# «SPENCER UNGER Abstract. From large cardinals we obtain the consistency of the existence of a singular cardinal κ of coﬁnality ω at which the ...»

4.2. Complex Properties of R. In this subsection we prove that in V R β the forcing R/R β is equivalent to a forcing with a deﬁnition similar to R. This will provide a key component in our proof that the tree property holds at κ++ in the extension.

First we make note of a slightly diﬀerent, but equivalent (in the sense of forcing) deﬁnition of R and its restrictions. Note that instead of the condition (a, p) ∈ A ∗ P in the ﬁrst two coordinates of R, we could have just taken a condition in RO(A ∗ P).

The projections needed in the deﬁnition of the ordering are just the projections σβ from Remark 3.8. If we call this new poset R, it is easy to see that R is a dense subset of R. Similarly we deﬁne R β as above by replacing Aβ ∗ Pβ with RO(Aβ ∗ Pβ ) in the deﬁnition of R β.

Having deﬁned these auxiliary posets it is easy to see that the map from R to R β given by (a, p, f ) → (πβ (a, p), f β) is a projection. In V R β we will deﬁne a poset R∗ which is equivalent to the poset R/R β and which resembles R. To deﬁne the ordering we will need suitable projections from (A ∗ P)/(Aβ ∗ Pβ ) to RO(Aγ ∗ Pγ )/(Aβ ∗ Pβ ) for γ β in B. These projections will be given by the following proposition. (The posets P, Q and R in the following proposition have no relation to the posets we’ve deﬁned above.) Proposition 4.7. Let P, Q and R be posets and assume that there are projections π : P → Q and σ : Q → R. If G is R-generic, then in V [G] π P/G is a projection from P/G to Q/G.

8 SPENCER UNGER Proof. Clearly the restriction of π is order preserving and sends the top element of P/G to the top element of Q/G. For the moment we work in V. Let p ∈ P and q ≤ π(p). Let D be the set of r ∈ R such that there is p ∈ P with (1) p ≤ p, (2) π(p ) ≤ q and (3) σ(π(p )) = r.

We claim that D is dense in R below σ(q). Suppose that r ≤ σ(q) since σ is a projection there is q ≤ q such that σ(q ) ≤ r. Since π is a projection there is a p ≤ p such that π(p ) ≤ q. Clearly σ(π(p )) ∈ D. Suppose that p ∈ P/G and q ≤ π(p) is in Q/G. Then since σ(q) ∈ G, D ∩ G = ∅ where D is deﬁned as above.

Let p witness that some r ∈ D ∩ G. Then p ∈ P/G, p ≤ p and π(p ) ≤ q as required.

Working in V R β, we deﬁne the forcing R∗ as follows. We let (a, p, f ) ∈ R∗ if and only if (a, p) ∈ (A ∗ P)/(Aβ ∗ Pβ ) and f is a partial function with domain a subset of B of size µ such that for each γ ∈ B, f (γ) is an (Aγ ∗ Pγ )/(Aβ ∗ Pβ )-name for a condition in Add(µ, 1). The ordering is deﬁned in a similar way to that for R, but for each γ using the restriction of πγ to (A ∗ P)/(Aβ ∗ Pβ ). These restrictions are projections by the previous proposition applied with πγ in place of π and σβ,γ (from Remark 3.8) in place of σ.

We can now state the main technical lemma of this section.

˙ β ∗ R∗ such that x ≤ y if and only Lemma 4.8. There is a map i from R to R if i(x) ≤ i(y) and the range of i is dense.

The proof is very similar to the proof of Lemma 2.12 in [1]. As with R we can show that R∗ is the projection of a product (see Deﬁnition 4.4 and Lemma 4.5).

** Lemma 4.9.**

In V R β, there is a µ-closed forcing Q∗ such that the identity map is a projection from (A ∗ P)/(Aβ ∗ Pβ ) × Q∗ to R∗.

The proof is straightforward.

5. The tree property at κ++ The proof in this section is somewhat diﬀerent from the proof in the paper of Cummings and Foreman [3]. We rely on the same analysis in terms of projections, but the forcings involved are no longer as nice. Our task is further complicated by a mistake on the very last page the Cummings and Foreman paper. In particular they attempt to prove that the quotient forcing in their paper corresponding to (A ∗ P)/(Aβ ∗ Pβ ) in our paper has the Knaster property. The key point in the argument is to show that conditions which witness the compatibility of conditions in A ∗ P are forced in to the quotient. A careful read of the paper shows that Cummings and Foreman have not done enough work to show that such conditions are forced in to the quotient. To ﬁx this problem we provide a further analysis of the quotient forcing. Our task is made a little easier, since in light of Lemma 2.4 we only need to show that the quotient squared has chain condition. The proof given below adapts easily to give the proof the analogous fact about the forcing in the Cummings and Foreman paper.

** Lemma 5.1.**

The tree property holds at κ++ in V R.

## ARONSZAJN TREES AND THE SUCCESSORS OF A SINGULAR CARDINAL 9

Our ﬁnal preliminary fact is about the chain condition of A ∗ P in V.

Proposition 5.9. In V, (A ∗ P)2 × (Aβ ∗ Pβ ) is µ-cc The proof is straight forward. Conditions with the same stem are compatible and κ+ of the A parts can be formed in to a ∆-system. We are now ready to prove that the quotient squared has chain condition.

This ﬁnishes the proof of Lemma 5.3 and with it the proof of Lemma 5.1

6. No special κ+ -trees In this section we give two proofs that there are no special κ+ -trees. The ﬁrst proof applies to the version of R where we follow Gitik and Sharon and take κn = κ+n. The second proof applies to the version of R where we follow Neeman and let the κn ’s be an increasing sequence of supercompact cardinals.

6.1. R with Gitik-Sharon. For this section we assume that κn = κ+n and so

**ν = κ+ω. We show the following:**

** Theorem 6.1.**

In V R there is a bad scale on κ+.

Proof. By arguments from [4] there is a bad scale f at κ in V A∗P which is witnessed by a stationary set S ⊆ κ+ω+1 from V. Using Lemma 4.5 and Easton’s Lemma, we see that V A∗P and V R have the same κ-sequences. It follows that f is still a scale in V R and every bad point of f in V A∗P remains bad in V R. It remains to see that the set S is still stationary in V R. It is enough to show that it is still stationary in the outer model V Q×(A∗P). Since Q is κ+ω+1 -closed in V, S is stationary in V Q.

By Easton’s Lemma, A ∗ P is κ+ω+1 -cc in V Q. The result follows.

** Remark 6.2.**

It should be noted that with a some extra work this argument applies for other choices of the κn ’s. A referee pointed out that arguments from [5] obtain the failure of approachability at κ for all choices of the κn ’s with no extra work required.

12 SPENCER UNGER

6.2. R with Neeman. For this section we assume that κn | n ω is an increasing sequence of supercompact cardinals. Further we assume that each κn is indestructible under κn -directed closed forcing. Under the above assumptions we prove the following theorem.

** Lemma 6.3.**

In the extension V R there are no special κ+ -trees Proof. It will be enough to show that the tree property holds at κ+ in an extension by A ∗ P × Q and that µ is preserved in this extension. If we had a special κ+ -tree in V R, then it would still be a special κ+ -tree in V (A∗P)×Q, which is impossible.

Recall that in the extension we have collapsed cardinals and µ has become κ+. By Lemma 4.5, we know that µ = κ+ is preserved in V (A∗P)×Q. It remains to show that the tree property holds in this model.

Recall that the term forcing Q was deﬁned in the ground model and is µ-closed.

So we consider the extension in question as an extension by Q and then by A ∗ P.

We want to show that in V Q, A ∗ P is Neeman’s forcing for some choice of measures and each of the κn ’s is still supercompact.

** Lemma 6.4.**

Q is µ-directed closed.

Proof. Let {fα : α η} be a set of conditions in Q for some η µ such that for any pair α0, α1 η there is a γ η such that fγ ≤ fα0, fα1. Deﬁne f to be a function such that dom(f ) = αη dom(fα ). The domain of f has size less than µ, because µ is regular. For each β ∈ dom(f ), let f (β) be a name for the union of fα (β) over all α η. We claim that for each β ∈ dom(f ), f (β) names a condition in Add(µ, 1)V Aβ ∗Pβ. Suppose that the claim fails. Then there is a β ∈ dom(f ), ρ µ, and condition (a, p) ∈ Aβ ∗ Pβ such that (a, p) (ρ, 0), (ρ, 1) ∈ f (β). By the deﬁnition of f, there are α0, α1 η, such that (a, p) fα0 (β)(ρ) = 0 and fα1 (β)(ρ) = 1. But this is impossible, because fα0 and fα1 are compatible in the ordering of Q.

˙ It follows that each Un is still an A-name for an appropriate measure and that each of the κn ’s is still supercompact since we made each of the κn ’s indestructible under κn -directed closed forcing. This ﬁnishes the proof since in V Q we have all of the conditions that we need to work the argument from [14].

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## ARONSZAJN TREES AND THE SUCCESSORS OF A SINGULAR CARDINAL 13

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Department of Mathematical Sciences, Carnegie Mellon University, Pittsburgh, PA 15213 E-mail address: sunger@cmu.edu