# Forcing, a technical detail

In the snippet below from Shelah's book P&I Forcing, in the definition 5.2(2) I do not follow why in this sentence [naturally extended to include $$N\prec (H(\mu^\dagger),\epsilon),\mu\in N$$] $$N$$ would really be in $$\mathbb{D}(N\cap H(\mu),P,p)$$

? Here, $$\mu<\mu^\dagger$$ so we may miss our $$N$$ there.

• I think "$\mathbb{D}$ ... include[s] $N$" should really be "the domain of $\mathbb{D}$ ... include[s] $N$:" the goal is for $\mathbb{D}(N,P,p)$ to be defined at all, even when $N$ might be too big for the original definition of $\mathbb{D}$. – Noah Schweber Dec 27 '20 at 21:52
• @NoahSchweber Could you be more precise and detailed : what is esp. the difference bewteen $\mathbb D$ and the domain of $\mathbb D$ ? Also, what do you mean by the last part of your comment ? I'm just looking at that page and cannot move on. – user2925716 Dec 27 '20 at 22:17
• Well, $\mathbb{D}$ is a function (from a certain set of triples to a certain set of filters), and it has a domain (the relevant set of triples) which is different from $\mathbb{D}$ itself. Unless I'm misreading it, the idea is that we can canonically extend an initial function $\mathbb{D}'$, which in an abuse of notation is conflated with $\mathbb{D}$ itself to be defined on a larger domain. – Noah Schweber Dec 27 '20 at 22:24

As Noah wrote in the comments, Shelah really means that $$N$$ is in the domain of $$\mathbb D$$. However some extra motivation for what's going on here might make things clearer. The point of the definition is to generalize $$\sigma$$-closed forcing. Shelah wants to iterate forcing notions that don't add reals. On the one hand, $$\sigma$$-closed forcing is the class of forcing notions par excellence that does not add reals but on the other hand $$\sigma$$-closed forcing is so non-destructive that there are limitations to what can be proved using it (for instance, $$\sigma$$-closed forcing does not specialize Aronszajn trees, it does not add branches to $$\omega_1$$-trees etc).
The relevant property Shelah wants to generalize is that if $$\mathbb P$$ is $$\sigma$$-closed, $$\mu$$ is sufficiently large and $$N \prec H_\mu$$ is a countable model containing $$\mathbb P$$ and some condition $$p$$ then $$any$$ generic $$p \in G \subseteq \mathbb P \cap N$$ over $$N$$ has a lower bound (it can be argued that this is what's really used in a lot of arguments involving $$\sigma$$-closed forcing). Roughly the idea for the generalization is to replace $$all$$ generics by $$most$$ generics (say a "measure one set") i.e. an element of a suitable filter of subsets of $$Gen(N, \mathbb P, p)$$.
To make this precise, we need to make sense of what types of filters we're going to use and this is what $$\mathbb D(N, \mathbb P, p)$$ is giving us. Now, the point is that if $$\mu$$ is sufficiently large (say $$(2^{|\mathbb P|})^+$$) then whether or not a generic $$G \subseteq \mathbb P \cap N$$ has a lower bound for $$N \prec H_{\mu^*}$$ for $$\mu^* > \mu$$ really only depends on $$N \cap H_\mu$$. Moreover, if $$\mu \in N$$ then an easy Tarski-Vaught argument tells you that $$N \cap H_\mu \prec H_\mu$$ so the filter of generics we get from $$\mathbb D (N, \mathbb P, p)$$ really need not be any different from the filter of generics we get from $$\mathbb D(N\cap H_\mu, \mathbb P, p)$$.
Uri Abraham's chapter in the handbook also discusses $$\mathbb D$$-complete forcing and the definitions and discussion there might help clarify things as well.