Earlier this semester, I worked on the following exercise:
Let $M$ be a transitive class model of ZFC and $\mathbb P=(P,<)\in M$ be a poset. Let $G$ be a filter on $\mathbb P$. Prove that the following are equivalent.
- For every $D\in M$, if $D$ is dense (in $\mathbb P$), $G\cap D\neq\varnothing$ (i.e. $G$ is $M$-generic).
- For every $A\in M$, if $A$ is a maximal antichain, then $G\cap A\neq \varnothing$.
- For all $r\in G$ and $D\in M$, if $D$ is dense below $r$, then $G \cap D\neq \varnothing$.
- For every $D\in M$, if $D$ is predense, then $G\cap D\neq \varnothing$.
- For every $D\in M$, if $D$ is dense and open, then $G\cap D\neq \varnothing$.
(This is essentially Jech ex. 14.2-14.6, among many other places I'm sure.)
After showing (1), (3), (4), and (5) are equivalent and that, for instance, (4) implies (2), it remained to show that (2) implies one of (1), (3), (4), or (5). I came up with the following proof (?) that (2) implies (4).
Suppose (2) holds, and let $D\in M$ be predense. We wish to show that $G\cap D\neq\varnothing$. By (2), it suffices to find $A\in M$ such that $A\subseteq D$ and $A$ is a maximal antichain.
To do this, work in $M$. Wellorder $D=\{p_0,p_1,\dots\}$, and define $A = \{p_\alpha \mid \forall \beta<\alpha, p_\alpha \perp p_\beta\}$. Clearly $A\subseteq D$ and $A$ is an antichain (any two conditions in $A$ are incompatible by construction).
It remains to see that $A$ is maximal, or equivalently that $A$ is predense. Suppose otherwise, i.e. there is some $q\in P$ such that $q\perp p$ for all $p\in A$. Since $D$ is predense, there is some $r\in D$ with $q\parallel r$. However, if this occurs then we should have $r\in A$, leading to the contradiction $q\perp r$. Thus $A$ is indeed maximal, as desired.
Jumping back out to $V$, we see that $A\in M$ (we defined it in $M$!), $A\subseteq D$, and that "$A$ is a maximal antichain in $\mathbb P$" is absolute between $M$ and $V$, so $A$ is still a maximal antichain when viewed from $V$. This completes the proof.
Does this work? All of my other proofs in this equivalence, say (a) implies (b), had me start with some object $D\in M$ satisfying the hypothesis in (b), and then expanding $D$ to some bigger set satisfying the hypotheses in (a), and then doing a further argument to show that the place where this bigger set meets $G$ is also in the original set. In contrast, this direction has me starting with $D$ and making a smaller set, which makes me worried it might not work for some reason I'm not seeing (my professor expressed concern about this, although we did not find an explicit flaw).
