2.7. METRIC SPACES AND THE HILBERT CUBE 81
Proof. Let A ∈ Expω(Y), that is, A ⊆ Y closed. Assume Expω(f)−1[{A}]is not connected, that is, Expω(f)−1[{A}] =B∪Cis the disjoint union of nonempty closed setsBandC. Since Expω(X) is normal,BandCcan be separated by disjoint open sets, and since they are com-pact, those disjoint open sets can be chosen such that they are unions of finitely many sets of the form
Vi = (U1i∪...∪Unii)∩♦U1i∩...∩♦Unii
withU1i,...,Uniirelatively open inB∪Cand i ∈ {1,...,m}. Then every Vi intersects either BorC, and wheneverVi∩B6= ∅andVj∩C6= ∅, thenViandVj are disjoint.
We will show that the setf−1[A]is inB. Since the situation is symmetric, it is then also inC, which is impossible.
LetS0 ∈Bbe arbitrary and recursively defineSlforl ∈ω. Namely, let Sel = f−1[A] \ [
{Uki | Sl∩Uki = ∅}.
Then everyVithat did not contain Sl also does not containSel, but sinceSl ⊆ Sel ⊆ f−1[A], Selis in Expω(f)−1[{A}]. So someVithat contains Sl must also containSel and thuseSl ∈ B.
IfSel = f−1[A], we are done, so assume thatf−1[A]contains a pointx /∈ Sel. Nowf−1(f(x)) intersectseSland its open supersetU1i∪...∪Unii, and since it is connected, there is a point
y ∈ f−1(f(x)) ∩ (U1i∪...∪Unii) \ Sel.
The set Sl+1 = Sel∪{y} then is still in Vi and thus in B, but it intersects strictly more of the open setsUk0i0 thanSl. Therefore this construction has to come to an end, where some eSl= f−1[A].
Now ifXis connected, the preimages of singletons underf : X → {p}are connected, so the same goes for Expω(f) : Expω(X) →{{p}}. Hence Expω(X)is connected.
Lemma 64. IfXis a locally connected metric space, then so is Expω(X).
Proof. It suffices to show that wheneverU1,...,Unare connected, then so isV =(U1∪...∪ Un)∩♦U1∩...∩♦Un, because the sets of that form are a basis. IfVwere the disjoint union of two nonempty open setsW1and W2, then each of them would have finite elements. Let m1,...,mn be such that there area1 ∈ W1anda2 ∈ W2 which intersect eachUi in at most mipoints. Since the map
f:Um1 1×...×Umnn → V, f(x) ={xi | i6m1+...+mn}
is continuous and theUiare connected, its image is connected, too. But it contains both a1 anda2, a contradiction.
Lemma 65. LethXn,fn,mibe an inverse system of compact, metrizable and connected spaces indexed byω. Let
hXω,fn,ωi = lim
←hXn,fn,mi.
ThenXω is compact, metrizable and connected.
Proof. The metrizability follows from Lemma 62, and the inverse limit is compact, because it is a closed subspace of the productQ
n∈ωXn.
Now assume that all the Xn are connected. Suppose Xω is the union of two nonempty clopen setsAandB; we have to show that they are not disjoint. As everyXn is connected, fn,m[A]andfn,m[B]always have a nonempty intersectionCn. Then the setsf−1n,ω[Cn]are a decreasing sequence of nonempty closed sets and hence have some pointxin common. Since everyfn,ω(x)is infn,ω[A],xis inA, and similarly forB.
In [Gsc75], G. R. Gordh and S. Mardeši´c prove that if in addition allXnare locally connected and all preimagesf−1n,m[{x}]of singletons are connected, thenXωis locally connected, too.
Summing up the results of this section, we now know that wheneverXis an object ofCmsuch thatVXis a Peano continuum and all(ΣX)−1[{x}]are connected, thenV∞Xis homeomorphic to the Hilbert cube.
As an example, letVX = SX = [0,1]andSΣX(A) = min(A). VXis a Peano continuum, and the sets SΣX−1
[{x}]are even path-connected: For any given elementA, F : [0,1]→ SΣX−1
[{x}], F(t) = {x+t(a−x) | a∈A}
is a path that connectsAto{x}. ThusVΣωXis homeomorphic toH. What is the set of autos-ingletons in this model? A ∈VωXis an autosingleton iffA= {x}, wherex =VΣωX(A), which means that for alln,VΣXn+1,ω(x) = VΣXn,ω[A] =
VΣXn,ω(x) . This recursive formula shows that the only autosingletons are the objects obtained by repeatedly taking singletons, and thus the set of autosingletons is homeomorphic toVX.
Analogously, we obtain for every natural numberna Hilbert cube model whose set of autosin-gletons is homeomorphic to[0,1]n, by settingVX= SX = [0,1]nandSΣX(A) =hx1,...,xni, wherexi =min{yi | hy1,...,yni ∈A}.
Open Questions
A topological characterization of positive set theory
We have seen thatGPK+(∞)+T3is equivalent toTS(∞)+ (A=∅∈V) +T3+Union. But we were unable to find such a formulation of GPK+(∞) alone and in fact we neither know whether GPK+(∞) implies the T3 separation axiom (or even just that V is a Hausdorff space), nor whether TS(∞)+Union implies the positive comprehension principle. Also, the union ax-iom is not really a topological statement. Is there a topological axax-iomatization ofGPK+(∞), that is, is GPF comprehension a topological property ofV?
Independence of the compactness axiom
In our proof that everyκ-hyperuniverse isκ-compact (Theorem 28), we made heavy use of the set theory external to that hyperuniverse, using for example cardinalsλwhich might be much larger thanκ. It is not clear that the theoryTSimplies that the universe isD-compact:
Without a choice principle (for classes) much stronger than the uniformization axiom the proof for hyperuniverses cannot be carried out within topological set theory. Does TS – or GPK+ orTS+Union+T4– imply compactness?
Normality and compactness
It follows from Lemmas 25 and 27 that whenever Expκ(Expκ(X))is normal,Xisκ-compact.
For the caseκ = ω, N. V. Velichko proved that in fact X is compact whenever Expω(X) is normal (cf. [Vel75])! Although not directly related to hyperuniverses and topological set theory, it would be interesting to know for whichκ > ωthis is true, too.
Cantor cubes
We were able to prove thatDκ is the universe space of a κ-hyperuniverse but no Dλ is the universe space of any clopen κ-hyperuniverse. Do the Dλ with λ > κ form (non-clopen) κ-hyperuniverses?
The duality of the limit constructions
The categoriesEHypandCHyp of hyperuniverses are canonically isomorphic: the mapsΣX are just reversed. We can extend this isomorphism to a functor defined on a larger subcat-egory ofCm∗: For each object X of Cm∗ such that ΣX is a homeomorphism, letRX be the corresponding object ofEx∗, that is,VRX = VX,ARX = AXandΣRX = (ΣX)−1. Iff:X → Y is a morphism, letRf be the morphism of Ex∗ defined by the same function Vf. In exactly the same way we defineRX if X is an object of Ex∗ such that VX is Hausdorff and locally κ-compact andSΣXis a homeomorphism.
Then for example,RVc = Ve, and for every objectXofEHypand dense embeddingf:Ve → X, there are unique horizontal arrows such that the following diagram commutes:
Ve //X //RVc
Ve
eVe
ii
f
OO
ReVc
55
Thus Ve really is the largest and Vc is the smallest hyperuniverse, in which Ve is dense.
But we do not know whether they are actually isomorphic, that is, whether the morphism Ve → RVc in the diagram is given by an injective map. If they are, then there is up to isomorphism only one hyperuniverse in whichVe can be densely embedded.
More generally, the same question poses itself for every suitable objectYinstead ofVe.
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Index
A-homeomorphism, 66
Cech-Stone compactification, 47ˇ
Cech-Stone compactification functor, 62ˇ additivity, 7
adjoint functors, 50
Alexandroff compactification, 42
Alexandroff compactification functor, 70 algebra, 44
atom, 2, 5 atom space, 39
atomless hyperuniverse, 39 axiom
additivity, 8 atoms, 6
comprehension, 6 exponential, 8 extensionality, 6 infinity, 15 nontriviality, 6 regularity, 29 topology, 8 union, 29 ball, 33, 55 base, 7
bounded positive formula, 26 BPF specification, 26
Cantor cube, 44
Cantor-Bendixson derivative, 9 category, 48
choice function, 31 class, 5
class axioms, 6
clopen hyperuniverse, 39 closed class, 6
closed subset of a partial order, 50 closure, 3, 6
coarse topology, 7
cocover, 7 codomain, 48 compact, 7 composite, 48 cone, 49 continuous, 7 deciding filter, 44 direct limit, 49 directed set, 49, 50 directed system, 49 discrete class, 7 domain, 48 epimorphism, 48 essential set theory, 8 exponential functor, 62 exponential space, 8 exponential ultrametric, 58 few, 7
filter, 44 fine topology, 7 finite class, 15 functor, 49
generalized positive formula, 2, 26 generated topology, 7
GPF specification, 26 Hausdorff space, 7 hyperspace, 8
hyperuniverse, 22, 39 hyperuniverses, 3 infinite class, 15 initial morphism, 49 initial object, 49 interior, 6 inverse limit, 50