A consistency result on long cardinal sequences

For any regular cardinal $\kappa$ and ordinal $\eta<\kappa^{++}$ it is consistent that $2^{\kappa}$ is as large as you wish, and every function $f:\eta \to [\kappa,2^{\kappa}]\cap Card$ with $f(\alpha)=\kappa$ for $cf(\alpha)<\kappa$ is the cardinal sequence of some locally compact scattered space.


Introduction
If X is a locally compact, scattered Hausdorff (in short: LCS) space and α is an ordinal, we let I α (X) denote the α th Cantor-Bendixson level of X. The cardinal sequence of X, CS(X), is the sequence of the cardinalities of the infinite Cantor-Bendixson levels of X, i.e. CS(X) = |I α (X)| : α < ht -(X) , where ht -(X), the reduced height of X, is the minimal ordinal β such that I β (X) is finite. The height of X, denoted by ht(X), is defined as the minimal ordinal β such that I β (X) = ∅. Clearly ht -(X) ≤ ht(X) ≤ ht -(X) + 1.
Let κ α denote the constant κ-valued sequence of length α. In [2] it was shown that the class C(α) is described if the classes C κ (β) are characterized for every infinite cardinal κ and ordinal β ≤ α. Then, under GCH, a full description of the classes C κ (α) for infinite cardinals κ and ordinals α < ω 2 was given.

2.
Proof of theorem 1.6 Graded posets. In [3], [6], [9] and in many other papers, the existence of an LCS space is proved in such a way that instead of constructing the space directly, a certain "graded poset" is produced which guaranteed the existence of the wanted LCS-space. From these results, Bagaria, [1], extracted the notion of s-posets and established the formal connection between graded posets and LCS-spaces. For technical reasons, we will use a reformulation of Bagaria's result introduced in [10].
If is an arbitrary partial order on a set X then define the topology τ on X generated by the family {U (x), X \ U (x) : x ∈ X} as a subbase, where U (x) = {y ∈ X : y x}.
So, instead of Theorem 1.6, it is enough to prove Theorem 2.4 below.
We say that I is an ordinal interval iff there are ordinals α and β with I = [α, β). Write I − = α and I + = β.
Note that I is a cofinal tree of intervals in the sense defined in [6]. So, the following conditions are satisfied: (i) For every I, J ∈ I, I ⊂ J or J ⊂ I or I ∩ J = ∅.
(ii) If I, J are different elements of I with I ⊂ J and J + is a limit ordinal, then I + < J + .
(iv) I n+1 refines I n for each n < ω.
Now if ζ < δ, we define the basic orbit of ζ (with respect to I) as We refer the reader to [6,Section1] for some fundamental facts and examples on basic orbits. In particular, we have that The underlying set of our poset will consist of blocks. The following set B below serves as the index set of our blocks: The underlying set of our poset will be To obtain a (κ, λ, δ, L δ κ )-good poset we take Y = B S and Define the functions π : X −→ δ and ρ : X −→ λ by the formulas π( α, ν ) = α and ρ( α, ν ) = ν.
. Finally we define the orbits of the elements of X as follows: To simplify our notation, we will write o(x) = o(π(x)) and o(x) = o(π(x)).
Forcing construction. Let Λ ∈ I and {x, y} ∈ X 2 . We say that Λ separates x from y if Definition 2.6. Now, we define the poset P = P, ≤ as follows: To complete the proof of Theorem 2.4 we will use the following lemmas which will be proved later: is dense in P Since λ <κ = λ , the cardinality of P is λ. Thus, Lemma 2.7 and Lemma 2.8 above guarantee that forcing with P preserves cardinals and 2 κ = λ in the generic extension.
Let G ⊂ P be a generic filter. Put A = {A p : p ∈ G}, i = {i p : p ∈ G} and = { p : p ∈ G}. Then A = X by Lemma 2.9(a).
Finally condition 2.1(c) holds by Lemma 2.9(b). So to complete the proof of Theorem 2.4 we need to prove Lemmas 2.7, 2.8 and 2.9.
Since κ is regular, Lemma 2.7 clearly holds.
Proof of Lemma 2.9. (a) Let p ∈ P be arbitrary. We can assume that (b) Let p ∈ P be arbitrary. By (a) we can assume that x ∈ A p . Write β = π(x). Let K be a finite subset of [α, β) such that α ∈ K and I(γ, n) + ∈ K ∪ [β, δ) for γ ∈ K and n < ω.
Let q = A q , q , i q . Next we check q ∈ P . Clearly (P 1), (P 2), (P 3) and (P 5) hold for q. (P4) also holds because if y ∈ A p and γ ∈ K then either b γ q y or they are q -incompatible. To check (P6) it is enough to observe that if Λ separates b γ and y, then z = b Λ + meets the requirements of (P6).
By the construction, q ≤ p.
The rest of the paper is devoted to the proof of Lemma 2.8.
In the first part of the proof, till Claim 2.16, we will find ν < µ < κ + such that r ν and r µ are twins in a strong sense, and r ν and r µ form a good pair (see Definition 2.15). Then, in the second part of the proof, we will show that if {r ν , r µ } is a good pair, then r ν and r µ are compatible in P.
For i ∈ σ let We put Z 0 = {δ i : i ∈ σ}. Since π ′′ A △ = {δ i : i ∈ K} we have π ′′ A △ ⊂ Z 0 . Then, we define Z as the closure of Z 0 with respect to I: By Claim 2.10(a), the sequence π(x ν,i ) : ν < κ + is strictly increasing for i ∈ D ∪M. Since |Z| < κ, and | o * (x ν,k )| ≤ κ for x ν,k ∈ B S ∩A △ , we can assume that Our aim is to prove that there are ν < µ < κ + such that the forcing conditions r ν and r µ are compatible. However, since we are dealing with infinite forcing conditions, we will need to add new elements to A ν ∪ A µ in order to be able to define the infimum of pairs of elements {x, y} where x ∈ A ν \ A µ and y ∈ A µ \ A ν . The following definitions will be useful to provide the room we need to insert the required new elements. Let Assume that i ∈ σ 1 ∪ σ 2 . Let For i ∈ σ 2 , since γ(δ i ) < δ i and δ i = lim{π(x ν,i ) : ν < κ + } by Claim 2.10(a) for all i ∈ D ∪ M, we can assume that (H) π(x ν,i ) ∈ J(δ i ) \ γ(δ i ), and so π(x ν,i ) / ∈ Z, for all i ∈ D ∪ M. We will use the following fundamental facts. Claim 2.11. If x ν,i ν x ν,j then δ i ≤ δ j .
Proof. Assume that i, j ∈ K and δ i = δ j . By Claim 2.11, we have δ i < δ j . Since i ∈ F ∪ M and x ν,i ν x ν,j imply x ν,i = x ν,j and so δ i = δ j , we have that i ∈ D, and so π(x ν,i ) < δ i , cf(δ i ) = κ + and J(δ i ) + = δ i by Proposition 2.5 .
Since x ν,k = x ν,j , we have x ν,k ∈ B S , and so k ∈ K ∪ D. But as π(x ν,k ) = δ i ∈ Z we obtain k / ∈ D by (H), and so k ∈ K, which implies , and x ν,i ∈ A △ ∩ B S , we have k / ∈ D ∪ M by (G). Thus k ∈ K, and so x ν,k ∈ A △ . Hence Claim 2.14. Assume that x ν,i and x ν,j are compatible but incomparable in r ν . Let x ν,k = i ν {x ν,i , x ν,j }. Then either x ν,k ∈ A △ or δ i = δ j = δ k .
To finish the proof of Lemma 2.8 we will show that ( †) If {r ν , r µ } is a good pair, then r ν and r µ are compatible.
So, assume that {r ν , r µ } is a good pair. Write In order to amalgamate conditions r ν and r µ , we will use a refinement of the notion of amalgamation given in [6,Definition 2.4]. Let be an order-preserving injective function for some ordinal θ < κ, and for x ∈ A ′ let Since cf(γ(δ x )) = κ and |A ′ | < κ we have So, for every x ∈ A ′ , y x ∈ B S with π(y x ) < π(x). Define functions g : Y −→ A ν andḡ : Y −→ A µ as follows: Now, we are ready to start to define the common extension r = A, , i of r ν and r µ . First, we define the universe A as Clearly, A satisfies (P1). Now, our purpose is to define . Extend the definition of g as follows: g : A −→ A ν is a function, We introduce two relations on A p ∪ A q ∪ Y as follows: Then, we put The following claim is well-known and straightforward.
The following straightforward claim will be used several times in our arguments.

Sublemma 2.19.
is a partial order on Proof. We should check that ν is transitive, because it is trivially reflexive and antisymmetric. So let s t u. We should show that s u. Since x z implies g(x) ν g(z), we have g(s) ν g(t) ν g(u) and so (⋆) g(s) ν g(u).
, then (⋆) implies s R1 u or s ν u or s µ u, which implies s u by (⋆). So we can assume that s ∈ A ν (the case s ∈ A µ is similar), and so u ∈ Y or u ∈ A µ .
Assume that t ∈ Y . Then s R2 t, and so there is a ∈ A △ such that g(s) ν a ν g(t). Since t u implies g(t) ν g(u), we have g(s) ν a ν g(u), and so s R2 u. Thus s u.
If t ∈ Y , then s R2 t, and so there is a ∈ A △ such that g(s) ν a ν g(t). Since t u implies g(t) ν g(u), we have g(s) ν a ν g(u), and so s R2 u. Thus s u.
Assume that t ∈ A ν ∪ A µ . Then t R2 u, and so there is a ∈ A △ such that g(t) ν a ν g(u). Then g(s) ν a ν g(u), and so s R2 u. Thus s u.
So, by the previous Sublemma 2.19 and by the construction, (P2) and (P3) hold for .
Next define the function i : A 2 −→ A ∪ {undef} as follows: Let i{s, t} = undef if s and t are not -compatible. If s and t are compatible, then so are g(s) and g(t) because x y implies g(x) ν g(y) by Claim 2.18. Moreover i ν {s, t} = i µ {s, t} for {s, t} ∈ A △ 2 by condition (C)(e), so the definition above is meaningful, and gives a function i.

If x /
∈ B S then x ∈ M and γ(δ x ) < π(x) < δ x by (H), and so Then x ∈ F and so
Proof. We should distinguish two cases.
To check (P4) we should prove that i{s, t} is the greatest common lower bound of s and t in A, .
Assume first that s and t are not twins. Note that by Claim 2.18, g(s) and g(t) are ν -compatible. Write v = i ν {g(s), g(t)}.
Since v = g(v) ν g(s) and v ∈ A △ , we have v R2 s. Similarly v R2 t. Thus v is a common lower bound of s and t.
To check that v is the greatest lower bound of s, t in A, let w ∈ A, w s, t. Then g(w) ν g(s), g(t). Thus g(w) ν i ν {g(s), g(t)} = v.
To check (P5) observe that g(s) and g(t) are incomparable in A ν . Indeed, g(s) ν g(t) implies v = g(s) ∈ A △ and so g(s) ν g(t) implies s R2 t, which contradicts our assumption that s and t are -incomparable. Thus, by applying (P5) in r ν , π(v) ∈ f{g(s), g(t)}.
If g(s) and g(t) are ν -comparable then δ g(s) = δ g(t) , because otherwise we would infer from Claim 2.12 that s, t are -comparable, which is impossible. Now assume that g(s) and g(t) are ν -incomparable. If δ v < δ g(s) , then there is a ∈ A △ ∩ B S with v ν a ν g(s) by Claim 2.12. Thus v = i ν {a, g(t)} and so v ∈ A △ by Claim 2.13. Thus δ v = δ g(s) , and similarly δ v = δ g(t) .
To check (P4) first we show that y v s, t. Indeed g(v) ν g(s) implies y v R1 s. We obtain y v R1 t similarly. Let w s, t. Assume first that δ g(w) < δ v . Since w s, t we have g(w) ν g(s), g(t) by Claim 2.18 and hence g(w) ν i ν {g(s), g(t)} = v. By Claim 2.12 there is a ∈ A △ such that g(w) ν a ν v. Thus w R2 y v .
If s and t are twins, then s ∈ A ′ implies that i{s, t} = y s and we can proceed as above in Case 2.2.
We should find v ∈ A such that s v t and π(v) = Λ + . Note that since s t, we have δ g(s) ≤ δ g(t) by Claim 2.11. We can assume that {s, t} / ∈ A ν 2 ∪ A µ 2 because r ν and r µ satisfy (P6).
Thus Λ separates a from g(t). Applying (P6) in r ν for a and g(t) and Λ we obtain b ∈ A ν such that a ν b ν g(t) and π(b) = Λ + .
Thus g(s) ν b ν g(t) implies s R2 b R2 t, and so s b t.
If Λ + = π(a), then we are done because g(s) ν a ν g(t) implies s a t.
We will see that this case is not possible.
As s t and [γ(δ s ), J(δ s ) + ) ∩ Z = ∅ we have that t / ∈ A µ . Since s ∈ A ν , s t and δ s = δ g(t) we have t / ∈ Y , and so t ∈ A ν , which was excluded.
By means of a similar argument, we can show that s ∈ A µ is also impossible.
Thus we proved that r is a common extension of r ν and r µ . This completes the proof of Lemma 2.8, i.e. P satisfies κ + -c.c.