The maps Expκ(gi) : Expκ(X) → Xeicommute with the mapsgfij, so we obtain a continuous map f :Expκ(X) → Xefrom the inverse limit property ofXeand we only have to show that it is bijective.
Firstly, let a,b ∈ Expκ(X) be distinct, wlog let x ∈ a\b. Then for everyy ∈ b, there is an iy ∈ I, such thatgiy(x)6= giy(y). SinceXiis Hausdorff, there are disjoint openUy 3 giy(x) andVy 3 giy(y) separating them. But κ-few sets g−1i
y[Vy] suffice to cover b, and if j is an upper bound of theirκ-few indicesiy, thengj(x)∈/ gj[b], because
giyj◦gj(x) = giy(x) ∈/ Vy = giyj◦gjh g−1i
y[Vy]i
for ally.
As a consequencegj[a]6=gj[b]. Thus the map Expκ(gj) : Expκ(X) →Xej, which equalsgej◦f, mapsaandbto distinct points, implying thatfalso does.
Secondly, letb∈ X. Then for everye i,gei(b)is aκ-compact subset ofXi. We claim that the set
a = \
i∈I
g−1i [gei(b)]
is an element of Expκ(X)andf(a) = b.
Wheneverj > i,
g−1j [gej(b)] ⊆ g−1j [g−1ij [geij◦egj(b)]] = g−1i [gei(b)].
So if aκ-small family of setshg−1i
α [geiα(b)]|α < γiis given, andjis an upper bound of theiα, everyg−1i
α [egiα(b)]is a superset ofg−1j gej(b)
, which is nonempty. Therefore these sets have noκ-small subcocover and becauseXisκ-compact,ais in fact nonempty, soa∈Expκ(X) To provef(a) = b, it suffices to verify thatgej◦f(a) = gej(b)for allj. Butgej◦f = Expκ(gj) and it follows from the definition ofathat Expκ(gj)(a) ⊆ gej(b). To prove the converse, let x /∈Expκ(gj)(a), that is, letg−1j [{x}]be disjoint froma. Sinceg−1j [{x}]isκ-compact there are κ-few setsg−1i [gei(b)]whose intersection withg−1j [{x}]is empty, and becauseI isκ-directed, there exists a single i > j, such that g−1j [{x}]∩g−1i [egi(b)] = ∅. Hence x /∈ gij[gei(b)] = egij(egi(b)) =gej(b).
2.4. ULTRAMETRIC SPACES AND GENERALIZED CANTOR CUBES 55
αfor eachα < κ. Then these are equivalence relations with the following properties:
\
α∈κ
∼α = ∆X
\
α∈B
∼α = ∼β wheneverβ =sup(B)andB⊆κis bounded.
In particular∼βis a subset of∼αwheneverβ>α, and∼0 = X2. In fact, any family of binary relations on X with these properties defines a κ-ultrametric via d(x,y) = sup{α | x ∼α y} (which is in fact max{α| x∼α y}unlessx=y).
We call [x]α = {y | x ∼α y} the α-ball of x. The topology induced by d is the coarsest κ-topology such that allα-balls forα < κare open. Then for everyα < κ, theα-balls consitute a partition ofXof clopen sets, and ifXisκ-compact, that partition has onlyκ-few members.
A space isκ-ultrametrizableif its topology is induced by aκ-ultrametric.
Lemma 38. Every regular space with weight6κisκ-ultrametrizable.
Proof. By Lemma 31, there is a basehBα|α < κiof clopen sets. Setting x∼α y iff ∀β < α. x∈Bβ ⇔ y∈Bβ
defines aκ-ultrametric inducing the topology ofX, because on the one hand, every[x]αis the intersection ofαclopen sets, and on the other hand, everyBαis a union ofα+1-balls.
A mapf :X → Y betweenκ-ultrametric spaces is auniformly continuousif for everyβthere is anαsuch that
∀x0,x1 ∈X. x0∼αx1 ⇒ f(x0)∼β f(x1).
Given a setC⊆κ, we callfaC-nonexpansivemap if for allα∈ C,
∀x0,x1 ∈X. x0 ∼αx1 ⇒ f(x0)∼α f(x1),
a C-isometryif it is bijective and bothfandf−1areC-nonexpansive, and simply anisometry if it is aκ-isometry.
Lemma 39. LetXandYbeκ-ultrametric spaces andf:X→ Y continuous.
1. IfXisκ-compact,fis uniformly continuous.
2. If f is uniformly continuous, there is a closed unbounded C ⊆ κ such that f is C-nonexpansive. In fact, the set Cof all αsuch that f is {α}-nonexpansive is closed un-bounded.
3. Iffis bijective and bothf−1andfare uniformly continuous, there is a closed unbounded C ⊆κsuch thatfis aC-isometry. In fact, the setCof allαsuch thatfis an{α}-isometry is closed unbounded. So in particular, for any twoκ-ultrametrics on a spaceXgiven by h∼α |α∈κiandh∼α0 |α∈κi, the set ofα ∈κwith∼α=∼α0 is closed unbounded.
Proof. (1): Letβ ∈ κ. Then theβ-balls constitute a partition ofY of clopen sets, so the set P = {f−1[[y]β]| y∈Y}of their preimages is a partition ofXof clopen sets. The set
{[z]γ | γ ∈κ, z∈ X, P contains a superset of[z]γ} is an open cover ofX, so it has a subcover{[zi]γi | i∈I}indexed by aκ-smallI. Let
α = sup
i∈I
γi.
Now wheneverx0 ∼α x1,x0 is in some[zi]γi. Sinceα > γi, the sameγi-ball must contain x1. But[zi]γi is a subset of some elementf−1[[y]β]ofP. In particular,f(x0)∼β y ∼βf(x1).
(2): First let us show thatCis closed, so assume thathβα|α<γiis an increasing sequence in Cfor some limit ordinalγ < κ, and letδ = supα<γ βα. Now ifx0 ∼δ x1, then in particular x0 ∼βα x1for allα < γ. Since allβαare inC, this impliesf(x0) ∼βα f(x1)for allα < γ. But since∼δ is the intersection of these∼βα, it follows thatf(x0) ∼δ f(x1), soδ∈ C.
To verify thatCis unbounded, we have to find for everyα∈ κan ordinalβ >αinC. To this end, letα0= αand proceed recursively: For everyn∈ω, there is anαn+1such that
∀x0,x1 ∈X. x0 ∼αn+1 x1 ⇒ f(x0) ∼αn f(x1).
If we defineαω = supn∈ω αn, thenα 6 αω ∈ C, becausex0 ∼αω x1implies∀n. x0 ∼αn+1
x1, which implies∀n.f(x0) ∼αn f(x1), which finally entailsf(x0) ∼αω f(x1).
(3) follows from (2), because if C0 is the closed unbounded set given by (2) and C1 is the corresponding closed unbounded set forf−1 instead off, thenC0∩C1 is closed unbounded, too. But that is exactly the set of allα ∈ κfor whichfmaps theα-balls ofXbijectively onto theα-balls ofY, as claimed.
Let X be a κ-ultrametric space and let SX be the set of all limit ordinals δ < κ such that every descending sequenceh[xα]α|α < δiinXhas a nonempty intersection, or equivalently, such that every descending sequenceh[xα]α|α∈ Ciwith unboundedC⊆ δhas a nonempty intersection. Let INS be the nonstationary ideal. Then we call SX/INS ∈ P(κ)/INS the solidityofX. Aκ-ultrametric space with solidityκ/INS is calledsolid.
Proposition 40. IfXandY are homeomorphicκ-compactκ-ultrametric spaces, then SX ≡SY mod INS.
In particular, the solidity of a κ-compact κ-ultrametrizable space does not depend on the choice of a specificκ-ultrametric.
Proof. Letf :X→ Ybe a homeomorphism and letC⊆κbe closed unbounded such thatfis aC-isometry, as in Lemma 39. We will show thatSX andSY agree on the closed unbounded set Lim(C) of the limit points of C. Since the situation is symmetric, it suffices to prove SX∩Lim(C) ⊇ SY∩Lim(C), so let δ ∈ SY∩Lim(C). Let h[xα]α|α < δi be a descending
2.4. ULTRAMETRIC SPACES AND GENERALIZED CANTOR CUBES 57
sequence inX. For everyα ∈ δ∩C, there is a yα, such that f[[xα]α] = [yα]α. Then since δ∈ SY, the sequenceh[yα]α|α ∈δ∩Cihas a nonempty intersection. Therefore
\
α<δ
[xα]α = \
α∈δ∩C
[xα]α = \
α∈δ∩C
f−1[[yα]α] = f−1
"
\
α∈δ∩C
[yα]α
# 6= ∅,
which proves thatδ ∈SX.
Theorem 41. Every perfect, κ-compact, solid, κ-ultrametrizable space is homeomorphic to the Cantor cubeDκ.
Every perfect,κ-compact,κ-ultrametrizable space is homeomorphic to a closeda⊆Dκ.
Proof. LetXbe perfect,κ-compact,κ-ultrametrizable.
IfX is in addition solid and h∼α |α ∈ κi is a κ-ultrametric, there is a closed unbounded C ⊇ SX. Let hβα|α < κi be a monotonously increasing enumeration of the elements of C. Then h∼βα |α ∈ κi is aκ-ultrametric onXof whichno decreasing sequence of balls has an empty intersection. So we can wlog assume that Xisκ-ultrametrized in such a way that SX =κ.
We recursively define an injective functiong from the set of balls to D<κ such that g(a) ⊂ g(b)iffa ⊃ b, whereby we letg(X)be the empty sequence, and for limit ordinals α, we let g([x]α) = S
β<αg([x]β).
Letα < κandx∈X. Letβbe minimal such that[x]α 6= [x]β. (Such an ordinal exists because xis not an isolated point.) Choose any enumerationh[yγ]β|γ < δ+2iof theβ-balls in[x]α. SinceXisκ-compact Hausdorff,δ < κ. Define
g([yγ]β) = g([x]α)_0γ_1 forγ < δ+1, and g([yδ+1]β) = g([x]α)_0δ_0.
Then rng(g) maps disjoint balls to incomparable sequences. The sequences g([x]α) with α < κ for a given x define a unique cofinal branch in D<κ and therefore a unique point f(x) = S
α<κg([x]α) ∈ Dκ. We claim that with this definition,f : X → Dκ is injective and continuous.
First of all,fis injective, because ifx6=y, there is anαsuch that[x]α 6= [y]αand thusg([x]α) andg([y]α)are incomparable. Butf(x)is an extension of the former andf(y)is an extension of the latter.
To show thatfis continuous, we have to verify that each preimagef−1(At)of an open basis setAt = {s ∈Dκ | t⊆ s}is open. f(x) ∈ At means thatg([x]α)is comparable tosfor allα, and in particular that someg([x]α)is an extension ofs. Sof−1(At)is the union of all[x]αfor whichs ⊆g([x]α), which is open.
To show thatfis surjective in the caseSX = κ, lets ∈Dκ. Since rng(g)is cofinal and order-reversing, the set of preimagesA= g−1[{s α |α < κ}]for allα < κconstitutes a decreasing sequence of balls. Ift⊂ sandt∈ rng(g), there is anxsuch thatg([x]α) = t. Ifβ > αandδ
are as in the definition ofg, there is either aγsuch thatt_0γ_1 ⊂sort_0δ_0 ⊂s. Then there is aβ-ball mapped to that extension oft, and in particular thatβ-ball is inA. SoAhas no least element. But ifAhasκ-few elements thenT
A = [x]αfor some α < κandx ∈ X, andg([x]α) ⊂ s, so[x]α ∈ A. It follows that Ahasκ-elements and by compactnessT
Ais nonempty. Thus it contains exactly onexandf(x) = s.
Since Expκ(Dκ)is perfect,κ-compact andκ-ultrametrizable, there is a homeomorphismΣX : a → Expκ(Dκ)for some closeda ⊆ Dκ by Theorem 41. Hence withVX = Dκ andSX = a, this defines a hyperuniverse. We shall see, however, thatSXcannot be clopen.
Since a similar reasoning also applies to κ = ωand the notion of solidity is not needed for the surjectivity of f, Dℵ0 is in fact the universe space of an atomless ω-hyperuniverse. S.
Sirota proves in [Sir68] thatDℵ1 is homeomorphic to its hyperspace, too! On the other hand, in [Š76] L. B. Šapiro shows that Expω(Dℵ2) is not the continuous image of any Dλ. Since Expω(Dℵ2)is the continuous image of every Expω(Dτ) for τ > ℵ2, this implies that noDτ withτ >ℵ2is an atomlessω-hyperuniverse.
So what about the generalized Cantor cubes with additivityκ > ω? For which λis Dλ an atomless κ-hyperuniverse? Is Dκ an example? Since if κ is weakly compact, Expκ(Dκ) is a perfect, κ-compact κ-ultrametrizable space, this amounts to the question of whether it is solid.
The exponentialκ-ultrametric on the hyperspace Expκ(X)of aκ-ultrametric spaceXis defined such thata ∼α biffaandbintersect the same< α-balls, that is:
a∼α b iff ∀β < α ∀x∈ X. a∩[x]β = ∅ ⇔ b∩[x]β =∅
Forκ-compact spaces, this is in fact a κ-ultrametric generating the exponentialκ-topology:
Givena ∈Expκ(X), theα-ball aroundais
[a]α = \
β<α
[
[x]β∩a6=∅
[x]β ∩ \
[x]β∩a6=∅
♦[x]β
, which is open in Expκ(X), because the intersections areκ-small. Conversely,
[x]α = [{x}]α+1 and ♦[x]α = [
[x]α∈B⊆{[y]α|y∈X}
h[Bi
α+1
are open with respect to the exponential ultrametric, and by Lemma 30, these sets are a subbasis of Expκ(X). Hence the topologies coincide.
For every regularλ < κlet Eκλ = {α < κ | cf(α) = λ}. Let W ⊆ κbe the set of all regular cardinals. Note that eachEκλis almost disjoint fromW.
2.4. ULTRAMETRIC SPACES AND GENERALIZED CANTOR CUBES 59
Theorem 42. Letκbe weakly compact and letXbe aκ-compactκ-ultrametric space.
1. SX∩Eκω 6SExp
κ(X) 6SX mod INS. 2. If X has a perfect subset, thenSExp
κ(X) 6W∪Eκω mod INS.
In particular, the solidity of everyκ-ultrametrizable clopenκ-hyperuniverseXis at most W∪Eκω/INS.
Proof. We endow Expκ(X)with the exponentialκ-ultrametric.
(1): Firstly, assume γ ∈ SX∩Eκω and let h[bα]α+1|α < γi be a decreasing sequence in Expκ(X). Sincebα ∼α+1 S
x∈b[x]α, we can wlog assume thatbαis a union ofα-balls. Then the bα constitute a decreasing sequence of subsets of X. We claim thatb = T
α<γbα is in every [bα]α+1. Since b ⊆ bα, it is clear that b intersects at most the α-balls intersected bybα, so we have to show that it in fact intersects every [x]α with x ∈ bα. To prove this, let α0 = α and let hαn|n ∈ ωi be an increasing sequence cofinal in γ. Let x0 = x. If [xn]αn intersectsbαn, then there must be some αn+1-ball[xn+1]αn+1 ⊆ [xn]αn intersecting bαn+1. Since γ ∈ SX, there exists anx ∈ X which is a member of every[xn]αn. And since [xn]αn ⊆bαn,xis a member ofb, proving thatbin fact intersects[x]α.
Now assumeγ ∈ SExp
κ(X)and let h[xα]α|α < γi a decreasing sequence of balls inX. Then h[{xα}]α+1|α < γiis a decreasing sequence of balls in Expκ(X), which by our assumption has a nonempty intersection. Ifais in that intersection andy∈a, theny∼α xαfor everyα < γ, soyis in the intersection of the[xα]α, which proves thatγ∈SX.
(2): We define a closed unboundedC ⊆ κ of cardinals such that the closed unbounded set Lim(C) of its limit points is disjoint fromSExp
κ(X)\(W∪Eκω). We giveC as the range of a normal functionh : κ→ κwithh(0) = 0: For eachα < κ, there areκ-fewh(α)-balls. Since Pis perfect, eachh(α)-ball[x]αthat intersectsPhas at leastκdistinct points in common with P, so there is a least cardinal β[x]α < κ, such that [x]α is the union of at leasth(α) distinct β[x]α-balls intersectingP. Let
h(α+1) = sup
x∈X
β[x]
α. Since there are onlyκ-fewα-balls,h(α+1) < κ.
Now letγ ∈ Lim(C)\(W∪Eκω)andτ = cf(γ). Thenω < τ < γand there is an increasing sequencehα|α < τiinC∩[τ,γ)converging toγ. Then by our construction ofC, each[x]α intersectingP consists of at leastτdistinctα+1-balls intersectingP.
Consider the treeT of all nondecreasing functions t : α → τ with α < τ, finite range and t(β)> βfor allβ < α, ordered by inclusion. Sinceτis regular and uncountable, this tree has no cofinal branch, but everyt ∈ T has a successor on every levelα < τ, and every element has exactlyτdistinct immediate successors.
We recursively define a mapg : T → P which maps each levelTα ofT injectively to P and in addition has the property thatg(t) = g(t α) whenever there is aβ < α such that tis
constant on[β,dom(t)). Firstly, letg(∅)be any point ofP. Ifg(t)has been defined andtis on levelα, map the immediate successors oftto any distinct elements ofP∩[g(t)]α which are in differentα+1-balls, such that g(t_max(rng(t))) = g(t) whenevert_max(rng(t)) ∈ T.
By our choice ofC, there are indeed enoughα+1-balls intersectingP. If the levelα= dom(t) oftis a limit, then by our construction,hg(t β)|α < βiis eventually constant and we can defineg(t)to be that final value, too.
Letbα = {g(t) | t ∈ Tα} for all α < τ. Then the sequenceh[bα]α+1|α < τi in Expκ(X) is descending: Ifα < β < τandg(t) ∈ bβ, theng(t) ∼α g(t α) ∈ bα. On the other hand, ifg(t) ∈ bα, then g(t) ∼α g(t_ββ\α) ∈ bβ. Thus bα and bβ in fact intersect the same α-balls and hencebα ∼α+1 bβ, which means[bα]α+1 = [bβ]α+1⊇ [bβ]β+1.
Now assume that a ∈ T
α<τ[bα]α+1, and let x ∈ a. Then for every α < τ, there is a g(tα) ∈ bα withg(tα) ∼α x. Thus for allα < β < τ,g(tα) ∼α g(tβ), which implies that tα ⊆tβ. Hence{tα | α < τ}is a cofinal branch inT, a contradiction.
We conclude that T
α<τ[bα]α+1 is a decreasing sequence with an empty intersection and thereforeγ /∈SExp
κ(X).
Finally, if X is a clopen κ-ultrametrizable κ-hyperuniverse, then SX is homeomorphic to Expκ(VX)and has solidity at mostW∪Eκω/INS. But if a descending sequence of balls has an empty intersection in the clopen subsetSX, then it has an empty intersection inVX, so the solidity ofVXcannot be greater:SVX 6SSX 6W∪Eκω mod INS.
With the canonicalκ-ultrametric defined byd(f,g) = min({κ} ∪ {α < κ | f(α) 6= g(α)}), the Cantor cube Dκ (as well as thePelczynski spaceD6κ) is solid, because ifh[fα]α+1|α < δi is a decreasing sequence, fβ βmust be an extension offα αfor allα < β < δ, and hence S
α<δ(fα α)is a function each extension of which is an element ofT
α<δ[fα]α. So it cannot be a clopen κ-hyperuniverse by Theorem 42. From that we can even conclude that none of the spacesDλ with greaterλis an atomless hyperuniverse:
Theorem 43. Letλ,τ > κand assume thatκ is mildlyν-ineffable for some regular cardinal ν>λ,τwithν<κ = ν. ThenDλ 6=∼ Expκ(Dτ).
Proof. By Theorem 32, these spaces areκ-compact. Dλ is the inverse limit of the system of projections πab : Db → Da with πab(f) = f a for f ∈ Db and a ⊆ b ∈ Pκ(λ), and by Lemma 37, Expκ(Dτ)is the inverse limit of the system of mapspa,b :Expκ(Db) → Expκ(Da), wherepab = Expκ(πab)fora⊆b∈ Pκ(τ).
By Lemma 36, if they were homeomorphic, then there would existκ-closed subsetsK⊆ Pκ(λ) andKe ⊆ Pκ(τ) of size κ, such that the inverse limits of the corresponding subsystems are homeomorphic. Now replacingK by its downwards closure in Pκ(λ) does neither increase its cardinality nor change the inverse limit nor affect its κ-closedness, so assume that K is downwards closed. Thus there exists a seta ⊆ λ of sizeκ with K = Pκ(a). For the same reason we can assume there is ab⊆Pκ(τ)of sizeκwithKe = Pκ(b).
But then the inverse limits of these subsystems would beDa and Expκ(Db), and that would implyDκ =∼ Da =∼ Expκ(Db) =∼ Expκ(Dκ), contradicting Theorem 42.