DIAMOND AND ANTICHAINS

JAMES CUMMINGS AND ERNEST SCHIMMERLING

Abstract. It is obvious that ♢ implies the existence of an antichain of stationary sets of cardinality 2^{ℵ_{1}}, which is the largest possible cardinality. We show that the obvious antichain is not maximal and find a less obvious extension of it by ℵ_{2} more stationary sets.

Let A = ⟨A_{α} | α < ω_{1}⟩ be a ♢-sequence. For X ⊆ ω_{1}, let

S_{X} = {α < ω_{1} | X ∩ α = A_{α}}.

Then each S_{X} is stationary. Moreover, if X ≠ Y, then S_{X} ∩ S_{Y} is bounded in ω_{1}. Thus ⟨S_{X} | X ⊆ ω_{1}⟩ is an antichain of stationary sets of cardinality 2^{ℵ_{1}}. Part of what we will show is that ⟨S_{X} | X ⊆ ω_{1}⟩ is not a maximal antichain.

First we record some easy observations. In the Boolean completion of P(ω_{1})/NS, S_{X} is the Boolean value of the sentence

If j_{G} : V ⟶ M is the generic ultrapower embedding (where the wellfounded part of M is is identified with its Mostowski collapse), then

j_{G}(A)_{ω_{1}^{V}} = X.

Moreover, the Boolean value of the sentence

j_{G}(A)_{ω_{1}^{V}} ∈ V

is precisely the join of the antichain ⟨S_{X} | X ⊆ ω_{1}⟩.

If X is a set of countable ordinals, then we write acc(X) for the set of accumulation points of X, that is {α < ω_{1} | sup(X ∩ α) = α}.

Definition 1. Let T = {α < ω_{1} | S_{A_{α}} ∩ acc(A_{α}) = {α}}.

In other words, T is the collection of β < ω_{1} such that β = sup(A_{β}) but there are no accumulation points α < β of A_{β} such that A_{α} = A_{β} ∩ α.

1991 Mathematics Subject Classification. 03E.

The research of the authors is partially supported by the NSF.

The second author was partially supported by IHES during his June 5-25, 2002, visit. He thanks Stevo Todorcevic and Boban Velickovic for conversations during the visit on the main question answered in this report.

Lemma 2. T is stationary and |T ∩ S_{X}| ≤ 1 for all X ⊆ ω_{1}. In particular, ⟨S_{X} | X ⊆ ω_{1}⟩ is not a maximal antichain in NS.

Proof. Consider an arbitrary club C ⊆ ω_{1}. Then acc(C) is a club subset of C and S_{C} ∩ acc(C) is stationary. Let α be the least element of S_{C} ∩ acc(C). Then α is the unique element of T ∩ S_{C} ∩ acc(C). In particular, α ∈ T ∩ C. This shows that T is stationary.

Now let X ⊆ ω_{1} and α < β be elements of S_{X}. Then A_{α} = A_{β} ∩ α. It is enough to show that if β ∈ T, then α ∉ T. Assume β ∈ T. Because α ∈ β ∩ S_{A_{β}} and β ∈ T, we may conclude that α ∉ acc(A_{β}). So α ∉ acc(A_{α}). Hence α ∉ T. □

Let us record the easy observation that T is the Boolean value of the sentence

j_{G}(A)_{ω_{1}^{V}} is unbounded in ω_{1}^{V} but if α < ω_{1}^{V} is an accu-
mulation point of j_{G}(A)_{ω_{1}^{V}}, then A_{α} ≠ j_{G}(A)_{ω_{1}^{V}} ∩ α.

We have shown that our ♢-sequence is also a ♢(T^{c}) sequence where T^{c} = ω_{1} − T. Many useful examples are obtained by the standard trick of redefining A_{α} for α ∈ T. For example, if α ∈ T, then let A′_{α} be an unbounded subset of α of order type ω, while if α ∈ T^{c}, then keep A′_{α} = A_{α}. Observe that T forces

j_{G}(A′)_{ω_{1}^{V}} is unbounded in ω_{1}^{V} and has order type ω. Note that in this example, if T′ is computed from A′ the same way that T was computed from A, then T ⊆ T′.

Our next goal is to define sets T_{δ,ρ} so that

⟨T_{δ,ρ} | δ < ω_{2} & ρ < ω_{1}⟩^{⌢}⟨S_{X} | X ⊆ ω_{1}⟩

is an antichain of stationary sets. For ω_{1} ≤ δ < ω_{2}, let e_{δ} map ω_{1} onto δ with e_{δ}(0) = 0. This choice will remain fixed for the rest of our note.

Definition 3. By recursion on δ < ω_{2}, define ∂^{δ}X for X ⊆ ω_{1} as follows. Let

∂^{0}X = acc(X).

If δ = γ + 1, then let

∂^{δ}X = {α < ω_{1} | ot(α ∩ S_{X} ∩ ∂^{γ}X) = α}.

If δ is a countable limit ordinal, then let

∂^{δ}X = ⋂_{γ<δ}∂^{γ}X.

If δ is an uncountable limit ordinal, then let

∂^{δ}X = ▽_{ι<ω_{1}}∂^{e_{δ}(ι)}X.

We remark that there is an alternative approach in which the countable limit stages in the definition of ∂^{δ}X are handled using diagonal intersections instead of intersections. There is very little difference between the two approaches, and so we simply picked one randomly here.

Lemma 4. Let δ < ω_{2}.

(1) For all X ⊆ ω_{1} and α < ω_{1},

α ∩ ∂^{δ}X = α ∩ ∂^{δ}(α ∩ X).

If, in addition, α ∈ acc(X) and A_{α} = X ∩ α, then

α ∈ ∂^{δ}X ⟺ α ∈ ∂^{δ}A_{α}.

(2) For all unbounded X ⊆ ω_{1}, ∂^{δ}X is club in ω_{1}.

(3) For all ε such that δ ≤ ε < ω_{2}, there exists ι < ω_{1} such that

(∂^{ε}X) − ι ⊆ ∂^{δ}X

whenever X ⊆ ω_{1}.

(4) If C is a club subset of ω_{1}, then

∂^{δ}C ⊆ C.

The proof of Lemma 4 is standard. Let us write ι_{δ,ε} for the least ι as in Part 3 of Lemma 4. Our requirement that e_{δ}(0) = 0 was to ensure that ι_{0,δ} = 0.

Definition 5. For δ < ω_{2} and ρ < ω_{1}, let

T_{δ,ρ} = {α < ω_{1} | α ∈ ∂^{δ}A_{α} & ot (α ∩ S_{A_{α}} ∩ ∂^{δ}A_{α}) = ρ} .

In other words, T_{δ,ρ} is the set of α < ω_{1} such that S_{A_{α}} ∩ ∂^{δ}A_{α} has order type ρ + 1 and greatest element α. Note that T_{0,0} = T.

Theorem 6. ⟨T_{δ,ρ} | δ < ω_{2} & ρ < ω_{1}⟩^{⌢}⟨S_{X} | X ⊆ ω_{1}⟩ is an antichain of stationary sets. Moreover, if P and Q are distinct elements of this antichain, then P ∩ Q is bounded in ω_{1}.

Theorem 6 follows immediately from Lemmas 7 – 12 below.

Lemma 7. For δ < ω_{2} and ρ < ω_{1}, T_{δ,ρ} is stationary in ω_{1}.

Proof. Let C be an arbitrary club subset of ω_{1}. Then ∂^{δ}C is a club subset of C and S_{C} ∩ ∂^{δ}C is stationary. Let α be the ρ-th element in the increasing enumeration of S_{C} ∩ ∂^{δ}C. Then α is the unique element of T_{δ,ρ} ∩ S_{C} ∩ ∂^{δ}C. In particular, α ∈ T_{δ,ρ} ∩ C. □

Lemma 8. Let δ < ω_{2} and ρ < σ < ω_{1}. Then

T_{δ,ρ} ∩ T_{δ,σ} = ∅.

Proof. Otherwise, there would be an ordinal α such that α is both the ρ-th and the σ-th element of S_{A_{α}} ∩ ∂^{δ}A_{α}. This is absurd. □

Lemma 9. Let δ < ω_{2} and ρ < ω_{1}. Then

|T_{δ,ρ} ∩ S_{X}| ≤ 1

whenever X ⊆ ω_{1}.

Proof. Let α < β be elements of S_{X}. Then A_{α} = A_{β} ∩ α. For contradiction, suppose that both α and β are elements of T_{δ,ρ}. Then, for some σ < ρ, α is the σ-th element of β ∩ S_{A_{β}} ∩ ∂^{δ}A_{β}. By Part 1 of Lemma 4 and the fact that A_{α} = A_{β} ∩ α, we have that that α ∈ T_{δ,σ}. This contradicts Lemma 8. □

Lemma 10. Let δ < ε < ω_{2} and ρ, σ < ω_{1}. Suppose that ε is a limit ordinal. If ε is countable, then

T_{δ,ρ} ∩ T_{ε,σ} ⊆ (δ + 1) ∪ {ρ}.

If, on the other hand, ε is uncountable and e_{ε}(ι) = δ + 1, then

T_{δ,ρ} ∩ T_{ε,σ} ⊆ ι ∪ {ρ}.

Proof. First suppose that ε is uncountable. Let

α ∈ T_{δ,ρ} ∩ T_{ε,σ}

such that α > ι. Because α ∈ T_{ε,σ},

α ∈ ∂^{δ+1}A_{α}.

So

ot(α ∩ S_{A_{α}} ∩ ∂^{δ}A_{α}) = α.

Because α ∈ T_{δ,ρ},

ot(α ∩ S_{A_{α}} ∩ ∂^{δ}A_{α}) = ρ.

Thus α = ρ.

The proof when ε is countable is essentially the same, so we omit it. □

Lemma 11. Let γ < ω_{2} and ρ, σ < ω_{1}. Then

T_{γ,ρ} ∩ T_{γ+1,σ} ⊆ {ρ}.

Proof. Let α ∈ T_{γ,ρ} ∩ T_{γ+1,σ}. Then α is both the ρ-th and the α-th element of S_{A_{α}} ∩ ∂^{γ}A_{α}. So α = ρ. □

Lemma 12. Suppose that δ < ε and ρ, σ < ω_{1}. Then T_{δ,ρ} ∩ T_{ε+1,σ} is bounded in ω_{1}.

Proof. Assume for contradiction that T_{δ,ρ} ∩ T_{ε+1,σ} is unbounded in ω_{1}. Consider an arbitrary α ∈ T_{δ,ρ} ∩ T_{ε+1,σ}. Then α is the ρ-th element of

S_{A_{α}} ∩ ∂^{δ}A_{α}

and the σ-th element of

S_{A_{α}} ∩ ∂^{ε+1}A_{α}.

So α is the α-th element of

S_{A_{α}} ∩ ∂^{ε}A_{α}

by the definition of ∂^{ε+1}A_{α}.

Recall that by Part 3 of Lemma 4, there is an ordinal ι = ι_{δ,ε} such that for all α > ι,

∂^{ε}A_{α} − ι ⊆ ∂^{δ}A_{α}.

Let us assume that α is large enough that α > ι and ι + α = α. Then α is the α-th element of

(S_{A_{α}} ∩ ∂^{ε}A_{α}) − ι.

Hence α is the α-th element of

S_{A_{α}} ∩ ∂^{δ}A_{α}.

Therefore α = ρ. □

That completes the proof of Theorem 6. We close with some remarks and a conjecture.

It is natural to consider the sets T_{δ,ρ} as being ordered lexicographically according to the index (δ, ρ). The reader may wonder why we have been listing sets of the form S_{X} after all those of the form T_{δ,ρ}. The reason is as follows. Let X be unbounded in ω_{1}. Recall that forcing below S_{X} produces a generic ultrapower in which

X = j_{G}(A)_{ω_{1}^{V}} .

Observe that ω_{1} is the ω_{1}-st element of ∂^{δ}X for all δ < ω_{2}. Thus, in the obvious sense,

ω_{1} ∈ ∂^{ω_{2}}j_{G}(A)_{ω_{1}^{V}} ,

which goes beyond what happens if we force below any of the sets T_{δ,ρ}. We conjecture that the join of the antichain

⟨T_{δ,ρ} | δ < ω_{2} & ρ < ω_{1}⟩

is the Boolean value of the sentence j(A)_{ω_{1}^{V}} ∉ V and j(A)_{ω_{1}^{V}} is unbounded in ω_{1}^{V}, perhaps under one of the familiar hypotheses on the canonical functions.

It is easy to see that ⟨T_{δ,ρ} | δ < ω_{2} & ρ < ω_{1}⟩^{⌢}⟨S_{X} | X ⊆ ω_{1}⟩ is not necessarily a maximal antichain. The example we have in mind involves redefining A on T again as follows. Let A′′_{α} be a subset of ω that codes a wellorder of type α whenever α ∈ T, while if α ∈ T^{c}, then let A′′_{α} = A_{α}. Then T forces

j(A′′)_{ω_{1}^{V}} ∉ V and j(A′′)_{ω_{1}^{V}} ⊆ ω. Let R_{β} = {α < ω_{1} | sup(A_{α}) = β} for β < ω_{1}. It is obvious that

⟨R_{β} | β < ω_{1}⟩ is an antichain the join of which is the the Boolean value of

j(A)_{ω_{1}^{V}} is bounded in ω_{1}^{V}.

Thus our conjecture above says exactly that

⟨T_{δ,ρ} | δ < ω_{2} & ρ < ω_{1}⟩^{⌢}⟨S_{X} | X ⊆ ω_{1} & sup(X) = ω_{1}⟩^{⌢}⟨R_{β} | β < ω_{1}⟩

is a maximal antichain. Even if the conjecture turns out to be true, we still would not know how to describe the Boolean value of

j(A)_{ω_{1}^{V}} ∉ V and j(A)_{ω_{1}^{V}} is bounded in ω_{1}^{V}. as the join of a specific antichain.

  E-mail address: jcumming@andrew.cmu.edu
  E-mail address: eschimme@andrew.cmu.edu

Department of Mathematical Sciences, Carnegie Mellon University, Pittsburgh, PA 15213-3890, USA