w23-logic-2/inputs/lecture_20.tex

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2024-01-15 15:47:24 +01:00
\lecture{20}{2024-01-15}{}
\begin{idea}
We want to add a new object that satisfies certain condition.
The elements of the forcing notion correspond to approximations
of this object.
A filter picks some information which we want to be true.
Being a filter ensures that this information does not contradict itself.
\end{idea}
\begin{definition}
Assume that $M$ is a transitive model of $\ZFC$,
and $\bP \in M$ a poset.
$G \subseteq \bP$ is said to be \vocab{$M$-generic for $\bP$}
if whenever $D \subseteq \bP$ is dense
and in $M$, then $G \cap D \neq \emptyset$.
\end{definition}
\begin{remark}
That is the same as being $\{D \subseteq \bP \text{ dense } : D \in M\}$-generic
with generic defined as in \yaref{def:forcingwords}.
\end{remark}
% \begin{remark}
% We can not prove that such a transitive model actually exists (we need inaccessible cardinals for that).
% However there are transitive models of finite fragments of $\ZFC$.
% % cf. Kuhnen, Chapter 7
% So proofs using tansitive models can be made more general by
% talking only about finite fragments, i.e.~a finite subset of the axioms of $\ZFC$ (since every proof can only
% use finitely many axioms).
% \end{remark}
\begin{definition}[Cohen Forcing]
\yalabel{Cohen Forcing}{Cohen Forcing}{def:cohenf}
Let $\bP$ be the set of finite partial function $p$ from $\omega$ to $2$,
i.e.~$\cP = 2^{<\omega}$.
The order on $\bP$ is described by $q \le p \colon\iff q \supseteq p$.
$\bP$ is called the \vocab{Cohen forcing}.
\end{definition}
\begin{fact}
Assume $X \subseteq 2^\omega$ is countable,
Then there is $x \in 2^\omega \setminus X$.
\end{fact}
Of course we already know that, but let's use it
to test our machinery:
\begin{proof}
Assume that $X = \{x_n \colon n \in \omega\}$
is an enumeration of $X$.
Let $D_n = \{ p \in \bP : \exists i \in \dom(\bP).~x_n(i) \neq p(i)\}$.
This makes sure that we get a ``new'' element
not belonging to $X$.
\begin{claim}
$D_n$ is dense in $\bP$.
\end{claim}
\begin{subproof}
Assume $q \in \bP$.
Let $i = 1 + \max(\dom(q))$.
Note that $i \not\in \dom(q)$.
Let $p = q \cup \{(i, 1-x_n(i))\}$.
Then $p \in D_n$.
\end{subproof}
Let $E_i = \{p \in \bP : i \in \dom(\bP)\}$.
This makes sure that our ``new'' element
is defined everywhere.
\begin{claim}
$\forall i < \omega.~E_i \subseteq \bP \text{ is dense}$.
\end{claim}
\begin{subproof}
Assume $q \in \bP$.
If $ i \in \dom(q)$ pick $p = q \in E_i$.
If $i \not\in \dom(q)$,
let $p = q \cup \{(i,0)\} \in E_i$.
\end{subproof}
Let $\cD = \{D_n : n < \omega\} \cup \{E_i : i < \omega\}$.
This is a countable subset of dense sets.
By \yaref{lem:egenfilter}
there is a $\cD$-generic filter $G$.
Let $y = \bigcup G$.
Note that $y$ is a function, since any two elements of $G$ are compatible.
\end{proof}
Note that the ``new'' element did already exists, so we used forcing language
to find it but didn't actually do anything.
\begin{lemma}
Let $M$ be a transitive model of $\ZFC$
and let $\bP = (P, \le ) \in M$.
Let $D \subseteq \bP$, $D \in M$, $p \in \bP$.
Then
\begin{enumerate}[(1)]
\item $\bP$ is a partial order iff $M \models \text{``$\bP$ is a partial order''}$.
\item $D$ is dense in $\bP$ iff $M \models \text{``$D$ is dense in $\bP$''}$.
\item $D$ is dense below $p$ iff $M \models \text{``$D$ is dense below $p$''}$
(this only makes sense if $p \in M$).
\end{enumerate}
\end{lemma}
\begin{proof}
All the definitions are $\Delta_0$, so we can apply
\yaref{lem:d0absolute}.
\end{proof}
\begin{definition}
Assume that $M$ is a transitive model of $\ZFC$
and $\bP \in M$ is a poset.
$G \subseteq \bP$ is called a
\vocab{$\bP$-generic filter over $M$}
or \vocab{$M$-generic filter for $\bP$}
if
\[
\forall D \in M.~((D \subseteq \bP \text{ is dense}) \implies G \cap D \neq \emptyset).
\]
\end{definition}
\begin{corollary}
If $M$ is a countable transitive model of $\ZFC$,
$\bP \in M$ is a poset
and $p \in \bP$,
then there is an $M$-generic filter $G \subseteq \bP$
with $p \in G$.
\end{corollary}
\begin{remark}
The filter usually exists outside of $M$.
$M$ itself does not think that $M$ is countable,
since $M \models \ZFC$.
But from the outside, we see that $M$ is countable,
so we can find a filter.
\end{remark}
\begin{definition}
Assume that $\bP$ is a poset.
$\bP$ is said to be \vocab{atomless}
if for all $p \in \bP$ there are $q,r \in \bP$
such that
\begin{enumerate}[(1)]
\item $q \le p$, $r \le p$,
\item $q \bot r$.
\end{enumerate}
\end{definition}
\begin{example}
The \yaref{def:cohenf} is atomless.
\end{example}
Usually we are only interested in atomless partial orders.
\begin{lemma}
Assume that $M$ is a transitive model of $\ZFC$,
$\bP \in M$ an atomless poset
and let $G \subseteq \bP$ be $M$-generic for $\bP$.
Then $G \not\in M$.
\end{lemma}
\begin{proof}
Towards a contradiction assume $G \in M$.
Define $D\coloneqq \bP \setminus G$.
We'll show that $D \subseteq \bP$ is dense,
which is a contradiction, since $G$ was assumed to be $M$-generic.
Let $q \in \bP$ and
let $r,s$ be two extensions of $q$
such that $r \bot s$.
These exist because $\bP$ is atomless.
Since $G$ is a filter,
it can contain at most one of $\{r,s\}$,
wlog.~$s \not\in G$.
In particular, $s \in D$ and $s \le q$.
Hence $D$ is dense in $\bP$.
\end{proof}
\begin{lemma}
Assume that $M$ is a transitive model of $\ZFC$,
$\bP \in M$ a poset, $G \subseteq \bP$ an $M$-generic filter
and $p \in G$.
If $D$ is dense below $p$,
then $G \cap D \neq \emptyset$.
\end{lemma}
\begin{proof}
Let $E = D \cup \{ q \in \bP : q \bot p\}$.
$E \subseteq \bP$ is dense:
Let $r \in \bP$.
\begin{itemize}
\item If $r || p$ let $s \le r,p$.
Since $D$ is dense below $p$,
there exists $\overline{s} \in D$ such that $\overline{s} \le s$.
Since $D \subseteq E$, $\overline{s} \in E$.
\item If $r \bot p$, then it is obvious that $r \in E$.
\end{itemize}
Since $E \in M$, $G \cap E \neq \emptyset$.
\begin{IEEEeqnarray*}{rCl}
G \cap (D \cup \{q \in \bP : q \bot p\}) &\neq & \emptyset\\
\implies (G\cap D) \cup \underbrace{(G \cap \{q \in \bP : q \bot p\})}_{\emptyset} & \neq & \emptyset\\
\end{IEEEeqnarray*}
\end{proof}
\begin{definition}
Assume that $\bP$ is a poset.
\begin{enumerate}[(1)]
\item $A \subseteq \bP$ is said to be an \vocab{antichain}
iff for all $p \neq q$ in $A$,
$p \bot q$.
\item An antichain $A \subseteq \bP$ is a
\vocab{maximal antichain}
iff $\forall r \in \bP$, there exists $a \in \bP$
such that $p || r$.
\item $X \subseteq \bP$ is said to be \vocab{open}
if $\forall p \in X.~\forall q \le p.~ q \in X$.
\end{enumerate}
\end{definition}
\begin{remark}
Note that if $A$ is a maximal antichain
in $\bP$, then it is maximal in $(\{A \subseteq \bP : A \text{ is an antichain}\}, \subseteq )$.
Using $\AxC$, every antichain can be extend to a maximal antichain.
The statement ``$A$ is an antichain'' is $\Delta_0$.
Note that ``every antichain of $\bP$ is countable'' is not necessarily
absolute between transitive models of $\ZFC$.
\end{remark}
\begin{lemma}
Assume that $M$ is a transitive model of $\ZFC$,
$\bP \in M$ a poset and $G \subseteq \bP$ a filter.
Then the following are equivalent:
\begin{enumerate}[(1)]
\item $G$ is $\bP$-generic over $M$.
\item $G \cap A \neq \emptyset$ for every
maximal antichain $A \in M$.
\item $G \cap D \neq \emptyset$ for every dense
open $D \in M$ with $D \subseteq \bP$
\end{enumerate}
\end{lemma}
We'll prove this next time.