Several authors have recently attempted to show that the intersection of three simply connected subcontinua of the plane is simply connected provided it is non-empty and the intersection of each two of the continua is path connected. In this note we give a very short complete proof of this fact. We also confirm a related conjecture of Karimov and Repovš

E. D. Tymchatyn

Vesko Valov

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GLASNIK MATEMATIČKIVol. 41(61)(2006), 159 – 163ON INTERSECTION OF SIMPLY CONNECTED SETS INTHE PLANEE. D. Tymchatyn and Vesko ValovUniversity of Saskatchewan, Canada and Nipissing University, CanadaAbstract. Several authors have recently attempted to show that theintersection of three simply connected subcontinua of the plane is simplyconnected provided it is non-empty and the intersection of each two of thecontinua is path connected. In this note we give a very short completeproof of this fact. We also confirm a related conjecture of Karimov andRepovš.1. IntroductionA homology (resp., singular) cell is a compact metric space whose Vietoris(resp., singular) homology groups are trivial. Helly [6] proved the followingresult which is now known as the Topological Helly Theorem:Theorem 1.1. Let S = {S0, ..., Sm}, m ≥ n, be a finite family of ho-mology cells in Rn such that the intersection of every subfamily H of S isnonempty if the cardinality |H| ≤ n+ 1 and it is a homology cell if |H| ≤ n.Then ∩i=mi=0 Si is a homology cell.Versions of Theorem 1.1 for singular homology have been proved by De-brunner [5]. Alexandroff and Hopf [1, p. 295] also established a simple proofof a combinatorial version of the Helly theorem.A topological space is said to be simply connected if it is path connectedand has trivial fundamental group. It is known [4] that a compact subspaceof the plane is a singular cell if and only if it is simply connected.2000 Mathematics Subject Classification. 54C55, 55M15, 54F15.Key words and phrases. Helly theorem, plane continua, absolute retracts.The first author was partially supported by NSERC grant NPG 0005616.The second author was partially supported by NSERC grant RGPIN 261914.159160 E. D. TYMCHATYN AND V. VALOVIn Section 2 of the paper [6] Helly proved that if Si, i = 1, . . . , 4, aresingular cells in R2 such that all intersections Si1 ∩Si2 ∩Si3 are singular cells,then ∩i=4i=1Si is not empty. Hence to prove the Topological Helly Theorem forsingular cells in R2, it suffices to prove the following:Proposition 1.2. Let S0, S1 and S2 be three simply connected compactain the plane such that the intersection of any two of them is path connectedand ∩i=2i=0Si 6= ∅. Then ∩i=2i=0Si is simply connected.Bogatyi [2] has pointed out that no complete proof of this propositioncan be found in the literature. He proved the proposition in the special casethat Si are Peano continua. Karimov and Repovš [7], established that, withthe hypotheses of Proposition 1.2, ∩i=2i=0Si is cell-like connected (i.e., everytwo points can be connected by a cell-like continuum). We prove Proposition1.2 by showing that ∩i=2i=0Si is path connected. We also give an affirmativeanswer to a conjecture of Karimov and Repovš [7] by proving the followingproposition:Proposition 1.3. If X and Y are compact AR’s in the plane, then so iseach component of X ∩ Y .2. Proof of Proposition 1.2Since the intersection of any family of simply connected sets in the planehas a trivial fundamental group with respect to each of its points, it sufficesto show that ∩i=2i=0Si is path connected. Let 0, 1 ∈ ∩i=2i=0Si and I ⊂ S0 ∩ S1,J ⊂ S0 ∩ S2 and K ⊂ S1 ∩ S2 be arcs from 0 to 1. Consider the componentsJn, n = 1, 2, .., of J\(I ∪K)which are not in S1. If the family {J1, J2, ..} isinfinite, then limi→∞diamJi = 0 because it is the family of components of anopen set in the arc J . Since 0 and 1 are end-points of J , it follows that no Jiseparates I ∪ J ∪K. Suppose Ji lies in a bounded component of R2\(I ∪K).Since the locally connected continuum I ∪K separates Ji from ∞ in R2, somesimple closed curve in I∪K ⊂ S1 does so as well [8, Chapter X, Section 61, II,Theorem 5, p. 513]. Since S1 is simply connected, this would imply Ji ⊂ S1,a contradiction. Thus, no Ji lies in a bounded component of R2\(I ∪K).We are going to construct for every n ≥ 1 an arc Jn ⊂ S0 ∩ S2 ∩(I ∪ J ∪K)\(J1 ∪ .. ∪ Jn)from 0 to 1. Let L be a half line irreducible from J1 to ∞in R2\(I ∪K) and ε = min{d(x, y) : x ∈ L, y ∈ I ∪K}. Then ε > 0 becauseI ∪ K is compact and disjoint from the closed set L. Hence, L meets onlyfinitely many, say {J1 = Ji1 , .., Jim}, of the sets {J1, J2, ..}. We may supposeL meets Jij in exactly one point for each j ∈ {1, ..,m}. We may also supposethat Jij ∪ I ∪K separates Jij−1 from ∞ in R2 for j = 2, ..,m if m > 1. GiveJ its natural linear order with initial point 0. If x, y ∈ J let xy denote thearc in J irreducible from x to y. Now, let x0 and x1 be the end-points of JimON INTERSECTION OF SIMPLY CONNECTED SETS IN THE PLANE 161with x0 < x1 and let y0 = max I ∩ 0x0, y1 = min I ∩ x11. Denote by M thearc in I irreducible from y0 to y1. Then y0y1 ∪M ⊂ I ∪ J is a simple closedcurve containing Jim . By the Jordan curve theorem, y0y1 ∪M bounds a diskD with boundary y0y1 ∪M . Let L∗ be the unbounded component of L\Jimand Dm ⊂ D be the bounded component of R2\(I ∪ J ∪K) whose boundarycontains Jim .Then Dm ⊂ DI ∩DK , where DI (resp., DK) is the component of R2\(I ∪J) (resp., R2\(J ∪K)) containing Dm. Note that, as in the first paragraphof the proof, DI ⊂ S0 because I ∪ J ⊂ S0. Similarly, DK ⊂ S2. Thus,Dm ⊂ S0 ∩ S2 and Dm ⊂ S0 ∩ S2 since S0 ∩ S2 is closed.Moreover, Fr(Dm) ⊂ I ∪ J ∪K. It is well known [9, Theorem 2, p. 39]that each continuum contained in the union of finitely many arcs is locallyconnected. So Fr(Dm) is locally connected. As above, let C ⊂ Fr(Dm) bethe simple closed curve that separates Dm from ∞ in R2. Note that Jim ⊂ Cbecause L∗ ⊂ R2\(I∪J∪K)joins Jim ⊂ Dm to ∞. Let J1,m ⊂(J∪C)\Jim ⊂S0 ∩ S2 be an arc from 0 to 1. If m = 1 let J1 = J1,m. If m > 1 repeat theabove arguments with J1,m in place of J and Jim−1 in place of Jim to obtainan arc J1,m−1 in S0∩S2∩((J1,m\Jim−1)∪ (I ∪K))from 0 to 1. After m suchsteps we obtain an arc J1 = J1,1 ⊂ (I ∪ J ∪K)∩S0 ∩S2\(Ji1 ∪ Ji2 ∪ ..∪ Jim)from 0 to 1.Suppose we have already constructed an arc Jn ⊂ S0 ∩ S2 in I ∪ J ∪Kfrom 0 to 1 such that Jn∩(J1 ∪ ..∪Jn)= ∅. If Jn ∩Jn+1 = ∅, let Jn+1 = Jn.If Jn ∩ Jn+1 6= ∅, we repeat the above arguments with Jn in place of J andJn+1 in place of J1 to obtain an arc Jn+1 ⊂ S0 ∩ S2 ∩((Jn ∪ I ∪K)\Jn+1)from 0 to 1. By induction, we construct a sequence of arcs {Jn}∞n=1 from 0to 1 withJn+1 ⊂ S0 ∩ S2 ∩((I ∪ J ∪K)\n+1⋃i=1Ji).Let J∗ = lim sup Jn. ThenJ∗ ⊂(S0 ∩ S2) ∩((I ∪ J ∪K)\∞⋃i=1Ji)⊂ S1is a continuum from 0 to 1. As above, J∗ is locally connected. So, there is anarc in J∗ ⊂ S0 ∩ S1 ∩ S2 from 0 to 1.3. Proof of Proposition 1.3Let C be a component of X ∩ Y . If K is the topological hull of C, thenK ⊂ X and K ⊂ Y since neither X nor Y separates R2. So, K = C.By unicoherence of R2 it follows that Fr(C), the boundary of C in R2, isconnected.162 E. D. TYMCHATYN AND V. VALOVBy the well-known result of Borsuk [3] (that every locally connected planecontinuum not separating the plane is an AR), it remains to prove that C islocally connected. Since C is a continuum in the plane, it suffices to provethat Fr(C) is locally connected. To prove this it suffices to show that everypair of points of Fr(C) is separated by a finite set (see [10, p. 99]).SinceX is simply connected, locally connected subcontinuum in the plane,by [10, Chapter IV], all true cyclic elements of X are topological disks Di suchthat the cardinality of Di ∩ Dj is at most 1 for i 6= j and, if the sequence{Di} is infinite, then lim diamDi = 0 . Hence, each Fr(Di) is a simple closedcurve and Fr(X) = X\⋃ int(Di) is a locally connected continuum with aparticularly simple structure. Let x and y be distinct points in Fr(C) ⊂Fr(X) ∪ Fr(Y ). If x and y do not both lie in any one cyclic element of X ,then an one point set separates x and y in X and, hence, in C. Thus, we maysuppose that there are cyclic elements D in X and E in Y with x, y ∈ D∩E.Now x in int(D) implies there is a neighborhood W of x in Fr(X) ∪ Fr(Y )with W ⊂ int(D). Then a finite set P separates Fr(Y )\W from x in Fr(Y )since Fr(Y ) is rim-finite. Hence, P separates x from Fr(X) ∪ Fr(Y )\W . Sowe may suppose x, y ∈ Fr(D) ∩ Fr(E) (see [8, 49.V, Theorem 3, p. 244]).Let F be a two-point set in Fr(E) which separates x and y in Fr(E).Then F separates x and y in Fr(Y ) [10, IV.3.1, p. 67]. So F ∪ (D\Fr(Y ))separates x and y in D. Since D is hereditarily normal, a closed subset A ofF ∪ (D\Fr(Y )) separates x and y in D. Since D is unicoherent, a componentA′ of A separates x and y in D. We can construct an arc A′′ ⊂ D suchthat A′′ separates x and y in D and A′′ ∩ Y ⊂ F . Indeed, let N be an openneighbourhood of A′\F . So N is the union of a null collection, locally finiteexcept at points of F , of open balls Bi each centered at a point of A′\Fand having closure in R2\Fr(Y ). Then the closure of N and, hence, Fr(N)separates x and y in D. Hence, a component G of Fr(N) separates x and y inD. As above, G ⊂ Fr(N) ⊂ ∪Fr(Bi) ∪ F is locally connected at each pointof G\F . Since F is finite and a continuum cannot fail to be locally connectedonly at points of a zero dimensional set, G is locally connected. Since x andy are on the boundary of D, there is an arc A′′ in G which separates x and yin D.If we also take A′′ to be irreducible with respect to separating x and yin D (see [8, V.49, Theorem 3, p. 244]), then A′′ ∩ Fr(D) will contain justtwo points c and d. As above, A′′ separates x and y in X because D is acyclic element of X . So A′′ ∩(Fr(X) ∪ Fr(Y ))⊂ F ∪ {c, d} separates x andy in Fr(C) ⊂(Fr(X) ∪ Fr(Y ))⊂ X . So, Fr(C) is rim-finite, hence, locallyconnected.Acknowledgements.We would like to thank the referee for his/her many valuable remarks.ON INTERSECTION OF SIMPLY CONNECTED SETS IN THE PLANE 163We also thank Dr. U. Karimov for pointing out an error in the first versionof this note and Dr. S. Bogatyi for several fruitful discussions and comments.References[1] P. Alexandroff and H. Hopf, Topologie, Chelsea, New York, 1972.[2] S. Bogatyi, The topological Helly theorem, Fundam. Prikl. Mat. 8 (2002), 365-405 (inRussian).[3] K. Borsuk, Sur les retracts, Fund. Math. 17 (1931), 152-170.[4] J. W. Cannon, G. R. Conner and A. Zastrow, One-dimensional sets and planar setsare aspherical, Topology Appl. 120 (2002), 23-45.[5] H. E. Debrunner, Helly type theorems derived from basic singular homology, Amer.Math. Monthly 77 (1970), 375-380.[6] E. Helly, Über Systeme von abgeschlossenen Mengen mit gemeinschaftlichen Punkten,Monatsh. Math. Phys. 37 (1930), 281-302.[7] U. Karimov and D. Repovš, On the topological Helly theorem, Topology Appl. 153(2006), 1614-1621.[8] K. Kuratowski, Topology, II, Academic Press, New York, 1968.[9] N. Steenrod, Finite arc-sums, Fund. Math. 23 (1934), 38-53.[10] G. Whyburn, Analytic Topology, Amer. Math. Soc. Colloq. Publ. 28, Providence,1942.[11] R. Wilder, Topology of Manifolds, Amer. Math. Soc. Colloq. Publ. 32, Providence,1949.E. D. TymchatynDepartment of Mathematics and StatisticsUniversity of SaskatchewanMcLean Hall, 106 Wiggins RoadSaskatoon, SK, S7N 5E6CanadaE-mail : tymchat@math.usask.caDepartment of Computer Science and MathematicsNipissing University100 College Drive, P.O. Box 5002North Bay, ON, P1B 8L7CanadaE-mail : veskov@nipissingu.caReceived : 3.6.2004.Revised : 20.12.2004. & 15.6.2005.

English

On intersections of simply connected sets in the plane

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GLASNIK MATEMATICKIVol. 41(61)(2006), 159 { 163ON INTERSECTION OF SIMPLY CONNECTED SETS INTHE PLANEE. D. Tymchatyn and Vesko ValovUniversity of Saskatchewan, Canada and Nipissing University, CanadaAbstract. Several authors have recently attempted to show that theintersection of three simply connected subcontinua of the plane is simplyconnected provided it is non-empty and the intersection of each two of thecontinua is path connected. In this note we give a very short completeproof of this fact. We also conrm a related conjecture of Karimov andRepovs.1. IntroductionA homology (resp., singular) cell is a compact metric space whose Vietoris(resp., singular) homology groups are trivial. Helly [6] proved the followingresult which is now known as the Topological Helly Theorem:Theorem 1.1. Let S = fS0; :::; Smg, m  n, be a nite family of ho-mology cells in Rn such that the intersection of every subfamily H of S isnonempty if the cardinality jHj  n+ 1 and it is a homology cell if jHj  n.Then \i=mi=0 Si is a homology cell.Versions of Theorem 1.1 for singular homology have been proved by De-brunner [5]. Alexandro and Hopf [1, p. 295] also established a simple proofof a combinatorial version of the Helly theorem.A topological space is said to be simply connected if it is path connectedand has trivial fundamental group. It is known [4] that a compact subspaceof the plane is a singular cell if and only if it is simply connected.2000 Mathematics Subject Classication. 54C55, 55M15, 54F15.Key words and phrases. Helly theorem, plane continua, absolute retracts.The rst author was partially supported by NSERC grant NPG 0005616.The second author was partially supported by NSERC grant RGPIN 261914.159160 E. D. TYMCHATYN AND V. VALOVIn Section 2 of the paper [6] Helly proved that if Si, i = 1; : : : ; 4, aresingular cells in R2 such that all intersections Si1 \Si2 \Si3 are singular cells,then \i=4i=1Si is not empty. Hence to prove the Topological Helly Theorem forsingular cells in R2, it suces to prove the following:Proposition 1.2. Let S0; S1 and S2 be three simply connected compactain the plane such that the intersection of any two of them is path connectedand \i=2i=0Si 6= ;. Then \i=2i=0Si is simply connected.Bogatyi [2] has pointed out that no complete proof of this propositioncan be found in the literature. He proved the proposition in the special casethat Si are Peano continua. Karimov and Repovs [7], established that, withthe hypotheses of Proposition 1.2, \i=2i=0Si is cell-like connected (i.e., everytwo points can be connected by a cell-like continuum). We prove Proposition1.2 by showing that \i=2i=0Si is path connected. We also give an armativeanswer to a conjecture of Karimov and Repovs [7] by proving the followingproposition:Proposition 1.3. If X and Y are compact AR's in the plane, then so iseach component of X \ Y .2. Proof of Proposition 1.2Since the intersection of any family of simply connected sets in the planehas a trivial fundamental group with respect to each of its points, it sucesto show that \i=2i=0Si is path connected. Let 0; 1 2 \i=2i=0Si and I  S0 \ S1,J  S0 \ S2 and K  S1 \ S2 be arcs from 0 to 1. Consider the componentsJn, n = 1; 2; ::, of Jn I [K which are not in S1. If the family fJ1; J2; ::g isinnite, then limi!1diamJi = 0 because it is the family of components of anopen set in the arc J . Since 0 and 1 are end-points of J , it follows that no Jiseparates I [ J [K. Suppose Ji lies in a bounded component of R2n I [K.Since the locally connected continuum I [K separates Ji from 1 in R2, somesimple closed curve in I[K  S1 does so as well [8, Chapter X, Section 61, II,Theorem 5, p. 513]. Since S1 is simply connected, this would imply Ji  S1,a contradiction. Thus, no Ji lies in a bounded component of R2n I [K.We are going to construct for every n  1 an arc Jn  S0 \ S2 \ I [ J [K)n J1 [ :: [ Jn from 0 to 1. Let L be a half line irreducible from J1 to 1in R2n(I [K) and " = minfd(x; y) : x 2 L; y 2 I [Kg. Then " > 0 becauseI [ K is compact and disjoint from the closed set L. Hence, L meets onlynitely many, say fJ1 = Ji1 ; ::; Jimg, of the sets fJ1; J2; ::g. We may supposeL meets Jij in exactly one point for each j 2 f1; ::;mg. We may also supposethat Jij [ I [K separates Jij 1 from 1 in R2 for j = 2; ::;m if m > 1. GiveJ its natural linear order with initial point 0. If x; y 2 J let xy denote thearc in J irreducible from x to y. Now, let x0 and x1 be the end-points of JimON INTERSECTION OF SIMPLY CONNECTED SETS IN THE PLANE 161with x0 < x1 and let y0 = max I \ 0x0, y1 = min I \ x11. Denote by M thearc in I irreducible from y0 to y1. Then y0y1 [M  I [ J is a simple closedcurve containing Jim . By the Jordan curve theorem, y0y1 [M bounds a diskD with boundary y0y1 [M . Let L be the unbounded component of LnJimand Dm  D be the bounded component of R2n(I [ J [K) whose boundarycontains Jim .Then Dm  DI \DK , where DI (resp., DK) is the component of R2n(I [J) (resp., R2n(J [K)) containing Dm. Note that, as in the rst paragraphof the proof, DI  S0 because I [ J  S0. Similarly, DK  S2. Thus,Dm  S0 \ S2 and Dm  S0 \ S2 since S0 \ S2 is closed.Moreover, Fr(Dm)  I [ J [K. It is well known [9, Theorem 2, p. 39]that each continuum contained in the union of nitely many arcs is locallyconnected. So Fr(Dm) is locally connected. As above, let C  Fr(Dm) bethe simple closed curve that separates Dm from 1 in R2. Note that Jim  Cbecause L  R2n I[J[K joins Jim  Dm to1. Let J1;m   J[CnJim S0 \ S2 be an arc from 0 to 1. If m = 1 let J1 = J1;m. If m > 1 repeat theabove arguments with J1;m in place of J and Jim 1 in place of Jim to obtainan arc J1;m 1 in S0\S2\ (J1;mnJim 1)[ (I [K)from 0 to 1. After m suchsteps we obtain an arc J1 = J1;1  (I [ J [K)\S0 \S2n Ji1 [ Ji2 [ ::[ Jimfrom 0 to 1.Suppose we have already constructed an arc Jn  S0 \ S2 in I [ J [Kfrom 0 to 1 such that Jn\ J1[ ::[Jn = ;. If Jn\Jn+1 = ;, let Jn+1 = Jn.If Jn \ Jn+1 6= ;, we repeat the above arguments with Jn in place of J andJn+1 in place of J1 to obtain an arc Jn+1  S0 \ S2 \ (Jn [ I [K)nJn+1from 0 to 1. By induction, we construct a sequence of arcs fJng1n=1 from 0to 1 withJn+1  S0 \ S2 \ (I [ J [K)nn+1[i=1Ji:Let J = lim sup Jn. ThenJ   S0 \ S2) \  (I [ J [K)n 1[i=1Ji  S1is a continuum from 0 to 1. As above, J is locally connected. So, there is anarc in J  S0 \ S1 \ S2 from 0 to 1.3. Proof of Proposition 1.3Let C be a component of X \ Y . If K is the topological hull of C, thenK  X and K  Y since neither X nor Y separates R2. So, K = C.By unicoherence of R2 it follows that Fr(C), the boundary of C in R2, isconnected.162 E. D. TYMCHATYN AND V. VALOVBy the well-known result of Borsuk [3] (that every locally connected planecontinuum not separating the plane is an AR), it remains to prove that C islocally connected. Since C is a continuum in the plane, it suces to provethat Fr(C) is locally connected. To prove this it suces to show that everypair of points of Fr(C) is separated by a nite set (see [10, p. 99]).SinceX is simply connected, locally connected subcontinuum in the plane,by [10, Chapter IV], all true cyclic elements of X are topological disks Di suchthat the cardinality of Di \ Dj is at most 1 for i 6= j and, if the sequencefDig is innite, then lim diamDi = 0 . Hence, each Fr(Di) is a simple closedcurve and Fr(X) = XnS int(Di) is a locally connected continuum with aparticularly simple structure. Let x and y be distinct points in Fr(C) Fr(X) [ Fr(Y ). If x and y do not both lie in any one cyclic element of X ,then an one point set separates x and y in X and, hence, in C. Thus, we maysuppose that there are cyclic elements D in X and E in Y with x; y 2 D\E.Now x in int(D) implies there is a neighborhood W of x in Fr(X) [ Fr(Y )with W  int(D). Then a nite set P separates Fr(Y )nW from x in Fr(Y )since Fr(Y ) is rim-nite. Hence, P separates x from Fr(X) [ Fr(Y )nW . Sowe may suppose x; y 2 Fr(D) \ Fr(E) (see [8, 49.V, Theorem 3, p. 244]).Let F be a two-point set in Fr(E) which separates x and y in Fr(E).Then F separates x and y in Fr(Y ) [10, IV.3.1, p. 67]. So F [ (DnFr(Y ))separates x and y in D. Since D is hereditarily normal, a closed subset A ofF [ (DnFr(Y )) separates x and y in D. Since D is unicoherent, a componentA0 of A separates x and y in D. We can construct an arc A00  D suchthat A00 separates x and y in D and A00 \ Y  F . Indeed, let N be an openneighbourhood of A0nF . So N is the union of a null collection, locally niteexcept at points of F , of open balls Bi each centered at a point of A0nFand having closure in R2nFr(Y ). Then the closure of N and, hence, Fr(N)separates x and y in D. Hence, a component G of Fr(N) separates x and y inD. As above, G  Fr(N)  [Fr(Bi) [ F is locally connected at each pointof GnF . Since F is nite and a continuum cannot fail to be locally connectedonly at points of a zero dimensional set, G is locally connected. Since x andy are on the boundary of D, there is an arc A00 in G which separates x and yin D.If we also take A00 to be irreducible with respect to separating x and yin D (see [8, V.49, Theorem 3, p. 244]), then A00 \ Fr(D) will contain justtwo points c and d. As above, A00 separates x and y in X because D is acyclic element of X . So A00 \  Fr(X) [ Fr(Y )  F [ fc; dg separates x andy in Fr(C)   Fr(X) [ Fr(Y )  X . So, Fr(C) is rim-nite, hence, locallyconnected.Acknowledgements.We would like to thank the referee for his/her many valuable remarks.ON INTERSECTION OF SIMPLY CONNECTED SETS IN THE PLANE 163We also thank Dr. U. Karimov for pointing out an error in the rst versionof this note and Dr. S. Bogatyi for several fruitful discussions and comments.References[1] P. Alexandro and H. Hopf, Topologie, Chelsea, New York, 1972.[2] S. Bogatyi, The topological Helly theorem, Fundam. Prikl. Mat. 8 (2002), 365-405 (inRussian).[3] K. Borsuk, Sur les retracts, Fund. Math. 17 (1931), 152-170.[4] J. W. Cannon, G. R. Conner and A. Zastrow, One-dimensional sets and planar setsare aspherical, Topology Appl. 120 (2002), 23-45.[5] H. E. Debrunner, Helly type theorems derived from basic singular homology, Amer.Math. Monthly 77 (1970), 375-380.[6] E. Helly, Uber Systeme von abgeschlossenen Mengen mit gemeinschaftlichen Punkten,Monatsh. Math. Phys. 37 (1930), 281-302.[7] U. Karimov and D. Repovs, On the topological Helly theorem, Topology Appl. 153(2006), 1614-1621.[8] K. Kuratowski, Topology, II, Academic Press, New York, 1968.[9] N. Steenrod, Finite arc-sums, Fund. Math. 23 (1934), 38-53.[10] G. Whyburn, Analytic Topology, Amer. Math. Soc. Colloq. Publ. 28, Providence,1942.[11] R. Wilder, Topology of Manifolds, Amer. Math. Soc. Colloq. Publ. 32, Providence,1949.E. D. TymchatynDepartment of Mathematics and StatisticsUniversity of SaskatchewanMcLean Hall, 106 Wiggins RoadSaskatoon, SK, S7N 5E6CanadaE-mail : tymchat@math.usask.caDepartment of Computer Science and MathematicsNipissing University100 College Drive, P.O. Box 5002North Bay, ON, P1B 8L7CanadaE-mail : veskov@nipissingu.caReceived : 3.6.2004.Revised : 20.12.2004. & 15.6.2005.

On intersections of simply connected sets in the plane

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