# Equivalent conditions for $M$-genericity

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.

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

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.