OF SPACES THAT LOOK LIKE [0, ∞)
우리는 CH 하에서 space X가 H와 유사하다고 가정하고, 그 나머지가 H∗과 homeomorphic 인지 증명한다. 이는 두 개의 공간이 quotient mapping에 의해 동일한 집합으로 매핑되는 것을 보여주는 Lemma 2.2를 사용하여 증명된다.
우리는 또한 CH가 모든 표준 하위 연속체 (standard subcontinua) 가 H∗와 homeomorphic 인 경우와 동치임을 보인다. 여기서, 표준 하위 연속체는 Iu = ∩{In : n ∈ U} 형태의 closed interval sequence ⟨In⟩n에 의한 것이다.
또한, 우리는 특정 하위 연속체 (subcontinua)가 H∗와 homeomorphic 인 경우를 보인다. 이러한 연속체는 natural candidate로 여겨지며, Lemma 2.1과 Wallman duality을 사용하여 증명된다.
이 논문은 Čech-Stone 나머지를 연구하는 데 중요한 기여를 한다. 또한 CH의 성질에 대한 새로운 조건을 제시한다.
한글 요약 끝:
OF SPACES THAT LOOK LIKE [0, ∞)
arXiv:math/9305202v1 [math.LO] 18 May 1993ˇCECH-STONE REMAINDERSOF SPACES THAT LOOK LIKE [0, ∞)ALAN DOW AND KLAAS PIETER HARTAbstract. We show that many spaces that look like the half line H = [0, ∞)have, under CH, a ˇCech-Stone-remainder that is homeomorphic to H∗.
Wealso show that CH is equivalent to the statement that all standard subcontinuaof H∗are homeomorphic.The proofs use Model-theoretic tools like reduced products and elementaryequivalence.IntroductionThe purpose of this note is to answer (partially) some natural questions aboutthe ˇCech-Stone remainder of the real line or rather the remainder of the space H =[0, ∞) as the remainder of R is just a sum of two copies of H∗.Our first result says that under CH the space H∗is, to a certain extent, unique:if X is a space that looks a bit like H then X∗and H∗are homeomorphic. To ‘looka bit like H’ the space X must be a connected ordered space with a first element,without last element, of countable cofinality and of weight at most c. The weightrestriction is necessary, because if the weight of X is larger than c then so is theweight of X∗and therefore X∗cannot be homeomorphic with H∗.As a consequence various familiar connected ordered spaces have a ˇCech-Stoneremainder that is homeomorphic to H∗.
So the remainders of the lexicographicordered square (minus the vertical line on the right) and of any Suslin line arehomeomorphic to H∗.The second result is concerned with the so-called standard subcontinua of H∗:take a discrete sequence ⟨In⟩n of closed intervals in H and put, for any u ∈ω∗,Iu = TU∈u clSn∈U In. Then Iu is a standard subcontinuum of H∗.We show that CH is equivalent to the statement that all standard subcontinuaare homeomorphic.
This solves Problem 264 from Hart and van Mill [1990].Our final result shows that certain subcontinua of the standard subcontinua arehomeomorphic to H∗; the precise statement is in Section 3, here suffice it to saythat these continua are natural candidates for being homeomorphic to H∗.As may be expected we shall not directly construct homeomorphisms betweenthe spaces in question—it’s too hard to take care of 2ω1 points in ω1 steps—but1991 Mathematics Subject Classification. Primary 54D40; Secondary 54F15 03C20.Key words and phrases.
ˇCech-Stone remainder, continuum, real line, Wallman spaces, elemen-tary equivalence, saturated models.1
2ALAN DOW AND KLAAS PIETER HARTwe show that the spaces have isomorphic bases for the closed sets (isomorphic aslattices). That this works follows from the results of Wallman from [1938], to bedescribed in Section 1 below.A few words on how we show that the bases are isomorphic as lattices: Weimplicitly and explicitly use a powerful result from Model Theory which says thatunder quite general circumstances various structures are isomorphic.
In each casethe bases are identified as reduced products of families of easily described lattices.The factors of these products are pairwise elementary equivalent and hence so arethe products themselves. Furthermore these products satisfy an certain saturationproperty.
The combination of elementary equivalence and this saturation propertyimplies that the lattices are isomorphic. A more detailed explanation can be foundin Section 3.The paper is organized as follows.
Section 1 contains some preliminary remarks.In Section 2 we prove the first result, the proof is self-contained (i.e., requires nomodel theory). In Section 3 we prove the results about the standard subcontinua,here we appeal to standard fact from Model Theory to keep the proofs pleasantlyshort.
The final Section 4 deals with a special case of Theorem 2.1 that can beproved under weaker assumptions.1. Preliminaries1.1.
Sums of compact spaces. We shall be dealing with sums of compact spacesa lot, so it’s worthwhile to fix some notation.
So let X = Ln∈ω Xn be a topologicalsum of compact spaces; we always take Sn∈ω{n} × Xn as the underlying set of thespace. The map q : X →ω defined by q(n, x) = n extends to βq : βX →βω.
Weshall always denote the fiber of u ∈ω∗under the map βq by Xu.1.2. The half line.
Our main objects of interest are the half line H = [0, ∞) andits ˇCech-Stone remainder H∗. The space H∗is a quotient of another space—thatis somewhat easier to handle—by a very simple map.Indeed, consider the space M = ω × I—the sum of ω many copies of the unitinterval I.
The map πH : M →H defined by πH(n, x) = n + x maps M onto H andthe map π∗H = βπH ↾M∗maps M∗onto H∗.A key point in our proof is to see what kind of identifications are made by π∗H, sowe take a better look at the components of M∗. Because ω∗is zero-dimensional andbecause Iu is connected for every u we know exactly what the components of M∗are: the sets Iu.Furthermore, each Iu has a natural top and bottom: we call the point 0u =u- lim⟨n, 0⟩the bottom point and 1u = u- lim⟨n, 1⟩the top point.
The continuum Iuhas many cut points: for every sequence ⟨xn⟩n in (0, 1) the point xu = u- lim⟨n, xn⟩is a cut point of Iu and this set of cut points is dense. It follows that Iu is irreduciblebetween 0u and 1u, which means that there is no proper subcontinuum of Iu thatcontains 0u and 1u.We can put a preorder on Iu: say x ≤u y iff every subcontinuum of Iu thatcontains 0u and y also contains x.
The layer of the point x is the set {y : y ≤ux and x ≤u y}. This order is continuous in the sense that {y : y ≤u x} is theclosure of {y : y
We shall use this order in the proof of Theorem 3.2.
ˇCECH-STONE REMAINDERS OF SPACES THAT LOOK LIKE [0, ∞)3We turn back to the map π∗H; using standard properties of the ˇCech-Stone com-pactification one can easily prove the next lemma (if u ∈ω∗then u + 1 is theultrafilter generated by {U + 1 : U ∈u}).Lemma 1.1. For every u ∈ω∗the map π∗H identifies the points 1u and 0u+1 andthese are the only identifications made.The continua Iu govern most of the structure of H∗; they are known as thestandard subcontinua of H∗.
More information on H∗can be found in the surveyHart [1992].1.3. Wallman spaces.
As mentioned above we construct the homeomorphismsindirectly via isomorphisms between certain lattices of closed sets of the spaces inquestion.This is justified by the results of Wallman from [1938]; Wallman generalizedthe familiar Stone duality for Boolean algebras and zero-dimensional spaces to aduality for lattices and compact spaces. We briefly describe this ‘Wallman duality’.If L is a lattice then a filter on L is a subset F such that 0 ̸∈F, if x1, x2 ∈Fthen x1 ∧x2 ∈F and if x1 ∈F and x1 ≤x2 then x2 ∈F.
An ultrafilter on L isjust a maximal filter. The set XL of ultrafilters on L is topologized by taking thefamily of all sets of the form x+ = {F : x ∈F} with x ∈L as a base for the closedsets of XL.
The space XL is always compact, it is Hausdorffiff L satisfies a certaintechnical condition.If B is a base for the closed sets of a compact Hausdorffspace then B satisfiesthis condition. Thus, X = XB whenever B is a (lattice) base for the closed setsof B.
It is now easy to see that two compact Hausdorffspaces with isomorphic(lattice) bases for the closed sets are homeomorphic.2. Remainders of spaces that look like HThis section is devoted to a proof of the result mentioned in the introduction,namelyTheorem 2.1 (CH).
Let X be a connected ordered space with a first element,with no last element, of countable cofinality and of weight c. Then X∗and H∗arehomeomorphic.We shall construct the homeomorphism indirectly, via spaces that are mappedonto H∗and X∗respectively.Remember from 1.2 that H∗is the quotient of M∗obtained by identifying1u and 0u+1 for every u ∈ω∗and that the map is called πH.We can construct a similar situation for X∗: take a strictly increasing and cofinalsequence ⟨an⟩n in X with a0 = min X. For every n let Jn = [an, an+1] and considerthe sum Y = Ln Jn.The map π : Y →X defined by π(n, x) = x identifies⟨n, an+1⟩and ⟨n + 1, an+1⟩for every n.As in the case for H∗and M∗the only identifications made by π∗= βπ ↾Y ∗areof u- limn⟨n, an+1⟩and u- limn⟨n + 1, an+1⟩= u + 1- limn⟨n, an⟩for every u ∈ω∗.In other words, for every u ∈ω∗the top point of Ju is identified with the bottompoint of Ju+1.
We denote the top point of Ju by tu and the bottom point by bu.This gives rise to the following lemma.
4ALAN DOW AND KLAAS PIETER HARTLemma 2.2. If h : M∗→Y ∗is a homeomorphism that maps Iu to Ju and moreovermaps 1u to tu for every u then h induces a homeomorphism from H∗onto X∗.□The maps π ◦h and πH have exactly the same fibers.
Both are closed, beingcontinuous between compact spaces, hence quotient mappings.Hence H∗(thequotient of M∗by πH) and X∗(the quotient of M∗by π ◦h) are homeomorphic. □Our efforts then will be directed towards constructing a homeomorphism betweenM∗and Y ∗that satisfies the assumptions of Lemma 2.2.Rather than constructing a homeomorphism we shall construct two bases B and Cfor the closed sets of M∗and Y ∗respectively and an isomorphism between themthat will induce the desired homeomorphism.To construct B we consider the lattice generated by the closed intervals in I. Itis a base for the closed sets of I.We let Ln be the corresponding lattice for In.
The product lattice L = Qn Lncorresponds in a natural way to a base for the closed sets of M. The reduced productL∗= Qn Ln/fin—obtained by identifying x and y whenever {n : x(n) ̸= y(n)} isfinite—will then correspond in a natural way to a base for the closed sets of M∗.This will be the base B.In a similar way we find C: let Kn be the lattice generated by the closed intervalsof Jn and consider K = Qn Kn and the reduced product K∗= Qn Kn/fin. Thelattice corresponds to a base C for the closed sets of Y ∗.Finding an isomorphism between L∗and K∗is the same thing as finding a bijec-tion ϕ between L and K such that for all x, y ∈L we have x ≤∗y iff ϕ(x) ≤∗ϕ(y),where x ≤∗y means that {n : xn ≤yn} is cofinite.To ensure that the induced homeomorphism maps Iu to Xu for every u, it sufficesto ensure that whenever y = ϕ(x) the sets {n : xn = ∅} and {n : yn = ∅} as wellas the sets {n : xn = In} and {n : yn = Xn} differ by a finite set only.Furthermore, to get h(0u) = bu and h(1u) = tu for every u we simply mapthe closed set bM =⟨n, 0⟩: n ∈ωto bX =⟨n, an⟩: n ∈ωand the settM =⟨n, 1⟩: n ∈ωto tX =⟨n, an+1⟩: n ∈ω.
We leave it to the reader tocheck that this will indeed suffice.We shall construct a bijection ϕ from L to K that satisfies the following condi-tions:(α) ϕ(bM) = bX and ϕ(tM) = tX, and(β) for every x and y in K there is an N ∈ω such that for every n ≥N the setsof endpoints of ϕ(x)(n) and ϕ(y)(n) have the same configuration as the setsof endpoints of x(n) and y(n). By this we mean the following.
(1) The closed sets x(n) and ϕ(x)(n) have the same number of intervals andthe families of intervals are similar in that if the ith interval of x(n)consists of one point then so does the ith interval of ϕ(x)(n) and viceversa. The same is demanded of y(n) and ϕ(y)(n).
(2) If {ai : i < k}, {bj : j < l}, {ci : i < k} and {dj : j < l} are the setsof endpoints of x(n), y(n), ϕ(x)(n) and ϕ(y)(n) respectively (all sets inincreasing order) then for all i < k and j < l we have ai <, =, > bj iffci <, =, > dj.Condition (α) is one of the demands made at the outset; in combination with (β)it ensures that for example the sets {n : x(n) = In} and {n : ϕ(x)(n) = Xn} differ
ˇCECH-STONE REMAINDERS OF SPACES THAT LOOK LIKE [0, ∞)5by a finite set.Condition (β) also readily implies that x ≤∗y iff ϕ(x) ≤∗ϕ(y) for all x, y ∈K.By CH we can construct ϕ in an induction of length ω1; but rather than settingup the whole bookkeeping apparatus we show how to perform a typical inductivestep. So assume we have a bijection ϕ : A →B that satisfies (α) and (β), whereA and B are countable subsets of K and L respectively with tM, bM ∈A andtX, bX ∈B.Let ⟨xi⟩i be an enumeration of A and let yi = ϕ(xi) for all i.
We show how tofind ϕ(x) for an arbitrary x ∈K \ A (the task of finding ϕ−1(y) for y ∈L \ B isessentially the same).First find an increasing sequence ⟨nk⟩k of natural numbers such that wheneveri, j < k and n ≥nk the endpoints of xi(n) and xj(n) and those of yi(n) and yj(n)are in the same configuration. Using the fact that the intervals In and Xn aredensely ordered it is now an easy matter to find y ∈L such that the endpoints ofx(n) and xi(n) and those of y(n) and yi(n) have the same configuration whenevernk ≤n < nk + 1 and i < k. We put ϕ(x) = y of course.This completes the proof of Theorem 2.1.Remark 2.3.
Lemma 2.2 brings up an interesting question.If h : M∗→Y ∗were just any homeomorphism then it would have to map components of M∗tocomponents of Y ∗and thus would induce a map ϕ from ω∗to ω∗by h[Iu] = Jϕ(u).It is readily seen that ϕ is an autohomeomorphism of ω∗: Note that the set C = {u :h(0u) = tϕ(u)} is clopen so that we may change h by first turning the Iu with u ∈Cupside-down. But then ϕ merely mirrors the action of h on the set {0u : u ∈ω∗}and hence it is an autohomeomorphism.The problem is now to find an autohomeomorphism of M∗that permutes the Iuin the same way as ϕ−1 permutes the points of ω∗for then we could simply say: ifM∗and Y ∗are homeomorphic then H∗and X∗are homeomorphic.
We formulatethis as an explicit question.Question 2.4. Is there for every autohomeomorphism ϕ of ω∗an autohomeomor-phism h of M∗such that h[Iu] = Iϕ(u) for all u ∈ω∗?3.
More homeomorphic continuaThe argument given in Section 2 is actually a careful proof of a special case ofa general Model-Theoretic result. We shall give a brief sketch of this result andthen show how it may be used to show that a few more continua of interest arehomeomorphic.The result says “elementary equivalent and countably saturated models of size ω1are isomorphic”.Two models for a theory are said to be elementary equivalent if they satisfythe same sentences, where a sentence is a formula without free variables.Thismay be rephrased in a more algebraic way; two models A and B are elementaryequivalent iff the following holds: if {x1, .
. .
, xn} ⊆A and {y1, . .
. , yn} ⊆B aresuch that for every formula ϕ with n free variables ϕ(x1, .
. .
, xn) holds in A iffϕ(y1, . .
. , yn) holds in B then for every formula ψ for which there is an x ∈A such
6ALAN DOW AND KLAAS PIETER HARTthat ψ(x1, . .
. , xn, x) holds there is also a y ∈B such that ψ(y1, .
. .
, yn, y) holds(and vice versa of course).By way of example consider dense linear orders with first and last points. Anytwo such sets are elementary equivalent: if F = {x1, .
. .
, xn} and G = {y1, . .
. , yn}are as in the previous paragraph then we simply know that xi ≤xj iff yi ≤yj andxi is the first (last) element iff yi is.
The conclusion will then be: for every x thatis in a certain position with respect to F then there is a y in the same position withrespect to G.A countably saturated model is one in which, loosely speaking, every countablesystem of equations has a solution iff every finite subsystem of it has a solution.A countably saturated dense linear order is generally known as an η1-set: if Aand B are countable and a < b for every a ∈A and b ∈B then there is an x suchthat a < x < b for all a and b.The well-known theorem of Hausdorff from [1914] that under CH any two η1-sets of cardinality c are isomorphic can now be seen as a special case of the generalisomorphism theorem.The ‘typical inductive step’ from Section 2 may be modified to show that thatthe reduced product modulo the finite sets is countably saturated: we were lookingfor an element of L∗that satisfied the same equations as x and we used the factthat we could always satisfy any finite number of these equations.We refer to the book Chang and Keisler [1977] for the necessary backgroundon Model Theory.We shall now use this Model-Theoretic approach to show that many more con-tinua are homeomorphic, under CH. As noted in the introduction, the first resultsolves Problem 264 from Hart and van Mill [1990].Theorem 3.1.
The Continuum Hypothesis is equivalent to the statement that allstandard subcontinua of H∗are homeomorphic.□One direction was done by Dow in [1984]: under ¬CH there are u and v forwhich Iu and Iv are not homeomorphic.For the other direction we note that we can obtain a base for the closed sets of Iusimply by taking the ultraproduct L/u. This product is actually an ultrapowerbecause the Ln are all the same.The proof is finished by noting that L/u and L/v are elementary equivalent (bothare elementary equivalent to L0) and countably saturated (Chang and Keisler[1977, Theorem 6.1.1.
]); by the general isomorphism theorem the ultrapowers areisomorphic.□The next result shows that, again under CH, all layers of countable cofinalityare homeomorphic. Indeed the following, stronger, theorem is true.Theorem 3.2 (CH).
Let ⟨an⟩n be an increasing sequence of cut points in some Iuand let L be the ‘supremum’ layer for this sequence. Then L is homeomorphicto H∗.□To begin we note that, because Iu is an F-space, the closed interval [a0, L] is theˇCech-Stone compactification of the interval [a0, L).We now follow the proof of Theorem 2.1.
ˇCECH-STONE REMAINDERS OF SPACES THAT LOOK LIKE [0, ∞)7Form the intervals Jn = [an, an+1] and the topological sum Y = Ln Jn. Themap π : Y →[a0, L) that identifies ⟨n, an+1⟩and ⟨n+1, an+1⟩for every n induces amap from Y ∗onto L (the restriction of βπ).
This map is of the same nature as π∗H:it identifies the top point of Ju and the bottom point of Ju+1 for every u ∈ω∗.Our aim is to find a homeomorphism h : M∗→Y ∗that satisfies the assumptionsof Lemma 2.2. We shall do this, again, via an isomorphism between bases for theclosed sets of M∗and Y ∗respectively.We shall use the lattice L∗as a base for M∗and we make a base for Y ∗asfollows: For each n the interval Jn is homeomorphic with Iu and hence it hasa base Kn for the closed sets that is elementary equivalent to Ln.
The reducedproduct K∗= Qn Kn is then a base for the closed sets of Y ∗.We may now copy the inductive construction of a bijection from L to K fromthe proof of Theorem 2.1. The only difference is that we can no longer rely on thelinear order of Jn when we are constructing the images coordinatewise.
Insteadwe enumerate the countably many formulas from lattice theory with parametersfrom A and use elementary equivalence to produce for every x a y such that, asn gets bigger, there are more and more formulas that x(n) and y(n) both satisfy orboth do not satisfy (for the y’s we replace the parameters from A with their imagesunder ϕ of course).□Remark 3.3. The lattice Kn is not the lattice generated by the intervals of Jn;indeed, the Wallman space of the latter lattice is a linearly ordered continuum andin fact the continuum that one gets by collapsing the layers of Jn to points.4.
A special caseConsider the long line L of length ω1 × ω; that is, we take the ordinal ω1 × ωand stick an open unit interval between α and α + 1 for every α < ω1 × ω.Apparently Eric van Douwen raised the question whether L∗and H∗could behomeomorphic. Theorem 2.1 implies that the answer is yes, under CH.A slight modification of the methods in Section 2 will show that the answer iseven yes if d = ω1.Theorem 4.1.
If d = ω1 then L∗and H∗are homeomorphic.□We shall find, of course, a homeomorphism h : M∗→Y ∗of the familiar kind,where Y = ω × [0, ω1] and [0, ω1] denotes the long segment of length ω1.Now, because d = ω1, we may take a sequence ⟨xα : α < ω1⟩of points in Iω withthe following properties:(1) For all α and all n we have 0 < xα(n) < 1. (2) If β < α then xβ <∗xα.
(3) If x is such that x(n) < 1 for all n then there is an α such that x <∗xα.It is then an easy matter to define a sequence ⟨hα : α < ω1⟩of homeomorphisms,wherehα :[n{n} × [0, xα(n)] →ω × [0, α],such that• hα{n} × [0, xα(n)]= {n} × [0, α] for all α and n, and• if β < α then hα extends hβ except on a finite number of vertical lines.
8ALAN DOW AND KLAAS PIETER HARTIt is then straightforward to check that this sequence induces the desired homeo-morphism from M∗onto Y ∗.□Question 4.2. Is d = ω1 equivalent to the statement that L∗and H∗are homeo-morphic?We note that d = ω1 iff M∗and Y ∗are homeomorphic; this is so because d = ω1iff the character of the set of top points of M∗is ω1.
We have just seen that thisimplies that M∗and Y ∗are homeomorphic; on the other hand if M∗and Y ∗arehomeomorphic then clearly the set of top points of M∗has character ω1.References1938Chang, C. C. and H. J. Keisler.
[1977]Model Theory. Studies in Logic and the foundations of mathematics 73.North-Holland, Amsterdam.Dow, A.
[1984]On ultra powers of Boolean algebras. Topology Proceedings, 9, 269–291.Hart, K. P.[1992]The ˇCech-Stone compactification of the Real Line.
In Recent Progress in GeneralTopology, M. Huˇsek and J. van Mill, editors, chapter 9, pages 317–352.North-Holland, Amsterdam.Hart, K. P. and J. van Mill. [1990]Open problems on βω.
In Open Problems in Topology, J. van Mill and G. M. Reed,editors, chapter 7, pages 97–125. North-Holland, Amsterdam.Hausdorff, F.[1914]Grundz¨uge der Mengenlehre.
Chelsea Publishing Company, New York. Reprintfrom 1978 of original edition published in Leipzig.Wallman, H.[1938]Lattices and topological spaces.
Annals of Mathematics, 39, 112–126.Department of Mathematics, York University, 4700 Keele Street, North York, On-tario, Canada M3J 1P3E-mail address: dowa@nexus.yorku.caFaculty of Technical Mathematics and Informatics, TU Delft, Postbus 5031, 2600 GA Delft,the NetherlandsE-mail address: wiawkph@dutrun2.tudelft.nl
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