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@ -24,8 +24,8 @@ and $\forall x \in y.~\phi$ abbreviates $\forall x.~x \in y \to \phi$.
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If $\phi(x_0,\ldots,x_m) \in \Sigma_n$,
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then $(\forall x_0.~\ldots\forall x_m.~\phi(x_0,\ldots,x_m)) \in \Pi_{n+1}$.
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If $\phi(x_0,\ldots,x_n) \in \Pi_n$,
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then $(\exists x_0.~\ldots\exists x_n.~\phi(x_0,\ldots,x_n) \in \Sigma_{n+1}$.
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If $\phi(x_0,\ldots,x_m) \in \Pi_n$,
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then $(\exists x_0.~\ldots\exists x_m.~\phi(x_0,\ldots,x_m) \in \Sigma_{n+1}$.
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$\Delta_n \coloneqq \Sigma_n \cap \Pi_n$.
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\end{definition}
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@ -47,6 +47,7 @@ and $\forall x \in y.~\phi$ abbreviates $\forall x.~x \in y \to \phi$.
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\end{notation}
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\begin{lemma}
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\label{lem:d0absolute}
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Let $M$ be transitive, $\phi \in \Delta_0$
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and $a_0,\ldots, a_n \in M$.
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Then $M \models\phi(a_0,\ldots,a_n)$
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@ -56,7 +57,6 @@ and $\forall x \in y.~\phi$ abbreviates $\forall x.~x \in y \to \phi$.
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Clearly $M \models a_i \in a_j \iff V \models a_i \in a_j$
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and $M \models a_i = a_j \iff V \models a_i = a_j$,
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i.e.~the lemma holds for atomic $\phi$.
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% TODO transitivity needed?
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It is clear that if $M \models \phi_i \iff V \models \phi_i, i = 1,2$,
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then also $M \models \lnot \phi_i \iff V \models \lnot \phi_i$
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@ -65,8 +65,8 @@ and $\forall x \in y.~\phi$ abbreviates $\forall x.~x \in y \to \phi$.
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Assume that the lemma holds for $\phi$.
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Then it also holds for $\exists a_i \in a_j.~\phi$:
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We have that $a_i \in a_j$ is atomic and by the assumption that the lemma holds for $\phi$
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so since $M$ is transitive, a witness can be transferred from $V$ to $M$
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and vice versa.
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so since $M$ is transitive,
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a witness can be transferred from $V$ to $M$ and vice versa.
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The case of $\forall a_i \in a_j.~\phi$ can be treated similarly.
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\end{proof}
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A similar arguments yields \vocab{upwards absoluteness} for $\Sigma_1$-formulas
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@ -167,6 +167,7 @@ is a partially ordered set (\vocab{poset})
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if $\le $ is reflexive, symmetric nd transitive.
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\begin{definition}
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\label{def:forcingwords}
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A non-empty poset $\bP = (P, \le )$ is called a \vocab{forcing notion}.
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The elements of $P$ are called \vocab{conditions}.
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If $q \le p$ we say that $q$ is \vocab{stronger} than $p$.%
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@ -196,6 +197,7 @@ if $\le $ is reflexive, symmetric nd transitive.
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\end{definition}
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\begin{lemma}
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\label{lem:egenfilter}
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Let $\cP = (P, \le )$ be a poset,
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$\cD$ a countable family of dense subsets of $P$
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and $p \in P$.
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230
inputs/lecture_20.tex
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230
inputs/lecture_20.tex
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@ -0,0 +1,230 @@
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\lecture{20}{2024-01-15}{}
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\begin{idea}
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We want to add a new object that satisfies certain condition.
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The elements of the forcing notion correspond to approximations
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of this object.
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A filter picks some information which we want to be true.
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Being a filter ensures that this information does not contradict itself.
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\end{idea}
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\begin{definition}
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Assume that $M$ is a transitive model of $\ZFC$,
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and $\bP \in M$ a poset.
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$G \subseteq \bP$ is said to be \vocab{$M$-generic for $\bP$}
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if whenever $D \subseteq \bP$ is dense
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and in $M$, then $G \cap D \neq \emptyset$.
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\end{definition}
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\begin{remark}
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That is the same as being $\{D \subseteq \bP \text{ dense } : D \in M\}$-generic
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with generic defined as in \yaref{def:forcingwords}.
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\end{remark}
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% \begin{remark}
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% We can not prove that such a transitive model actually exists (we need inaccessible cardinals for that).
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% However there are transitive models of finite fragments of $\ZFC$.
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% % cf. Kuhnen, Chapter 7
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% So proofs using tansitive models can be made more general by
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% talking only about finite fragments, i.e.~a finite subset of the axioms of $\ZFC$ (since every proof can only
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% use finitely many axioms).
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% \end{remark}
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\begin{definition}[Cohen Forcing]
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\yalabel{Cohen Forcing}{Cohen Forcing}{def:cohenf}
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Let $\bP$ be the set of finite partial function $p$ from $\omega$ to $2$,
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i.e.~$\cP = 2^{<\omega}$.
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The order on $\bP$ is described by $q \le p \colon\iff q \supseteq p$.
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$\bP$ is called the \vocab{Cohen forcing}.
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\end{definition}
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\begin{fact}
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Assume $X \subseteq 2^\omega$ is countable,
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Then there is $x \in 2^\omega \setminus X$.
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\end{fact}
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Of course we already know that, but let's use it
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to test our machinery:
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\begin{proof}
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Assume that $X = \{x_n \colon n \in \omega\}$
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is an enumeration of $X$.
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Let $D_n = \{ p \in \bP : \exists i \in \dom(\bP).~x_n(i) \neq p(i)\}$.
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This makes sure that we get a ``new'' element
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not belonging to $X$.
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\begin{claim}
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$D_n$ is dense in $\bP$.
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\end{claim}
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\begin{subproof}
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Assume $q \in \bP$.
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Let $i = 1 + \max(\dom(q))$.
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Note that $i \not\in \dom(q)$.
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Let $p = q \cup \{(i, 1-x_n(i))\}$.
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Then $p \in D_n$.
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\end{subproof}
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Let $E_i = \{p \in \bP : i \in \dom(\bP)\}$.
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This makes sure that our ``new'' element
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is defined everywhere.
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\begin{claim}
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$\forall i < \omega.~E_i \subseteq \bP \text{ is dense}$.
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\end{claim}
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\begin{subproof}
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Assume $q \in \bP$.
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If $ i \in \dom(q)$ pick $p = q \in E_i$.
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If $i \not\in \dom(q)$,
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let $p = q \cup \{(i,0)\} \in E_i$.
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\end{subproof}
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Let $\cD = \{D_n : n < \omega\} \cup \{E_i : i < \omega\}$.
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This is a countable subset of dense sets.
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By \yaref{lem:egenfilter}
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there is a $\cD$-generic filter $G$.
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Let $y = \bigcup G$.
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Note that $y$ is a function, since any two elements of $G$ are compatible.
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\end{proof}
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Note that the ``new'' element did already exists, so we used forcing language
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to find it but didn't actually do anything.
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\begin{lemma}
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Let $M$ be a transitive model of $\ZFC$
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and let $\bP = (P, \le ) \in M$.
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Let $D \subseteq \bP$, $D \in M$, $p \in \bP$.
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Then
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\begin{enumerate}[(1)]
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\item $\bP$ is a partial order iff $M \models \text{``$\bP$ is a partial order''}$.
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\item $D$ is dense in $\bP$ iff $M \models \text{``$D$ is dense in $\bP$''}$.
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\item $D$ is dense below $p$ iff $M \models \text{``$D$ is dense below $p$''}$
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(this only makes sense if $p \in M$).
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\end{enumerate}
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\end{lemma}
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\begin{proof}
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All the definitions are $\Delta_0$, so we can apply
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\yaref{lem:d0absolute}.
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\end{proof}
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\begin{definition}
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Assume that $M$ is a transitive model of $\ZFC$
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and $\bP \in M$ is a poset.
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$G \subseteq \bP$ is called a
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\vocab{$\bP$-generic filter over $M$}
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or \vocab{$M$-generic filter for $\bP$}
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if
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\[
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\forall D \in M.~((D \subseteq \bP \text{ is dense}) \implies G \cap D \neq \emptyset).
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\]
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\end{definition}
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\begin{corollary}
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If $M$ is a countable transitive model of $\ZFC$,
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$\bP \in M$ is a poset
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and $p \in \bP$,
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then there is an $M$-generic filter $G \subseteq \bP$
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with $p \in G$.
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\end{corollary}
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\begin{remark}
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The filter usually exists outside of $M$.
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$M$ itself does not think that $M$ is countable,
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since $M \models \ZFC$.
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But from the outside, we see that $M$ is countable,
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so we can find a filter.
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\end{remark}
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\begin{definition}
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Assume that $\bP$ is a poset.
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$\bP$ is said to be \vocab{atomless}
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if for all $p \in \bP$ there are $q,r \in \bP$
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such that
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\begin{enumerate}[(1)]
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\item $q \le p$, $r \le p$,
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\item $q \bot r$.
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\end{enumerate}
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\end{definition}
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\begin{example}
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The \yaref{def:cohenf} is atomless.
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\end{example}
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Usually we are only interested in atomless partial orders.
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\begin{lemma}
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Assume that $M$ is a transitive model of $\ZFC$,
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$\bP \in M$ an atomless poset
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and let $G \subseteq \bP$ be $M$-generic for $\bP$.
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Then $G \not\in M$.
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\end{lemma}
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\begin{proof}
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Towards a contradiction assume $G \in M$.
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Define $D\coloneqq \bP \setminus G$.
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We'll show that $D \subseteq \bP$ is dense,
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which is a contradiction, since $G$ was assumed to be $M$-generic.
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Let $q \in \bP$ and
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let $r,s$ be two extensions of $q$
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such that $r \bot s$.
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These exist because $\bP$ is atomless.
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Since $G$ is a filter,
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it can contain at most one of $\{r,s\}$,
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wlog.~$s \not\in G$.
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In particular, $s \in D$ and $s \le q$.
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Hence $D$ is dense in $\bP$.
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\end{proof}
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\begin{lemma}
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Assume that $M$ is a transitive model of $\ZFC$,
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$\bP \in M$ a poset, $G \subseteq \bP$ an $M$-generic filter
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and $p \in G$.
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If $D$ is dense below $p$,
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then $G \cap D \neq \emptyset$.
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\end{lemma}
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\begin{proof}
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Let $E = D \cup \{ q \in \bP : q \bot p\}$.
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$E \subseteq \bP$ is dense:
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Let $r \in \bP$.
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\begin{itemize}
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\item If $r || p$ let $s \le r,p$.
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Since $D$ is dense below $p$,
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there exists $\overline{s} \in D$ such that $\overline{s} \le s$.
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Since $D \subseteq E$, $\overline{s} \in E$.
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\item If $r \bot p$, then it is obvious that $r \in E$.
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\end{itemize}
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Since $E \in M$, $G \cap E \neq \emptyset$.
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\begin{IEEEeqnarray*}{rCl}
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G \cap (D \cup \{q \in \bP : q \bot p\}) &\neq & \emptyset\\
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\implies (G\cap D) \cup \underbrace{(G \cap \{q \in \bP : q \bot p\})}_{\emptyset} & \neq & \emptyset\\
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\end{IEEEeqnarray*}
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\end{proof}
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\begin{definition}
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Assume that $\bP$ is a poset.
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\begin{enumerate}[(1)]
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\item $A \subseteq \bP$ is said to be an \vocab{antichain}
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iff for all $p \neq q$ in $A$,
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$p \bot q$.
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\item An antichain $A \subseteq \bP$ is a
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\vocab{maximal antichain}
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iff $\forall r \in \bP$, there exists $a \in \bP$
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such that $p || r$.
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\item $X \subseteq \bP$ is said to be \vocab{open}
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if $\forall p \in X.~\forall q \le p.~ q \in X$.
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\end{enumerate}
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\end{definition}
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\begin{remark}
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Note that if $A$ is a maximal antichain
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in $\bP$, then it is maximal in $(\{A \subseteq \bP : A \text{ is an antichain}\}, \subseteq )$.
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Using $\AxC$, every antichain can be extend to a maximal antichain.
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The statement ``$A$ is an antichain'' is $\Delta_0$.
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Note that ``every antichain of $\bP$ is countable'' is not necessarily
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absolute between transitive models of $\ZFC$.
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\end{remark}
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\begin{lemma}
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Assume that $M$ is a transitive model of $\ZFC$,
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$\bP \in M$ a poset and $G \subseteq \bP$ a filter.
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Then the following are equivalent:
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\begin{enumerate}[(1)]
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\item $G$ is $\bP$-generic over $M$.
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\item $G \cap A \neq \emptyset$ for every
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maximal antichain $A \in M$.
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\item $G \cap D \neq \emptyset$ for every dense
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open $D \in M$ with $D \subseteq \bP$
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\end{enumerate}
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\end{lemma}
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We'll prove this next time.
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