Higher-dimensional forcing -------------------- Bernhard Irrgang

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3 D forcing

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Morasses

A simplified (κ,1)-morass is a structure M =<< θα | α ≤ κ >,< Fαβ | α < β ≤ κ >> satisfying the following conditions:
(P0) (a) θ0 = 1, θκ = κ+, ∀α < κ  0 < θα < κ.
(b) Fαβ is a set of order-preserving functions f : θα → θβ.
(P1) |Fαβ| < κ for all α < β < κ.
(P2) If α < β < γ, then Fαγ = {f ∘ g | f ∈ Fβγ,g ∈ Fαβ}.
(P3) If α < κ, then Fα,α+1 = {id ↾ θα,fα} where fα is such that fα ↾ δ = id ↾ δ and fα(δ) ≥ θα for some δ < θα.
(P4) If α ≤ κ is a limit ordinal, β1,β2 < α and f1 ∈ Fβ1α, f2 ∈ Fβ2α, then there are a β1,β2 < γ < α, g ∈ Fγα and h1 ∈ Fβ1γ, h2 ∈ Fβ2γ such that f1 = g ∘ h1 and f2 = g ∘ h2.
(P5) For all α > 0, θα = ⋃ {f[θβ] | β < α,f ∈ Fβα}.

Lemma
Let α < β < κ, τ1,τ2 < θα, f1,f2 ∈ Fαβ and f1(τ1) = f2(τ2). Then τ1 = τ2 and f1 ↾ τ1 = f2 ↾ τ2.

A simplified morass defines a tree < T,≺>.
Let T = {< α,ν >| α ≤ κ,ν < θα}.
For t =< α,ν >∈ T set α(t) = α and ν(t) = ν.
Let < α,ν >≺< β,τ > iff α < β and f(ν) = τ for some f ∈ Fαβ.
If s ≺ t, then f ↾ (ν(s) + 1) is uniquely determined by the lemma. So we may define πst := f ↾ (ν(s) + 1).

Lemma
The following hold:
(a) ≺ is a tree, htT (t) = α(t).
(b) If t0 ≺ t1 ≺ t2, then πt0t1 = πt1t2 ∘ πt0t1.
(c) Let s ≺ t and π = πst. If π(ν′) = τ′, s′ =< α(s),ν′ > and t′ =< α(t),τ′ >, then s′≺ t′ and πs′t′ = π ↾ (ν′ + 1).
(d) Let γ ≤ κ, γ ∈ Lim. Let t ∈ Tγ. Then ν(t) + 1 = ⋃ {rng(πst) | s ≺ t}.

Example


As example, we force ω2 ⁄→ (ω : 2)ω2 .
More precisely, we want to add a function f : [ω2]2 → ω such that {ξ < α | f(ξ,α) = f(ξ,β)} is finite for all α < β < ω2.
For a,b ⊆ ω2 set [a,b] := {< α,β >| α ∈ a, β ∈ b, β < α}.
Let P := {p : [ap,bp] → ω | ap,bp ⊆ ω2 finite }.
We set p ≤ q iff q ⊆ p and ∀α < β ∈ aq ∀ξ ∈ (bp - bq) ∩ α p(α,ξ)≠p(β,ξ).
Let π : λ → θ be an order-preserving map. Then π : λ → θ induces maps π : λ2 → θ2 and π : λ2 × ω → θ2 × ω in the obvious way:
π : λ2 → θ2, < γ,δ > ↦→ < π(γ),π(δ) >
π : λ2 × ω → θ2 × ω, < x,ε > ↦→ < π(x),ε > .

Base Case: β = 0
Then we only need to define P(1).
Let P(1) := {p ∈ P | ap,bp ⊆ 1}.

Successor Case: β = α + 1
We first define P(φβ). Let it be the set of all p ∈ P such that
(1) ap,bp ⊆ φβ
(2) fα-1[p], (id ↾ φα)-1[p] ∈ P(φα)
(3) p ↾ ((φβ \ φα) × (φα \ δ)) is injective
where fα and δ are like in (P3) in the definition of a simplified gap-1 morass.
For all ν ≤ φα P(ν) is already defined. For φα < ν ≤ φβ set P(ν) = {p ∈ P(φβ) | ap,bp ⊆ ν}. Set σst : P(ν(s) + 1) → P(ν(t) + 1),p↦→πst[p].

Limit Case: β ∈ Lim
For t ∈ Tβ set P(ν(t) + 1) = ⋃ {σst[P(ν(s) + 1)] | s ≺ t} and P(λ) = ⋃ {P(η) | η < λ} for λ ∈ Lim where σst : P(ν(s) + 1) → P(ν(t) + 1),p↦→πst[p].

Motivation

Two-dimensional applications

Three-dimensional application

Open problems