diff --git a/inputs/lecture_19.tex b/inputs/lecture_19.tex index 6d5fe9d..6a22304 100644 --- a/inputs/lecture_19.tex +++ b/inputs/lecture_19.tex @@ -24,8 +24,8 @@ and $\forall x \in y.~\phi$ abbreviates $\forall x.~x \in y \to \phi$. If $\phi(x_0,\ldots,x_m) \in \Sigma_n$, then $(\forall x_0.~\ldots\forall x_m.~\phi(x_0,\ldots,x_m)) \in \Pi_{n+1}$. - If $\phi(x_0,\ldots,x_n) \in \Pi_n$, - then $(\exists x_0.~\ldots\exists x_n.~\phi(x_0,\ldots,x_n) \in \Sigma_{n+1}$. + If $\phi(x_0,\ldots,x_m) \in \Pi_n$, + then $(\exists x_0.~\ldots\exists x_m.~\phi(x_0,\ldots,x_m) \in \Sigma_{n+1}$. $\Delta_n \coloneqq \Sigma_n \cap \Pi_n$. \end{definition} @@ -47,6 +47,7 @@ and $\forall x \in y.~\phi$ abbreviates $\forall x.~x \in y \to \phi$. \end{notation} \begin{lemma} + \label{lem:d0absolute} Let $M$ be transitive, $\phi \in \Delta_0$ and $a_0,\ldots, a_n \in M$. Then $M \models\phi(a_0,\ldots,a_n)$ @@ -56,7 +57,6 @@ and $\forall x \in y.~\phi$ abbreviates $\forall x.~x \in y \to \phi$. Clearly $M \models a_i \in a_j \iff V \models a_i \in a_j$ and $M \models a_i = a_j \iff V \models a_i = a_j$, i.e.~the lemma holds for atomic $\phi$. - % TODO transitivity needed? It is clear that if $M \models \phi_i \iff V \models \phi_i, i = 1,2$, then also $M \models \lnot \phi_i \iff V \models \lnot \phi_i$ @@ -65,8 +65,8 @@ and $\forall x \in y.~\phi$ abbreviates $\forall x.~x \in y \to \phi$. Assume that the lemma holds for $\phi$. Then it also holds for $\exists a_i \in a_j.~\phi$: We have that $a_i \in a_j$ is atomic and by the assumption that the lemma holds for $\phi$ - so since $M$ is transitive, a witness can be transferred from $V$ to $M$ - and vice versa. + so since $M$ is transitive, + a witness can be transferred from $V$ to $M$ and vice versa. The case of $\forall a_i \in a_j.~\phi$ can be treated similarly. \end{proof} A similar arguments yields \vocab{upwards absoluteness} for $\Sigma_1$-formulas @@ -167,6 +167,7 @@ is a partially ordered set (\vocab{poset}) if $\le $ is reflexive, symmetric nd transitive. \begin{definition} + \label{def:forcingwords} A non-empty poset $\bP = (P, \le )$ is called a \vocab{forcing notion}. The elements of $P$ are called \vocab{conditions}. If $q \le p$ we say that $q$ is \vocab{stronger} than $p$.% @@ -196,6 +197,7 @@ if $\le $ is reflexive, symmetric nd transitive. \end{definition} \begin{lemma} + \label{lem:egenfilter} Let $\cP = (P, \le )$ be a poset, $\cD$ a countable family of dense subsets of $P$ and $p \in P$. diff --git a/inputs/lecture_20.tex b/inputs/lecture_20.tex new file mode 100644 index 0000000..15c5775 --- /dev/null +++ b/inputs/lecture_20.tex @@ -0,0 +1,230 @@ +\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.