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Topology and its Applications 153 (2006) 1302–1308

www.elsevier.com/locate/topo

Dimensions of strongn-point sets

John Cobb

Mathematics Department, University of Idaho, Moscow, ID 83844-1103, USA

Received 18 June 2004; received in revised form 24 March 2005; accepted 29 March 2005

Abstract

n-point sets (plane sets which hit each line inn points) and strongn-point sets (in addition hit eaccircle in n-points) exist (forn � 2, n � 3 respectively) by transfinite induction, but their propertotherwise are difficult to establish. Recently forn-point sets the question of their possible dimensihas been settled: 2- and 3-point sets are always zero-dimensional, while forn � 4, one-dimensionan-point sets exist. We settle the same question for strongn-point sets: strong 4- and 5-point seare always zero-dimensional, while forn � 6, both zero-dimensional and one-dimensional strn-point sets exist. 2005 Elsevier B.V. All rights reserved.

MSC:54G99; 54F45

Keywords:n-point set; Strongn-point set; Planar set

1. Introduction

An n-point [8] set is a set in the plane which hits each line in exactlyn points, anda strongn-point set [1] hits each line and each circle in exactlyn points; it is ann-pointset that has additional properties. No explicit examples are known: constructionstransfinite induction, hence properties other than existence are not immediately apRecently progress on their possible dimensions (zero or one) has been made. Theresults may be summarized as:Every2-point set[5], 3-point set[3], and strong3-point set[1] is zero-dimensional, while for alln � 4 there exist[1] one-dimensionaln-point sets.

E-mail address:johncobb@uidaho.edu.

0166-8641/$ – see front matter 2005 Elsevier B.V. All rights reserved.doi:10.1016/j.topol.2005.03.013

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We will settle all the remaining cases:There exist zero-dimensional sets of both tyfor all n; strongn-point sets forn = 4,5 must be zero-dimensional, while for eachn � 6,one-dimensional strongn-point sets exist.

2. Dimension zero

One reason for introducing strongn-point sets in [1] was to be able to prove somzero-dimensional results [1, Theorem 6.4]; in this spirit we generalize a little more.

Definition. For positive integersk � 2, � � 3, andm � 5, a (k, �,m)-point setis a setwhich hits each line in exactlyk points, each circle in exactly� points, and each noncirculaellipse in exactlym points; apartial (k, �,m)-point set has each ‘exactly’ replaced bymost’.

Note that strongn-point sets are dense in the plane [1, 6.2], whilen-point sets can benowhere dense (examples later).

Theorem 1. For k � 2, � � 3, andm � 5, there exist(k, �,m)-point sets.

The proof of Theorem 1 is by transfinite induction (e.g., [1, 6.1]), iterating the followlemma as often as needed at each step.

Lemma 1. If E is a partial (k, �,m)-point set with|E| < c, and G is a line/circle/noncircular ellipse having|E ∩ G| less than its target value, then there is a pointp ∈ G \ E suchthatE ∪ {p} is a partial (k, �,m)-point set.

Proof. LetH be the union of all lines, circles, and noncircular ellipses (excludingG itself)uniquely determined by subsets ofE (two points for lines, three non-colinear points fcircles, and five points no three colinear for noncircular ellipses);H ∩ G is the union offewer thanc sets each having at most four points, so pickp ∈ G \H. If E ∪ {p} were nota partial set, there would be aG′ with p ∈ G′ ∩ G and |E ∩ G′| already equal its targevalue; but thenG′ would be part ofH andp would not have been chosen inG′. �Theorem 2. There exist zero-dimensionaln-point and strongn-point sets for all possiblen.

This will follow from two lemmas.

Lemma 2. Noncircular ellipses depend continuously on sets of five points in the follosense: if G is a noncircular ellipse andP = {p1, . . . , p5} consists of five distinct pointof G, then ifP ′ = {p′

1, . . . , p′5} are points of the plane withdist(pi,p

′i ), 1 � i � 5, suf-

ficiently small, the setP ′ determines a(unique) noncircular ellipseG′ whose center andfour cardinal points(‘vertices’) are close to those ofG. (This assures thatG′ is close toGin the Hausdorff distance sense, in the focus-directrix sense, and in the foci-and-sdistances sense.)

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Proof. Let the plane have an arbitrary coordinate system (some restrictions will bposed later). For setsP ′ of five points, letQ(P ′) denote the linear system of fivequations obtained by plugging the coordinates of the points ofP ′ into the equationAx2 + Bxy + Cy2 + Dx + Ey + F = 0; the coefficientsA, . . . ,F are the ‘unknowns’By hypothesisQ(P ) has a unique solution, and at least one ofA, . . . ,F is nonzero; pickone of these and fix it, so thatQ(P ′) is a linear system of five equations in five unknowThe Cramer’s Rule determinants for solvingQ(P ′) are continuous inP ′; sinceQ(P ) hasa unique solution, its denominator determinant is nonzero, and forP ′ close toP this deter-minant will be bounded away from zero. HenceQ(P ′) has a unique solution and, since tnumerator determinants are continuous inP ′, the solution coefficientsA′, . . . ,F ′ are closeto those of the equation forG. SinceG is an ellipse, its discriminantB2 −4AC is negative,and forP ′ close toP the systemQ(P ′) determines an ellipseG′ with coefficients close tothose ofG.

The angleθ of rotation to eliminate thexy-term fromQ is given by cos2θ = (A −C)/B; sinceG is noncircular, imposing on the original coordinate system the condthat neither axis ofG should make an angle of 0 orπ/4 with either coordinate axis assurthat B �= 0 andA �= C for G, so we may assume that both are true forG′ also; henceG′ is noncircular. Finally, to find the centers and cardinal points ofG andG′, we wouldseparately rotate each and complete the squares; choosingP ′ close toP makes the resultfor G′ close to those forG. �Lemma 3. Each(k, �,5)-point set is zero-dimensional.

Proof. LetS be such a set andz ∈ S. Working inside a small circular neighborhood ofz, letI be a horizontal open interval with centerz andI ∩S = z. LetJ be a vertical open intervamissingS with center onI , and letG be a noncircular ellipse with centerz, major axisvertices inI , and hittingJ in two points. LetP ⊂ G consist of five points on the opposiside ofJ from z; sinceS is dense, by Lemma 2 there is a set of five pointsP ′ ⊂ S closeto P , on the opposite side ofJ from z, which determines a noncircular ellipse withz in itsinterior and hittingJ in two points. Then the component of (interiorP ′) \ J containingz

is a small open set whose boundary missesS. HenceS is zero-dimensional. �Proof of Theorem 2. By Lemma 3, an(n,n,5)-point set is a zero-dimensionaln-point setand strongn-point set. �

3. Strong 4- and 5-point sets

By [4,6,7], ann-point set is one-dimensional if and only if it contains an arc.

Theorem 3. Every strong4-point set and every strong5-point set is zero-dimensional.

The proof will follow immediately from Lemmas 4 and 6.

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Definition. An arcA pierces a circleC at p ∈ A ∩ C if there is an arcA′ ⊂ A with p

a nonendpoint ofA′ such thatp = A′ ∩ C and the two components ofA′ \ {p} lie onopposite sides ofC.

Lemma 4. No strong4-point set can contain an arc.

Proof. Suppose to the contrary that some strong 4-point setS contains an arcA. Let C bea circle whose center is a point ofA and having the endpoints ofA in its exterior. ThenA ∩ C consists of 2, 3, or 4 points, andA piercesC at two (or more) of them; letx andy

two of these piercing points, and letp andq be the two other points ofC ∩ S. The centerM of C lies on the perpendicular bisectorL of the segment connectingp andq. For somesmall open subintervalJ of L aboutM , if M ′ ∈ J andC′ is the circle with centerM ′passing throughp andq, we haveC′ ∩ S = {p,q, x′, y′}, wherex′ andy′ are containedin small disjoint compact sets aboutx andy. Since

⋃{C′: m′ ∈ J } consists ofp, q, andtwo thin open lines, it contains a small open set which missesS. But S is dense, and thicontradiction completes the proof.�Lemma 5. If A is an arc that is not a line segment, then some circleC is pierced byA inat least three points.

Proof. Let x, y, z ∈ A be three non-collinear, non-endpoints, and letC be the circle passing through them. If all three of{x, y, z} are piercing points, we are done. Otherwise,say,x is a piercing point andy is a non-piercing point, then aty the arcA ‘reflects back’into one side, say the exterior, ofC; increasing the radius ofC slightly yields three piercingpoints. Lastly, if none of{x, y, z} is a piercing point, thenA ‘reflects back’ at two of theminto the same side ofC; again, increasing or decreasing the radius ofC produces three(actually four) piercing points. �Lemma 6. No strong5-point set can contain an arc.

Proof. Repeat the proof of Lemma 4, using Lemma 5’s circle and three piercing poiplace ofx andy. �

4. Strong n-point sets for n ��� 6

Theorem 4. For eachn � 6, there exist one-dimensional strongn-point sets.

Lemma 7. For n � 3 andk � 3 each partial strongn-point set can be extended to a stro(n + k)-point set.

The proof of Lemma 7 follows that of Theorem 5.2 of [1].

Lemma 8. The setV = {(x, x2), x > 0} is a partial strong3-point set.

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Proof. Let p1,p2,p3 be three distinct points ofV , and letC be the circle through themThe centerM of C is the intersection of the perpendicular bisectors of the segmentsp1p2andp2p3; routine analytic geometry shows that thex-coordinate ofM is negative, soMis on the opposite side of they-axis from V . If p1’s coordinates are(x1, y1), let q bethe point(−x1, y1); q is closer toM thanp1 is, soq is insideC. Then the half parabol(y = x2) \ V contains a pointp4 onC. Since a circle and a parabola can have at mostpoints in common,C cannot hitV in a fourth point. �

The proof of Theorem 4 follows immediately from Lemmas 7 and 8.

5. Further results

Example 1 (Nowhere dense, zero-dimensionaln-point sets). Let K be a Cantor set inR,and letX = (K × R) ∪ (R × K) in R

2; X is a nowhere dense set which hits each linecpoints. In the construction of ann-point setS, we may choose all of the points fromX. Theresulting set is zero-dimensional, since every arc inX contains a vertical or a horizontsegment, and hence cannot be contained inS.

Example 2 (Nowhere dense, one-dimensionaln-point sets, forn � 4). Let C be a circlemissingX; in the extension of the partial 2-point setC to ann-point set choose all the nepoints fromX.

The (small inductive) definition of one-dimensional is a point-wise definition, wour one-dimensional sets are so because they are constructed to contain a knownmight wonder if they could be one-dimensional at a ‘significantly larger’ subset, peat every point. The following theorem shows that the answer is:NO. (Note that ifS is one-dimensional atp ∈ S, then every relative neighborhood ofp is a one-dimensional partian-point set, and hence by [4,6,7] must contain an arc.)

Theorem 5. The union of all the arcs in ann-point set is nowhere dense in the plane; hencethe set of points at which a strongn-point set or a densen-point setS is one-dimensionais nowhere dense inS.

Proof. Suppose to the contrary thatS is ann-point set whose arcs are dense in the inteof a circleC0. Then there is a circleC1 insideC0 and an open horizontal bandH1 betweentwo horizontal lines such that each horizontal line inH1 hits S in at least one point insidC1. SinceH1 ∩ intC0 ∩ extC1 is a non-empty open set insideC0, it contains a circleC2andH1 contains a bandH2, with the property that each horizontal line insideH2 hits S atleast once inside each ofC1 andC2. Continuing, we get circlesC1,C2, . . . ,Cn+1 insideC0 with disjoint interiors, and bandsH1 ⊃ H2 ⊃ · · · ⊃ Hn+1 such that each horizontal linin Hn+1 hitsS in n + 1 points. �

Note that in Example 2, the circleC is a relatively open set, hence is not nowhere dein S, but is nowhere dense in the plane.

J. Cobb / Topology and its Applications 153 (2006) 1302–1308 1307

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This leads to the question: How many disjoint arcs can ann-point set contain?

Example 3 (For n � 4, there aren-point sets which contain a countably infinite collectiof pairwise disjoint arcs). Each circle contains such a collection of arcs, so letA be theunion of one such collection in the circleC of Example 2.A is a partial 2-point set, sby 5.2 of [1] may be extended to ann-point set by adding points which we may choofrom X; sinceC andX are disjoint closed sets, this will not create any new arcs.

Theorem 6. Non-point set can contain uncountably many pairwise disjoint arcs.

Proof. Suppose to the contrary that someS does contain such a collectionF . For rationalmultiplesθi andθj of 2π with θi < θj , letW(θi, θj ) be the open cone with vertex the origconsisting of all rays from the origin whose counter-clockwise angle from the pox-axis is betweenθi andθj . For eachF ∈ F , there existsW(θi, θj ) such that each ray iW(θi, θj ) intersectsF ; sinceF is uncountable and the collection{W(θi, θj )} is countable,some line through the origin hitsS in an uncountable set.�

The proofs of Theorems 5 and 6 do not use the full force ofn-point sets; they can bextended slightly:

Theorem 7. LetX be a plane set which intersects each line in a finite, nonempty set.the union of all the arcs contained inX is nowhere dense in the plane, andX cannotcontain uncountably many pairwise disjoint arcs.

(Continue theH ’s of Theorem 5 infinitely, and then use⋂

Hn.)While we have introduced(k, �,m)-point sets as tools in the construction ofn-point and

strongn-point sets, we can also say a little about their dimensions.

Theorem 8. There exist zero-dimensional(k, �,m)-point sets of all orders, and fork � 4,� � 6, andm � 9 there exist one-dimensional(k, �,m)-point sets.

Proof (Sketch). For one-dimensional, note that the half-parabola used earlier is a p(2,3,4)-point set; use a slight generalization of Lemma 7 to extend it as desired.

For zero-dimensional, let a(k, �,m,5)-setbe a(k, �,m)-set which in addition hits eachyperbola in exactly five points; existence proof as in Theorem 1. Since the discrimof a hyperbola is positive, the ‘five-point continuity’ of Lemma 2 holds for hyperbowith the modification ‘. . . Hausdorff distance sense on round disks centered at the oof sufficiently large radius’. About each point of a(k, �,m,5)-set we may construct sma‘concave rectangles’ with hyperbolic sides missing the set.�

Some other sorts of generalizations ofn-point sets are in [2].

References

[1] K. Bouhjar, J.J. Dijkstra, J. van Mill, Three point sets, Topology Appl. 112 (2) (2001) 215–227.

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[2] J. Cobb, Are there generalizations ofn-point sets? Questions Answers Gen. Topology 22 (2004) 81–90.[3] D.L. Fearnley, L. Fearnley, J.W. Lamoreaux, Every three-point set is zero-dimensional, Proc. Amer

Soc. 131 (2003) 2241–2245.[4] D.L. Fearnley, L. Fearnley, J.W. Lamoreaux, On the dimension ofn-point sets, Topology Appl. 129 (2003

15–18.[5] J. Kulesza, A two-point set must be zero-dimensional, Proc. Amer. Math. Soc. 116 (1992) 551–553.[6] J. Kulesza, J. Schweig, Rim-finite, arc-free subsets of the plane, Topology Appl. 124 (2002) 475–485[7] A. Le Donne, Partialn-point sets and zero-dimensionality, Topology Appl. 128 (2003) 169–172.[8] S. Mazurkiewicz, Sur un ensemble plan qui a avec chaque droite deux et seulement deux points c

(1914), in: K. Borsuk, et al. (Eds.), Traveaux de Topologie et Ses Applications, PWN, Warsaw, 1969, p47.