The previous two posts was dedicated to stating, explaining and applying a certain result in core model theory, the PD dichotomy, without using any inner model theory at all. This post is then the final post in this short series in which we’ll actually prove the dichotomy. This blog series, and especially the following proof, grew out of some work with Stefan Mesken.

The following is the dichotomy we’ve been working with for the last couple of posts.

The PD dichotomy. Let be either a -fixed point or . Then

- The core model exists for some ; or
- , projective determinacy, holds in an inner model.

*Technically speaking*, this is not really a dichotomy, as stated here. A dichotomy would also have the property that the two outcomes are mutually exclusive, which isn’t the case here. This was mostly done to minimise the number of definitions and complications, and the disjunctive conclusion is *still correct*. A more “correct” dichotomy, for the experts, would be the following.

The “correct” PD dichotomy.Let be either a -fixed point or . Assume that has many Woodins (say ), for every . Then either

- The core model exists for some and is -iterable; or
- is total on and is -iterable, for every . In particular, holds.

Note here that *if* had a limit of Woodins then would satisfy , so we still get in an inner model. So the first dichotomy is really quite simplified. I’ll provide our sketch of the proof below, and the full proof can be found **here**.

**Sketch of proof.** Assume that (2) fails, so that there’s a least and such that either doesn’t exist or it exists and isn’t -iterable. If then all we’re saying is that doesn’t exist, but then it’s a well-known theorem of Jensen that in that case. So we may assume that for some .

We may further assume that all premice are tame, as otherwise (2) holds. By Corollary 2.11 of Jensen et al. (2009) we get that exists and is countably iterable. We now have two cases.

**Case 1.** has a Woodin cardinal.

Let be the strict sup of all the Woodin cardinals of , which exists as has many Woodins. In this case we want to compare with , and for this to be possible it suffices to show that they’re both -iterable above . By the choice of we get that is -iterable, so we have to show that is -iterable above as well. A technical lemma ensures that this is actually the case, so we have a coiteration

.

By universality of we get that . We can further show that neither branch move, so that . But is countably iterable, so is as well, and one can show that this also implies that is -iterable as well.

**Case 2. ** has no Woodin cardinals.

Let be any uncountable cardinal and let . Note that by our choice of . Define now to be if is a limit cardinal of , and otherwise let . By Lemma 3.3 of Jensen et al. (2009) it holds that is countably iterable. In particular, is not Woodin in , as it’s trivial when , and otherwise it’s because

.

Now, by Fernandes () and Jensen & Steel (2013) this means that we can build the core model , as the only places they’re using that there’s no inner model with a Woodin cardinal are to guarantee that has no Woodin cardinals, which is simply our case assumption, and in Lemma 4.27 of Jensen & Steel (2013) in which they require that isn’t Woodin in .

As was arbitrary we get that the core model exists. Since the core model doesn’t have any Woodins either in this case, -structures trivially exist, making -iterable as well. **QED**