Geometriae Dedicata75: 177–186, 1999.© 1999Kluwer Academic Publishers. Printed in the Netherlands.

177

Unions of Sets Starshaped via Staircase Paths or viaPaths of Bounded Length

MARILYN BREEN?

Department of Mathematics, University of Oklahoma, 601 Elm Avenue, Norman, OK 73019-0315,U.S.A. e-mail: mbreen@ou.edu

(Received: 10 February 1998)

Abstract. Let S be an orthogonal polygon in the plane,S simply connected, and letk = 2, 3. SetSis a union ofk sets starshaped via staircase paths if and only if for everyF finite,F ⊆ bdryS, thereis a setG ⊆ bdryS arbitrarily close toF and pointssi ,1 6 i 6 k, (depending onG) such that eachpoint ofG is clearly visible from somesi . An analogous result holds for a union of 2 sets starshapedvia α-paths whenS is a closed simply connected set in the plane. Each result is best possible.

Mathematics Subject Classifications (1991):Primary: 52A30, 52A35.

Key words: unions of starshaped sets, staircase paths,α-paths.

1. Introduction

We begin with some definitions. LetS be a nonempty set in the plane. SetS iscalled anorthogonal polygon(rectilinear polygon) if and only if S is a connectedunion of finitely many convex polygons (possibly degenerate) whose edges areparallel to the coordinate axes. Letλ be a simple polygonal path inR2 whoseedges[vi−1, vi], 1 6 i 6 m, are parallel to the coordinate axes. Pathλ is calleda staircase path if and only if the associated vectors alternate in direction. That is,for i odd the vectorsvi−1vi have the same direction, and fori even the vectorsvi−1vi have the same direction, 16 i 6 m. We use the termsnorth, south, east,west, northeast, northwest, southeast, southwestto describe the relative positionof points. For pointsx andy in setS, we sayx seesy (x is visible from y) viastaircase pathsif and only if there is a staircase path inS which contains bothxandy. Similarly,x clearly seesy (y is clearly visiblefrom x) via staircase pathsifand only if for some neighborhoodN of y, x sees via staircase paths each point ofN ∩ S. Finally, setS is starshaped via staircase pathsif and only if for some pointp in S, p sees each point ofS via staircase paths. The set of all such pointsp is thestaircase kernelof S,KerS.

We may adapt the notions of visibility and starshapedness toα-paths as well.Let S be a set in the plane. A continuous functionλ with domain some compact

? Supported in part by NSF Grant DMS-9504249

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interval and with range inS will be a curve in S. For convenience, we identify thecurve with its range. For some fixedα > 0 and for pointsx, y in S, we sayx seesy (x is visible from y) via α-paths if and only if there is a curveλ in S whichcontains bothx andy and whose length is at mostα. Similarly, we may defineclear visibility viaα-paths, sets starshaped viaα-paths, and theα-kernel of a set.

A rich variety of theorems exist for sets which are starshaped via segments,ranging from Krasnosel’skii’s famous characterization theorem [7] to art galleryresults [10]. Furthermore, some of these results have been adapted and exten-ded to accomodate other notions of visibility and starshapedness, including star-shapedness via staircase paths (see [2–3]) and starshapedness viaα-paths (see [1]).Continuing such studies, this paper will use alternate concepts of starshaped setsto examine their unions, obtaining appropriate analogues of a characterization oftwo starshaped sets given in [4]. Interestingly, certain differences distinguish theanalogues which emerge, for the staircase path result will hold for unions of threeor fewer starshaped sets, while theα-path result will work for unions of at mosttwo such sets. In each case, the bound is best possible.

Throughout the paper, we use convS, cl S, int S, rel intS, and bdryS to denotethe convex hull, closure, interior, relative interior, and boundary, respectively, forset S. The distance between pointsx and y will be dist(x, y), and whenx 6=y,L(x, y) will represent their line. The reader may refer to Valentine [11], to Lay[9], to Danzer, Grünbaum, Klee [5], and to Eckhoff [6] for discussions concerningvisibility and starshapedness via segments.

2. Unions of Sets Starshaped via Staircase Paths

The following lemma will be useful.

LEMMA 1. LetS be an orthogonal polygon in the plane. If for every finite subsetF of bdryS there existk points ofS (depending onF) such that each point ofFsees one of thesek points via staircase paths inS, then there existk points ofSsuch that every point ofbdryS sees one of thesek points via staircase paths inS.

Proof. By a theorem of Lawrence, Hare, and Kenelly [8, Th. 1], there is ak-partition {B1, . . . , Bk} of bdryS such that forF finite, F ⊆ bdryS, all points ofF ∩ Bi see a common point, 16 i 6 k. For A ⊆ Bi,A finite, let VA be thesubset ofS seeing all points ofA via staircase paths inS. Using an argument in [3,Lem. 1], setVA is a finite union of boxes and hence is compact. Clearly for eachi,16 i 6 k, the family{VA: A finite,A ⊆ Bi} has the finite intersection property,so∩{VA: A finite,A ⊆ Bi} 6= ∅. Forpi in this intersection,pi sees all points ofBivia staircase paths inS, and pointsp1, . . . , pk satisfy the lemma. 2THEOREM 1. Let S be an orthogonal polygon in the plane. If every boundarypoint ofS sees pointp via staircase paths inS, thenp is in the staircase kernelof S.

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UNIONS OF SETS STARSHAPED VIA STAIRCASE PATHS 179

Figure 1.

Proof. Let x belong toS to show thatp seesx via staircase paths inS. Ifx ∈ bdryS, there is nothing to show, so assumex ∈ int S. Consider the horizontaland vertical lines atx. Each intersectsS in a segment of maximal length, and welet n, e, s, w denote the associated endpoints north, east, south, and west ofx,respectively. The horizontal and vertical lines atx determine four closed quadrantsatx, and without loss of generality we assume thatp is southeast ofx, in quadrantIV. By our hypothesis,p sees each ofs, e via staircase pathsλ(p, s), µ(p, e),respectively.

Consider the vertical and horizontal lines atp. If the vertical line atp meets[x, e], say atp′, then[p′, x] ⊆ S. (See Figure 1.) Also,[p, p′] cannot meet∼ Ssince this would produce a boundary point ofS not visible fromp, impossible.Hence[p, p′] ⊆ S, and[p, p′] ∪ [p′, x] is a staircase path inS from p to x, thedesired result.

If the horizontal line atp meets[x, s], a parallel proof shows thatp seesx viastaircase paths inS. In case neither situation occurs, thenp must lie outside therectangle with verticesx, s, e. If p is east of the rectangle, thenµ(p, e) ∪ [e, x]provides ap − x staircase path inS. If p is south of the rectangle, thenλ(p, s) ∪[s, x] provides such a path. We conclude thatp seesx via staircase paths, andp ∈ KerS. 2The result in Theorem 1 fails for two or more points. That is, every boundary pointof S may be visible (or even clearly visible) from one of two points withoutS beinga union of two starshaped sets. Consider the following example.

EXAMPLE 1. Let S be the set in Figure 2, with shaded regions in∼S. Observethat every boundary point ofS is clearly visible via staircase paths either fromp1

or from p2. However,S cannot be a union of two sets which are starshaped viastaircases: no point ofS sees via staircases bothb1 andb2. Parallel statements holdfor paira1 andb2 and for paira2 andb1. Hence forS to be a union of two starshapedsetsS1 and S2, we must havea1, b1 ∈ S1 and a2, b2 ∈ S2 (for an appropriatelabeling). But then KerS1 must lie on the vertical segment atp1,KerS2 on thehorizontal segment atp2. Unfortunately, no such points see pointx via staircasepaths inS.

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Figure 2.

Figure 3.

However, if we require setS to be simply connected, the result in Theorem 1may be extended tok points, 16 k 6 3.

THEOREM 2. Let S be a simply connected orthogonal polygon in the plane. Ifevery boundary point ofS sees one ofp1, p2, p3 via staircase paths inS, thenevery point ofS sees somepi via staircase paths,1 6 i 6 3, andS is a union ofthree sets which are starshaped via staircase paths.

Proof. Let x ∈ int S to show thatx sees somepi via staircase paths, 16 i 6 3.As in Theorem 1, consider the horizontal and vertical lines atx, with n, e, s, w theendpoints of the corresponding maximal length segments inS north, east, south,west ofx, respectively. Then some pair fromn, e, s, w will be visible via staircasepaths from the same pointpi , 16 i 6 3. Suppose thatn, e are visible via staircasepaths inS from p1. Sinceλ = [n, x] ∪ [x, e] is ann-e staircase inS, by [3, Lem.2], it follows thatp1 sees all points ofλ via staircase paths inS, sop1 seesx. Theargument may be applied to any two ofn, e, s, w visible from the same pointpi ,so this finishes the proof. 2The following example shows that the result in Theorem 2 fails fork points,k > 4.

EXAMPLE 2. LetS be the set in Figure 3. Every boundary point ofS is visible (infact, clearly visible) via staircase paths from one of the pointspi , 16 i 6 4. How-ever, no two of the pointsx1, x2, x3, x4, y can be assigned to the same starshapedsubset ofS, soS cannot be a union of four sets which are starshaped via staircasepaths.

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UNIONS OF SETS STARSHAPED VIA STAIRCASE PATHS 181

The next result provides a characterization for certain unions of orthogonallystarshaped sets.

THEOREM 3. Let S be an orthogonal polygon in the plane, and letk > 1. If Sis a union ofk orthogonally starshaped sets, then for everyF finite,F ⊆ bdryS,there exist a setG ⊆ bdryS arbitrarily close toF and pointss1, . . . , sk (dependingonG) such that each point ofG is clearly visible via staircase paths inS from oneof thesi points. The converse holds fork = 1 and, if S is simply connected, for26 k 6 3. Moreover, clearly visible may be replaced by visible in each statement.

Proof. To establish the first statement, letS = ∪{Si: 1 6 i 6 k}, where eachSi is starshaped via staircase paths andsi ∈ KerSi, 1 6 i 6 k. By results in [3,Lem. 1], the family of lines containing edges ofS gives rise to a finite collectionAof closed subsetsA of S, each of which is either a segment or a rectangular region.ForF a finite subset of bdryS andx in F , there correspond setAx in A and edgeex such thatx ∈ ex ⊆ Ax ∩bdryS. Forw ∈ rel intAx , some pointsi , says1, seeswvia staircase paths, and hence by [3, Ob. 1],s1 sees all points ofAx via such paths.We may choose a pointpx of rel intex arbitrarily close tox. Thenpx is clearlyvisible froms1, and the set of points{px: x ∈ F } satisfies the theorem.

We establish the converse using the weaker term ‘visible’ in place of ‘clearlyvisible.’ ForF any finite subset of bdryS and forn any natural number, there existsetGn within 1

nof F and associated pointssin,1 6 i 6 k, with each point of

Gn visible from somesin. Standard convergent subsequence arguments producecorresponding limitssi , 16 i 6 k. Moreover, for eachi, setVi = {y: si seesy viastaircase paths inS} is closed, and hence each point ofF is visible from somesi ,16 i 6 k. Using Lemma 1, there existk points ofS such that each point of bdrySis visible from one of thesek points. Ifk = 1, then by Theorem 1,S is orthogonallystarshaped. IfS is simply connected and 26 k 6 3, then by Theorem 2,S is aunion ofk orthogonally starshaped sets.

Previous examples show that the converse above fails forS not simply con-nected and fork ≥ 4, even using the notion of clear visibility. Moreover, it isinteresting to observe that subsetsF of bdryS may be visible from one ofk pointswithout the clear visibility condition holding for anyG sufficiently close toF .Consider the following examples, one forS not simply connected andk = 2, theother forS simply connected andk = 4.

EXAMPLE 3. LetS be the set in Figure 4, adapted from Example 1. Every bound-ary point ofS is visible fromp1 or p2. Yet for F = {a1, b1, a2, b2, c} and forGsufficiently close toF , no q1, q2 exist such that each point ofG is clearly visiblefrom q1 or q2.

EXAMPLE 4. LetS be the simply connected set in Figure 5, withTi = [bi, ci],16i 6 4. Each boundary point ofS is clearly visible from a point of{ti : 16 i 6 4} if

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Figure 4.

Figure 5.

and only if (for an appropriate labeling)ti ∈ Ti,16 i 6 4. However, no boundarypoint of the rectangular regionR is clearly visible from anyti in Ti .

3. Unions of Sets Starshaped viaα-Paths

In this section we develop analogues of the previous results for sets which arestarshaped viaα-paths. We begin with an adaptation of Lemma 1.

LEMMA 2. LetS be a nonempty closed set in the plane,S 6= R2, and letα > 0.Assume that for every finite subsetF of bdryS there existk points ofS (dependingonF) such that each point ofF sees one of thesek points viaα-paths inS. Thenthere existk points ofS such that each boundary point ofS sees one of thesekpoints viaα-paths inS. Moreover, ifS is simply connected, thenS is compact.

Proof. By [1, Cor.], forx in S, setVx = {y: x seesy viaα-paths inS} is closed,soVx is compact. Thus the argument in Lemma 1 may be adapted to prove the firststatement.

To establish the second statement, notice that sinceS 6= ∅ andS 6= R2,bdryS 6=∅. Also, it is easy to see that bdryS is bounded: otherwise, we could select pointsb1, . . . , bk+1 in bdryS with dist(bi , bj ) > 2α for 1 6 i < j 6 k + 1. But thensetF = {b1, . . . , bk+1} would violate our hypothesis. Hence bdryS is compact.To show that setS is compact as well, we prove thatS is a subset of the compactset conv(bdryS): for y in R2 ∼ conv(bdryS), select a simple closed curveλcontainingy and disjoint from conv(bdryS), with conv(bdryS) in the boundedregion determined byλ. Sinceλ is disjoint from bdryS, λ is a subset either of intSor of R2 ∼ S. If λ ⊆ int S, then sinceS is simply connected, conv(bdryS) ⊆ int S,

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UNIONS OF SETS STARSHAPED VIA STAIRCASE PATHS 183

Figure 6.

and bdryS ⊆ int S, clearly impossible. Therefore,λ ⊆ R2 ∼ S, soy ∈ R2 ∼ S.We conclude thatS ⊆ conv(bdryS). SinceS is closed, it follows thatS is compact.

THEOREM 4. Let S be a nonempty closed, simply connected set in the plane,S 6= R2, and letα > 0. If every boundary point ofS sees one ofp, q via α-pathsin S, then every point ofS sees one ofp, q via such paths, andS is a union of twosets which are starshaped viaα-paths.

Proof. Let x ∈ S to show thatx sees one ofp, q via α-paths. Ifx ∈ bdryS, theresult is immediate, so assumex ∈ int S, and letN be a spherical neighborhood ofx in S. By Lemma 2,S is compact. Hence for every lineL throughx, there existpointsyL, zL in bdryS such thatx ∈ [yL, zL] ⊆ L∩ S. Further,yL sees eitherp orq (or both) viaα-paths inS. A parallel statement holds forzL. LetAp denote the setof all yL andzL points which seep via α-paths inS, and letBq be the set of suchpoints which seeq via α-paths. Finally, defineA = {w: {w} = [x, y] ∩ bdryN fory in Ap}. Similarly, defineB for Bq . Using [1, Cor.], it is easy to see thatAp,Bqare compact and henceA,B are compact, too.

Certainly, bdryN = A ∪ B. Moreover, since bdryN is connected, setsA andB cannot be separated. Thus either (1) one ofA or B is empty or (2)A ∩ B 6= ∅.Either way, there exists some lineL throughx such that bothyL andzL see thesame pointp or q, sayp, via α-paths inS. Since[yL, zL] is ayL − zL geodesic inS, by [1, Th. 1] it follows thatp sees each point of[yL, zL] via α-paths, andp seesx via such paths. Thus each point ofS sees one ofp, q via α-paths, finishing theproof. 2It is interesting to observe that analogues of Theorem 4 fail for 3 or more points(see Example 5) and forS not simply connected, even whenS is compact (seeExample 6).

EXAMPLE 5. Let T be the planar region bounded by an equilateral triangle ofside length 2

√3. At each vertexvi of T , letDi be the closed disk of radius

√3, i =

1,2,3, and letS = T ∪D1∪D2∪D3. (See Figure 6.) Forα = √3, every boundarypoint of S sees somevi via anα-path, yetS is not a union of 3 sets which arestarshaped viaα-paths.

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Figure 7.

EXAMPLE 6. Let S be the nonsimply connected set in Figure 7, with shaded(open) regions inR2 ∼ S. Segments[p, p′], [q, q ′] have length3

2 while othersegments have length 1. Forα = 3

2, every boundary point ofS sees one ofp, q viaα-paths, yetS is not a union of 2 sets which are starshaped viaα-paths.

The final theorem is anα-path analogue of Theorem 3.

THEOREM 5. LetS be a nonempty closed simply connected set in the plane,S 6=R2, and letα > 0. SetS is a union of two sets which are starshaped viaα-pathsif and only if for every finite setF ⊆ bdryS, there exist a finite setG ⊆ bdrySarbitrarily close toF and pointspG, qG (depending onG) such that each point ofG is clearly visible viaα-paths from one ofpG, qG. An analogous result holds withclearly visible replaced by visible.

Proof. We begin with the sufficiency of the condition. Suppose the propertyabove holds for every finite setF ⊆ bdryS. Then for each natural numbern,there exist setGn within 1/n of F and pointspn, qn satisfying our hypothesis.Furthermore, by an argument like the one in Lemma 2, setS is compact. Hencestandard arguments produce subsequences of{pn},{qn} which converge, say topF , qF in S, respectively. Each point ofF will be visible via α-paths from oneof pF , qF . Hence by Lemma 2 there exist pointsp, q of S such that each boundarypoint of S is visible viaα-paths from one ofp, q, and by Theorem 4S is a unionof two sets which are starshaped viaα-paths. An identical argument holds withclearly visible replaced by visible.

To prove the necessity, assume thatS is a union of two sets which are starshapedvia α-paths, withp, q in the respective kernels. DefineVp, Vq to be the subsets ofS seen byp, q, respectively. LetF be a finite subset of bdryS. It suffices to showthat for x in F andN any open spherical neighborhood ofx, N contains somepoint of bdryS clearly visible from one ofp, q. In case eitherp or q fails to seesome point ofN ∩ bdryS, the result is immediate. Hence assume that bothp andq see viaα-paths inS all points of(clN) ∩ bdryS.

LetK be a component ofN∩ bdryS. We assert that fors in S∩N∩(conv clK),bothp andq sees via α-paths: sinceK is connected, certainlys lies on a segmentwhose endpoints are in clK. The intersection of this segment withS producesa segment inS which containss and whose endpoints are in(clN) ∩ bdryS ⊆

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UNIONS OF SETS STARSHAPED VIA STAIRCASE PATHS 185

Vp ∩ Vq . Thuss lies on a geodesic inS whose endpoints are inVp ∩ Vq, and by [1,Th. 1], s ∈ Vp ∩ Vq .

Choose any pointv in K. If v is clearly visible fromp or q, the argument isfinished. Otherwise,v is not clearly visible fromp. Hence there is some pointw inS ∩N such thatp cannot seew. By previous comments,w ∈ int S ∼ (conv clK).Select a spherical neighborhoodM of w with clM ⊆ (int S) ∩ N ∼ (cl convK)such thatp sees no point of clM. Define T = conv(M ∪ K). Clearly T =∪{[k,m]: k in convK,m in M} ⊆ N . We assert that fort in T ∩ S, q seest :point t lies on a segment of maximal length inS ∩ [kt ,mt ] for appropriatekt inconvK,mt in M. Any endpoint of this segment is either in(bdryS) ∩ N ⊆ Vq orin (clM) ∩ S ⊆ Vq or in cl convK ⊆ Vq . Hencet lies on a geodesic inS whoseendpoints are inVq , and again by [1, Theorem 1],t ∈ Vq. We conclude thatqsees viaα-paths inS all points ofT ∩ S. In case intT contains a point of bdryS,such a point will be clearly visible fromq, finishing the proof. Otherwise, sinceM ⊆ int T ∩ int S 6= ∅, all points of intT will belong to intS, andT ⊆ S. Thenqwill see all points ofT via S. Also,K ⊆ (bdryS) ∩ T ⊆ bdryT .

Observe that sinceT = conv(M ∪ K),bdryT contains distinct segments[m1, k1] and[m2, k2], each of maximal length, withmi ∈ clM andki ∈ clK, i =1,2. SinceM is fully two-dimensional,m1 6= m2. For the moment, assumek1 6=k2, and choose a pointk0 of K on neither lineL(mi, ki), i = 1,2. LetB be the(open) component ofR2 ∼ (L(m1, k1) ∪ L(m2, k2) ∪ clM) containingk0. If qclearly seesk0, the argument is finished. Otherwise, there must exist some pointw′ in B ∩ S ∩ N which q cannot see. As in our previous argument,w′ ∈ int S ∼(conv clK). Moreover, sinceq sees all points ofT ,w′ /∈ cl T . Select a sphericalneighborhoodM ′ of w′ with clM ′ ⊆ (int S) ∩ N ∩ B ∼ (cl T ) such thatq seesno point of clM ′. Hencep sees each point of clM ′. DefineT ′ = conv(K ∪M ′).By our earlier argument, if intT ′ contains a point of bdryS, thenp clearly seesthis point, finishing the proof. Otherwise, intT ′ ⊆ int S andp sees all ofT ′ via S.Also,K ⊆ bdryT ′. Form ∈ M andm′ ∈ M ′, [m,m′] meetsK at some pointk inK ∩ conv(M ∪M ′) ⊆ K ∩ (T ∪ T ′). However,T ∪ T ′ contains a neighborhood ofk in S, contradicting the fact thatk ∈ bdryS.

The only other possibility is thatk1 = k2. ThenK ⊆ [m1, k1] ∪ [m2, k2], andsinceK is not a singleton set, we may assume thatK meets(m1, k1). Selectk0

relatively interior toK ∩ (m1, k1). If q clearly seesk0, the argument is finished.Otherwise, selectw′ not visible fromq such thatw′ and intT are on oppositesides of lineL(m1, k1), w

′ ∈ (intS) ∩N . Label the open halfplanes determined byL(m1, k1) so thatw′ belongs to halfplaneL1. Choose a spherical neighborhoodM ′of w′ with clM ′ ⊆ (int S)∩N ∩L1, such thatp sees all points of clM ′. As in ourearlier argument, defineT ′ = conv(K∪M ′). If int T ′ meets bdryS, the argument isfinished. This must occur, for otherwiseT ∪T ′ ⊆ S would contain a neighborhoodof k0, impossible sincek0 ∈ bdryS. This completes the proof of the theorem.2

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Previous examples show that Theorem 5 fails for unions of 3 sets and forS notsimply connected.

In conclusion, it would be interesting to determine whether analogous resultshold in higher dimensions, either for these or for other types of starshapedness.

References

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2. Breen, M.: An improved Krasnosel’skii-type theorem for orthogonal polygons which arestarshaped via staircase paths,J. Geom.51 (1994), 31–35.

3. Breen, M.: Staircase kernels in orthogonal polygons,Arch. Math. 59 (1992),588–594.

4. Breen, M. and Zamfirescu, T.: A characterization theorem for certain unions of two starshapedsets inR2, Geom. Dedicata22 (1987), 95–103.

5. Danzer, L., Grünbaum, B. and Klee, V.: Helly’s theorem and its relatives, In:Convexity, Proc.Sympos. Pure Math. 7, Amer. Math. Soc., Providence, R.I., 1962, pp. 101–180.

6. Eckhoff, J.: Helly, Radon, and Carathéodory type theorems, In: P. M. Gruber and J. M. Wills(eds),Handbook of Convex Geometry, Vol. A, North Holland, New York, 1993, 389–448.

7. Krasnosel’skii, M. A.: Sur un critère pour qu’un domaine soit étoilé,Math. Sb.(61) 19 (1946),309–310.

8. Lawrence, J. F., Hare, W. R. Jr. and Kenelly, J. W.: Finite unions of convex sets,Proc. Amer.Math. Soc.34 (1972), 225–228.

9. Lay, S. R.:Convex Sets and Their Applications, Wiley, New York, 1982.10. O’Rourke, J.:Art Gallery Theorems and Algorithms, Oxford University Press, New York, 1987.11. Valentine, F. A.:Convex Sets, McGraw-Hill, New York, 1964.

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