Someone commented under one of my previous posts that, intuitively, a generic set isn't supposed to have any "conspicuous properties". I wonder what the precise meaning of that comment was and whether we could provide some kind of proof of this.
For reference, I'm working with these definitions:
A dense set $D$ is any subset of the forcing notion $P$ such that every $p \in P$ has an extension in $D$. A generic ideal $G \subseteq P$ is an ideal that intersects every dense subset of $P$ that is also an element of the model $\mathbf{M}$.
I have been thinking about this myself, but haven't gotten anywhere concrete. Here's what I was working with: consider a specific forcing notion, say the set $P \in \mathbf{M}$ of all finite partial functions from $\mathbb{N}$ to $\mathbb{N}$.
Suppose $D \subseteq P$. I will call some $(a_1, a_2, \dots, a_n) \in \mathbb{N}^n$ "determined in $D$" if the sentence
$$(\exists b_1 \exists b_2 \dots \exists b_n) (\forall f \in D) ((f(a_1) \neq b_1) \lor (f(a_2) \neq b_2) \lor \dots \lor (f(a_n) \neq b_n))$$
is true. If there exists a determined tuple in $D$, $D$ can't be dense. This is because the partial function $f$ determined by $f(a_1) = b_1, f(a_2) = b_2, \dots f(a_n) = b_n$ won't have an extension in $D$. It's also fairly easy to see the converse from this, i.e. any set $D$ that has no determined tuples is dense. If we consider an "infinite" tuple, then we can have a dense set in which that tuple is determined, but this can't be done for finite tuples. So, in this case, a set isn't dense iff it "determines" a finite tuple, which can be thought of intuitively as carrying a positive amount of information. I like this perspective and find it quite intuitive, so I'd be satisfied with a generalization of this.
This also seems like it should translate into a definability argument (something like "iff $D$ is a definable subset of $P$, it's not dense", but that exact phrasing doesn't work). I know some model theory and generally find it very intuitive, so I'd be very happy if I could connect it to something like that. I haven't been able to actually translate it, though, or to generalize it in any way for other forcing notions. I also don't see an easy way to connect it to genericity. So my question is: can this approach be generalized? If so, how exactly? If not, is there another intuitive interpretation of what a dense set is and how can we show that rigorously?
The statement about $G$ not having any "conspicuous properties" means that any first-order property of $M[G]$ depends only on the facts that
(1) $G$ is $M$-generic on $P,$
and (2) some particular condition $p$ belongs to $G.$
In detail:
Let $\varphi$ be a first-order sentence in the language of set theory. Then there exists some $p\in G$ with the following property:
For every $H\subseteq P$ such that $H$ is $M$-generic on $P$ and $p\in H,$ we have that $$M[H]\models\varphi\;\text{ iff }\;M[G]\models\varphi.$$
You can even allow $\varphi$ to contain constants from $M$ or names representing members of the generic extension. (Any names must be interpreted as appropriate in $M[G]$ or $M[H].)$
Moreover, if the partial ordering $P$ is homogeneous and if we just allow $\varphi$ to contain constants from $M$ (but no mentions of other names), then all generic $G$ agree on whether $\varphi$ is true in $M[G].$ (In other words, we don't need $p.)$