Suppose $G$ is $M$ generic over some model $M$ of $ZFC$, for some poset $P$. We have the following equivalences:

  1. $G$ is $M$-generic, i.e, for every $D \in M$ such that $D$ is dense in $P$, $G \cap D \not = \emptyset$

  2. $G$ meets every maximal antichain in $M$

  3. $G$ meets every predense set in $M$

  4. $G$ meets every open and dense set in $M$

  5. For every $r \in G$, if $D$ is dense below $r$ (i.e, for every $p \leq r$ there is a $q \in D$ such that $q \leq p$), and $D \in M$, $G \cap D \neq \emptyset$

I wanted to prove these equivalences. Most of these are pretty easy, but some of them I've blanked on.

$1 \implies 4$ is obvious. $4 \implies 1$ is: given a dense $D \in M$, take the downward closure $D' = \{q \in P: \exists p \in D q \leq p \}. D' \in M$ because $M$ satisfies $ZFC$ and is transitive and this is a simple definition. $D'$ is then dense and open so $G \cap D' \not = \emptyset$ and then just use the upwards closure of filters.

$3 \implies 1$ is obvious. $1 \implies 3$ is: given any $D \in M$ predense we have that $(D$ is predense$)^M$ and thus $M$ knows that for any $p \in P$ there is a $q \in D$ such that $p$ and $q$ are comparable, i.e, there is $r_{pq} \in P$ such that $r_{pq} \leq p, q$. Now define $D' = \{r_{pq} : p \in P$ and $q \in D$ such that $p, q$ comparable $\}$ and then I was thinking $D' \in M$ since $(Choice)^M$ (so a random $r_{pq}$ ca nbe chosen if there are many. $D'$ is dense so $G$ meets it at which point you just use upwards closure again. I feel less comfortable with this argument.

$5 \implies 1$ clearly, and $3 \implies 2$ clearly (as maximal antichains are predense). So all that's left is $2 \implies 1,4,5$ and $1,3,4 \implies 5$.

Perhaps I'm being an idiot but I've been having trouble with these ones.

Does this work for $1 \implies 5$? Suppose $D \in M$ dense below some $r$. Then define $D' = \{q \in P:q$ is incompatible with $r$, or $q \in D\}$. This is dense: given any $p \in P$, if $p$ incompat. with $r$, $p \in D'$ so assume not. So that means $p$ is compatible with $r$, which means there is $q$ such that $q \leq p, r$. By density below $r$ there is $q' \in D$ such that $q' \leq q \leq p, r$ so $q' \in D'$ and $q' \leq p$; therefore $D'$ is dense. Then $G \cap D \not = \emptyset$, and it is clear that intersection has to be in $p$, due to the directedness of filters. Does this work?

Finally $2 \implies 1,3,4,5$: this one I'm just stuck with. I may be missing something obvious.

  • $\begingroup$ Try looking at Theorem III.3.60 in Kunen's Set Theory (the newer edition). $\endgroup$
    – Reveillark
    Feb 19, 2021 at 2:26

1 Answer 1


There are multiple questions here; let me address the question of applying $(2)$ (which seems the most unclear point).

The key to applying $(2)$ is to note that every dense set contains a maximal antichain. Specifically, fix a dense set $D$ and let $A$ be a set with the following properties:

  • $A\subseteq D$,

  • $A$ is an antichain, and

  • there is no antichain $B\subseteq D$ with $A\subsetneq B$.

The existence of such an $A$ is guaranteed, as usual, by Zorn's Lemma (consider the partial order of antichains which are subsets of $D$). Now I claim that this $A$ is in fact a maximal antichain.

For suppose otherwise. Let $a\not\in A$ such that $A\cup\{a\}$ is an antichain. Then no extension of $a$ can meet $D$, since if $b\le a$ with $b\in D$ then $A\cup\{b\}$ would be an antichain strictly containing $A$ which is a subset of $D$.

The idea, then, is the following: "If $G$ meets every maximal antichain, then $G$ meets every dense set since every dense set contains a maximal antichain." Do you see how to appropriately formulate this to get $(2)\rightarrow(1)$?

Note that unlike the "maximal-antichain-to-dense-open" translation $$A\leadsto\{p: \exists a\in A(p\le a)\},$$ there is in general no canonical way to find a maximal antichain inside a given dense set. Indeed, it is consistent with $\mathsf{ZF}$ that there is a partial order $P$ with top element $\mathbb{1}_P$ which is separative (= nowhere-trivial, from the forcing perspective) but which has no maximal antichains other than $\{\mathbb{1}_P\}$; in such a poset, the dense set $P\setminus\{\mathbb{1}_P\}$ does not contain a maximal antichain.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .