Some checks failed
Build latex and deploy / checkout (push) Failing after 15m58s
131 lines
4.5 KiB
TeX
131 lines
4.5 KiB
TeX
\lecture{23}{2024-01-25}{Negation of CH is consistent to ZFC}
|
|
|
|
\begin{goal}
|
|
We want to construct a model of $\ZFC$
|
|
such that $2^{\aleph_0} \ge \aleph_2$.
|
|
% = ?
|
|
\end{goal}
|
|
|
|
Let $M$ be a countable transitive model of $\ZFC$.
|
|
Suppose that $M \models \CH$
|
|
(otherwise we are done).
|
|
|
|
Let $\alpha = \omega_2^M$.
|
|
|
|
Let $\bC(\alpha) \coloneqq \{p : p\colon \alpha \to \bC \text{ is a function such that } \{\xi < \alpha : p(\xi) \neq \emptyset\} \text{ is finite} \}$,
|
|
ordered by $p \le_{\bC(\alpha)} q$ iff
|
|
$p(\xi) \le_{\bC} q(\xi)$ for all $\xi < \alpha$.
|
|
|
|
Recall that $\bC$ is the set of finite sequences of natural numbers
|
|
ordered by $p \le_{\bC} q$ iff $p \supseteq q$.
|
|
|
|
Let $g $ be $\bC(\alpha)$-generic over $M$.
|
|
For $\xi < \alpha$ let $x_\xi = \bigcup \{ p(\xi) : p \in g\}$.
|
|
We have already seen that $x_\xi\colon \omega \to \omega$
|
|
is a function
|
|
and $x_\xi \neq x_\eta$ for $\xi \neq \eta$.
|
|
|
|
We have $M[g] \models\ZFC$.\footnote{We only handwaved this step.}
|
|
As $g \in M[g]$,
|
|
we have $\langle x_\xi : \xi < \alpha\rangle \in M[g]$.
|
|
Therefore $M[g] \models \text{``$2^{\aleph_0} \ge \alpha$''}$.
|
|
Also $\alpha = \omega_2^M$.
|
|
However the proof is not finished yet,
|
|
since we need to make sure,
|
|
that $M[g]$ does not collapse cardinals.
|
|
|
|
We only have $M[g] \models 2^{\aleph_0} \ge \aleph_2^M$,
|
|
i.e.~we need to see $\aleph_2^{M[g]} = \aleph_2^M$.
|
|
|
|
\begin{claim}
|
|
Every cardinal of $M$ is still
|
|
a cardinal of $M[g]$.
|
|
\end{claim}
|
|
This suffices, because
|
|
then $\aleph_0^M = \aleph_0^{M[g]}$,
|
|
$\aleph_1^M = \aleph_1^{M[g]}$,
|
|
$\aleph_2^{M}= \aleph_2^{M[g]}$, $\ldots$
|
|
|
|
\begin{definition}
|
|
Let $(\bP, \le )$ be a partial order.
|
|
We say that $\bP$ has the
|
|
\vocab{countable chain condition}
|
|
(\vocab{c.c.c.})%
|
|
\footnote{it should really be the ``countable antichain condition''}
|
|
iff there is no uncountable
|
|
antichain,
|
|
i.e.~every uncountable $V \subseteq \bP$
|
|
contains compatible $p \neq q$.
|
|
\end{definition}
|
|
|
|
We shall prove:
|
|
\begin{claim}
|
|
\label{l23:c:1}
|
|
For all $\beta$, $\bC(\beta)$ has the c.c.c.
|
|
\end{claim}
|
|
\begin{claim}
|
|
\label{l23:c:2}
|
|
If $\bP \in M$ and
|
|
$M \models \text{`` $\bP$ has the c.c.c.''}$
|
|
and $h$ is generic over $M$,
|
|
then all $M$-cardinals are still
|
|
$M[h]$ cardinals.
|
|
\footnote{Being a cardinal is $\Pi_1$,
|
|
so $M[h]$ cardinals are always $M$ cardinals.}
|
|
\end{claim}
|
|
\begin{yarefproof}{l23:c:2}
|
|
Suppose not.
|
|
Let $\kappa$ be minimal such that $M \models \text{``$\kappa$ is a cardinal''}$,
|
|
but $M[h] \models \text{``$\kappa$ is not a cardinal''}$.
|
|
Then $\kappa = (\lambda^+)^M$ for some unique $M$-cardinal $\lambda < \kappa$.
|
|
By minimality, $\lambda$ is also an $M[h]$-cardinal.
|
|
|
|
Let $f \in M[h]$ be such that $M[h] \models \text{``$f$ is a surjection from $\lambda$ onto $\kappa$''}$.
|
|
There is a name $\tau \in M^{\bP}$ with $\tau^h = f$.
|
|
|
|
We then have some $p \in h$
|
|
with $p \Vdash_M^\bP \text{``$\tau$ is a surjection from $\check{\lambda}$ onto $\check{\kappa}$''}$.
|
|
|
|
Let $\xi < \lambda$.
|
|
Consider $X_\xi \coloneqq \{\eta < \kappa: \exists q \le_{\bP} p .~q \Vdash \tau(\check{\xi}) = \check{\eta}\} \in M$.
|
|
|
|
$X_\xi$ is countable in $M$ by the following argument (in $M$):
|
|
For every $\eta \in X_\xi$,
|
|
let $q_\eta \le p$ be such that $q_\xi \Vdash^{\bP}_M \tau(\check{\xi}) = \check{\eta}$.
|
|
The set $\{q_\eta : \eta \in X_\xi\}$ is an antichain
|
|
as for $\eta_1 \neq \eta_2$ we have that $q_{\eta_i} \Vdash \tau(\check{\xi}) = \check{\eta_i}$,
|
|
so they are not compatible.
|
|
So $\{q_\eta : \eta \in X_\xi\}$ is countable by the c.c.c.
|
|
Thus $X_\xi$ is countable.
|
|
|
|
|
|
|
|
Therefore we may define a function in $M$
|
|
\begin{IEEEeqnarray*}{rCl}
|
|
F\colon \lambda \times \omega &\longrightarrow & \kappa
|
|
\end{IEEEeqnarray*}
|
|
such that for all $\xi < \lambda$
|
|
\[
|
|
\{F(\xi,n) : n < \omega\} = X_\xi.
|
|
\]
|
|
|
|
$F$ is surjective since $f$ is surjective:
|
|
For $\eta < \kappa$,
|
|
there is some $\xi < \lambda$ such that $M[h] \models \text{``$f(\xi) = \eta$''}$,
|
|
there is some $\overline{q} \in h$ with $\overline{q} \Vdash ^{\bP}_M \tau(\check{\xi}) = \check{\eta}$.
|
|
Pick $q \le \overline{q},p$.
|
|
This shows $\eta \in X_\xi$
|
|
hence $\eta = F(\xi, n)$ for some $n$.
|
|
But $|\lambda \times \omega| = |\lambda| = \lambda$,
|
|
so in $M$ there is a surjection $F' \colon \lambda \to \kappa$,
|
|
but $\kappa$ is a cardinal in $M$ $\lightning$.
|
|
\end{yarefproof}
|
|
|
|
\begin{yarefproof}{l23:c:1}
|
|
Omitted.
|
|
% TODO combinatorial argument
|
|
\end{yarefproof}
|
|
|
|
|
|
|