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Can anyone explain what is the intuition behind the following definition?

Definition 4.25 Let $\Bbb P$ be a poset. Let $\phi(x_1,\ldots,x_n)$ be a formula, $p\in\Bbb P$, and let $\tau_1,\ldots,\tau_n$ be $\Bbb P$-names. We define $p\Vdash^*\phi(\tau_1,\ldots,\tau_n)$ by recursion on the complexity of $\phi$ as follows.

  1. $p\Vdash^*\tau_1=\tau_2$ if and only if the following hold.

    1. For all $\langle\pi_1,s_1\rangle\in\tau_1$, the set $$\{q: q\leq s_1\rightarrow\exists\langle\pi_2,s_2\rangle\in\tau_2(q\leq s_2\land q\Vdash^*\pi_1=\pi_2)\}$$ is dense below $p$.

    2. For all $\langle\pi_2,s_2\rangle\in\tau_2$, the set $$\{q:q\leq s_2\rightarrow\exists\langle\pi_1,s_1\rangle\in\tau_1(q\leq s_1\land q\Vdash^*\pi_1=\pi_2)\}$$ is dense below $p$.

  2. $p\Vdash^*\tau_1\in\tau_2$ if and only if the set $$\{q:\exists\langle\pi,s\rangle\in\tau_2(q\leq s\land q\Vdash^*\tau_1=\pi)\}$$is dense below $p$.
  3. $p\Vdash^*\phi(\tau_1,\ldots,\tau_n)\land\psi(\tau_1,\ldots,\tau_n)$ if and only if $$p\Vdash^*\phi(\tau_1,\ldots,\tau_n)\text{ and }p\Vdash^*\psi(\tau_1,\ldots,\tau_n).$$

I know that the sign $p \Vdash \phi(x_1,...,x_n)$ somehow suppose to tell me that for any generic filter which contains $p$, $M[G] \models \phi(x_1,...,x_n)$. But, what is the connections to the definition here above?

Thank you

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    $\begingroup$ You can show that $p\Vdash^* \varphi(x_1,\dots,x_n)$ if and only if for any generic filter $G$ containing $p$, $M[G]\models\varphi(x_1,\dots,x_n)$ using a simple induction argument. $\endgroup$ – Alex Kruckman Mar 2 '15 at 19:02
  • $\begingroup$ Three notes: 1) There's a typo in the definition. Case B should read "For all $\langle \pi_2,s_2\rangle\in \tau_2$..." 2) The definition as given only covers formulas which are conjunctions of atomics, though it can easy be extended to cover all formulas. 3) For the record, I was not the one who cast the vote to close. $\endgroup$ – Alex Kruckman Mar 2 '15 at 19:04
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    $\begingroup$ There is no point asking this questions until you have read the proof of definability + truth lemma if you really care. I never actually read the details of these boring proofs from a book although I did see my instructor do the proof and we all wished it ended soon. By the way, your notes look like Kunen's 1980 book with typos but cuter font. $\endgroup$ – hot_queen Mar 2 '15 at 19:28
  • $\begingroup$ @hot_queen: Not quite: there are a few minor changes from Definition $3.3$ in the $1980$ book. At a guess this is from Ken’s $2011$ book. $\endgroup$ – Brian M. Scott Mar 2 '15 at 20:01
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Since you already know that $\Vdash$ is defined using generic filters, and you know that a filter is generic if and only it it meets every dense set of the ground model, this becomes quite an obvious choice of definition.

We can prove by induction on the complexity of the formula, that if $p\Vdash^*\varphi(\tau)$ then $p\Vdash\varphi(\tau)$. The reason is that the atomic formulas are defined using dense sets, and as it turns out these are dense open sets. And the intersection of dense open sets [below a condition] is a dense open set [below the condition].

This means that $p\Vdash^*\varphi(\tau)$ if and only if every generic filter $G$ such that $p\in G$, has some condition below $p$ (and therefore $p$ itself) which ensures that $\varphi(\tau)$ is true. So the intuition, if I had to give any for the definition, would be that $\Vdash^*$ is defined as a condition "which occurs on dense open sets [below a condition]", and therefore it is equivalent to saying that it is realized by every generic filter [containing said condition].

Once you see the proof of the equivalence between the two relations, this becomes quite obvious. So let me add a bit of motivation instead. The reason to do that is that given a formula $\varphi(\tau)$ in the language of forcing, the statement $p\Vdash^*\varphi(\tau)$ is definable internally to the ground model. So we can ask whether or not some condition forces a formula or not, internally. So we don't have to use a generic filter in order to find out whether or a statement is independent of $\sf ZFC$.

Namely, if we have a forcing notion such that $p\Vdash^*\varphi$ and $q\Vdash^*\lnot\varphi$, then we know that $\sf ZFC$ neither proves nor disproves $\varphi$. And this can be "pulled" to arithmetical meta-theories. So instead of using $\sf ZFC$ as a meta-theory and countable transitive models of [fragments of] $\sf ZFC$, you can use something as weak as $\sf PA$, and even weaker, as your meta-theory and still prove these sort of consistency results.

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