models \neg \varphi_\emptyset(e')$.

So $\varphi_\emptyset$ is not \gls{abs} between
substructures.

Instead of substructures, we will restrict out
attention to \emph{transitive substructures}: in
particular, to $M$ transitive ($\forall x, x \in M
\implies x \subseteq M$ or equivalently $x \in M
\wedge y \in x \implies y \in M$) such that $M
\models \ZFC$.

Next time: theorems about absoluteness for
transitive substructures.