ω_{1}-compactness in Type I manifolds

Sina Greenwood^{∗}and Peter Nyikos^{†}

November 19, 2003

Abstract

In this paper we establish that any well-pruned ω_{1}-tree, T, admits an ω_{1}-compact Type I manifold if T does not contain an uncountable antichain. If T does contain an uncountable antichain, it has been shown that whether or not T admits an ω_{1}-compact manifold is undecidable in ZFC.

AMS classification: Primary 54D99; 57N15; Secondary 03E65; 05C05. Keywords: Type I manifold; Υ-tree; ω_{1}-compact.

1 Introduction

In [G] the question of whether or not an arbitrary tree admits a Type I manifold was examined. First, an arbitrary tree admits a Type I manifold if and only if it is an ω_{1}-tree and well-pruned. It was found that under the axiom ♢, any well-pruned ω_{1}-tree admits an ω_{1}-compact manifold, whereas under the axiom (∗), if a well-pruned tree T admits an ω_{1}-compact manifold, then T does not contain an uncountable antichain. It was also shown, in ZFC, that any wellpruned ω_{1}-tree which did not contain an uncountable antichain or a Souslin subtree admits an ω_{1}-compact manifold. We now show that it does not matter whether T contains a Souslin subtree.

A tree, ⟨T, ≤⟩, which we will denote simply as T, is a partially ordered set such that for each t ∈ T the set of predecessors of t, {s ∈ T : s < t}, is wellordered. We will assume throughout that T has a single least element called the root of T.

Definition 1.1. Let T be a tree.

(a) If t ∈ T, t̂= {s ∈ T : s < t}.

(b) For each ordinal α, the α-th level of T, or T (α), is the set {t ∈ T (α) : t̂ has order type α}.

∗University of Auckland, Private Bag 92019, Auckland, New Zealand. Funded by: New Zealand Science and Technology Foundation

†University of South Carolina, Columbia, SC 29208, USA

(c) T_{α} is the subtree ⋃_{β<α} T (β).

(d) If t ∈ T (α) for some ordinal α, t^{+} = {s ∈ T (α + 1) : s > t}.

(e) The height of T is the least ordinal α such that T (α) = ∅.

An ω_{1}-tree T has height ω_{1} and each level T (α) is countable.

A space X is of Type I if it is the union of an ω_{1}-sequence Σ = {U_{α} : α < ω_{1}} of open subspaces such that \overline{U_{α}} ⊂ U_{β} whenever α < β, and such that \overline{U_{α}} is Lindelöf for all α. If in addition U_{α} = ⋃_{β<α} U_{β} for any limit ordinal α, then Σ is a canonical sequence for X.

We will assume from now on that M is a nonmetrisable manifold. Given any canonical sequence Σ for a manifold M, it is possible to relate Σ to an ω_{1}-tree. Nyikos [N] defines such a tree, Υ(Σ), as follows:

Definition 1.2. Let M be a Type I manifold, and let Σ = ⟨U_{α} : α < ω_{1}⟩ be canonical for M. The tree of non-metrisable-component boundaries associated with Σ, denoted Υ(Σ), is the collection of all sets of the form ∂C such that C is a nonmetrisable component of M ∖ \overline{U_{α}} for some α, with the following order: if τ, σ ∈ Υ(Σ), then τ < σ iff σ is a subset of a component whose boundary is τ.

We usually denote this tree by just Υ if Σ is clear, and refer to it as an Υ-tree. We say that a tree admits a manifold with property P, if there exists a Type I manifold with property P, and with an Υ-tree isomorphic to T.

2 ω_{1}-compactness

Definition 2.1. A space is ω_{1}-compact if it does not contain an uncountable closed discrete subset.

Theorem 2.2. If T is a well-pruned ω_{1}-tree which does not contain an uncountable antichain, then there exists an ω_{1}-compact Type I manifold whose Υ-tree is T.

Proof. We will use induction to construct a manifold M by defining an open Lindelöf subset U_{α} for each α ∈ ω_{1}, so that ⋃_{α∈ω_{1}} U_{α} is the underlying set for M, and Σ = ⟨U_{α} : α ∈ ω_{1}⟩ is a canonical sequence.

Let U_{0} = (−1, 1) × (−1, 0) with the usual topology.

We will use the following notation:

• For each t ∈ T let I_{t} be a copy of (−1, 1) × [0, 1) with the usual topology, and denote the point in I_{t} corresponding to ⟨x, y⟩ by I_{t}⟨x, y⟩. (For each α the underlying set for U_{α} is ∪{I_{s} : s < t, t ∈ T (α)} ∪ U_{0}, and (−1, 1) × {0} will be the boundary component in Υ(Σ) corresponding to t.)

• For each α ∈ ω_{1} and t ∈ T (α), if α is a successor ordinal define i_{t} : I_{t} → (−1, 1) × [α, α + 1) by i_{t}(I_{t}⟨x, y⟩) = ⟨x, α + y⟩. If α is a limit ordinal define j_{t} : I_{t} → (−1, 1) × [1, 2) by j_{t}(I_{t}⟨x, y⟩) = ⟨x, 1 + y⟩.

• For each t ∈ T let B_{t} = (⋃{I_{s} : s ≤ t} ∪ U_{0}). In each case B_{t} will be associated with an embedding h_{t} : B_{t} → (−1, 1) × (−1, α + 1), where t ∈ T (α), and will therefore have a coordinate system determined by this embedding. We will denote h_{t}^{−1}(⟨x, y⟩) by ⟨x, y⟩_{t}.

• For each t ∈ T let {t^{n} : n < |t^{+}|} denote the successors of t.

• For each successor ordinal α, pick a homeomorphism

f_{α} : (−1, 1) × (−1, α + 1) → (−1, 1) × (−1, α),

satisfying the following properties:

(i) f_{α}((−1, 1) × (α, α + 1)) = A, where

A = \{⟨x, α − 1 + y⟩ : (−1 < x < 0) ∧ (\frac{1}{2}(x^{2} + 1) < y < \frac{1}{2} ( \sqrt[+]{|x|} + 1))\};

(ii) f_{α}((−1, 1) × ({α}) = ∂A, where

∂A = \{⟨x, α − 1 + y⟩ : (−1 < x < 0) ∧ (\frac{1}{2}(x^{2} + 1) = y) ∨ (y = \frac{1}{2} ( \sqrt[+]{|x|} + 1))\};

(iii) f_{α}((−1, 1) × (α − 1, α) = (−1, 1) × (α − 1, α) ∖ (A ∪ ∂A);

(iv) ∀x ≤ 0, f_{α}(\{x\} × (α − 1, α)) = \{x\} × (α − 1, α − 1 + \frac{1}{2}(x^{2} + 1));

(v) ∀x > 0, f_{α}({x} × (α − 1, α)) ⊂ (−x, x) × (α − 1, α); and

(vi) f_{α} ↾ (−1, 1) × (0, α − 1] is the identity.

Denote the root of T by r and define h_{r} : B_{r} → (−1, 1) × (−1, 1) to be the identity on U_{0} and let h_{r}(I_{r}⟨x, y⟩) = ⟨x, y⟩ for each I_{r}⟨x, y⟩.

Suppose α is a successor ordinal, U_{α} has been defined and for each t ∈ T (α − 1), h_{t} has been defined. Define h_{t^{0}} : B_{t^{0}} → (0, 1) × (0, α + 1) by letting h_{t^{0}} ↾ B_{t} = h_{t} and h_{t^{0}} ↾ I_{t^{0}} = i_{t^{0}}. If t has more than one successor, we will now define for each n > 0 a function

φ_{t^{n}} : (B_{t} ∪ ⋃_{m≤n} I_{t^{m}} ) → (−1, 1) × (−1, α + 1)

and h_{t^{n}} is then φ_{t^{n}} ↾ B_{t^{n}}.

Let φ_{t^{0}} = h_{t^{0}}. If φ_{t^{n}} has been defined, let φ_{t^{n+1}} ↾ (B_{t} ∪ ⋃_{m≤n} I_{t^{m}}) = f_{α} ∘ φ_{t^{n}} and φ_{t^{n+1}} ↾ I_{t^{n+1}} = i_{t^{n+1}}.

Topologise U_{α+1} so that each φ_{t^{n}} is a homeomorphism. Note that this is well defined and consistent with the topology on B_{t}. Let U_{α+1} = ⋃_{n<|t^{+}|} B_{t^{n}} have the direct limit topology.

If α is a limit ordinal, let U_{α} = ∪_{β<α}U_{β} with the direct limit topology. It remains to define U_{α+1}.

Consider the equivalence classes: [t] = {s ∈ T (α) : ŝ = t̂}. For each equivalence class pick a member t = t^{0} and denote the other members t^{1}, t^{2} etc. Pick a homeomorphism

g_{α} : (−1, 1) × (−1, α + 1) → (−1, 1) × (−1, 2)

which preserves the first coordinate of each point, preserves horizontal lines, and such that g_{α}((−1, 1) × (−1, α]) = (−1, 1) × (−1, 1], and define e_{[t]} : ⋃_{s<t} B_{s} → (−1, α) by

e_{[t]}(I_{s}⟨x, y⟩) = h_{s}(I_{s}⟨x, y⟩).

For each equivalence class [t], define h_{t} : B_{t} → (0, 1) × (0, α + 1) by letting h_{t} ↾ ⋃_{s<t} B_{s} = e_{[t]} and h_{t} ↾ I_{t} = i_{t}.

If |[t]| > 1 we define for each n > 0 a function

ψ_{t^{n}} : (∪_{s<t}B_{s}) ∪ (∪_{m≤n}I_{t^{m}}) → (−1, 1) × (−1, 2)

and h_{t^{n}} will be (g_{α}^{−1} ∘ ψ_{t^{n}}) ↾ B_{t^{n}}. Let ψ_{t} = g_{α} ∘ h_{t} and if ψ_{t^{n}} has been defined let

ψ_{t^{n+1}} ↾ ∪_{s<t}B_{s} ∪ (∪_{m≤n}I_{t^{m}}) = f_{1} ∘ ψ_{t^{n}}

and ψ_{t^{n+1}}|I_{t^{n+1}} = j_{t^{n+1}}. Topologise U_{α+1} so that each ψ_{t^{n}} is a homeomorphism.

For each α ∈ ω_{1}, U_{α} is Lindelöf since T (α) is countable and for each t ∈ T (α), B_{t}∖I_{t} is homeomorphic to (−1, 1)×(−1, α). U_{α} is connected since U_{0} ⊂ B_{t}∖I_{t}.

Clearly M is a manifold with Υ(Σ) = T.

We now show that M is ω_{1}-compact. For any p ∈ M, if s = min{t ∈ T; p ∈ B_{s}}, let p = ⟨p(x), p(y)⟩_{s}. Then for every t > s, if p = ⟨x′, y′⟩_{t}, |x′| ≤ |p(x)| by properties (iv) and (v) of f_{α}. Suppose X is an uncountable discrete set, then there exist a, b ∈ (−1, 1) such that Y = {p ∈ X : a < p(x) < b} is uncountable.

In the order inherited from T, Y is a tree in which every antichain is countable because it is an antichain of T. Since Y is uncountable, it is an ω_{1}-tree. Hence there exists α < ω_{1} and t ∈ T (α) such that B_{t} ∩ Y is infinite, and since h_{t}^{−1}((a, b) × (1, α + 1)) has countably compact closure, B_{t} ∩ Y has a limit in M. Thus M does not contain an uncountable closed discrete set.

If T contains a Souslin subtree but is not Souslin, then T has at most countably many uncountable paths which cease to branch at some level α say. The manifold constucted in the proof above ensures that for each path P, P ∖ T_{α+1} is ω_{1}-compact, as well as the Souslin subtree.

♢ implies that any well-puned ω_{1}-tree, T, admits an ω_{1}-compact manifold. ♢ is needed for the case when T contains an uncountable antichain. We will consider one case in which ♣ is adequate to obtain an ω_{1}-compact manifold from a tree with an uncountable antichain.

Let ♣_{2} denote the following axiom: there is a family {λ_{α} : α < ω_{1}} of countable limit ordinals, and a family {A_{α} : α < ω_{1}} such that:

1. both dom(A_{α}) and ran(A_{α}) are cofinal subsets of λ_{α}, of order type ω; and,

2. if X is a subset of ω_{1} × ω_{1} that meets each subset of the form [α, ω_{1}) × [α, ω_{1}), then there exists α such that A_{α} ⊂ X.

We will call such a family {A_{α} : α < ω_{1}}, a ♣_{2}-sequence.

Theorem 2.3. ♣ implies ♣_{2}.

We use the following formulation of ♣: there exists a family

{L_{γ} : γ < ω_{1}, γ is a limit ordinal}

such that each L_{γ} is a cofinal subset of γ of order type ω, and such that if X is an uncountable subset of ω_{1}, then there exists γ such that L_{γ} ⊂ X.

We will make use of the lemma below, [D, Theorem 4.3].

Lemma 2.4. If

{L_{γ} : γ < ω_{1}, γ is a limit ordinal}

witnesses ♣, and X is an uncountable subset of ω_{1}, then S = {γ : L_{γ} ⊂ X} is stationary.

Proof of Theorem 2.3. Let φ : ω_{1} → ω_{1} × ω_{1} be a bijection, and let

C = {γ ∈ ω_{1} : φ([0, γ)) = [0, γ) × [0, γ)}.

A standard leapfrog argument shows C is a club. Let C′ be the derived set of C and for each γ ∈ C′ let B_{γ} = φ(L_{γ}). Since L_{γ} is cofinal in γ of order type ω, at most finitely many members of B_{γ} can be in [0, α) × [0, α) for any α < γ; hence at least one projection of B_{γ} is unbounded in [0, γ). Let {λ_{α} : α < ω_{1}} list in order all members of C′ such that both projections are unbounded ω-sequences in λ_{α}. Let A_{α} = B_{λ_{α}} for all α. Claim. A = {A_{α} : α < ω_{1}} witnesses ♣_{2}.

Proof of Claim. Suppose X meets every set of the form [α, ω_{1})×[α, ω_{1}). Let Y = {(β_{α}, δ_{α}) : α ∈ ω_{1}} be a subset of X such that

α ≤ min{β_{α}, δ_{α}} ≤ max{β_{α}, δ_{α}} < min{β_{ε}, δ_{ε}}

whenever α < ε.

Since φ^{−1}(Y) is uncountable, there is a stationary set S such that L_{σ} ⊂ φ^{−1}(Y) for all σ ∈ S. Let γ ∈ C′ ∩ S. Then B_{γ} ⊂ Y, and at least one projection of B_{γ} is unbounded in γ. This forces both dom(B_{γ}) and ran(B_{γ}) to be ω-sequences with supremum γ. So B_{γ} ∈ A and B_{γ} ⊂ X, as desired.

The following example demonstrates that the full strength of ♢ in Theorem [G, Theorem 5.4], need not be necessary.

Example 2.5. Assume ♣. There exists an ω_{1}-compact Type I manifold M, such that if Σ is any canonical sequence for M, Υ(Σ) contains an uncountable antichain.

We will define a Type I manifold M with the underlying set:

{⟨α, β⟩ ∈ 𝕃^{+} × 𝕃^{+} : α, β ≠ 0} ∖ {⟨α, β⟩ : α ∈ [1, ω_{1}), β ∈ ω_{1}}.

Let T be the topology on M as a subspace of 𝕃^{+} × 𝕃^{+} with the usual topology. For each α < ω_{1} let U_{α} = (0, α)^{2} ∩ M. We will define a topology on each U_{α} such that ∂U_{α} = {α} × [(0, α) ∩ M] ∪ (0, 1) × {α}, U_{α} ≈ ℝ^{2}, and ⟨U_{α} : α ∈ ω_{1}⟩ is canonical.

Let M ∖⋃_{α<ω_{1}} ∂U_{α} have the topology inherited from T . Note that U_{0} = ∅, and U_{1} has the usual topology, and hence U_{1} ≈ ℝ^{2}.

Suppose α is a successor ordinal, the topology on U_{α} has been defined, and U_{α} ≈ ℝ^{2}. Then let U_{α+1} be the connected sum of (0, α − \frac{1}{2}]^{2} ∩ M, and U_{α+1} ∖ (0, α − \frac{1}{2})^{2} with the topology inherited from T .

If α is a limit, U_{β} has been defined for each β < α, and U_{β} ≈ ℝ^{2} for each β, let U_{α} = ∪_{β<α}U_{β} with the direct limit topology. Since each U_{β} ≈ ℝ^{2}, U_{α} ≈ ℝ^{2} [B].

To complete the construction it remains to define the topology on U_{α+1} when α is a limit. We need only define a neighbourhood base for each x ∈ ∂U_{α}.

Let A = {A_{α} : α ∈ L} be a ♣_{2}-sequence. Choose a function ϕ_{α} : U_{α+1} → U_{α+1} such that:

(a) ϕ_{α}|U_{α} is an embedding;

(b) ϕ_{α} is the identity on the sets \{(0, 1)×\{β\} : β ≤ α+1\} and (0, \frac{1}{4})×(0, α+1];

(c) for each ⟨x, y⟩ ∈ U_{α+1}, ϕ_{α}(⟨x, y⟩) = ⟨x′, y⟩ for some x′ ≤ x;

(d) if β ∈ domA_{α}, A_{α} ∈ A, then ϕ_{α}({1} × (β, β + 1)) and ϕ_{α}({α} × (β, β + 1)) are circular arcs in (0, 1) × (β, β + 1), such that ϕ_{α} (⟨1, β + \frac{1}{2}⟩) = ⟨\frac{1}{2}, β + \frac{1}{2}⟩, ϕ_{α}(⟨α, β + \frac{1}{2}⟩) = ⟨\frac{3}{4}, β + \frac{1}{2}⟩, ϕ_{α}(\{α + 1\}, (β, β + 1)) = (\{2\} × (β, β + 1)), lim_{x→0}ϕ_{α}(⟨α, β + x⟩) = ⟨1, β⟩, and lim_{x→1}ϕ_{α}(⟨α, β + x⟩) = ⟨1, β + 1⟩.

Such a function ϕ_{α} exists since domA_{α} is countable. Topologise U_{α+1} so that ϕ_{α} is an embedding. Clearly U_{α+1} ≈ ℝ^{2}.

This completes the construction.

M is a manifold, since if x ∈ M then x ∈ U_{α} for some α, and since U_{α} ≈ ℝ^{2}, x has a Euclidean neighbourhood.

Clearly M is Type I and Σ = ⟨U_{α} : α ∈ ω_{1}⟩ is a canonical sequence for M. Υ(Σ) contains an uncountable antichain. Υ(Σ) consists of a path P, which

may be thought of as the main trunk of Υ(Σ), with a path emanating from P at each successor level. These paths do not themselves branch. By choosing a member of Υ(Σ) from each of the paths off P, we may obtain an uncountable antichain.

We now establish that M is ω_{1}-compact. Suppose Z is a closed uncountable discrete subset of M. Then Z ⊄ U_{α} for any α. We consider 3 cases.

1. Z ∩ (0, 1) × 𝕃^{+} is uncountable.

For some a, b ∈ (0, 1), the set Z′ = {⟨x, y⟩ ∈ Z : a < x < b} is uncountable. If a ∈ (0, \frac{1}{4}), then Z′ must have a limit point on [a, \frac{1}{4}]×\{α\} for some limit ordinal α. If not, Z′ must have a limit point on [\frac{1}{4}, b] × \{α\} for some limit ordinal α.

2. X ∩ ((0, ω_{1}) × (β, β + 1)) is uncountable for some β ∈ ω_{1}. For some a, b ∈ (0, 1), the set Z′ = {⟨x, y⟩ ∈ X : β + a < y < β + b} is uncountable. This set has a limit point on {α} × [β + a, β + b] for some limit ordinal α.

3. Z ∩(0, 1)×𝕃^{+} is countable, and for each β ∈ ω_{1}, X ∩((0, ω_{1}) × (β, β + 1)) is countable. Hence B = {β : Z ∩([1, ω_{1})×(β, β+1)) ≠ ∅} is uncountable.

For each β ∈ B, pick one point ⟨x_{β}, y_{β}⟩ ∈ Z ∩ ((0, ω_{1}) × (β, β + 1)). Let Z′ = {⟨x_{β}, y_{β}⟩ : β ∈ B}. Then for some a, b ∈ (0, 1), Y = {⟨x_{β}, y_{β}⟩ ∈ Z′ : β + a < y_{β} < β + b} is uncountable. Let B′ = {β : ⟨x_{β}, y_{β}⟩ ∈ Y }. Note that for each α ∈ ω_{1}, if β ∈ domA_{α} and x ∈ (0, 1), then the first coordinate of ϕ_{α}(⟨α, β + x⟩) is a fixed value, call it x(ϕ_{α}), since each point in ϕ_{α}(\{α\} × (β, β + 1)) lies on a circular arc through ⟨\frac{3}{4}, β + \frac{1}{2}⟩, ⟨1, β⟩ and ⟨1, β + 1⟩. Now let λ^{β} = min{α : ⟨x_{β}, y_{β}⟩ ∈ U_{α}} for each β ∈ B′, and observe that sup{λ^{β} : β ∈ B′} = ω_{1}. Let A = {⟨β, λ^{β}⟩ : β ∈ B′}. Choose α such that A_{α} ∈ A and A_{α} ⊂ A. Since for each β ∈ dom(A_{α}), ⟨x_{β}, y_{β}⟩ ∈ ((0, ω_{1}) × (β, β + 1))∩U_{α}, 1 < x_{β} < α_{β} < α and β +a < y_{β} < β +b, if the first coordinate of ϕ_{α}(⟨x_{β}, y_{β}⟩) is x, then \frac{1}{2} < x < max\{a(ϕ_{α}), b(ϕ_{α})\} and hence Z has a limit point on [\frac{1}{2}, m] × \{α\}, where m = max\{a(ϕ_{α}), b(ϕ_{α})\}.

References

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[N] Peter Nyikos The Theory of Nonmetrizable Manifolds, in Handbook of Set-theoretic Topology, K. Kunen and Jerry Vaughan, Eds, 1984, 633-684.