Inverse Limit Models

• a continuous mapSΣ^X :Exp^c_κ(VX)→ SX.

We callAXtheatom spaceofXand defineAΣ^X = id_AXand

VΣ^X = AΣ^X∪SΣ^X : AX∪Exp^c_κ(V_X) → VX,

the comprehension map of X. The morphisms f : X → Y of Cm^∗ are continuous mapsVf : VX→ VY such that the image ofAf = Vf AXis a subset ofAY, the image ofSf = Vf SX is a subset ofSYand the diagram

Exp^c_κ(VX)

SΣ^X

Exp^cκ(Vf) //Exp^c_κ(VY)

SΣ^Y

SX ^{Sf //}SY

commutes. Note that as the hyperspace of a locally κ-compact Hausdorff space, Exp^c_κ(VX)is also locallyκ-compact Hausdorff by Lemma 23. And by Lemma 30 it has weight6 λ.

We define the functor:Cm^∗ → Cm^∗ as follows:

AX = AX SX = Exp^c_κ(VX) Af = Af Sf = Exp^c_κ(Vf) AΣ^X = idAX SΣ^X = Exp^c_κ(VΣ^X)

Then Σ^X : X → Xis a morphism and the following diagram commutes, which shows that X 7→ Σ^Xis a natural transformation fromto the identity:

X

Σ^X

f //Y

Σ^Y

X ^{f //}Y

Thecomprehension maps categoryCmis the full subcategory ofCm^∗given by those objectsX whose universe spaceVX(or equivalentlySX) isκ-compact. And thecategory of (comprehen-sion map) hyperuniversesCHypis the full subcategory ofCmgiven by those objectsXwhere Σ^X is bijective and thus a homeomorphism. The restriction Cmis a functor from Cmto Cm. Finally, letPCm^∗ be the subcategory ofCm^∗ of only those objectsXwithκ-properSΣ^X and only the κ-proper morphisms.

We define theκ-Alexandroff compactification functorωfromPCm^∗ toCmas follows:

AωX = AX Aωf = Af

SωX = ωSX Sωf = ωSf,

that is, Sωf(p) = p, and using the universal property of the κ-Alexandroff compactification, letSΣ^ωX :Exp^c_κ(VωX) → SωXbe the unique map such that the diagram

Exp^c_κ(VX)

SΣ^X

Exp^cκ(ι_VX) //Exp_κ(VωX)

SΣ^X

SX ^{ιSX //}SωX

2.6. INVERSE LIMIT MODELS 71

commutes and everything not in rng(Exp^c_κ(ι_VX)) is mapped to p, that is, such that Vι_X = idAX∪ι_SX defines a morphism ι_X : X → ωX. This is possible by Lemma 29 because as SX is open inSωX, Exp^c_κ(VX) is open in Exp_κ(VωX), and SΣ^X is assumed to beκ-proper. For a κ-proper morphismf :X→ Y,ωfis in fact a morphism, because every cell in the diagram

Exp_κ(VωX) ^{Expκ(Vωf) //}

SΣ^X

Exp_κ(VωY)

SΣ^Y

Exp^c_κ(VX)

Exp^cκ(VιX)

ii

SΣ^x

Exp^c_κ(Vf) //Exp^c_κ(VY)

Exp^cκ(Vι_Y) 55

SΣ^Y

SX

SιX

tt

Sf //SY

SιY

**SωX ^{Sωf //}SωY

is commutative, soSΣ^ωY◦Exp_κ(Vωf)◦Exp^c_κ(Vι_X) = Sωf◦SΣ^ωX◦Exp^c_κ(Vι_X)and Exp^c_κ(Vι_X) can be cancelled on the right because ifa ∈ Exp_κ(VωX) is not in the image of Exp^c_κ(Vι_X), then it containsp, and in that case,p∈Exp_κ(Vωf)(a)and hence

SΣ^ωY(Exp_κ(Vωf)(a)) = p = Sωf(p) = Sωf(SΣ^ωX(a)).

The mapX7→ ι_Xis a natural transformation from the identity onPCm^∗toω.

Lemma 51. WheneverYis inCHyp,Xis inPCm^∗,Σ^Xis surjective andf:X→ Y homeomor-phically embedsVXas an open subset inVY, then there is a unique morphismg : Y → ωX such thatg◦f = ι_Xandg(x) =pfor everyx ∈VY\rng(Vf).

Proof. Applying Lemma 29 to (Vf)⁻¹, we obtain a map Vg : VY → VωXwith the required properties, and we only have to show that it defines a homeomorphism. Firstly, leta =f(b).

Then by definition

Σ^ωX◦Exp_κ(Vg)(a) = Σ^ωX◦Exp^c_κ(Vg◦f)(b) = Σ^ωX◦Exp^c_κ(ι_X)(b) = ι_X◦Σ^X(b)

= Vg◦f◦Σ^X(b) = g◦Σ^Y◦f(b) = Vg◦Σ^Y(a)

If, on the other hand, a ∈ Exp_κ(VY)\rng(f), then a contains an element not in rng(f) and hencep ∈ Exp_κ(Vg)(a)and Σ^ωX◦Exp_κ(Vg)(a) = p. Therefore it suffices to show that Σ^Y(a) ∈/ rng(f), because that entailsVg(Σ^Y(a)) = p, too. So assume Σ^Y(a) = f(x). Then Σ^Y(a) = f(Σ^X(b))for someb, because Σ^Xis surjective. It follows that Σ^Y(a) = Σ^Y◦f(b) and thusa= f(b), contradicting our assumption.

Lemma 52. Inverse limits exist in the categoryCm. LethX_i,fijibe an inverse system inCm and lethX,f_i :X→ X_iibe its inverse limit inCm. Then:

hVX,Vfii= lim

←

TophVX_i,Vfiji and hSX,Sfii= lim

←

TophSX_i,Sfiji andSΣ^X is the unique map such that for alli,SΣ^Xi◦Exp(Vf_i) =Sf_i◦SΣ^X.

Proof. This is exactly dual to Lemma 46 about direct limits in Ex: We show that the object X with the morphismsf_i defined by those topological limits is in fact an inverse limit of the given system inCm.

For each i, let h_i : Z → X_i be a morphism and assume that for all j > i, h_i = f_ij◦h_j. We then defineVhusing the inverse limit property ofVX. This is the unique candidate for a suitable morphism h : Z → X, and we just have to show that it is in fact a morphism. For eachi, every path from Exp^c_κ(VZ)toSX_iis equal in the diagram

Exp^c_κ(VX)

Exp^c_κ(Vfi) **

SΣ^X

Exp^c_κ(VZ)

Exp^c_κ(Vh)

oo

Exp^c_κ(Vhi)

tt

SΣ^Z

Exp^c_κ(VX_i)

SΣ^Xi

SXi

SX

Sfi

33

Sh //SZ

Shi

kk

and thus Sfi◦Sh◦SΣ^Z = Sfi◦SΣ^X◦Exp_κ(Vh). Since hSX,Sfii is the inverse limit of the SX_i, it follows thatSh◦SΣ^Z =SΣ^X◦Exp_κ(Vh).

Also dually to the situation in the categoryEx, we recursively define functors^β :Cm→ Cm and morphismsΣ^X_α,β:^βX→ ^αXfor ordinalsα 6β:

⁰X = X Σ^X_α,α = id^α_X ⁰f = f ^α+1X = ^αX Σ^X_α,β+1 = Σ^X_α,β◦Σ^{βX α+1}f = ^αf For limit ordinalsγ, we define

h^γX,Σ^X_α,γi = lim

←

Cmh^αX,Σ^X_α,βi_α6β<γ

and iff :X → Y is a morphism, we use the inverse limit property of^γX: ^γfis defined as the unique morphism, such that^αf◦Σ^X_α,γ = Σ^Y_α,γ◦^γffor allα < γ. ThenX 7→ Σ^X_α,βis a natural transformation from^βto^α, that is, for every morphismf : X→ Y, the following diagram commutes:

^αX

^αf

^βX

Σ^X_α,β

oo

^βf

^αY ^βY

Σ^Y_α,β

oo

The construction will stop at^κX, becauseSΣ^X_κ+1,κ = SΣ^κX will already be a homeomor-phism: Sinceκis regular, the systems

hVX_α,VΣ^X_α,βi and hExp_κ(VX_α),Exp_κ(VΣ^X_α,β)i = hSX_α+1,SΣ^X_α+1,β+1i

2.6. INVERSE LIMIT MODELS 73

areκ-directed, so by Lemma 37,SΣ^X_κ,κ+1is a homeomorphism from the hyperspace Exp_κ(VX_κ) of the limit of the former to the limitSX_κ of the latter.

We define^∞ =^κ. Then^∞is a functor fromCmto the category of hyperuniversesCHyp.

Note that ifΣ^Xis surjective,Σ^X_0,∞also is, so thatXis a quotient of the resulting hyperuniverse.

Finally, we show thatΣ^X_0,∞is terminal among the morphisms from objects ofCHyptoX: Theorem 53. LetXbe an object ofCm,Y an object ofCHypandf :Y → X. Then there is a unique morphismg :Y → ^∞Xsuch thatf= Σ^X_0,∞◦g, namelyg =^∞f◦(Σ^Y_0,∞)⁻¹.

X Y

f

oo

g

ww^∞X

Σ^X_0,∞

gg

The functor^∞ :Cm → CHypis a right adjoint to the inclusion functor.

Proof. Again, this proof is mostly dual to that of Theorem 48, although the situation is a bit easier because instead ofγ_Xwe can always useκalias∞. SinceΣ_0,κ is a natural transforma-tion,

Σ^X_0,κ◦g = Σ^X_0,κ◦^κf◦(Σ^Y_0,κ)⁻¹ = f, so^κf◦(Σ^Y_0,κ)⁻¹has the required property.

It remains to show that for every morhism g, the equation f = Σ^X_0,κ◦g implies g = ^κf◦ (Σ^Y_0,κ)⁻¹. By induction onα 6κ, we will show that the following diagram commutes:

^αX ^κX

Σ^X_α,κ

oo

^αY

^αf

OO

Σ^Y_0,α

//Y

g

OO

Then the caseα =κwill imply our claim, proving uniqueness:

^κf = Σ^X_κ,κ◦g◦Σ^Y_0,κ = g◦Σ^Y_0,κ The caseα= 0 is just our assumptionf= Σ^X_0,κ◦g.

Now letα = β+1. Applying to the induction hypothesis forβ, we obtain the left cell in the diagram

^αX ^κ+1X

Σ^X_α,κ+1

oo

Σ^X_κ,κ+1

//^κX

^αY

^αf

OO

Σ^Y_1,α

//Y

g

OO

Σ^Y_0,1

//Y

g

OO

and the right cell commutes becauseΣis a natural transformation. SinceΣ^X_κ,κ+1 = Σ^κX is an isomorphism, this proves the caseα.

Finally, letαbe a limit ordinal and assume the induction hypothesis for allβ < α. Then every cell in the diagram

^αX

Σ^X_β,α **

^κX

Σ^X_α,κ

oo

Σ^X_β,κ

tt^βX

^βY

^βf

OO

^αY

Σ^Y_β,α 44

Σ^Y_0,α

//

^αf

OO

Y

Σ^Y_0,β

jj

g

OO

commutes and hence

Σ^X_β,α◦^αf◦ Σ^Y_0,α−1

= Σ^X_β,α◦Σ^X_α,κ◦g : Y →^βX for allβ < α. So by the inverse limit property of^αX, it follows that

^αf◦ Σ^Y_0,α−1

= Σ^X_α,κ◦g.

To prove that^κis a right adjoint, we show that the bijection

Φ_Y,X :hom(Y,X)→ hom(Y,^κX), f7→ ^κf◦(Σ^Y_0,κ)⁻¹ has the required property:

Φ

Y,eXe(h₁◦f◦h₀) = ^κ(h₁◦f◦h₀)◦(Σ^Y_0,κ^e )⁻¹ = ^κh₁◦^κf◦^κh₀◦(Σ^Y_0,κ^e )⁻¹

= ^κh₁◦^κf◦(Σ^Y_0,κ)⁻¹◦h₀ = ^κh₁◦Φ_Y,X(f)◦h₀

for all morphismsh₀ :Ye→ Y,h₁ :X→ Xeandf:Y → X.

A difference with the categoryExis that proving nontriviality of^∞Xis a lot easier here: It suffices thatΣ^X is surjective andXis nontrivial. And in fact, every^∞Xarises also from an objectYwith surjectiveΣ^Y:

Proposition 54. Let Xbe an object of Cm. LetVY = rng(Σ^X_0,∞) and SY = SX∩VY. Then Σ^Y = Σ^X (AX∪Exp_κ(VY))defines an objectY and^∞Xis isomorphic to^∞Y.

Proof. Givena∈ Exp_κ(VY), letb= (Σ^X_0,∞)⁻¹[a]∈Exp_κ(V^∞X). Then

Σ^X(a) = Σ^X(Σ^X_0,∞[b]) = Σ^X_0,1(Σ^X_1,κ+1(b)) = Σ^X_0,∞(Σ^X_κ,κ+1(b)) ∈ rng(Σ^X_0,∞) ThereforeΣ^Xactually maps Exp_κ(VY)toVYand the restriction defines an object ofCm.

2.6. INVERSE LIMIT MODELS 75

VΣ^X_0,∞ also defines a surjective morphism h : ^∞X → Y, such that h◦f = Σ^X_0,∞ for the canonical inclusionf :Y → X. By Theorem 53, there is a uniqueeh :^∞X→ ^∞Ysuch that f◦Σ^Y_0,∞◦eh =Σ^X_0,∞. But then

Σ^X_0,∞◦^∞f◦eh = f◦Σ^Y_0,∞◦eh = Σ^X_0,∞ = Σ^X_0,∞◦id∞X.

Since again by Theorem 53, id∞Xis the unique morphism^∞X→ ^∞Xwith that property, ^∞f◦he = id∞X. Analogously,

f◦Σ^Y_0,∞◦he◦^∞f = Σ^X_0,∞◦^∞f = f◦Σ^Y_0,∞ = f◦Σ^Y_0,∞◦id∞Y

implieseh◦^∞f = id∞Y, becausef, being injective, cancels on the left, so Theorem 53 can be applied again.

Proposition 55. There exists aκ-hyperuniverse of weight>λ

Proof. Let AX = ∅,VX discrete with |VX| = λ. Then Exp^c_κ(VX) is discrete, too: Everya ∈ Exp^c_κ(VX)isκ-small and

{a} = a ∩ \

x∈a

♦{x}

is open. Sinceλ^<κ = λ, Exp^c_κ(VX)also has sizeλand there is a bijectionVΣ^X : Exp^c_κ(VX) → VX(which in particular isκ-proper). ThenXis inPCm^∗and^∞ωXis aκ-hyperuniverse.Σ^ωX_0,∞ is a surjection fromV^∞ωXontoVωX. ButVωXhasλisolated points and their preimages in V^∞ωXmust be disjoint open sets, soV^∞ωXhas at least weightλ.

Another very general example of a surjective mapΣ^Xis the union map: Take anyκ-compact Hausdorff spaces Z 6= ∅ and AX, setSX = Exp_κ(Z). Then the mapSΣ^X : Exp(VX) → SX defined as

SΣ^X(b) = S

(b∩SX) ifb /∈ AX Z ifb∈ AX

is continuous, because the preimage of V is ♦V∪(V∪AX) and the preimage of ♦V is♦♦V∪AX for every proper open subsetV ⊂ Z. The set of singletons in SX is homeo-morphic toZ, and so is the set of singletons of singletons {{x}} ∈ SX withx ∈ Z, and so on. These subspaces also survive limit steps and it turns out that^∞Xhas a closed subset of autosingletons homeomorphic toZ.⁸

A subsetA⊆VXis calledtransitiveif(Σ^X)⁻¹[A∩SX]⊆A.Aispersistentif it is transitive, open andΣ^Xmaps(Σ^X)⁻¹[A]homeomorphically ontoA.

Lemma 56. LetXbe an object ofCm^∗andA⊆VX.

1. If A is transitive in X, then f⁻¹[A]is transitive in Y for every morphism f : Y → X.

Moreover,A∪(A∩AX)is transitive inXandι_X[A]is transitive inωX.

8A similar hyperuniverse, constructed in a different way, is described extensively in [FH96b].

2. IfAis persistent inX, then(Σ^X)⁻¹[A]andA∪AXboth are persistent inX, andι_X[A]

is persistent inωX.

3. If hX_α|f_α,βi_α,β<γ is an inverse system in Cm, f_0,α maps f⁻¹_0,α[A] homeomorphically ontoAfor everyα < γandf⁻¹_0,α[A]is persistent for eachα < γ, then if

hX_γ|f_α,γi_α<γ= lim

←

CmhX_α|f_α,βi_α,β<γ

is the inverse limit,f_0,γ f⁻¹_0,γ[A]is a homeomorphism andf⁻¹_0,γ[A]is persistent.

Proof. (1): These claims are easily verified by direct calculation:

(Σ^Y)⁻¹[f⁻¹[A]∩SY] = (Σ^Y)⁻¹[f⁻¹[SX∩A]] = (f)⁻¹[(Σ^X)⁻¹[SX∩A]]

⊆ (f)⁻¹[A] =(f⁻¹[A])

(Σ^X)⁻¹[A] ⊆ ((Σ^X)⁻¹[A]) = ((Σ^X)⁻¹[A∩SX]∪(A∩AX))

⊆ (A∪(A∩AX))⊆(A∪AX)

(Σ^ωX)⁻¹[ι_X[A]∩SωX] = ι_X[(Σ^X)⁻¹[A∩SX]⊆ι_X[A] =(ι_X[A])

Since (A∪(A∩AX))∩SX = A, the second calculation proves that A∪(A∩AX) is transitive.

(2): In the light of (1) and the fact that(Σ^X)⁻¹[A],A,A∩AXandι_X[A]are open, we only have to worry about whether their preimages are mapped homeomorphically onto the sets in question. Since (Σ^X)⁻¹[A] is a subset of A∪(A∩AX) and AΣ^X is a homeomorphism by definition, it suffices to consider the mapΣ^X (Σ^X)⁻¹[A]. By definition that equals Exp^c_κ(Σ^X (Σ^X)⁻¹[A]), and the exponential of a homeomorphism is a homeomorphism itself.

Forι_X[A], the statement is even more trivial, becauseΣ^XandΣ^ωXagree on Exp^c_κ(VX)and in particular onA.

(3): From (1) we know thatf⁻¹_0,γ[A]is transitive. It follows from the construction of the inverse limit in the category of topological spaces that f⁻¹_0,γ Ais a homeomorphism because every f⁻¹_0,α Ais. But then Exp^c_κ(f⁻¹_0,γ A)is a homeomorphism, too. Since(Σ^X0)⁻¹[A∩SX] ⊆ A, the left, top and bottom arrow in the diagram

Exp^c_κ(VX₀)

SΣ^X0

Exp^c_κ(VXγ)

Exp^cκ(f_0,γ)

oo

SΣ^Xγ

SX₀ SXγ

Sf0,γ

oo

mapA homeomorphically to its preimage, so the right one must be a homeomorphism be-tween these preimages, too, implying thatf⁻¹_0,γ[A]is persistent.

Theorem 57. If X is an object of Cm^∗ and A ⊆ VX is persistent, then (Σ^ωX_0,α)⁻¹◦ι_X[A] is persistent in^αωX, and(Σ^ωX_0,α)⁻¹◦ι_X Ais a homeomorphism for every ordinalα.

2.6. INVERSE LIMIT MODELS 77

In particular,(Σ^ωX_0,∞)⁻¹◦ι_X[A]is persistent in the hyperuniverse^∞ωXand mapped homeo-morphically ontoA.

Proof. The proof is by induction on α and follows immediately from Lemma 56: The κ-compactification step at the beginning is (2) and the limit step is (3).

Proposition 58. LetXbe an object ofPCm^∗ with non-κ-compactVXand letΣ^Xbe a homeo-morphism. Thene_X = (Σ^ωX_0,∞)⁻¹◦ι_Xis a morphism which embedsVXas a dense open subset inV^∞ωX.

IfYis inCHypand ˜e :X→ YembedsVXas a dense open subset inVY, then there is a unique morphismg :Y → ^∞ωXsuch thate_X = g◦˜e.

Proof. IfΣ^Xis a homeomorphism, thenΣ^ωXmaps rng(ι_X) = (Σ^ωX)⁻¹[rng(ι_X)] homeomor-phically onto rng(ι_X), which is dense and open. It follows inductively that everyΣ^ωX_0,αmaps the dense and open preimage of rng(ιX)homeomorphically onto rng(ιX). Thus the definition ofVe_Xmakes sense. Ve_Xalso defines a morphism: Let a ∈ Exp_κ(X). Then since Exp_κ(Σ^ωX_0,κ) also maps the preimage of rng(Exp^c_κ(ι_X))homeomorphically onto rng(Exp^c_κ(ι_X)),

Se_X◦SΣ^X = (SΣ^ωX_0,∞)⁻¹◦Sι_X◦SΣ^X = (SΣ^ωX_0,∞)⁻¹◦SΣ^ωX◦Exp^c_κ(ι_X)

= (SΣ^ωX_0,κ)⁻¹◦SΣ^ωX◦Exp_κ(Σ^ωX_0,κ)◦(Exp_κ(Σ^ωX_0,κ))⁻¹◦Exp^c_κ(ι_X)

= (SΣ^ωX_0,κ)⁻¹◦SΣ^ωX_0,κ◦SΣ^∞ωX◦(Exp_κ(Σ^ωX_0,κ))⁻¹◦Exp^c_κ(ι_X)

= SΣ^∞ωX◦Exp^c_κ(e_X)

Given ˜e, by Lemma 51 there is an f : Y → ωX with f◦˜e = ι_X, and since the image of ˜e is dense, this f is unique. Theorem 53 yields a unique g : Y → ^∞ωX with f = Σ^X_0,∞◦g.

Thus g is unique such that Σ^X_0,∞◦g◦e˜ = ι_X. But that means that the image of g◦˜e is in (Σ^ωX_0,∞)⁻¹[rng(ι_X)]and by composing with (Σ^ωX_0,∞)⁻¹ on the left, we obtain the equivalent equation e_X =g◦e.˜

As in the case ofEx^∗, letVV_c = V_κwith the discrete topology,AV_c = {∅}andSΣ^Vc the identity – which makes sense because Exp^c_κ(VV_c) = SV_c. ThenV_c = ^∞ωV_c is a hyperuniverse in which the set e_Vc[VVc]of well-founded sets is dense and the homeomorphic image of VVc. By Proposition 58, it is the smallest hyperuniverse with that property. In fact, V_c turns out to beκ-ultrametrizable in a canonical way which shows that it is isomorphic to the structure originally described by R. J. Malitz in [Mal76] and E. Weydert in [Wey89]:

Theorem 59. Letx∼0 yfor allx,y∈ VV_c. For everyα, letx∼α+1 ywhenever

∀˜x∈ (Σ^Vc)⁻¹(x) ∃˜y∈(Σ^Vc)⁻¹(y) ˜x∼α y˜ and vice versa.

At limit steps, take the intersection. Then the sequenceh∼α |α < κidefines aκ-ultrametric inducing the topology ofVVc.

Proof. Firstly we show that∼α actually is a superset of∼β for allα < β < κ. We do this by induction onβ. Assume x ∼β y. If βis a limit ordinal, thenx ∼α yfor everyα < βby definition. Now assumeβ =γ+1. Since for all ˜x ∈(Σ^Vc)⁻¹(x)and ˜y∈ (Σ^Vc)⁻¹(y), ˜x∼γ y˜ implies ˜x ∼δ y˜ for all δ < γ by the induction hypothesis, it follows thatx ∼δ+1 y. If γis a limit, this impliesx ∼γ y, because ∼γ is the intersection of these ∼δ. Otherwise, it follows from the case whereδis the immediate predecessor ofγ.

It also follows inductively that every∼α is an equivalence relation: Intersections of equiva-lences are equivaequiva-lences, the definition is symmetric, and whenever x ∼α+1 y ∼α+1 z, just choose a ˜ywith ˜x ∼α y, and a ˜˜ zwith ˜y ∼α ˜zfor every ˜x as in the definition – then by the induction hypothesis ˜x∼α ˜z, proving thatx∼α+1 z.

If all[x]_αare open, then byκ-compactness, there are onlyκ-few of them. Hence every

[x]_α+1 = [

˜ x∈x

[˜x]_α ∩ \

˜ x∈x

♦[˜x]_α is open, too. Ifαis a limit, every[x]_αequalsT

β<α[x]_β. Thus it follows inductively, that there areκ-few[x]_αfor everyα < κ, and they are all open.

Next we show that ∼κ, which is the intersection of all∼α for α < κ, equals∼κ+1, that is, it is not a proper superset of it: Let x 6∼κ+1 y. Wlog assume that there is an ˜x ∈ (Σ^Vc)⁻¹(x) such that there is no ˜y ∈ (Σ^Vc)⁻¹(y) with ˜x ∼κ ˜y. Thus for every ˜y ∈ (Σ^Vc)⁻¹(y), there is anα_y˜ such that ˜xis not in[˜y]_αy˜. Sinceyisκ-compact, there is a family of such[˜y_i]_αi with aκ-small index setI, which covers y. Letβ < κbe an upper bound to these α_i. This means that ˜x∼βy˜ holds fornoy˜ ∈(Σ^Vc)⁻¹(y). Thusx6∼β+1 yand in particularx6∼κ y.

Now let q : VV_c → VMbe the quotient map, where VM = VV_c/ ∼κ. Then q◦Σ^Vc(a) = q◦Σ^Vc(b) iff Σ^Vc(a) ∼κ Σ^Vc(b), which is equivalent to Σ^Vc(a) ∼κ+1 Σ^Vc(b), which in turn means that a and b intersect the same equivalence classes [x]_κ. Thus Exp_κ(q)(a) = Exp_κ(q)(b). We have shown thatΣ^Vc factors through∼κin the sense that it induces a home-omorphismVΣ^M from {∅}∪Exp_κ(VM) to VMand thus a quotient hyperunverse M, with a quotient morphismq :V_c → M.

On the other hand, it follows recursively from the definition thatxα ywheneverx6=yand x,y∈ e_Vc[V_α]. In particular,q◦e_Vc is still injective and embedsV_c as an open subset inM.

Since V_c is minimal with that property, q must therefore be an isomorphism. That implies that∼κis the diagonal and thus the familyh∼α |α < κiis aκ-ultrametric onVVcitself. Since every[x]_αis open, it induces the topology.

Thetree model, a hyperuniverse presented by E. Weydert in [Wey89], is isomorphic to^∞X, whereAXandSXare one-point spaces.⁹ He conjectured that the isolated points are dense in V^∞X(in his terms, that V^∞X isperfect). Theorem 61 is a more general criterion for this property and proves Weydert’s conjecture.

9Asequential tree T is a tree of sequences closed with respect to restrictions. For sets of sequences A let Aα= {xα|x∈A}. Define recursively: Bα ={x∈Q

ξ<αP(Bξ) |∀ζ<ξ<α xζ =xξζ}. Thend(x,y) =min({κ}∪ dom(x∆y))defines aκ-ultrametric onVY = Bκ. WithAY = {∅}, the mapΣ^Y : Bκ → AY∪Exp_κ(Bκ),Σ^Y(x) = {y|∀α yα∈x}is a homeomorphism. This is E. Weydert’stree model. But in factB_α+1corresponds exactly to the hyperspace ofBα, and for limit ordinalsα,Bαis the inverse limit of its predecessors. Hence these sets correspond to the spacesV^αXin a canonical way.

2.6. INVERSE LIMIT MODELS 79

Given an object Xof Cm, we call a point x ∈ V^αX simpleif at least one of the following conditions is met:

• αis not a successor’s successor.

• x∈ A^αX.

• α =β+2,x∈ S^αX,Σ^X_β,β+1 xis injective and everyy∈ xis simple.

Asimple sequenceis a sequencehx_γ,γ ∈ [α,β)isuch that for everyγ,x_γ ∈V^γXis a simple point andΣ^X_γ1,γ2(x_γ2) = x_γ1 wheneverγ₂ > γ₁.

Lemma 60.LetXbe an object ofCmsuch thatVXis discrete andΣ^Xis surjective,α < β < κ, and lethx_γ,γ∈[α,β)ibe a simple sequence. Then the sequence can be extended to[α,β+1) and ifβ >α+ω,x_βis the unique point such thatΣ^X_γ,β(x_β) = x_γ for allγ ∈[α,β).

Proof. First of all, everyV^αXwithα < κis discretely small and hence discrete. The proof goes by induction onβ.

Ifβis a limit ordinal, it follows from the definition of the inverse limit thatx_βis unique (and it is simple by definition).

Next assume that β = δ+1. Choose any z ∈ (Σ^X_δ,β[{x_δ}])⁻¹. Thenx_γ+1 = Σ^X_γ+1,β(z) = Σ^X_γ,δ[z] for every γ ∈ [α,δ). Thus for every y ∈ z, the sequence hΣ_γ,δ(y),γ ∈ [α,δ)i is a simple sequence. By the induction hypothesis, it has a simple extension w_y. Then the set x_β = {w_y | y∈z}is simple itself and since

Σ^X_γ+1,β(x_β) = Σ^X_γ,δ[z] = Σ^X_γ+1,β(z)

for everyγ ∈[α,δ), we haveΣ^X_δ,β(x_β) = x_δ(either by the inverse limit property ifδis a limit, or simply becauseδis of the formγ+1).

Ifβ > α+ω, the sequenceshΣ_γ,δ(y),γ ∈ [α,δ)i by the induction hypothesis have a unique extension. Hencew_y =yfor ally∈zand thusz= x_β.

Theorem 61. LetXbe an object ofCmsuch thatVXis discrete. Then the isolated points are dense inV^∞X.

Proof. Wlog assume thatΣ^X is surjective. The nonempty sets of the form(Σ^X_α,∞)⁻¹[U]with limit ordinalsαconstitute an open base ofV^∞X, so it suffices to show that each contains an isolated point. Givenx_α ∈ U, recursively choose simple pointsx_γ according to Lemma 60, such thathx_γ|γ ∈ [α,κ)iis a simple sequence. Then there is exactly one pointx_κ ∈ V^∞X such thatΣ^X_γ,κ(x_κ) = x_γ for eachγ. And by Lemma 60,x_κis in fact the only preimage of the pointx_α+ω. But{x_α+ω}is open, so{x_κ}is open, too, and hencex_κ is isolated.

Im Dokument Topological set theories and hyperuniverses (Seite 75-86)