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If $P$ is a notion of forcing in $M$, then $G$ is a $P$-generic filter over $M$ if $G\subseteq P$ is a filter, and for every $D\in M$ which is a dense subset of $P$, $G\cap D\neq\varnothing$.

Equivalently we can replace $D$ being dense by being pre-dense, open dense, or a maximal antichain.

Given such a generic filter, and $\dot x$ which is a $P$-name, we can define the interpretation of $\dot x$ by the filter $G$ by recursion, $$\dot x^G=\{\dot y^G\mid\exists p\in G:\langle p,\dot y\rangle\in\dot x\}.$$

Clearly, we don't need genericity in order to interpret names. We can talk about interpretation using arbitrary filters.

Question. Given a filter $G$, is there some reasonable condition stating that $G$ is generic if and only if it interprets certain names (e.g. names which are forced to be ordinals) "properly"?

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  • $\begingroup$ I suppose saying that $G$ is generic iff it interprets the canonical name $\dot{G}$ to be a filter generic over the ground model isn't helpful. $\endgroup$ – Miha Habič Jan 20 '14 at 13:10
  • $\begingroup$ Miha, yeah, it's not quite what I was aiming for. I was looking more for something like "every name of an ordinal is interpreted as an ordinal". $\endgroup$ – Asaf Karagila Jan 20 '14 at 13:11
  • $\begingroup$ In that direction you get some pretty underwhelming results. For example, if $x$ is some arbitrary nonempty set, $G$ is generic iff it interprets every name for $x$ as $x$: for any maximal antichain $A$ build a name $\tau_A$ by mixing check names $\check{x}$ over $A$. This will be interpreted to be $x$ iff $G$ meets $A$ and will be empty otherwise. $\endgroup$ – Miha Habič Jan 20 '14 at 13:33
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    $\begingroup$ I see (and probably you force upwards as well!). The idea of the usual convention is that names are like characteristic functions, but with truth values in the forcing notion. $\endgroup$ – Joel David Hamkins Jan 20 '14 at 15:05
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    $\begingroup$ @Joel: I have never forced upwards. :-) $\endgroup$ – Asaf Karagila Jan 20 '14 at 16:47
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Your question is fairly open-ended, but here is one way to do it. For any antichain $A\subset P$, consider the name $\dot a_A=\{\langle\check a,a\rangle\mid a\in A\}$, which is a mixture of the elements of $A$ on the antichain $A$. This is the name that is trying to be an element of $A$, with values determined by $A$ itself.

Theorem. The following are equivalent, for a filter $G\subset P$.

  1. $G$ is $V$-generic.
  2. $(\dot a_A)^G\in A$ for each maximal antichain $A\subset P$.
  3. $(\dot a_A)^G$ is nonempty for every maximal antichain.

Proof. The filter $G$ is generic if and only if it meets every maximal antichain, and in this case, 2 will hold, and clearly 2 implies 3. If 3 holds, then there must be some $a\in A\cap G$, and so $G$ will be $V$-generic. QED

Miha suggests a similar idea in the comments, namely, take any nonempty set $x$, and consider the names $\dot x=\{\langle\check y,a\rangle \mid y\in x\text{ and }a\in A\}$, for any maximal antichain $A$. Miha's observation is that although every condition forces $\check x=\dot x$, nevertheless if $G$ is not generic and misses the antichain $A$, then $\dot x^G$ will be empty. This idea arrives at the following, which seems to meet the requirement of your question.

Theorem. (Miha Habič, in comments) The following are equivalent:

  1. $G$ is $V$-generic.
  2. $G$ interprets all names for non-zero ordinals as non-zero ordinals. That is, if $\mathbb{1}\Vdash(\tau$ is a nonzero ordinal), then $\tau^G$ is a nonzero-ordinal.
  3. $G$ interprets all names for the number $1$ correctly.

Proof. Clearly 1 implies 2 and 3. Conversely, if 2 or 3 holds, let $A$ be any maximal antichain and let $\tau=\{\langle 0,a\rangle \mid a\in A\}$. This is a name forced to be the ordinal $1$, and it is easy to see that $\tau^G$ is either $0$ or $1$, depending on whether $G$ meets $A$. So $1$ holds. QED

One can use dense sets instead of antichains, which will avoid a dependence on AC.

Lastly, let me add then when you are entertaining the idea of interpreting names via a filter that is not generic, then you should not be using the value as you have defined it, but rather you should be using the Boolean ultrapower instead. In this method, you use the Boolean algebra rather than just a partial order, and define the equivalence on names $\sigma=_G\tau\iff[\![\sigma=\tau]\!]\in G$, and similarly for $\sigma\in_G\tau$. Now, the ultrapower map is $x\mapsto [\check x]_G$, meaning the equivalence class under this equivalence relation. The result is an elementary embedding $$j:V\to\check V_G\subseteq V^{\mathbb{B}}/G,$$whose target is the ground model of the corresponding quotient $V^{\mathbb{B}}/G$. The following theorem appears in the Boolean ultrapower paper:

Theorem. The following are equivalent:

  1. $G\subset P$ is $V$-generic.
  2. The Boolean ultrapower by $G$ is an isomorphism of $V$ with $\check V/G$.

This theorem is part of the explanation that "non-generic" is the right generalization of "non-principal" when considering the ultrapower on a complete Boolean algebra instead of merely on a powerset.

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  • $\begingroup$ Thank you Joel. It's been helpful, and it also means that I'll have to do all the definitions and proofs the hard way... :-) $\endgroup$ – Asaf Karagila Jan 20 '14 at 14:53
  • $\begingroup$ I'm not sure what you mean... $\endgroup$ – Joel David Hamkins Jan 20 '14 at 14:54
  • $\begingroup$ Oh, of course you don't. It's a work in progress... I'll keep you posted. :) $\endgroup$ – Asaf Karagila Jan 20 '14 at 14:55

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