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A simple closed curve which fails to pierce a disc at only one point Goff, William Sidney 1972

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A SIMPLE CLOSED CURVE WHICH FAILS TO PIERCE A,DISC AT ONLY ONE POINT by WILLIAM SIDNEY GOFF B.Sc . j U n i v e r s i t y of V i c t o r i a , 1969 A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF SCIENCE i n the Department of Mathematics We accept t h i s t h e s i s as conforming to the requ i r e d standard THE UNIVERSITY OF BRITISH COLUMBIA September, 1972 In p resen t ing t h i s t h e s i s in p a r t i a l f u l f i l m e n t o f the requirements f o r an advanced degree at the U n i v e r s i t y o f B r i t i s h Columbia, I agree tha t the L i b r a r y sha l l make i t f r e e l y a v a i l a b l e f o r reference and s tudy . I f u r t h e r agree t h a t permiss ion f o r ex tens ive copying o f t h i s t h e s i s f o r s c h o l a r l y purposes may be granted by the Head o f my Department or by h i s r e p r e s e n t a t i v e s . I t is understood tha t copying o r p u b l i c a t i o n o f t h i s t h e s i s f o r f i n a n c i a l ga in sha l l not be al lowed w i thou t my w r i t t e n pe rm iss ion . Department o f The U n i v e r s i t y o f B r i t i s h Columbia Vancouver 8, Canada i i ABSTRACT A simple closed curve J i s constructed i n R which f a i l s to p i e r c e a d i s c at only one p o i n t . The i n t e r s e c t i o n s of various surfaces i n R are i n v e s t i g a t e d to prove that J has the d e s i r e d property. Heavy use i s made of approximation of surfaces i n R with p o l y h e d r a l surfaces. J i s used to construct simple closed curves which f a i l to p i e r c e d i s c s at various f i n i t e and i n f i n i t e sets of p o i n t s . The most i n t e r e s t i n g of these i s a simple closed curve which p i e r c e s a d i s c at one dense set of p o i n t s while f a i l i n g to p i e r c e a d i s c at another dense set of points which has measure 0 . TABLE OF CONTENTS INTRODUCTION . . . . NOTATIONS AND DEFINITIONS CONSTRUCTION . . . . PRELIMINARY SET UP PROPERTIES OF S f l D SOME USEFUL LEMMAS OUTLINE . . . . . AN INVESTIGATION OF CERTAIN 2-SPHERES EXAMPLES RELATED TO J BIBLIOGRAPHY . . . . i v LIST OP FIGURES FIGURE 1 . . . . . . 2 FIGURE 2 . . . . . . . 2 FIGURE 3 . . . . . . 2 FIGURE 4 . . . . . . . 5 FIGURE 5 . . . . . . 5 FIGURE 6 . . . . . . . 8 FIGURE 7 . . . . . . l x FIGURE 8 . . . . . . 1 5 FIGURE 9 . . . . . . ]_6 FIGURE 10 . . . . . . 2.6 FIGURE 11 . . . . . . ]_8 FIGURE 12 . . . . . . . . 2 3 FIGURE 13 . . . . . . . . 2 5 FIGURE 14 . . . . . . . 2 8 FIGURE 15 . . . . . . < 3 1 FIGURE 16 . . . . . . . 3 2 FIGURE 17 ' . . . . . . . . 3 4 FIGURE 18 . . . . . . . 3 6 FIGURE 19 . . . . . . . 3 6 V ACKNOWLEDGMENT I wish to thank Dale Rolfsen f o r h i s many comments and suggestions which kept me on a s u c c e s s f u l course i n my i n v e s t i g a t i o n s . 1 INTRODUCTION. A simple closed curve J i n Euclidean 3-space i s s a i d to p i e r c e a d i s c at a point x i n J i f and only i f there e x i s t s a t o p o l o g i c a l 2 - d i s c D i n 3-space such that x i s the only point i n the i n t e r s e c t i o n of D and J and the boundary of D l i n k s J . An example of a simple closed curve that p i e r c e s a d i s c at a point x i s given i n Figure 1. A l l tame simple closed curves (that i s , simple closed curves which are equivalent to f i n i t e polygons) i n 3-space p i e r c e d i s c s at each of t h e i r p o i n t s . Even a w i l d simple closed curve (that i s , a simple closed curve that i s not tame) may p i e r c e a d i s c at each of i t s points as I l l u s t r a t e d i n Figure 2. In [ 1 ] Bing gave an example of a simple closed curve which f a i l e d to p i e r c e a d i s c at each of i t s p o i n t s . To the other extreme, i t i s the purpose of t h i s paper to give an example of a simple closed curve (sec) which f a i l s to p i e r c e a d i s c at only one point and then to use t h i s sec to construct examples of scc's which f a i l to p i e r c e d i s c s at various f i n i t e , countable and countable dense sets of p o i n t s . To t h i s end one could take a w i l d sec such as the one i n Figure 2 and l i n k each corresponding p a i r of t r e f o i l s i n an attempt to obstruct d i s c p i e r c i n g at the w i l d point (that i s , the point to which the t r e f o i l s converge). While t h i s i s reasonable i t i s not enough since as i n d i c a t e d i n Figure 3 a d i s c can be constructed, a s n a k e - l i k e cone, FIGURE 3 3 which f o l l o w s the curve down to i t s w i l d point without i n t e r s e c t i n g the curve except at the w i l d p o i n t . Next one could t r y to create o b s t r u c t i o n s to a tube, such as the one i n Figure 3, f o l l o w i n g the w i l d curve down to i t s w i l d p o i n t . I t i s with these ideas i n mind that the curve i n t h i s paper i s constructed. NOTATIONS AND DEFINITIONS. We w i l l denote Euclidean n-space by R n. Let B £ = 3 i 2 2 2 {(x,y,z)e-R | x + y + z < e} where e > 0, be the e p s i l o n b a l l . Let i t s closure be denoted by B . Let S = B ^ -~B • e —e e E be the e p s i l o n sphere. Note that by sphere we always mean the 2-sphere and by d i s c we always mean the 2-disc. I f A and B are closed sets i n R and p_ i s the standard metric i n R^ then define d i s t (A,B) = inf{p(a,b) | a s A & b e B } . 3 3 I f S i s a sphere i n R then R~^^ S i s the union of two d i s j o i n t connected components. Let i n t S be the bounded component and l e t extS be the unbounded component. I f D i s a d i s c and K i s a sec i n D then D—-"K i s the union of two d i s j o i n t connected components and . We w i l l use su b s c r i p t e d V's to denote the components of a d i s c minus a sec. I f S i s a sphere and K i s a sec on S then S-^ " K i s the union of two d i s j o i n t connected components and . We w i l l use sub s c r i p t e d U's to denote the components of a sphere minus a sec. 4 n A set X i n R i s a f i n i t e polyhedron i f there i s a f i n i t e s i m p l i c i a l complex L such that |L| = X. A set Y i n R n i s l o c a l l y p o l y h e d r a l at a point p i n Y i f there i s n a neighbourhood of p i n R i n t e r s e c t i n g Y i n a f i n i t e polyhedron. A set W i n R n has i t s points i n general p o s i t i o n i f no k+2 of them l i e i n a k-plane f o r k = 0 ,1,2,...,n-l. F i n a l l y , i f M i s a manifold we denote i t s boundary by 3M. CONSTRUCTION. In [3, pp. 166] R. H. Fox gives an example of a p a i r of arcs that are u n s p l i t t a b l e . A p a i r of d i s j o i n t arcs X and Y i n R are u n s p l i t t a b l e i f any 3 - c e l l which contains X i n i t s i n t e r i o r i n t e r s e c t s Y. These arcs are i l l u s t r a t e d i n Figure 4. Using these arcs as b u i l d i n g blocks the f o l l o w i n g sec i s constructed. Let T be the t r i a n g l e i n R with v e r t i c e s (0,0), (-1,1) and (1,1). Let X^U Y^ be a copy of the above p a i r of u n s p l i t t a b l e arcs i n R such that the boundary p o i n t s of X.^  are (-1,1,0) and (-1/2,1/2,0) and the corresponding boundary points of Y.^  are (1,1,0) and (1/2,1/2,0). We also want X-j^Y-^ cr: B i / y 2 ' L e t I x = { ( x , y ) e T | -1 < x < -1/2, 1/2 < y < 1} and l e t I Y = ' { ( x , y ) e T | 1/2 < x < 1, 1/2 < y < 1}. FIGURE 4 FIGURE 5 6. Let f i : xiUYi be a homeomorphism which i s the " i d e n t i t y " on the endpoints. See Figure 5 f o r an explanatory diagram. Let I x = {(x,y)<=:T | -1 / 2 < x < -1/4, 1/4 < y < 1 / 2 } and l e t I Y = {(x,y)e=:T | 1/4 < x < 1 / 2 , 1/4 < y < 1 / 2 } . Let F 2: R 2 -* R 2 be defined by F 2 ( x , y ) = ( 2 x , 2 y ) . Let F- |_/2 : r 3 r 3 b e defined by P 1 / 2 ( x , y , z ) = ( ( l / 2 ) x , ( l / 2 ) y , ( l / 2 ) z ) . Now define f_: I v U I v •* R 3 by d A 2 I 2 f 2 ( x , y ) = F 1 / 2 ( f 1 ( F 2 ( x , y ) ) ) . Let X„ = f 0 ( I v ) and Y„ = f„(I v ). Notice that d d A 2 d d 1 2. f Q : I (J I v -» X„[) Y„ i s a homeomorphism and X„U Y„ i s a copy d X 2 1 2 d d d d of our f i r s t p a i r of u n s p l i t t a b l e arcs U Y^ which i s i n f a c t attached to X-^ U Y 1 but now X 2 U Y^c=. ^iL/^2'^^'l/ (2 «V2) ' In general f o r 1 < i < 0 0 l e t I = {(x,y)e T | - 1 / 2 1 " 1 < x < - 1 / 2 1 , 1 / 2 1 < y < 1 / 2 1 " 1 } 7 and l e t I Y = {(x 3y)e:T | 1/21 < x < l / 2 1 - \ Define f. : I v U I v -»• R 3 by 1 A l I 1 1/21 < y < 1/21"1} f ± ( x , y ) = P 1 / 2 ( f 1 _ 1 ( P 2 ( x , y ) ) ) . Let X.= f (I ) and Y.= f (I ) . Then f : I UI + X.U Y 1 X A i X X I j _ X A i I i X X i s a homeomorphism where X^ y 1 S a c o P y °f the p a i r of u n s p l i t t a b l e arcs X j _ _ ] _ U Y^_^ and i s attached to the p a i r x i - i « Y i - r W e a l s o h a v e x i u Y i c = V ( 2 1 - 2 . ^ ) ^ B i / ( 2 1 - 1 - V 2 ' ) Let C = {(x,y)e=:R2 | y = l j -1 <. x < 1). Notice CcrT. Let C'= {(x,y,0)e:R 3 | x 2 + y 2 = 2, y > 0, -1 < x < 1} Define a homeomorphism g: C -»- C" such that g ( - l , l ) = (-1,1,0) and g ( l , l ) = (1,1,0). We can now define the d e s i r e d sec. OO 00 Let J = ± y i X ± (J 1^l1Y± U {(0,0,0)} U C . Define f: T -*• J by g ( x , y ) , ( x , y ) s c f ( x , y ) = -cJf.(x,y), (x,y)€=.l I j l X A i I i (0,0,0), (x,y) = (0,0) . f i s a homeomorphism hence J i s a sec which i s represented i n Figure 6. 8 FIGURE 6 9 Let p. = f ( - l / 2 1 - 1 , l / 2 1 1 ) and l e t p j = f d / 2 1 1 J l / 2 1 " 1 ) f o r 1 <_ i < °° and l e t p = f ( 0 , 0 ) = ( 0 , 0 , 0 ) . Notice l i m p.= l i m pT= p. j_-»-oo 2.-¥-oo We s p l i t the proof that J p i e r c e s a d i s c everywhere except at p i n t o the f o l l o w i n g p a r t s . PRELIMINARY SET UP. I t i s c l e a r that J p i e r c e s a d i s c at every point i n J-^{p}. Let D be a t o p o l o g i c a l disc i n R such that: (1) D fl J = (p> (2) 3D (1 J = 0 . We w i l l prove that 3D can not l i n k J . To do t h i s we w i l l prove that the 3D can be continuously shrunk to a point i n the compliment of J . This w i l l then imply that J f a i l s to p i e r c e a d i s c at p. R e c a l l that f o r 1 < i < « we have X 1 U Y ± c : B 1 / ( 2 1 - 2 . y 10 with B 1 / ( 2 l - 2 . v ^ ) f l X.={ P l}and B±/( 2±-2 . f l Y ±= (p^J. Consider the tangent plane to B-^y^±-2ti^^ at p^ ( r e s p e c t i v e l y P T ) . C a l l t h i s plane the tangent plane at p_ ( r e s p e c t i v e l y pT). The tangent plane contains a r b i t r a r i l y small 2-simplexes x ( r e s p e c t i v e l y x"*) such that: (1) T f ] J ={pi] ( T ' f l J =(p£J). (2) 3T(1 J = 0 OT' fl J = 0) . ( 3 ) 3T l i n k s J (3T' l i n k s J ) . Figure 7 i l l u s t r a t e s t h i s c o n s t r u c t i o n . C a l l these 2-simplexes which J pie r c e s at p^ (p^) s i m p l i c i a l d i s c s at £K ( P £ ) . Let d = d i s t ( p , 3 D ) . Since p and 3D are compact and p i s not i n 3D, d > 0. Let S be a po l y h e d r a l sphere i n R such that: (1) S fl J = { P l,pC} f o r some i > 0. (2) The face of S that contains p^ ( r e s p e c t i v e l y p^) i s a s i m p l i c i a l d i s c at p^ (pT). ( 3 ) Sex B d / 2 . (4) p e i n t S . Notice that ( 3 ) i m p l i e s 3D cz extS. Choose e > 0 such that B cr i n t S . Let e A = r v B . e 1 1 FIGURE 7 12 Let s = d l s t ( B ,S) and s^= d i s t ( A j J ) . Notice and are both p o s i t i v e . Let s = minimum(s /2,s^/2). We use Urysohn's Lemma to construct a f u n c t i o n T^: D -»- R such that: (1) f Q = 0 on B e / 2 . (2) f„(w) > 0 f o r we D B = A. 0 e (3) 0 <_ f Q(w) < s f o r a l l weD. Now we need the f o l l o w i n g theorem due to Bing [2] THEOREM (Approximation theorem f o r s u r f a c e s ) . I f M i s a 2-manifold (with or without boundary) i n R and i s a continuous non-negative f u n c t i o n defined on M, there i s . - 3 a 2-manifold M' i n R and a homeomorphism h: M -»• M^such that: (1) M' i s l o c a l l y p o l y h e d r a l at a l l points a such that f 0 ( h - 1 ( a ) ) > 0. (2) The distance between b and h(b) Is no more than f Q ( b ) f o r a l l b s H . 1 13 By t h i s theorem there e x i s t s a 2-disc D" such that (1) and (2) of the Approximation theorem hold. By property (1) of the d e f i n i t i o n of f Q and property (2) of the Approximation theorem Dfl B £ / 2 c Dfl D". Let A'= h(A) . By property (2) of f ^ and property (1) of the Approximation theorem A"* i s l o c a l l y p o l y h e d r a l . Since A' i s compact i t i s a f i n i t e polyhedron. By property (2) of the Approximation theorem, by the choice of s and by property (3) of fg we have D"fl Scr k'<*~ 3A" where 3A' = h( 3A) = h( ( 3B £ f| D) U 3D) . Notice that D" l i k e D has the property that D'fl J = {P> and BD"n J = 0 but D" a l s o has the d e s i r a b l e property that A" (crD") i s a f i n i t e polyhedron. Notice a l s o that 3D and 3D" are homotopic i n R 3-~-J. In f a c t DfjB / ? c = Df]D' so the boundaries of 14 the d i s c s can be shrunk to the same sec i n Dfl B e / 2 S h i s s i n g J . Therefore i t i s enough to show 3D Y does not l i n k J . For ease of n o t a t i o n l e t us assume that D and Acr D already had the important p r o p e r t i e s of and A"cz D^ , that i s , assume A i s a f i n i t e polyhedron and Dp S^r: A s - r 3 A . We may a l s o assume the v e r t i c e s of AU S are i n general p o s i t i o n . This could b accomplished by an a r b i t r a r i l y small adjustment without destroying the other p r o p e r t i e s mentioned. See Figure 8 f o r an i l l u s t r a t i o n of the set up to t h i s point.. PROPERTIES OF 'SOD. Notice that S fl D = Sfl A. Since S and A are i n general p o s i t i o n there are two p o s s i b i l i t i e s with respect to the i n t e r s e c t i o n of a 2-simplex of S and a 2-simplex of A. I f T i s a 2-simplex of S and a i s a 2-simplex of A then e i t h e r T(1 O = 0 or xf| c i s a s t r a i g h t l i n e segment. I f x f) a / 0 we can e s s e n t i a l l y have the two types of i n t e r s e c t i o n s In Figure 9. Hence Sf|A i s a 1-dimensional manifold w i t h a f i n i t e number of components since there are only a f i n i t e number of 2-simplexes i n S f l A . No component of S fl A can have a boundary since i f i t d i d have a boundary point i t would have to meet a free edge of a 2-simplex i n e i t h e r S or A. Now S i s without boundary so possesses no free edges and 3Afl S = 0 so there are no free edges of A i n t e r s e c t i n g S. Therefore FIGURE 8 16 FIGURE 9 FIGURE 10 17 S P A = L X U L 2 U . . . U L where each L. Is a sec. There e x i s t s at l e a s t one of 1 these L^, when considered as l y i n g on D, which d i v i d e s D i n t o 2 components and such that p e and 3Dc: . I f we could not f i n d an L. l i k e the one i n Figure 10 then the space l e f t a f t e r we cut out the d i s c s that the L^'s bound would be path connected and we could then construct a path on D from p to 3D which does not i n t e r s e c t S. But p o i n t s and 3 D c r e x t S th e r e f o r e such a path i s impossible. Let {K.}. m n be the subset of which has the above property of sepa r a t i n g p from 3D on D. Let D-^ K .= v] U V 2 J J J 1 2 1 where V. and V. are the components of D—"K. with p e V . J 2 ^ 1 ^ 1 ^ and 3Dcr: V.. There i s e x a c t l y one V., say V. corresponding J J m J o to the sec K. , which does not i n t e r s e c t .U K.. That i s , Jo J=l J no other K. i s " i n t e r i o r t o " K. on D. Of course we may J Jo have other L. 's i n t e r s e c t i n g V"!" . In a l a t e r part of the i Jo proof we w i l l remove these i n t e r s e c t i o n s where they .are i n e s s e n t i a l . Figure 11 i s an example of the s i t u a t i o n that might a r i s e In Sfl A at K. . J o 18 FIGURE 11 19 SOME USEFUL LEMMAS. LEMMA 1. I f S Is a 2-sphere In R which contains a s i m p l i c i a l d i s c at p^ ( r e s p e c t i v e l y pT) then: (1) J fl i n t S ? 0 . ( 2 ) J fl extS j* 0 . Proof: Let T ( r e s p e c t i v e l y x") be a s i m p l i c i a l d i s c at p^ ( p T ) . I f J f l i n t s = 0 we could shrink 3x Ox") to a point i n i n t S . I f J f| extS = 0 we could shrink 3x ( 3 x " ) to a point i n extS. But 3x ( 3 x " ) l i n k s J so we would get a c o n t r a d i c t i o n . Therefore we have J fl i n t S ^ 0 and J f] extS ? 0 . LEMMA 2 . There does not e x i s t a 2-sphere S i n R such that: ( 1 ) S'fl J = tp>. ( 2 ) i n t S ' f l J / 0 . (3) extS'fl J fi 0 . Proof: Any sec which i n t e r s e c t s i n t S " and extS" must i n t e r s e c t S" i n at l e a s t 2 p l a c e s . LEMMA 3- There does not e x i s t a 2-sphere S" i n R such that: ( 1 ) S ' f l J = {p±,P>. ( 2 ) i n t S " fl J 7* 0 . (3) extS ' f ] J ¥ 0 . 20 Proof: J — S " = I-^U I 2 where I and 1^ are the open arcs of J r e s u l t i n g from the removal of p and p. with p. , ne=: I , Without l o s s of g e n e r a l i t y assume I-^ 4"" i n t S ' and ^ 00 I„crrextS . I., contains {p.}. . ,-, . Therefore we can 2 1 * j j = i + l f i n d a p a i r of p o i n t s {p^.,p^} cr J such that 1-^  and P ^ ^ ^ 2 * H e n c e the arc from p^ to p ^ + ^ ( t h a t i s , X. ) i s contained i n I, while the arc from p." to p ' . k' 1 ^k+1 (that i s , Y. ) i s not contained i n I , . Therefore X,czr 5 k 1 k i n t S " and Y^err extS". By the approximation theorem f o r surfaces we can f i n d a p o l y h e d r a l 2-sphere S " such that X, cr i n t S ^ a n d Y, cr, e x t S " . Since X, and k k k Y, are u n s p l i t t a b l e we have a c o n t r a d i c t i o n so S^ k ^ can not e x i s t . OUTLINE. Before continuing with the proof i t w i l l be informative to give an o u t l i n e of the proof so f a r and of the parts yet to come. We have: A. Defined J . •3 B. Let D be a d i s c i n R J such that D fl J = (p) and 3D fl J = 0 . C. Proposed to show 3D can not l i n k J and hence J can not p i e r c e D at p. D. Constructed a p o l y h e d r a l sphere S i n R such that p s i n t S and 3DcrrextS. 21-E. Applied approximation theorem to get a d i s c D" close to D but pol y h e d r a l outside of a neighbourhood of p (assumed D already had the p r o p e r t i e s of D"). F. Proved S fl D = L^U L 2 U ... U L n where the L±'s were G. Proved a subset of the L^'s, {Kj}™_^, separated p from 9D and one of the K.'s, K. , was innermost J J o on D. H. Proved 3 lemmas to be used i n the f o l l o w i n g p a r t s . We w i l l now prove: ( d e t a i l s w i l l f o l l o w ) I. There are 2 cases: Case 1. Both p. and pT are i n the same component of S—-K. ( r e c a l l S f| J = J o {p^jpT}). In t h i s case the proof i s f i n i s h e d . Case 2. p. i s i n one component of S-*"K. — i Jo and p^ i s i n the other component. J . I f we l e t U be the component of K. which contains p ^ 3 V be the component of D<-*"Kj which contains p, H = K. and S'= HU UUV then Case 2 leads to a J o c o n t r a d i c t i o n i n part K below. K. There are again 2 cases: Case 1. S' i s a sphere. This leads to a c o n t r a d i c t i o n by the lemmas. 22 Case 2_. S' i s not a sphere. We make i n t o a sphere by c u t t i n g o f f i n e s s e n t i a l parts of V. This leads to a c o n t r a d i c t i o n by the lemmas. AN INVESTIGATION OP CERTAIN 2-SPHERES. Consider K. on S where S i s the p o l y h e d r a l sphere J o we constructed i n the p r e l i m i n a r y set up. K. d i v i d e s S J o i n t o 2 components. Suppose both p^ and p^ ( r e c a l l J fl S = { p i > P T } ) are i n the same component. Then, K.^ can be shrunk continuously to a point on the other component which l i e s i n R-i—- J . Now since 3D can be shrunk continuously on D i n R ~ J to K. we have that 3D can be shrunk to a J o point In R^ -- J and hence 3D can not l i n k J . The only other p o s s i b i l i t y i s i f p. i s i n one component of S—K. i Jo and p.7 i s i n the other. We w i l l show that t h i s can not happen and hence 3D can be shrunk to a point i n R—-J. Assume p. l i e s i n one component of S—K. and pT l i e s 1 J o 1 i n the other. To reduce n o t a t i o n l e t H = K. J o and v = v^ J o ( r e c a l l D--K. = V1. U V 2 and' p e v ! ). Let U be the component Jo Jo Jo Jo 23 FIGURE 12 24 of S—•-H which contains p.. 1 Consider s ' = H U V U U . I f U flV = 0 then S" i s the union of 2 di s c s with common boundary which touch nowhere e l s e . Hence, S" i s a sphere such that S'fl J = { p , p ± } since u f l J = ivJ and V fl J = {p}. U contains a s i m p l i c i a l d i s c at p^ ther e f o r e Lemma 1 im p l i e s i n t S ' f l J t 0 and extS'fl J ? 0 . We have a l l the hypotheses of Lemma 3 hence S" can not e x i s t so p^ and p;T must be i n the same component of S^-H. Suppose U flv ^ 0 , that i s , suppose Ufl V = H1U H 2U ... UHk where {H.}. , i s a subsequence of { L . } . = 1 ( r e c a l l that 3 3 J J the L. ' s are the components of S fl D) . Notice that the H.'s are scc' s . We w i l l show how to amputate c e r t a i n 3 parts of S" i n order to obtain a sphere such that Lemma 2 or Lemma 3 a p p l i e s . See Figure 13. Consider each H. as on U . Let J U . J be the component of U—' H. (U = UU H) which does not contain J H. Let V. 25 FIGURE 13 26 be the component of V—'H. (V = VU H) which does not contain H. Define H. to be innermost on U i f U. does not contain 3 3 any Hh<= ufl V. I t i s impossible f o r p_^  to be i n U. f o r any innermost H. f o r suppose H. was innermost on U and p U . V J U 3 3 1 j j j 0 since H. i s innermost on U so we can form the sphere 3 S"= v.U H.UD.. 3 3 3 See Figure 13- Notice S" fl Jcz { p ± , p } and S" fl J ¥• 0 since we are assuming p .e U.. C l e a r l y U. contains a s i m p l i c i a l d i s c at p so i n t S " f l J ^  0 and extS " " f l J ^ 0 by Lemma 1. S^ ' s a t i s f i e s a l l the hypotheses of e i t h e r Lemma 2 or Lemma 3 so S^" can not e x i s t hence p.^U.. Since there are only f i n i t e l y many H.'s In Ufl V there e x i s t s at l e a s t one, say H^, which i s innermost on U. We w i l l now construct a new sec on U. Let be a pol y h e d r a l sec In U such that the component of U - — w h i c h does not contain H has the f o l l o w i n g p r o p e r t i e s . (1) contains H 1 (H^ i s outside H^). (2) U' does not contain any other H . cr Ufl V ( a l l other H.'s are outside of H"). J 27 (3) does not contain p. (p. i s outside H')« See Figure 14 . Let i s a d i s c i n U. We f a t t e n an a r b i t r a r i l y small amount to form a 3 - c e l l F' such t h a t : (1) D ' c z 3p'. (2) V-^ n F'= 0 ( r e c a l l i s the component of V — w h i c h does not contain H). (3) 3P '0 V = H1IJ where H ~ i s a sec. (4) P ' f l J = 0 . Figure 14 c l a r i f i e s t h i s c o n s t r u c t i o n . Let be the component of 9 F " — w h i c h does not contain . We are f a t t e n i n g e i t h e r i n t o i n t S or i n t o extS depending on where V 1 i s s i t u a t e d and i n such a way that we get the n i c e s t p o s s i b l e i n t e r s e c t i o n with V. V - ^ ( 3 F ' ^ ' ) c o n s i s t s of 2 components. One component contains V . C a l l the other component 28 s F I G U R E 12 29 Consider vijv£-u.Hr-Again see Figure 14. In e f f e c t what we are doing i s c u t t i n g a p a r t i c u l a r hole i n V that contains to get the annulus V'. This hole we f i l l w i th the d i s c V£''(J H£^ which does not i n t e r s e c t S. For convenience continue to c a l l our modified d i s c (that i s , V 'y Vj'[J H { ' ) v- W e h a v e "amputated a f i n g e r " of V which contained an innermost H. on U. I t i s c l e a r that t h i s c u t t i n g procedure has no e f f e c t on the l i n k i n g of 3D and J . I t i s al s o s t i l l true that V fl J ='{p>, V i s bounded by H and V i s a d i s c . Now at l e a s t one of {H.}^_p i s innermost on U and we J 3 ^-do a s i m i l a r surgery on i t . In t h i s way a l l of {H.}._, may be removed. Hence again we have a sphere S^ = U U V \j H such that U f l V = 0 . We have already proved that t h i s case can not happen so we can never have p^ i n one component of S—H and pf i n the other. The case where both p. and pT are i n the same component of S—H always a p p l i e s and ther e f o r e 3D can be shrunk to a point i n R — J . F i n a l l y , we can conclude that J does not p i e r c e a di s c at p. 30 EXAMPLES RELATED T O .J. We w i l l construct scc's which f a i l to p i e r c e d i s c s at c e r t a i n f i n i t e and i n f i n i t e sets of p o i n t s . R e c a l l from Figure 6 that C i s the part of J which extends from p^ to p^ without passing through p. Let be the arc we get when we remove minus i t s endpoints from J . Let n be any i n t e g e r . 1 J o i n n copies of J " together at t h e i r endpoints i n such a way that we get a sec. Figure 15 i l l u s t r a t e s t h i s c o n s t r u c t i o n f o r n = 4 . C a l l t h i s sec ' J . n I t i s c l e a r that J n f a i l s to p i e r c e a d i s c at only n p o i n t s where each "bad" point of J n corresponds to a "bad" point on one of the copies of J ' . Therefore we have shown how to construct a sec which f a i l s to p i e r c e a d i s c at any f i n i t e number of p o i n t s . Again consider our example J . R e c a l l t h a t , i n p a r t , i t was composed of p a i r s of arcs X^ and f o r i ^ { l , 2 , . . . } . Suppose we e x t r a c t a small closed i n t e r v a l from each and each Y^. I f i n place of these closed i n t e r v a l s we i n s e r t copies of which decrease i n s i z e as i increases we again o b t a i n a sec which we w i l l c a l l J . FIGURE 15 FIGURE 16 33 Figure 16 explains t h i s c o n s t r u c t i o n . I t i s c l e a r that f a i l s to p i e r c e a d i s c at a l l the "bad" points of the attached copies of J ' . J a l s o f a i l s to p i e r c e a d i s c at CO the point which corresponds to p on the o r i g i n a l example J . To prove t h i s we can use the same argument we used to show that J f a i l s to p i e r c e a d i s c at p with only minor m o d i f i c a t i o n s . So we have constructed a sec which f a i l s to p i e r c e a d i s c at a countably i n f i n i t e number of p o i n t s . A l s o , t h i s countably i n f i n i t e c o l l e c t i o n of "nonpiercing" points has only one accumulation point and J r o f a i l s to p i e r c e a d i s c at t h i s point too. I f we take a countably i n f i n i t e number of copies of which decrease i n s i z e and attach them together as i n Figure 17 we again obtain a sec which f a i l s to p i e r c e a d i s c at' a countably i n f i n i t e number of p o i n t s . We w i l l c a l l t h i s sec OO This countably i n f i n i t e c o l l e c t i o n of "nonpiercing" points a l s o has only one accumulation p o i n t . Howerer, c l e a r l y p i e r c e s a d i s c at t h i s p o i n t . Our f i n a l example i s a sec which contains a countably i n f i n i t e dense set where the sec f a i l s to p i e r c e a d i s c and which also contains a countably i n f i n i t e dense set where i t p i e r c e s a d i s c . To construct t h i s sec we w i l l take a c i r c l e i n R and by a t t a c h i n g an i n f i n i t e number of FIGURE 17 35 copies of J " i n a s p e c i a l way we w i l l add the r e q u i r e d non-p i e r c i n g p o i n t s and p i e r c i n g points to make up both countably i n f i n i t e dense set s . The r e s t of t h i s e x p o s i t i o n concerns the s p e c i a l way we a t t a c h the copies of 3" i n order to get a sec which has the d e s i r e d p r o p e r t i e s . Let us assume that J had two more p r o p e r t i e s . C l e a r l y J i s l o c a l l y p o l y h e d r a l except at {p }?_-,U {p" }?_-,U{p} so j_ 1—J. 1 i—i. we w i l l consider J as an i n f i n i t e c o l l e c t i o n of s t r a i g h t l i n e segments which we w i l l c a l l edges. Also J could have been constructed so that J " was contained i n C where C i s the s o l i d c y l i n d e r i n R with c e n t r a l a x i s that part of the y a x i s from ( 0 , 0 3 0 ) to ( 0 , 1 , 0 ) and of diameter 2 . See Figure 18 . R e c a l l how we attached copies of J " to J o t and note how the s i z e of the i n t e r v a l removed (distance b'etween the end-points by the standard metric) where the copy of J ' was to be attached was d i r e c t l y p r o p o r t i o n a l to the s i z e of the copy of J'. We w i l l be a t t a c h i n g copies of J " i n a s i m i l a r f a s h i o n . We are going to a s s o c i a t e with each copy of J ' that we a t t a c h a set we c a l l J ^ j _ s c which i s j u s t J " with the a d d i t i o n of s i m p l i c i a l d i s c s , as described e a r l i e r , at each point i n OO CO tP-K -i U IP-/, -i • These d i s c s w i l l be the a c t u a l d i s c s * i i = l u ^ i i = l which our l a s t example w i l l p i e r c e . We are now ready to construct the f i n a l example. Take a copy of the u n i t i n t e r v a l and by i d e n t i f y i n g 0 and 1 form 36 FIGURE 18 FIGURE 19 37 a c i r c l e i n R w i t h (0,0,0) as the center. C a l l i t V Assume KQ has i t s p o i n t s l a b e l l e d as the u n i t i n t e r v a l w i t h 0(=1) on the negative x a x i s , 1/4 on the negative y a x i s , 1/2 on the p o s i t i v e x a x i s and 3/4 on the p o s i t i v e y a x i s . Now choose an i n t e r v a l on K Q which contains 1/2 such that the distance between the endpoints of the I n t e r v a l i s l e s s than 1. Attach a copy of 3" to KQ at t h i s I n t e r v a l so that we get a new sec which we w i l l c a l l K l -Now define a homeomorphism s r Ko - K i which i s the i d e n t i t y outside the chosen i n t e r v a l on and which takes the i n t e r v a l onto the attached copy of J ' i n such a way that the p oints on the copy of J " which correspond to {p^ u ' P r i - i a n <^ ' i : h e e n ^ p o i n t s of the edges have inverse images which are not dyadic r a t i o n a l s . We a l s o want g-^(l/2) to be the point on the copy of J ' which corresponds to p 3". We w i l l c a l l t h i s f i r s t attached copy of <T, J i r Notice that. J i n has a s o l i d c y l i n d e r c o n t a i n i n g i t . C a l l 38 t h i s c y l i n d e r c i r Let us al s o assume that was attached so that does not i n t e r s e c t KQ minus the i n t e r v a l . Figure 19 i l l u s t r a t e s t h i s f i r s t stage of the c o n s t r u c t i o n . We w i l l describe the 1^ stage of the c o n s t r u c t i o n f o r i > 1. Suppose we have a sequence of scc's { K 0 , K 1 , . . . , K 1 _ 1 } and a sequence of homeomorphisms {f~>-^>§>2 > ' ' ' J ^ i — l ^ w i t h g,: K, , -»- K, toh h-1 h f o r 1 <_ h < i . We w i l l a t t a c h 2 i _ 1 copies of J" to K^^, a l l of uniform s i z e , to form a new sec K. l and we w i l l do t h i s according to the f o l l o w i n g p l a n . F i r s t we determine where the points i n W = ( g i _ 1 o g 1 _ 2 o . ...og 1(j/2 1) | 0 < j < 2 1, j odd} are s i t u a t e d on ^. Next we w i l l choose closed i n t e r v a l s of uniform s i z e around these points i n K^_1 which are small enough to s a t i s f y some important c o n d i t i o n s . 39 F i r s t we want to ensure that the c o n s t r u c t i o n leads to a set which i s a sec. To accomplish t h i s we choose i n t e r v a l s so that newly attached copies of J" w i l l be smaller than p r e v i o u s l y attached copies. Therefore we choose non-overlapping i n t e r v a l s whose endpoints are a distance of l e s s than 1/2^^"""'"^ apart from each other. We al s o choose the i n t e r v a l s so that when new copies of J" are attached to the i-l'k* 1 stage we end up with a sec at the i t l r l stage. I f any of the po i n t s i n W l i e s i n a previous attachment of a copy of J " then we want our new i n t e r v a l s to l i e e n t i r e l y w i t h i n an edge i n such a way that when a copy of J " i s attached i t s c y l i n d e r w i l l not i n t e r s e c t any other c y l i n d e r except i f i t i s completely contained i n one. I f any of the points i n W does not l i e i n a previous attachment of J" then we want to choose an i n t e r v a l such that a newly attached copy of J " has a co n t a i n i n g c y l i n d e r which does not i n t e r s e c t any other c y l i n d e r . The l a s t property we need to ensure that our c o n s t r u c t i o n leads to a sec i s that the i n t e r v a l s be chosen so that the distance i n between the endpoints of the inverse images of the i n t e r v a l s v i a the map g 1 _ 1 g j _ _ 2 ••• §i i s l e s s than 1 / 2 2 1 . Next we want to ensure we end up with a set where our example p i e r c e s a d i s c . To do t h i s we choose i n t e r v a l s i n such a way that when a copy of J * i s attached i t s corresponding J C . does not i n t e r s e c t the J " of any other new or di s c d i s c J previous attachment of a copy of J " and does not i n t e r s e c t any other part of K. n minus the newly chosen i n t e r v a l s . 40 F i n a l l y , to ensure that a g e n e r a l i z a t i o n of our proof that J f a i l s to p i e r c e a d i s c at p works f o r each point i n a dense set contained i n our example we choose i n t e r v a l s so that where new copies of J ' are attached, arcs that were p r e v i o u s l y u n s p l i t t a b l e are not made s p l i t t a b l e . (Fox's argument i n [ 3 ] can be extended to the case where the arcs have s u f f i c i e n t l y small r e v i s i o n s . ) A f t e r f i n d i n g i n t e r v a l s which s a t i s f y these p r o p e r t i e s a t t a c h a copy of at each of these i n t e r v a l s . Label them J. . f o r 0 < j < 2 1 and j odd. Corresponding to these J . . we * 3 have t h e i r c o n t a i n i n g s o l i d c y l i n d e r s C. . We define a homeomorphism g. : K. n K. & i l - l 1 which i s the i d e n t i t y on K. n minus, the i n t e r v a l s and takes J l - l each i n t e r v a l to i t s corresponding attached copy of . Also we want the inverse images v i a the map g^og^_-^o . . . og-^ of a l l p o i n t s i n a l l J. . corresponding to {p.}? n i i ' { p T } ? , t- 1 j * i J i = l u * i i = l and a l l endpoints of edges i n a l l J . . not to be dyadic r a t i o n a l s . And we want g^og^_^o...og^(j/2 1) to be the point on J^. which corresponds to pG f o r each newly 41 defined J . . . C l e a r l y g ± : K± 1 ->- R 3 f o r 1 > 0. Define f ± = S 1og j__ 1o. . . og 1 f o r i > 0 and note f 1 : K Q R3. Let f = l i m f.. PROPOSITION 1. f e x i s t s and i s continuous Proof: Define d(g,g") = sup p ( g ( s ) , g " ( s ) ) . S€ where g,g": K + R 3. Notice d ( g i 3 i d K ) < 1/2 2^ 1 - 1^ t ^ - l by our choice of i n t e r v a l s at the i stage. Let i > 0 and j > i . Then d ( f . , f . ) < d ( f . , f 1 + 1 ) '+ d ( f . + 1 J f . + 2 ) +...+ d ( f j _ 1 , f j ) Since ^ ( f k _ 1 , f k ) = d ( g k , i d K ^ ) f o r k > 0 we have d ( f . , f . ) < 1 / 2 2 ( ( i + D - l ) + 1 / 2 2 ( ( i + 2 ) - l ) + _ + i / 2 2 ( J - 1 ) < ( 4 / 3 ) ( l / 2 2 1 ) Hence we have Cauchy uniform convergence which Implies f e x i s t s and, since each f. i s continuous, f i s a l s o . . PROPOSITION 2. f i s i n f e c t i v e . Proof: To each s E K ^ and f o r every i > 0 we w i l l be a s s o c i a t i n g the smallest c y l i n d e r to which f . ( s ) belongs, i f i t th 1 belongs to any, at the i stage of,the c o n s t r u c t i o n . 42 Suppose s^, K Q with s-^  ^ s 2 and f ( s 1 ) = f ( s 2 ) . There are 3 cases: Case 1. I f the images of s^ and s 2 do not both belong to the same c y l i n d e r at some stage then assume without l o s s of g e n e r a l i t y the image of s-^  i s i n a c y l i n d e r and the image of s 2 i s not i n that c y l i n d e r . By our con-s t r u c t i o n fCs.^) i s i n the c y l i n d e r and f ( s 2 ) i s not so f ( s ^ ) ^ f ( s 2 ) . C o n t r a d i c t i o n . Case 2. I f the images of s^ and s 2 are i n no c y l i n d e r s at a l l stages of the c o n s t r u c t i o n then f(s-^) = s-^  and f ( s 2 ) = s 2> C o n t r a d i c t i o n . Case 3. I f the images of s.^  and s 2 are i n the same c y l i n d e r s at a l l stages of the c o n s t r u c t i o n then by our choice of i n t e r v a l s s-^  and s 2 l i e i n an a r b i t r a r i l y small i n t e r v a l i n K Q. C o n t r a d i c t i o n . The f a c t that K Q i s compact and R i s Hausdorff now i m p l i e s that f i s a homeomorphism of Kg onto f ( K g ) . Hence f(Kg) i s a sec which we w i l l c a l l K . 00 Consider the set D = ' { s s L | s i s a dyadic r a t i o n a l } . Notice that f o r every dyadic r a t i o n a l s there i s an f^ which 43 takes s to a point i n where K_ f a i l s to p i e r c e a d i s c . I t i s c l e a r that each Kj f a i l s to p i e r c e a d i s c at t h i s point f o r j > i and by using methods s i m i l a r to the ones used i n proving that J f a i l s to p i e r c e a d i s c at p i t can be shown that K m f a i l s to p i e r c e a d i s c at f ( s ) . Notice that D i s dense i n KQ SO f(D) i s dense i n K^. Let U\ = ( s e K g | s = f T ^ ( t ) where t*= J _ c corresponds to a point i n {p^ }j°=i_ U ^ ) ± ^ ± = 2.c~ ^ * a n d J" ^ s 0(^d between 0 and 2 1 }. Let C O u = . U . u . . 1=1 i By our J ^ i s c c o n s t r u c t i o n above Kffl p i e r c e s a d i s c at each point i n f ( U ) . f(U) i s dense i n Kffl since c l e a r l y U i s dense i n K Q. SO K has the property that I t pi e r c e s a d i s c i n a dense set and, as noted above., f a i l s to p i e r c e a d i s c i n a dense s e t . Let D ' C K 0 be the inverse image v i a f of a l l non-piercing p o i n t s i n K^ Notice that f o r each i > 0 D" c o n s i s t s of the inverse image of the f i n i t e number of non-piercing points of K^ plus a set of points contained i n a countable union of i n t e r v a l s 44 C O whose t o t a l length i s l e s s than . E . 1 / 2 J . This i s j u s t the sum of the lengths of a l l i n t e r v a l s on which are the inverse images of i n t e r v a l s chosen a f t e r the i ^ * 1 stage. Since KQ i s j u s t the u n i t i n t e r v a l w i t h 0 and 1 i n d e n t i f i e d we can give i t the measure induced by the Lebesgue measure on the u n i t i n t e r v a l . Hence we can give K the measure ° 0 0 induced by f. Now i t i s c l e a r that D" has measure 0 i n hence f ( D " ) , the non-piercing points of K , has measure 0 . 45 BIBLIOGRAPHY 1. Bing, R. H., "A simple closed curve that p i e r c e s no d i s c , " J . Math. Pures Appl. , v o l . 3 5 ( 1 9 5 6 ) , 3 3 7 - 4 3 . 2. Bing, R. H., "Approximating surfaces with p o l y h e d r a l ones," Ann, of Math., v o l . 6 5 ( 1 9 5 7 ) , 4 5 6 - 8 3 . 3 . P o r t , M. K. J r . , ed, Topology of 3-ManlfoTds, P r e n t i c e -H a l l , Inc., Englewood C l i f f s , N. J . , 1962. 

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