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.

  1. For every $D\in M$, if $D$ is dense (in $\mathbb P$), $G\cap D\neq\varnothing$ (i.e. $G$ is $M$-generic).
  2. For every $A\in M$, if $A$ is a maximal antichain, then $G\cap A\neq \varnothing$.
  3. For all $r\in G$ and $D\in M$, if $D$ is dense below $r$, then $G \cap D\neq \varnothing$.
  4. For every $D\in M$, if $D$ is predense, then $G\cap D\neq \varnothing$.
  5. 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).

  • $\begingroup$ I’m not catching why $r$ needs to be in $A$? That said, I don’t see any issues with overall the structure of the proof. The “usual” proof of 2->1 is you use choice in $M$ find a maximal anti chain in $D$ then argue by denseness that it is maximal in $P$, which the idea you’re going for here. $\endgroup$ Mar 19 at 17:00
  • $\begingroup$ I'm realizing I was too fast and loose with my construction of $A$. I should have said that we build $A$ recursively by adding $p_\alpha$ if for all $\beta<\alpha$, if $p_\beta \in A$ then $p_\alpha \perp p_\beta$. In that case, if $A$ is not maximal and we have such $q$ and $r$, then $r$ is incompatible with all members of $A$, and thus those less than it in the wellordering, so it should have been included in $A$. $\endgroup$ Mar 19 at 17:10
  • $\begingroup$ All that being said, my actual question was more about whether the general structure of using choice to build a maximal antichain inside $D$ works (this was just my particular method of doing that), and you seem to suggest it does. Thanks! $\endgroup$ Mar 19 at 17:12
  • $\begingroup$ I had understood what you were doing despite the issue you mentioned… maybe I’m being dense but I still don’t understand why you can conclude $r$ is incompatible with all of $A$ just cause it’s compatible with something that is. $\endgroup$ Mar 19 at 18:31
  • $\begingroup$ Ah! You're right. It appears I forgot how posets work and was assuming that compatibility was transitive, but that's not true. I'll think for a little bit about whether or not this construction works, because in theory it should be equivalent to any other means of using choice to get a maximal antichain? $\endgroup$ Mar 19 at 18:40


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