Geometriae Dedicata 62: 107-114, 1996. 107 © 1996 KluwerAcademic Publishers. Printed in the Netherlands.

Nice Decompositions of R n Entirely into Nice Sets are Mostly Impossible

JOHN COBB Department of Mathematics, Unfi~ersity of ldaho, Moscow, Idaho 83843 e-mail:johncobb @ uidaho, edu.

(Received: 31 January 1995)

Abstract. While there are several interesting examples of partitions of R 3 into elements which individually are geometrically 'nice' - circles or segments - the partitions themselves fail to be 'nice', in the sense of forming continuous or upper semicontinuous decompositions. We show that this is no accident: R 3 has no continuous decomposition into circles, and no open subset of R '~ has an upper semicontinuous decomposition into convex compact nonsingleton sets.

Mathematical Subject Classifications (1991): Primary: 54B 15; Secondary: 52C22.

Key words: upper semicontinuous, continuous decomposition, circles

1. Introduction

Examples of decompositions of R 3 into circles (subsets isometric in the Euclidean metric to X 2 + if2 = 1,2 ) 0) Can be found in [CC], [S], [Kh], and [BM]. While the individual elements are 'nice' and most of these decompositions (_ partitions into disjoint sets) display in pieces some sort of continuity (definitions below), they in fact all fail to fit together 'nicely' ; this is unavoidable:

THEOREM 1. R 3 cannot be continuously decomposed into circles.

A second type of 'nice' elements are convex, compact, nonsingleton sets - convex cells of dimension > 0; examples are in [CC], [W], and [CW]. Again, such elements cannot fit together 'nicely' ; we have a stronger result:

THEOREM 2. There is no upper semicontinuous decomposition of R ~, or of any open subset of R n, into convex, compact, nonsingleton sets.

If G is a decomposition of R 3 into circles, a simple definition of continuity suffices: for x C R 3 let 9x be the (unique) element of G such that x E gz; G is said to be continuous if, whenever {x~}--+x, then the three sequences {center of gx~ }, {radius ofgz~ }, and {plane containing gz~ } all converge to the corresponding entities for g~.

More generally, a decomposition G of a space X is said to be upper semi-continuous if, for each open U C X , the set U{9 E G [ g C U} is also

108 JOHNCOBB

open in X. This is equivalent to the condition that, for each closed A C X, U{g E G ] g N A ~ (~} is closed in X. (Basic material on upper semicontinuous decompositions can be found in [Ku], [A], [CV], and [Da].) The decomposition space X / G has the elements of G as its points, and its topology is the smallest for which the decomposition (or projection) map, p" X- -+X/G defined by x E p(x), is continuous. G is upper semicontinuous if and only if p is a closed mapping. An upper semicontinuous decomposition G is continuous if in addition for each open set U C X, the set U{g E G [ g N U ¢ 0} is open.

For A C X, we denote its interior and boundary by intA and Fr(A), and its closure by A or cl A. If A is a cell, OA denotes its boundary in the manifold sense, irrespective of where A might be embedded. A is saturated (with respect to G) if A = U{g C G l g n A ¢ 0}.

2. Circles

Proof of Theorem 1. Suppose G were such a decomposition; let Y be the decom- position space Ra/G, and let p: R a c y be the decomposition projection.

• Y is connected and metrizable. Y is the continuous image of R 3, which is connected. A metric on Y can be defined by taking the distance between two elements of G to be the sum of three numbers: the distance between their centers, the difference of their radii, and the angular difference between their containing planes.

• The projection p is locally a product projection: disk × circle -~ disk, and hence Y is a 2-manifold. Let C be an element of G, let L be the line in R 3 through the center of C perpendicular to the plane of C, let D1 be a disk with center on C and lying in an open half-plane emanating from L, and let T1 be the torus in R 3 obtained by spinning D1 about L. Let U be the union of all elements of G contained entirely in intT1 and sufficiently close to C that each links L; since G is continuous, U is an open subset of R 3. Let D be a subdisk of D1 with the same center and lying entirely in U, and let T C U be the union of all elements of G which intersect D; each will intersect D in only one point. Parametrize the half-planes emanating from L by S 1, and define a function ~: D × SI -~T taking (x, 0) to the point of intersection of the half-plane corresponding to 0 with the element of G containing x. ~ is continuous, since G is continuous; it is one-to-one and onto by construction; hence it is a homeomorphism and T is a toms on which the p is a projection, p I D : D - ~ Y is an embedding, since it is one-to-one and D is compact; hence p Iint D is also an embedding. But p(int D) = p(intT), and intT is saturated by construction and open by Invariance of Domain; hence p(intD) is open in Y.

• Y is simply connected; in addition, if ~ : I--+Y is a path and, for i = O, 1, i f Pi E ~-1( i ) is given, then a may be lifted to/3 : I ~ R 3 with ~ = p o fl and /3(i) = Pi. Let P0 E C E G and let D and T be a disk and its associated canonical product neighborhood as above (with Po E D\OD)./3 = (p I intD) -1 lifts part

NICE DECOMPOSITIONS OF R n ENTIRELY INTO NICE SETS ARE MOSTLY IMPOSSIBLE 109

of a and is as desired at 0. Ignoring for the moment the prescribed Pa, extend this lifting as far as possible to some subinterval [0, i) C I. Let C = p - I ( a ( i ) ) , and let

and L be a canonical toms neighborhood of C and the line perpendicular to through its cente~ Some^half-plane emanating from L intersects the image of/3 at a point fl(t l) E T. Let D be the intersection of this half-plane with T. Replace/3 on [tl, t) by (p ] /3)-1 o c~; this extends/3 beyond t - thus a may be lifted on all of [0, 1], with prescribed image at one end. In case/3(1) ¢ P1, let C be the element of G containing/3(1) and P1, and let T be a canonical product neighborhood of C. Pick t2 such that/3([t2, 1]) C T. Let D be the cross-section o f T through/3(t2), and define a Tietze function on D that is zero on its boundary and at D f) C equals the angular distance between/3 (t2) and P1 determined by intersection with a half-plane as before. Replace/3 on [t2, 1] by rotating through the angles given by this Tietze function. This gives the desired lifting of a.

Now if c~ is a closed curve, it may be lifted to a closed curve. This lifted closed curve in R 3 can be contracted to a point, and the projection of this contraction into Y contracts ~.

• Y is homeomorphic to R 2. Y is a connected, simply connected 2-manifold, so it is either R 2 or S 2. Suppose Y is compact; then it can be covered by a finite collection of canonical neighborhoods, each of which can be lifted to a disk in R 3 whose closure is compact; let A be the compact set in R 3 which is the union of these closures of disks. Since G is continuous, the function ~5" R3--+R defined by 6( z ) = the diameter o f the element o f G which contains z is continuous; therefore there is a bound on the diameters of the elements of G which intersect A. But then there is some ball B C R 3 containing all these circles, and p - a ( y ) C B. This is a contradiction, since p is defined on all of R 3. Hence Y cannot be compact, so it must be homeomorphic to R 2.

• I f h • B--+Y is an embedding of a disk B into Y, then h can be lifted to H : B--+R 3 with h = p o H; in addition, i f A is an arc in OB, then H ] A can be prescribed beforehand. For any lifting of an embedding, each element of G which intersects H(int B) pierces it. First, suppose that h(B) is small enough that it lies in the image of a canonical toms neighborhood T. Pick a cross-section D of T, and let H = (p ] D) -1 o h; this lifts h except for the condition that H should extend a specified lifting ~" A-+R 3. T is canonically homeomorphic to D x S 1 by a homeomorphism that takes each {z} x S 1 to an element of G. D x / ~ covers D x S 1 by the map ~/= (z, t)--+{z} x e 2~i~. Let q~: A--+D x / ~ cover ~, and use Tietze's theorem to extend the second coordinate function of qb to if: B-+/L Then H = ~/o (H z if) is the desired lifting of h.

Now, in the general case suppose that B = [0, 1] x [0, 1] and that A = [0, 1] x {0}. Since each point of h(B) lies in the image of a canonical toms neighborhood, there is an integer N so large that the image under h of each of the squares of the subdivision of B formed by the lines z = i /N , y = i /N , 0 <_ i <_ N, will also lie in the image of a canonical toms neighborhood. Apply the special case first with B = [0, l / N ] x [0, l / N ] and A = [0, l / N ] z {0}; next apply it with

110 JOHNCOBB

B : [1/N, 2 /N] x [0, l / N ] and A : { l / N } x [0, l / N ] U [1/N, 2/N] x {0}; then on the next small square [Z/N, 3/N] x . . . over to [(N - 1) /N, 1] x . . . . Then repeat on the next r o w . . , x [ l /N , 2/N]; then o n . . . x [2/N, 3 / N ] ; . . . ; and finally to [(N - 1) /N, 1] x [(N - 1) /N, 1]. The piercing property follows automatically from the local product structure.

• There is a lifting ~: Y--+R 3 wi thp o ~ = id; each element of G hits d~(y) in exactly one point and pierces it there. This is similar to the last part of the previous step. Regard Y as R 2 and consider the ordinary partition into unit squares {Q}with the integer lattice points as comers. Order these squares Q1, Q 2 , . . . , carefully so that for each n, Qn intersects u{Qi I i < n} in an arc. (For example, take [0, 1] x [0, 1] first; then the eight squares intersecting it in cyclic order; then the 16 intersecting these; etc.) Define ff on Q1 anyway; once ff has been defined on u{Qi I i < n}, extend it to Q~, agreeing on the arc of intersection with the previous Q's.

• Contradiction. Let Co denote the element of G which contains ~(0, 0). In R2(= Y), for r > 0, let S~ denote the circle with radius r and center (0, 0), and consider the simple closed curves {~(S~) [ r > 0} in R 3. Each two are homo- topic in the complement of Co by a homotopy of the form {if(St) I a < t < b}; hence they either all link Co, or else none links it. Since Co pierces ~(Y), for r sufficiently small ff(S~) links Co. But consider the disk Do in R 3 bounded by Co : p(Do) is compact in R 2, so for r large enough, S~ misses it; then ~(S~) misses Do and can be contracted in its complement, hence cannot link Co. []

3. Convex Sets

We first prove several lemmas which will be needed in the proof of Theorem 2.

LEMMA 1 (Nonseparation Lemma). In l:g ~ suppose Y is a compact set for which there exist a plane L with 1 < dim L < n - 1, and an upper semicontinuous decomposition G of Y into convex sets, with the property that g f) L 7 ~ O, for each g E G .

Then Y does not separate R ~. Proof. We rely on Borsuk's Separation Theorem (see, for example, [Du]): A

compact set separates R n i f and only i f each map of the compact set into S n-1 is nullhomotopic. Let f • Y--+S ~-1 be given; since S ~-1 is an ANR, there is an extension f of f to an open neighborhood U of Y.

Let X = U N L; let dim L = k; and let R n-k be an (n - k)-plane orthogonal to L, so R n = L x R ~-k. Consider open sets V, with U D V D V D X, of the form V = N x M, where N C L is a compact neighborhood of X, and M C R n-k is a (n - h)-ball centered at the origin. Since V does not separate R ~, f I V-: V ~ S n - l is nullhomotopic. To show that f is nullhomotopic, we will construct a homotopy / / : Y x [ -+ U such that (writing Ht(y) ~ H(y , t)) H0 = identity and H1 (Y) C V for suitable V. Then f o / / ~ provides a homotopy between f and f o / / 1 , which is nullhomotopic since f I V i s .

NICE DECOMPOSITIONS OF R n ENTIRELY INTO NICE SETS ARE MOSTLY IMPOSSIBLE 1 1 1

It remains to show the existence of H and V. For each 9 C G there exist sets A and B, where A is a convex n-cell in R n and B is a saturated closed set in Y, satisfying A D int A D B D satinty B D g, where satinty B = U {g' C G I g' C intg B}. Using the compactness of Y, let B be a finite collection of such B's which covers Y, and let .A be the corresponding A's.

C m Let { i}i=1 be the collection of all (nonempty) intersections of subcollections of /3 , ordered by decreasing number of intersectands; hence if Ci M Cj ¢ O,

• 9 n then Ci f) Cj precedes both Ci and Cj. Let {D~}i= 1 be the intersections of the corresponding subcollections of .A.

For 1 < i < m, let Ei ×.Mi be a closed neighborhood of L N Ci in int Di, with Ei a convex k-cell in L and Mi an (n - k)-ball in R n-k centered at the origin, of radius so small that Ci misses (OEi) x Mi. Let M = Oim=lMi, N = Uim=lEi, and V = int(N × M).

{h }i=l of Y into U satisfying We will inductively construct homotopies i ,~

(i) h~(Uj<iCj) C V, 1 < i < m, (ii) h i = h~ -1, 1 < i < m (set h ° = identity),

(iii) h~ =- h~ -1 on Uj<<_i-lCj, 0 < t < 1, 1 < i < m, (iv) h~(Cj) C Dj, j >_ i, 0 < t < 1.

(Once this is done H is defined by performing the h~'s consecutively: Ht = himt_i for i /m <_ t <_ (i + 1)/m. By (ii) H will be continuous, and by (i) H i (Y) = h~(Y) C V.)

Suppose h i-1 has already been constructed. For y E Y, let/3(y) be the point h i - l / of Ei closest to 1 (y); since Ei is convex and compact this is well-defined and

fl" Y-+E~ is continuous. Let P = (Ci \ Uj<iCj) and Q = (uj<iCj) u {y e Y ] h~-l(y) ~ int Di}. Define ~" Y × I--+U by ~(y,t) = (1 - t)h~-l(y) + t/3(y)and let W = ~ - l ( i n t (E i × M)) ; since ~ is continuous, W is an open subset o f Y × I. We have P × { 1 } C W, and since ( P M Q) × {0} c W and int(Ei × M) is convex, it follows that (P M Q) × I c W. Hence there is a continuous 0: Y---+[0, 1) such that O(y) = 0 for y e Q and (y,O(y)) e W for y e P.

Set h~(y) = (1 - tO(y))h~i -1 (y) + tO(y)Z(y); this satisfies the inductive condi- tions and completes the proof of the lemma. []

LEMMA 2 (Contradiction Lemma). Suppose G is an upper semicontinuous decomposition of an open set 0 C R ~. Then there cannot exist a triple ('~,~, U), consisting of ~ E G, ~ C "g \ 0~, U an open subset of 0 containing ~, with the property that g M L 7~ ~ for every 9 E G and 9 C U, where L is the (n - dim~)- plane in R n through ~orthogonal to "~.

Proof Suppose there were; we may assume that U is bounded. Let k = dim if; let V = U{9 E G [ g C U }; let F be a closed (n - k)-disk in L Cl U with center ~ ; l e t X = OF; l e t Y = U{9 E G I g M X ~ ( ~ } . L e t Z = U{g E G 19 C Vand # N (L \ intL F ) ¢ (~}, and let W = V \ Z. Since L \ intL F is closed, Z is

112 JOHNCOBB

closed relative to V, and W is open in V and in R n. Since the elements of G are convex, W U Y = {g E G I g Cl F ¢ 0}, which is closed by upper semicontinuity. Hence Fr (W) C Y and, since ~ C W with ~ N Y = 0, Y separates R ~. But this contradicts the Nonseparation Lemma. []

LEMMA 3 (The Saturated Baire Category Theorem). Let G be an upper semicon- tinuous decomposition of a locally compact separable metric space 0 into compact sets, and let { Ai } be a countable collection of saturated closed sets with 0 = uAi. Then at least one Ai must contain a saturated open set.

Proof. Let 7r : O--+O/G be the decomposition projection; O/G is a locally compact separable metric space (see, for example, [CV] or [A]). {Tr(Ai)} is a countable collection of closed sets filling up O/G; since O/G is topologically complete, by the Baire Category Theorem some 7r(Ai) contains an open set, whose inverse image under 7r is saturated. []

LEMMA 4. Suppose that D is a convex k-cell in R k, and p E int D. Then there is an e > 0 so small that if D1 is a convex k-cell with OD1 C N~(OD), and if D2 is an isometric copy of D1 with D2 C D U N~(OD), then p E int D2.

Proof For e >_ 0, let D(e) = D U N~(OD); let g(e) = vol(D \ N~(OD)); and let f ( e ) = max{vol(Q) [ Q c D(e),Qconvex, p f~ intQ}. Since g(0) > f(0) , the continuity of f and g give g(e) > f (e ) for sufficiently small e. For such e, vol(D2) = vol(D1) _> g(e) > f (e) ,sop/E intD2. []

LEMMA 5. Let C be a convex k-cell in R k and e > O. Then there is a finite set {ql, q2, . . . , q,~} of points in N,(C) \ C such that, if D is a convex k-cell in [gk satisfying

(i) C C D, and (ii) qi ~ D, 1 < i < m,

then D C N~(C). Proof Let S = ON~(C) and T = ON~/2(C ). Fix q C intC. There is a

6 < e /2 such that for each x E S, if Yx is the intersection of T with the seg- ment joining x and q, we have that N,(yx) C int(convexhullof{x} U C). Chose {q l , . . . , q,~} C N,(C) \ C such that for each y E T, N6(y) contains at least one of the qi 's. []

Proof of Theorem 2. Suppose on the contrary that G were such a decomposition of an open set O C R '~. We may assume that O is bounded; for otherwise let V C U be a bounded open set which contains at least one element of G, and replace G by {g E G I g C V} and O by the union of the elements of this new G.

We may assume that all the elements of G have dimension less than the max- imum possible, n. For otherwise in R n+l = R n × R, consider G = {g × {t} [

NICE DECOMPOSITIONS OF R n ENTIRELY INTO NICE SETS ARE MOSTLY IMPOSSIBLE 113

g E G, t E R}; G is an upper semicontinuous decomposition of an open set and each of its elements have dimension less than the maximum, n + 1.

To apply the Contradiction Lemma, we need to find an open set of elements of G which are all 'about the same size and shape'. To this end, let C be the collection of all convex, compact, non-singleton sets in R '~- 1 such that each C E C satisfies

• C is the convex hull of a finite set of points, all of whose coordinates are rational

• if dim C = k, then C lies in the k-plane determined by the first k coordinates of R ~- 1

• origin E C \ OC

For C E C let A(C) = {9 E G ] 9 contains an isometric copy o f t } . By upper semicontinuity, A(C) is closed relative to O, and O = U{A(C) ] C E C}. Since C is countable, by the Saturated Baire Category Theorem at least one A(C) contains a saturated open set O(C). Fix k = max{dimC I C E C, O(C) exists } and let

= {C E C [ d imC = k, O(C) exists }. The elements of Care subsets o f R k and is partially ordered by set inclusion; hence C contains a maximal linearly ordered

subset C. Let C = U{C E C}; since O is bounded, C is a convex k-cell. (We do

not know whether O(C) exists; in fact, it is conceivable that A(C) = 0.) Apply

Lemma 4 to Cus ing the orig~ as the point p; let e be the result. Fix C' E C such that C \ N~(OC) C intC' C C.

Let (ql~...~ q,~) be the points given by Lemma 5 for C and e, and let Ci = convexhull({qi) U C), 1 _< i _< ra. Each A(Ci) contains no open set, since

is maximal and C C C C Ci for every C E C. By the Baire Category Theorem UA~_ 1 A(Ci) i s a closed set which contains no open set. Let t9 = O(C') \ LJ~=IA(Ci); O is a (nonempty) saturated open set.

Fix go E G with g0 C O and dimg0 = k. Such a go exi ts ; for if not, it would be the case that dim~ > k for each g E G with g C O. For r > 0 let I f(r) = U{9 E G I g C O and g contains a (k + 1)-baU of radius r). Each I ( ( r ) is closed and (9 = U{ K (r) I r rational}, so by the Saturated Baire Category Theorem some I f(r) would contain an open set, contradicting the maximality ofk.

Let ~: Co--+9o be an isometry, where Co is a convex k-cell with C ~ C Co C C; let p0 = ~y(origin); and let L be the (n - k)-plane in R ~ through Po orthogonal

to g0. There is an open set U, 9o C U C (), with the property that, for every 9 E G with 9 C U, 9 fl L ¢ 0. For otherwise, there would exist an infinite sequence (9i} of elements of G with 9i N L = 0; and an infinite sequence (K{} of convex k-cells with Ifi C 9i with each IQ isometric to the same K satisfying C' C I f C N~(~) and with the property that lim Ifi = Q c 9o. Now applying Lemma 4 with C, K, qo-l(Q) as D, D1, D2, respectively, we have that the origin is in i n t ~ - l ( Q ) , so po E Q \ OQ. But l im0If / = OQ, and for sufficiently large i, OKi and OQ are homotopic in R n \ L; hence 0 = 9{ N L D II'{ fl L ¢ 0, a

114 JOHNCOBB

contradiction. Thus such a U exists, and Lemma 2 supplies the final contradiction.[]

4. Questions

Can Theorem 1 (circles) be extended to upper semicontinuous decompositions? (It cannot be extended to open sets, as is shown by the interior of a standard toms.)

Can Theorem 1 be generalized to higher dimensions? (See [BM] and [BF] for noncontinuous examples.)

Can any of our results be extended to 'nice' decompositions whose elements are only topologically 'nice' rather than isometrically 'nice' ? Examples of decom- positions with elements less-than-isometrically 'nice' are found in [BM], [W], and [CW]; all of these decompositions fail to be upper semicontinuous. Even the simplest case, of R 3 into arcs, is unknown ([A], p. 9). In the opposite sense, it is unknown ([W]) whether R 3 has any sort of decomposition into squares.

References

[A] Armentrout, S.: Monotone decompositions of E 3, Topology seminar, Wisconsin, 1965, Princeton Univ. Press, 1966, 1-25.

[BF] Bankston, P. and Fox, R.: Topological partitions of Euclidean space by spheres, Amer. Math. Monthly 92 (1985), 423-424.

[BM] Bankston, P. and McGoveru, R. J.: Topological partitions, General TopologyAppL 10 (1979), 215-229.

[CW] Chew, J. and Wilker, J. B.: A tiling of R 3 by nearly congruent rhombi, C. R. Math. Rep. Acad. Sci. Canada 12 (1990), 37-40.

[CV] Christenson, C. O. and Voxman, W. L.: Aspects of Topology, Marcel Dekker, New York, 1977, pp. 469-476.

[CC] Conway, J. H. and Croft, H. T.: Covering a sphere with congruent great-circle arcs, Proc. Camb. Phil Soc. 60 (1964), 787-800.

[Da] Daverman, R. J.: Decompositions of Manifolds, Academic Press, Orlando, 1986. [Du] Dugundji, J.: Topology, Allyn and Bacon, Boston, 1966. [Kh] Kharazishvili, A. B.: Partition of a three-dimensional space into congruent circles, Soobshch.

Akad. Nauk Gruzin SSR 119 (1985), 57-60. [Ku] Kuratowski, C.: Topologie, PWN, Warsaw, 1961. [S] Szulkin, A.: R 3 is the union of disjoint circles, Amer. Math. Monthly 90 (1983), 640-641. [W] Wilker, J. B.: Tiling R 3 with circles and disks, Geom. Dedicata 32 (1989), 203-209.