255 lines
7.6 KiB
TeX
255 lines
7.6 KiB
TeX
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\lecture{22}{2024-01-22}{More Forcing}
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\begin{warning}+
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Forcing will not be relevant for the exam.
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Because of a lack of time, this is more of an outlook
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than a thorough presentation of the material.
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\end{warning}
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For the rest of the section, let us fix
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a transitive model $M$ of $\ZFC$
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a partial order $\mathbb{P}$
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and an $M$-generic filter $g$.
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\begin{definition}[$\mathbb{P}$-names]
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For an ordinal $\alpha \in M$%
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\footnote{Recall that $\Ord_M = \Ord \cap M$.},
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let $M^{\mathbb{P}}_\alpha$,
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the \vocab{$\mathbb{P}$-names} in $M$ of rank $\le \alpha$,
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be defined as follows:
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\[
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\tau \in M_{\alpha}^{\mathbb{P}} :\iff
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\tau \in M \land
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\tau \subseteq \mathbb{P} \times \bigcup \{M_\beta^{\mathbb{P}}: \beta < \alpha\},
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\]
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i.e.~the elements of $\tau \in M_\alpha^{\mathbb{P}}$
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are of the form $(p, \sigma)$,
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where $p \in \cP$ and $\sigma \in M^{\mathbb{P}}_\beta$
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for some $\beta < \alpha$.
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Finally $M^{\mathbb{P}} = \bigcup \{M_\alpha^{\mathbb{P}} : \alpha \in M\}$.
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\end{definition}
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Let $R$ be the relation on $M^{\mathbb{P}}$ defined by
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$\sigma R \tau$ iff $\exists p \in \mathbb{P}.~(p,\sigma) \in \tau$.
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If $\tau \in M^\mathbb{P}$
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and $(p, \sigma) \in \tau$,
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then $\sigma \in \{p, \sigma\} \in (p,\sigma) \in \tau$,
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so the relation $R$
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is well founded.
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\begin{definition}
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Let $\tau \in M_\alpha^{\mathbb{P}}$.
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Then $\tau^g$,
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the \vocab{$g$-interpretation of $\tau$},
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is defined to be
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\[
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\{\sigma^g : \exists p \in g .~(p,\sigma) \in \tau\}.
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\]
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\end{definition}
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\begin{definition}
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$M[g]$,
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the forcing extension of $M$
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given by $g$,
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is
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\[
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\{ \tau^g : \tau \in M^{\mathbb{P}}\} .
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\]
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\end{definition}
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\begin{lemma}
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$M[g]$ is transitive.
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\end{lemma}
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\begin{proof}
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Trivial!
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\end{proof}
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\begin{lemma}
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$M \cup \{g\} \subseteq M[g]$.
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\end{lemma}
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\begin{proof}
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For all $x \in M$ we need to find a name \vocab{$\check{x}$}
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such that $\check{x}^g = x$.
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We can recursively (along $\in$) define
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\[
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\check{x} = \{(p, \check{y}) : p \in \mathbb{P} \land y \in x\}.
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\]
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By induction, $\check{x} \in M$ for all $x \in M$.
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\begin{claim}
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$\check{x}^g = x$.
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\end{claim}
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\begin{subproof}
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Recall that $\mathbb{P} \neq \emptyset$.
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Inductively, we get
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\begin{IEEEeqnarray*}{rCl}
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\check{x}^g &=& \{\check{y}^g : \exists p \in g.~(p,\check{y}) \in \check{x}\}\\
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&\overset{\text{induction}}{=}& \{y : \exists p \in g.~(p,\check{y}) \in \check{x}\}\\
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&\overset{\text{definition of $\check{x}$}}{=}& \{y : y \in x\} = x.
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\end{IEEEeqnarray*}
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\end{subproof}
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So $M \subseteq M[g]$.
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We also need a name for $g$.
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Let \vocab{$\dot{g}$}$ \coloneqq \{(p, \check{p}) : p \in \mathbb{P}\}$.
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Indeed
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\begin{IEEEeqnarray*}{rCl}
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\dot{g}^g &=& \{\check{p}^g : \exists p \in g.~(p, \check{p}) \in \dot{g}\}\\
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&=& \{p : p \in g\} = g.
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\end{IEEEeqnarray*}
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\end{proof}
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\begin{lemma}
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\label{lem:mgmodelexfundinfpairunion}
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$M[g] \models \AxExt, \AxFund, \AxInf, \AxPair, \AxUnion$.
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\end{lemma}
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\begin{proof}
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\begin{itemize}
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\item \AxExt:
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The formula $\forall x.~\forall y .~((\forall z \in x.~z \in y \land \forall z \in y.~z \in x) \to x = y)$
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is $\Pi_1$, hence it is true in $M[g]$
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by %TODO REF downward absolutenes.
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\item \AxFund:
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Again,
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\[
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\forall x.~(\exists y \in x .~y = y \to \exists y \in x.~\forall z \in y.~z \not\in x)
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\]
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is $\Pi_1$.
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\item \AxInf
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can be written as
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\[
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\exists x .~(\underbrace{\neq \in x \land \forall y \in x.y \cup \{y\} \in x}_{\Sigma_0}).
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\]
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We have $ \omega \in M \subseteq M[g]$,
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so $M[g] \models \AxInf$.
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\item \AxPair:
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Let us assume $x,y \in M[g]$,
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say $x = \tau^g$ and $y = \sigma^g$.
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Let $\pi = \{(p,\tau) : p \in \mathbb{P} \} \cup \{(p,\sigma) : p \in \mathbb{P}\} \in M^{\mathbb{P}}$.
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Then $\pi^g = \{\tau^g, \sigma^g\} = \{x,y\}$,
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so $\{x,y\} \in M[g]$.
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As a $\cL_\in$-statement,
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$z = \{x,y\}$ is $\Sigma_0$,
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so $M[g] \models \text{``$\{x,y\}$ is the pair of $x$ and $y$''}$.
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Hence $M[g] \models \AxPair$.
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\item \AxUnion:
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Similar to \AxPair.\todo{Exercise}
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\end{itemize}
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\end{proof}
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Still missing are
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\begin{itemize}
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\item \AxPow,
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\item \AxAus,
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\item \AxRep,
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\item \AxC.
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\end{itemize}
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\begin{definition}[\vocab{Forcing relation}]
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Let $M$ be a countable transitive model of $\ZFC$
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and let $\mathbb{P} \in M$ be a partial order.
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Let $p \in \mathbb{P}$
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and let $\phi$ be a $\cL_{\in }$-formula.
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Let $\tau_1,\ldots, \tau_k \in M^{\mathbb{P}}$ be names.
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We say that
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$p$ \vocab[force]{forces} $\phi(\tau_1,\ldots, \tau_k)$,
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\[
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p \Vdash^{\mathbb{\cP}}_{M} \phi(\tau_1, \ldots, \tau_k),
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\]
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if for all $h \subseteq \mathbb{P}$
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which are $\mathbb{P}$-generic over $M$
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with $p \in h$,
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\[
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M[h] \models\phi(\tau_1^h, \ldots, \tau_k^h).
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\]
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\end{definition}
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\begin{theorem}
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Fix an $\cL_\in$-formula $\phi$.
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Then the relation
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\[
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R = \{(p,\tau_1,\ldots,\tau_k : p \Vdash ^{\mathbb{P}}_M \phi(\tau_1,\ldots,\tau_k)\}
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\]
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is definable over $M$
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(in the parameter $\mathbb{P}$).
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\end{theorem}
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\begin{proof}
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Omitted.
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\end{proof}
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\begin{theorem}[\vocab{Forcing Theorem}]
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Let $M$, $\mathbb{P}$, $g$,
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be as above,
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let $\phi$ be a formula,
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and let $\tau_1, \ldots, \tau_k \in M^{\mathbb{P}}$.
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Then the following are equivalent:
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\begin{enumerate}[(1)]
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\item $M[g] \models \phi(\tau_1^g, \ldots, \tau_k^g)$.
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\item There is some $p \in g$ with
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\[
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p \Vdash^{\mathbb{P}}_M \phi(\tau_1,\ldots, \tau_k).
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\]
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\end{enumerate}
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\end{theorem}
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\begin{proof}
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Omitted.
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\end{proof}
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\begin{theorem}
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$M[g] \models \ZFC$.
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\end{theorem}
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\begin{proof}
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We have already shown a part of this in
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\yaref{lem:mgmodelexfundinfpairunion}.
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Let us show that $M[g] \models \AxAus$,
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the rest is similar and left as an exercise.%
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\footnote{or done next semester in Logic IV!}
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Let $\phi$ be a formula,
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let $a, x_1,\ldots,x_k \in M[g]$.
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We need to see
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\[M[g] \models \exists y.~y = \{z \in a : \phi(z, x_1,\ldots,x_k)\}.\]
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If suffices to show that there is some $y \in M[g]$
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with $y = \{ z \in a : M[g] \models \phi(z, x_1,\ldots,x_k)\}$.
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For this, let us construct a name for $y$.
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Let $a = \tau^g$, $x_i = \sigma_i^g$.
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Let
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\[
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\pi = \{(p,\rho) : \exists \overline{p} > p.~(\overline{p}, \rho) \in \tau \land
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p \Vdash^{\mathbb{P}}_M \phi(\rho, \sigma_1, \ldots, \sigma_k)\}.
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\]
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We have $\pi \in M$, since the relation
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$ \Vdash^{\mathbb{P}}_M$ can be defined in $M$.
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Let $z \in a$ such that $M[g] \models \phi(z, x_1,\ldots,x_n)$.
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We have $z = \rho^g$
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for some $\rho$
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and there is $\overline{p} \in g$ with $(\overline{p}, \rho) \in \pi$.
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Now $M[g] = \phi(\rho^g, \sigma_1^g,\ldots\sigma_k^g)$.
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Let $p' \Vdash^{\mathbb{P}}_M \phi(\rho, \sigma_1,\ldots, \sigma_k)$,
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where $p' \in g$.
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We have $p', \overline{p} \in g$,
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so there is some $p \le p', \overline{p}$ with $p \in g$.
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Then $(p,\rho) \in \pi$,
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so $\rho^g \in \pi^g$.
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This shows that
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\[
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\{z \in a : M[g] \models \phi(z,x_1,\ldots,x_k)\} \subseteq \pi^g.
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\]
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The other inclusion is easy.
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\end{proof}
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