Compactness

toVbeing D-compact. And in fact,TSUwith a D-compact HausdorffV implies most of the additional axioms we have looked at so far, including the separation properties and the union axiom:

Leta ⊆ Tandx /∈ S

a. Then for every y ∈ a, there is absuch thaty ⊆ int(b)andx /∈ b.

The sets int(b) then cover a and by D-compact Hausdorffness, a discrete subfamily also does. But then the union of thesebis a superset ofS

anot containingx.

Another consequence of global D-compactness is that most naturally occurring topologies coincide: Point (2) of the following theorem not only applies to hyperspaces a, but also to products, order topologies and others. If the class of atoms is closed and unions of sets are sets, this even characterizes compactness (note that these two assumptions are only used in (3)⇒(1)):

Theorem 24(ESU+T₂+Union). IfAisT-closed, the following statements are equivalent:

1. Every set isD-compact, that is: IfT

A=∅, there is a discreted⊆AwithT d=∅.

2. Every HausdorffD-topologyT ∈ Vequals the natural topology: T = S T 3. For every seta, the exponentialD-topology onaequals the natural topology.

Proof. (1) ⇒ (2): Let A = S

T. Since A is T-closed in A, A ∈ T and thus A ∈ V. By definition, T ⊆A. For the converse, we have to verify that eachb ∈AisT-closed, so let y ∈ A\b. Consider the class Cof allu ∈ T, such that there is av ∈ T withu∪v = Aand y /∈ v. By the Hausdorff axiom, for everyx ∈ bthere is au ∈ Comittingx, sob∩T

C = ∅. ByD-compactness, there is a discreted ⊆Cwithb∩T

d= ∅. By definition ofC,y∈ intT(u) for every u, and sincedis discrete, the intersection T

u∈dint_T(u) is open. Therefore, every y /∈bhas aT-open neighborhood disjoint fromb.

(2)⇒(3) is trivial, because as aD-compact HausdorffD-topological space,aisT₃and hence ais Hausdorff by Lemma 22.

(3)⇒ (1): Lemma 22 also implies that ifaisT₂, thenaisT₃ andaisT₄, so it follows from the Hausdorff axiom that every set is normal.

Finally, we can proveD-compactness. LetA⊆a,T

A= ∅and letc =cl(A). ThenT c= ∅.

Since every set is regular andAis closed, the positive specification principle holds. Therefore B =

b∈c | \ b6= ∅

= {b∈c | ∃x∀y∈b x∈y} is a closed subset ofcnot containingc. In particular, there is an open base class

U ∩ \

i∈I

♦V_i

of the spaceccontainingcwhich is disjoint fromB. EveryU∩V_iis a relatively open subset ofc, so there is anx_i ∈ A∩U∩V_i, becauseAis dense inc. The set{x_i | i∈I}– and here we used the uniformization axiom – then is a discrete subcocover ofA.

Chapter 2

Models of Topological Set Theories

This chapter deals with hyperuniverses, that is, models ofTSwhose classes areallsubclasses of aκ-compact Hausdorff space, for some cardinal κ. Since it would be confusing to always have to discern the natural topology from the hyperuniverse’s topology, we will work inZFC now, although many of these constructions would be possible in ES_∞, too. But as we have established that ZFC is interpretable inES_∞, this is no limitation anyway. And because we have no need for “real” atoms or a universal set, we use the symbolsA,SandVto denote the atom, set and universe spaces of the models we construct.

2.1 Hyperuniverses

Aκ-hyperuniverseis aκ-topological Hausdorff spaceVXwith at least two points, together with a closed subsetSX ⊆VXand a homeomorphismΣ^X :SX→ Exp_κ(VX). We callAX = VX\SX theatom spaceofX,VXtheuniverse spaceandSXtheset space.

We omitted the condition that the set of atoms of the κ-hyperuniverse does not contain any subsets ofVX, because in the well-founded realm ofZFCit is irrelevant: Just replace AXby AX×{VX}, for example, to fulfill this additional requirement. Having done that it follows from Proposition 10 that I is an interpretation ofTS, where the domain of I isP(VX)∪AX, its atoms areA^I = AXanda∈^I bis defined as

b /∈AX ∧ a ∈ (b∩AX) ∪ Σ^X[b∩SX].

The axiom of infinity holds in Iiff κ > ω. In that case,κ must be inaccessible, as Theorem 28 will show. By definitionIinterprets the Hausdorff and uniformization axioms and, as will also follow from Theorem 28,D-compactness.

Let us call aκ-hyperuniverseXclopenifSXis clopen inVX, andatomlessifSX= VX. Clopen κ-hyperuniverses are of particular interest because by Proposition 15, they are models of GPF comprehension.

At first glance, these interpretationsInever satisfy ∅ ∈ V. But if we pick an elementa ∈ AX and defineA^I = AX\ {a}instead ofAX, then∅^I = a. In other words, the empty set is just an

atom that has been awarded set status. Therefore a clopenκ-hyperuniverse with exactly one atom provides a model ofGPK⁺.

Lemma 25. A spaceXisκ-compact iff there is no continuous¹strictly descending sequence of nonempty closed setshC_α|α < λi withT

α<λC_α = ∅, whose length is a regular cardinal λ>κ.

Proof. If such a sequence exists, then its members constitute a cocover of X. Every small subset’s indices are bounded by someα < λ, so its intersection is a superset ofC_α 6= ∅. Thus the sequence disprovesκ-compactness.

Assume conversely that Xis not κ-compact. Letλ be the least cardinal such that there is a cocover{A_β | β < λ} with noκ-small subcocover. For each α < λ, defineC_α = T

β<αA_β. Since noκ-small subset of{A_β | β < α}is a cocover andλis minimal,C_αmust be nonempty.

Every cofinal subsequence ofhC_α|α < λiis also a cocover without aκ-small subcocover, so it must have length λ. Hence λis regular. In particular, the sequence can be replaced by a strictly descending subsequence.

The following argument is based on an idea of J. Keesling ([Kee70]).

Lemma 26. LetXhave a dense subsetDof size µand a closed discrete subset Cof size ν, such that 2^µ < 2^ν. ThenXis not normal.

Proof. Ifµwere finite, thenX = Dwould have sizeµ, which is impossible becauseC ⊆ X has sizeν. Henceµ,ν >ℵ0.

SinceDis dense, each of the(2^ω)^µ = 2^µfunctions fromDtoRcan be continuously extended in at most one way to all ofX. Thus there are at most 2^µcontinuous functions fromXtoR.

IfXwere normal, by the Tietze extension theorem² each of the(2^ω)^ν = 2^ν functions from Cto the unit interval could be continuously extended toX, so there would exist at least 2^ν distinct continuous real-valued functions onX, a contradiction.

Lemma 27. Letλ > κbe a regular cardinal, endowed with a topology at least as fine as its order-κ-topology. Then Exp_κ(λ)is not normal.

Proof. First assume that λ = κ, that is, λ is a discrete space of size κ. By Lemma 26, it suffices to find a dense D ⊆ Exp_κ(λ) of size κ and a closed discrete C ⊆ Exp_κ(λ) of size

1In the sense thatCα=T

β<αCβfor each limit ordinalβ.

2Every bounded real-valued continuous function defined on a closed subset of a normal spaceXcan be con-tinuously extended to all ofX. (cf. [Kel68, Eng89])

2.1. HYPERUNIVERSES 41

2^κ. ForD, we can simply take the small subsets of X, because κ = κ^<κ. To construct C, first partitionλ into κ-large X₁, X₂, and let f_i : X → X_i be bijections. For each A ⊆ λ let F(A) =f₁[A]∪f₂[λ\A]. Then letC= {F(A)|A⊆λ}. To prove thatCis closed, letB /∈ C, so f⁻¹₁ [B∩X₁]6= {f⁻¹₂ [B∩X₂]:

• Either f⁻¹₁ [B∩X₁] and f⁻¹₂ [B∩X₂] have some point x in common: then ♦{f₁(x)}∩

♦{f₂(x)}is an open neighborhood ofBdisjoint fromC,

• or there is some point xnot contained in any of the two sets: then {{f₁(x),f2(x)} is such a neighborhood.

To see that it is also discrete, let F(A) ∈ C. ThenF(A)is a neighborhood disjoint from the rest ofC.

Now assume that λ > κ. Both Y = T

α<λ♦(λ\α) and Z = {{α} | α < λ} are closed in Exp_κ(λ), and they are disjoint. Assume that they can be separated by open sets. Then in particular, there is an openU⊇ Y, whose closure is disjoint fromZ.

We recursively defineκ-small setsA_α ∈ Ufor α < κ: Let γ_α = 1+sup S

β<αA_β . Then λ\γ_αis a clopen element ofYand therefore inU. ThusU∩(λ\γ_α)is nonempty and, being open, contains a κ-small elementA_α. Now the sequencehA_α|α < κi lies in theκ-compact set (1+γ)withγ = sup_α<κγ_αand hence has an accumulation pointB. For everyα < κ, allA_βwithβ > αare members of the closed set(λ\γ_α), soBmust be in their intersection (λ\γ). HenceB= {γ}, which is inZ, a contradiction.

Theorem 28. Everyκ-hyperuniverse isκ-compact. And if there exists aκ-hyperuniverse, then κis strongly inaccessible orω.

Proof. By Lemma 25, ifVXis notκ-compact, Exp_κ(VX)has a closed subset A = {C_α | α < λ} = C₀ ∩ \

α<λ



C_α+1 ∪ \

x∈C_α

♦{x}





from which there exists a continuous bijection C_α 7→ α to a regular cardinal λ > κ. By Lemma 27,Ais nonnormal. Hence Lemma 22 implies thatAis not regular andA is not Hausdorff. But

(Σ^X◦Σ^X◦Σ^X◦Σ^X)⁻¹[A]

is a closed subset of the Hausdorff spaceVXhomeomorphic toA, a contradiction. HenceVX isκ-compact.

If VX were discrete, then |VX|+1 = |Exp_κ(VX)|+1+|AX| = 2^|VX|+|AX|, which is only possible for|VX| = 1. But a κ-hyperuniverse has more than one point and hence VX is a nondiscreteκ-compactκ-topological space.

Ifκis singular, then every setBof closed sets with|B| = κis the union S

α<γB_α ofκ-small setsB_α ⊆ Bfor someγ < κ, and henceS

B = S

α<γ

SB_αis closed. Thus everyκ-additive

space is κ⁺-additive. So VX is a κ-compact κ⁺-additive nondiscrete Hausdorff space. In particular there exists a subsetA⊆ VXof sizeκ, which byκ⁺-additivity is discrete. But then {A\ {a} | a∈A} is a cocover with no κ-small subcocover, which contradicts κ-compactness.

Henceκmust be regular.

Now letγ < κbe infinite. Then there is a setA⊆ VXof sizeγandAis closed and discrete.

HenceAis a closed discrete subset of Exp_κ(VX)of size 2^γ, which is only possible if 2^γ < κ.

This proves thatκis a strong limit.

Im Dokument Topological set theories and hyperuniverses (Seite 43-48)