SOME PROBLEMS ON MOUNTAIN CLIMBING by PATRICK CHIA-LING HUNG B.S. Fu-Jen U n i v e r s i t y , T a i p e i . Taiwan, 1967 A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY i n the Department of MATHEMATICS We accept t h i s t h e s i s as conforming to the r e q u i r e d standard THE UNIVERSITY OF BRITISH COLUMBIA June, 1973 In presenting t h i s t h e s i s i n p a r t i a l f u l f i l m e n t of the requirements f o r an advanced degree at the U n i v e r s i t y of B r i t i s h Columbia, I agree that the L i b r a r y s h a 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 study. I f u r t h e r agree that permission f o r extensive copying of 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 of my Department or by h i s representatives. I t i s understood that copying or p u b l i c a t i o n of t h i s t h e s i s f o r f i n a n c i a l gain s h a l l not be allowed without my w r i t t e n permission. Department of MAV^M/\JlCS The U n i v e r s i t y of B r i t i s h Columbia Vancouver 8, Canada Date ii Supervisor: P r o f e s s o r J , V. Whittaker ABSTRACT Let f and f u n c t i o n s d e f i n e d on f(l) = g be two c o n t i n u o u s , r e a l - v a l u e d [0,1] with f ( 0 ) = g(0) and g ( l ) « The main r e s u l t o f t h i s c h a r a c t e r i z e the p r o p e r t y that the same connected component o f In Chapter I , we study (0,0) G(f,g), and (1,1) where f t h e s i s i s to (0,0) and G(f,g) = (1,1) are i n {(x,y)jf(x)=g(y)}. conditions implying that are i n the same connected component o f and. g are not n e c e s s a r i l y r e a l - v a l u e d f u n c t i o n s . We o b t a i n theorems to c h a r a c t e r i z e [0,1], In Chapter I I , we give a simple p r o o f o f a theorem by S i k o r s k i and Z a r a n k i e w i c z . In Chapter I I I , we o b t a i n our main result. In Chapter I V , we study pathwise connectedness i n G(f,g) and give some a p p l i c a t i o n s . In Chapter V, we study the q u e s t i o n o f s l i d i n g a chord o f g i v e n l e n g t h along a p a t h . An example i s g i v e n to show t h a t t h i s i s not always p o s s i b l e . iii TABLE OF CONTENTS Page INTRODUCTION CHAPTER CHAPTER 1 . . . . . . . . . . . . I : CHARACTERIZATIONS OF [0,1] . . . . . . 3 ; . . . . 17 I I : CONNECTEDNESS CHAPTER I I I : COMPATIBILITIES OF FUNCTIONS CHAPTER IV : SOME PROPERTIES OF MOUNTAIN CLIMBING AND APPLICATIONS CHAPTER 27 . . . . . V : CHORD SLIDING ALONG A PATH BIBLIOGRAPHY 49 59 70 ACKNOWLEDGEMENT I am deeply indebted his to P r o f e s s o r J , V. Whittaker f o r v a l u a b l e a s s i s t a n c e throughout the r e s e a r c h and p r e p a - ration of this The thesis. f i n a n c i a l support o f the U n i v e r s i t y o f B r i t i s h Columbia and the N a t i o n a l Research C o u n c i l o f Canada i s gratefully acknowledged. 1 INTRODUCTION Let d e f i n e d on The £ and g [0,1] with be two r e a l continuous f ( 0 ) = g(0) and f(l) =g(l). q u e s t i o n o f whether there e x i s t continuous and j from [0,1] h(l) = j(1) to such t h a t [0,1] with functions h(0) = j ( 0 ) fh(x) = gj(x) for a l l has been s t u d i e d by S i k o r s k i and Zarankiewicz Huneke functions h and x e [0,1] [ 1 ] , Homma [ 7 ] , [6] and Whittaker [ 2 ] , I f we c o n s i d e r the graphs o f mountains, then the e x i s t e n c e o f h f and and j g t o be t e l l s us t h a t c l i m b e r s can always climb the mountains m a i n t a i n i n g elevation ( h and j a common denote the h o r i z o n t a l progress o f the climbers). I f we c o n s i d e r the s e t G(f,g) {(x,y)| f(x) = g(y)}, then the e x i s t e n c e o f e q u i v a l e n t to the e x i s t e n c e o f a path (0,0) and d e f i n e d by in h and G(f,g) connectedness o f G(f,g) between t h e s i s i s to study the (0,0) and (1,1). i s important when we want to know i f the f u n c t i o n s exist. joining (1,1). The main purpose o f t h i s j j is h This and 2 The and s t a r t i n g p o i n t w i l l be a theorem of S i k o r s k i Zarankiewicz which s t a t e s that i f continuous f u n c t i o n s f(0) = g CO) (1,1) = 0 and from [0,1] and [0,1] f ( l ) » g(l') = 1, g then (0,0) [0,1] [0,1] get It turns out as w e l l as the p o i n t s 0 G(f,g). and get i s known, but Zarankiewicz. 1 In A c t u a l l y the the o r i g i n a l p r o o f r e q u i r e s a l o t of advanced t o p o l o g i c a l means. The p r o o f we have i s a very simple d i r e c t p r o o f . In Chapter I I I , we study the in i n more d e t a i l . G(f,g) between (0,0) can give a n e c e s s a r y and (1,1) (1,1) In Chapter V, we sliding discuss G(f,g) We (0,0) G(f,g) and i n the study some p r o p e r t i e s of and give some a p p l i c a t i o n s . the q u e s t i o n s t u d i e d by a chord o f f i x e d l e n g t h along o f view. An connectedness s u f f i c i e n t c o n d i t i o n that c a s e . In Chapter IV, we pathwise connectedness i n had and are i n the same connected component of most g e n e r a l we a theorem which g e n e r a l i z e s the above theorem proved by S i k o r s k i and r e s u l t we G(f,g). [0,1], i n terms of connectedness p r o p e r t i e s f o r Chapter I I , we and show that the above n i c e r e s u l t i s only true f o r f u n c t i o n s from characterize are with are i n the same connected component of In Chapter I , we can to f Fenn [3] of a path from our point i n t e r e s t i n g counter-example i s i n c l u d e d . A f t e r I the example, P r o f e s s o r R. 0. Davies showed me example i n terms of a simple curve. a better 3 CHAPTER I CHARACTERIZATIONS OF Let space Y. F(a,b;Y) a' and-.-, b Throughout be two. p o i n t s of a topological t h i s t h e s i s , we s h a l l denote by the c l a s s o f a l l continuous f u n c t i o n s closed unit i n t e r v a l f ( l ) = b. instead [0,1] I into Y with f(0) = a For s i m p l i c i t y , we s h a l l w r i t e of F(a,b;R), numbers), and belong to F(0,1;I) F(a,b;Y), •{(* ,x x ) 1 2 (where n n F(a,b) If and F f , f ^ ,«•«, f n G ( f , f , . . , , f ) i s the s e t : I f . U . ) = f.(x.) ' and i s the s e t o f a l l r e a l respectively. then I e R from the x i J j fora l l i,j}. Now we s t a t e two theorems which R. S i k o r s k i and K. Zarankiewicz proved i n [ 1 ] . Theorem 1.1: Let and (1,1) f, g belong to F. Then the p o i n t s are i n the same connected component o f (0,0) G(f,g). 4 Theorem 1.2: f ^ , £ 2 » • « « , . £ n belong to Let F with each c o n s i s t i n g of a f i n i t e number of monotone p i e c e s . there e x i s t ^ , y ,... , ^ for a l l x i n e F such [0,l]o to c h a r a c t e r i z e F i n a l l y , we £ and g are able to give theorems [0,1], 1.3: A space Y i s c a l l e d pathwise connected i f f o r each p a i r o f d i s t i n c t p o i n t s tinuous f u n c t i o n £(0) = a and £ from £ ( 1 ) = b. a homeomorphism, then we Lemma to c o n s i d e r i n the case where we do not assume are r e a l f u n c t i o n s . Remark Then that In t h i s c h a p t e r , we would l i k e Theorem 1.1 f^ a, b [0,1] I f we call in to Y, Y such can always Y there i s a conthat choose £ to be arcwise connected. 1.4: Let connected space a, b Y. be two d i s t i n c t p o i n t s of a pathwise I f f o r every £, g in F(a,b;Y), 5 (0,0) G(f,g), and (1,1) are i n the same connected component o f then every h in F(a,b;Y) i s surjective. Proof: If a point exists h ( I ) i s not equal to Y, then there exists c e Y - h ( I ) . By pathwise connectedness, there j e F(a,b;Y) with h(x) i s not e q u a l to j(1/2) = c. j (1/2) f o r any I t i s clear that x in [0,1], There- fore the s e t {(x,y) I 2 | x = 1/2} e and G(j,h) are d i s j o i n t . Hence (0,0) and (1,1) same connected component o f G(j,h). assumption t h a t (1,1) component o f (0,0) G(j,h). and Hence every are not i n the T h i s c o n t r a d i c t s our are i n the same connected h £ F(a,b;Y) is surjective. Lemma 1.5. Let Y be a H a u s d o r f f space and a, b be two d i s t i n c t p o i n t s i n the same path-connected component o f Suppose t h a t every f in F(a,b;Y) i s surjective. i s homeomorphic t o the c l o s e d u n i t i n t e r v a l I. Then Y. Y 6 Proof: Since a, b are i n the same path-connected com- ponent o f Y, g(I) = Y. By the Hahn-Mazurkiewiez Theorem Y there e x i s t g e F(a,b;Y). Clearly, [ 5 , p. 1 2 9 ] , i s a compact, l o c a l l y c o n n e c t e d , connected m e t r i c s p a c e , i.e., Y i s a Peano s p a c e , and so Hence there e x i s t s a homeomorphism with fl(0) = a 11(1) = Y. terval and Hence Y 11(1) = b. Y i s arcwise connected. fl from Since I into tl e F ( a , b ; Y ) , Y so i s homeomorphic t o the c l o s e d u n i t i n - I. Using the same argument as above, we immediately get a simple p r o o f of the f o l l o w i n g known r e s u l t . Theorem 1.6: If Y i s a H a u s d o r f f s p a c e , then connected i f and only i f Example Y i s pathwise Y i s arcwise connected. 1.7: The f o l l o w i n g example shows pathwise does not imply arcwise connectedness c o n s i s t o f the two p o i n t s {0,1} connectedness in general. Let w i t h the topology Y T = Y i.e., Y 0, and with 1 to c o n s i d e r . f([0,l/2)) - 0 continuous. Y We have only one p a i r o f d i s t i n c t p o i n t s , {4>f{0}, }« Hence and Define f f is Y. i s pathwise connected space. Since Y = 1. to [0,1] Then has only two p o i n t s , f([l/2,l]) from i s not arcwise c o n n e c t e d . Theorem 1.8: Let Y be a pathwise c o n n e c t e d , H a u s d o r f f space. Given two d i s t i n c t p o i n t s a, b o f Y, the f o l l o w i n g con- d i t i o n s are e q u i v a l e n t : (1) If f ,•g and then (0,0) (1,1) are i n the same connected component of (2) are i n F ( a , b ; Y ) , G(f,g). Y i s homeomorphic t o and b = 1 [0,1] with (or a =1, a = 0 b =0), Proof: (2) implies ( 1 ) . T h i s i s j u s t Theorem 1.1. Conversely i f we assume every h e F(a,b;Y) meomorphic t o a - 1, ction f [0,1]. b = 0). f i s surjective. Now we c l a i m ( 1 ) , then by Lemma 1.4, By Lemma 1.5 a ~ 0 and Y i s ho- b - 1 (or I f t h i s i s not t r u e , then a continuous can be c o n s t r u c t e d such t h a t i s not s u r j e c t i v e . f in F(a,b;Y) T h i s i s a c o n t r a d i c t i o n , s i n c e we funand •8 have proved t h a t a l l such and b = 1 (or a = 1, £ are s u r j e c t i v e . Hence a = 0 b =0). Lemma 1,9: Let f, g be two homeomorphisms from with f ( 0 ) = g(0) = a , then f ( l ) - g ( l ) i <j> and I to Y If f(I)i g(I), f ( l ) = g ( l ) = b. g ( l ) - f ( I ) i 4>. Proof: Assume subset of g ( I ) . Then there e x i s t s such t h a t of f ( I ) - g ( I ) = <{>, i . e . , f ( I ) i s a proper g(t) £ f ( I ) . a point It i s clear Note we have i n (0,1) g(t) i s a cut point g ( I ) , i . e . , g ( I ) - g ( t ) = U U V, are s e p a r a t e d . t a e U, where b e V. U and V Since f ( I ) i s contained i n g(I) - g(t) and i s a connected s e t , so f(I) U, i s c o n t a i n e d i n one o f f ( l ) = g ( l ) = b e V and possible. V. and not the o t h e r . F(0) = g(0) = a e U, f ( I ) - g ( I ) 1 <t>. Hence But which i s im- Similarly, g ( I ) - f ( I ) f <}» Lemma 1,10: Let given point of Y be an arcwise Y, connected space and Suppose t h a t f o r every p be a f e F(p,p;Y), 9 the p o i n t s (0,1) ponent o f and (1,0) are i n the same connected com- G ( f , f ) = {(x,y).-j f ( x ) = £ ( y ) } . (a) Y Then lias a unique a r c (Arc i s a homeomorphic image o f the c l o s e d u n i t i n t e r v a l tween any by (b) point x and p I.) be- which we denote px. For any or x^, px C 2 x2 in Y, we have px-^ Q px"2 P X J . Proof: (a) to Y with I f we have two homeomorphisms f ( 0 ) = g(0) = p , £ ( I ) f g(I)> [0,1] f, g f ( l ) = g(l) = x then by Lemma 1.9, there e x i s t from and t^, t 2 in with f ^ g C ^ ) = (j), g " 1 f C t 2 ) = <fr. Let f f(2t) j(t) Then gC2-2t) j e F(p,p;Y). for any t in for t in [0,1/2] for t in [1/2,1] By the c h o i c e of [1/2,1], t 2 , j(t) f j(t /2) 2 I Furthermore, j i s one to one .o.n j(t) t J(t /2) I t f o l l o w s that [0,1/2],. i f t f t /2 2 2 Similarly, j(t) Clearly, j j((2-t )/2) i f t f 1 G(j,j) S1 \j S , i s d i s j o i n t from S 1 = {(x,y) | x = t /2, 2 (2-^/2 where 2 t /2 < y < 1} 2 and S 2 = {(x,y) Hence | y = (2-t )/2, 1 (0,1) and ponent of Therefore 0 < x < (2-^/2} (1,0) are not i n the same connected com- G(j,j) and t h i s i s a contradiction. f o r any two homeomorphisms f(0) = g(0) = p and f, g with f ( l ) = g ( l ) = x, we-have that f(I) = g ( D . (b) Y with Let f, g be two homeomorphisms from f(0) = g(0) = p and f ( l ) = x-^ I to g(l)= x2< Define t1 Let to - sup {t | f ( t ) e g(I)> f(t}) = x3 = g ( t 2 ) . X3 and g([0,t2]) Then f(.[0tt^]) i s an a r c from i s an a r c from p to x3. By p a r t (a) f C [ o , t ] ) ' = g ( [ o , t 2 ] ) - Jx"3. i We c l a i m that x = x-^ or x = x 2 . I f this i s not t r u e , p 11 then t-^ f 1 and t 2 f 1. L e t f g(4x) l)x + 2 g(4(t2 h(x) = f ( 4 ( 1 - t 2 ) x + 3 t 1 - 2) ^ £ ( 4 - 4x) Clearly, h is in F(p,p;Y). for x in [0,1/4]. for x in [1/4,1/2], for x in [1/2,3/4], for x in [3/4,1], By the c h o i c e o f t ^ , f ( 4 ( l - t i ) x + 3 t i - 2) i g ( I ) f o r any x in (1/2,3/4], Therefore h(y) i h(x) f o r any (x.,y) S e 1 = [0,1/2] x (1/2,3/4], Similarly, h(x) i h ( y ) f o r any S i U S2 Hence and (1,0) G(h,h) x 3 (x,y) e S 2 = [1/4,1/2) x [1/2,1] =x and G(h,h) That means (0,1) are not i n the same connected component o f and t h i s c o n t r a d i c t s 1 are d i s j o i n t . or x Theorem 1.11: 3 our assumption. = x , i.e., px C P 2 1 x 2 Therefore or p x C 2 12 Let p be a g i v e n p o i n t o f a Peano space pose t h a t f o r every (1,0) Y f e F(p,p;Y), the p o i n t s are i n the same connected component o f i s homeomorphic to Y. (0,1) Supand G£f,f). Then [0,1], Proof: In order to prove t h a t [0,1], Y i s homeomorphic to we would l i k e to i n t r o d u c e an o r d e r i n g on Y first. Define i f px C x < y By p a r t py (b) o f Lemma 1.10, for (Y,<) x, y of becomes a Y. linearly ordered s e t . Remark: ing sense: < i s a l s o a dense o r d e r i n g i n the f o l l o w - Given any a and b Y with e x i s t an element c means so there i s a homeomorphism pa C £([0,1]) t in pb", = pF, (0,1). i s c l e a r that with such t h a t of f(0) = p, Take any paC a < c < b. a < c < b. f ( l ) = b. pb", Hence that e px z, e U, a < b with. f ( s ) = c. It i . e . , there i s an element c T h i s proves our remark. we want t o prove t h a t Take any f there f ( [ 0 , t ] ) = pa f o r some To prove the theorem, d e f i n e Y. Since s e ( t , l ) and l e t pc C a < b, A(x) e Y - A(x). A(x) = {y e Y | x i s a c l o s e d s e t f o r any Note t h a t Y - A(x) ="px and there i s an open s e t x f. U, U = px n V, x where U o f px V y}, in - {x}. such i s an open s e t 13 of Y. I t i s c l e a r that x : i s not i n Peano s p a c e , i n p a r t i c u l a r is an open connected s e t K V. Y V. Since Y i s l o c a l l y c o n n e c t e d , so there containing z-^ and c o n t a i n e d i n We know that a connected open subset o f a Peano space i s arcwise connected nected. z^, [ 5 , p . 118].. Hence I f there i s an element c a n z2 K i s arcwise con- z 2 e K Pi A(x) , b e J o i n e < i by an a r c T in K. then It i s clear T u pT-^ i s a pathwise c o n n e c t e d , H a u s d o r f f s p a c e . T U pz" 1 i s arcwise connected. x 4 T \j pzT^. - S i n c e Recall i.e., px C P2*2* Hence K n A(x) z2 e A(x) , A(x) Now we c l a i m that p z . 1 x < z2, x f. pz" 2 . | x e Y}, n (A(x) | x eY} Y implies inter- that has e x a c t l y one element. x^, then we have e i t h e r x^ < x 2 . and so has the f i n i t e | x e Y} i <j>. t h i s i s not t r u e , say we have x--, t A ( x 2 ) Therefore i s closed. The compactness o f n{A(x) us say we have that A(x) = <p. T h i s proves t h a t the complement o f section property. Let "T y T h i s i s i m p o s s i b l e because i s open and so ^(A(x) pz"? C ^ Hence Now the c l a s s o f a l l A(x) If isa By d e f i n i t i o n o f x-^ t H {A(x) x2 of x-^ < x 2 A(x2), | x e Y}. This or x 2 < x-^. we know contradicts our choice o f point If q. x-^. Hence <^ (A(x) j x e Y} Now we want to show pq = y» q < r, then This c o n t r a d i c t s our choice C pq, p h i c to and Take any r c Y. q £ A(r) and that i m p l i e s q f. H{A(x) | x pr has only one r £ pq". of Hence e Y}. q. Therefore pq = Y, so r <, q , i . e . Y . i s homeomor [0,1]. Lemma 1.12: Suppose to [0,1] and with (1,0) f i s a continuous f u n c t i o n f ( 0 ) = f ( l ) = 0. from Then the p o i n t s [0,1] (0,1) are i n the same connected component o f G ( f , f ) = {(x,y) | f ( x ) = f ( y ) } . Proof: Let c be a p o i n t where f a t t a i n s i t s maximum. Define: h-^x) = f ( c x ) / f ( c ) h2(x) Then h , (0,0), G(h = for x f((c-l)x + l)/f(c) h (1,1) h 2 ) . Let belong to F. in for x [0,1], in [0,1]. By Theorem 1.1, the p o i n t s are i n the same connected component o f 15 "G = {(x.y) e G ( f , f ) | 0 < x < c , c < y < 1} The mapping: (x,y) — — * (ex, d e f i n e s a homeomorphism from the f a c t t h a t the p o i n t s connected component o f (0,1) and G^ C GCh^,}^) (0,0) Gfh-pl^), and onto G-^. From (1,1) are i n the same we i n f e r t h a t the p o i n t s (c,c) are i n the same connected component o f G(f,f). Since to the d i a g o n a l and ( c - l ) y + 1) set G(f,f) i s symmetric with {(x,x) | x e [0,1]}, r e s p e c t to we get t h a t (1,0) ( c , c ) are i n the same connected component o f Therefore (1,0) and ponent o f G(f,f). (0,1) G(f,f). are i n the same connected com- Theorem 1.15: Let p be a g i v e n p o i n t o f a Peano space Y. Then the f o l l o w i n g c o n d i t i o n s are e q u i v a l e n t : (1) Y = [0,1] and (2) For every and (1,0) nent o f Proof: f p = 0 in ( o r p = 1) ; F(p:^p;Y), the p o i n t s (0,1) are i n the same connected compo- G(f,f). (1) implies Let (2) hold. morphic t o or 0. so that If p f 0 p3r2 subset o f Hence [0,1], Theorem 1.13, By Theorem 1.11, Y We c l a i m that and p p f 1, then take or This c o n t r a d i c t s p a r t p = 1. i s homeo- i s equal to e i t h e r i s not a subset o f p x , px*2. p = 0 ( 2 ) . T h i s i s j u s t Lemma 1.12. 0 < x 2 < p < x-^ and px"-^ i s not (b) o f Lemma 1.10. This completes the p r o o f o f 17 CHAPTER II CONNECTEDNESS In t h i s chapter we s h a l l prove the f o l l o w i n g rem theo- which i s a g e n e r a l i z a t i o n o f Theorem 1.1. Theorem 2.1; If (0,0,...,0) component o f The f ^ , f 2 , . •« , f and n (1,1,...,1) belong to F, then the p o i n t s are i n the same connected G ( f , f2 ,. .. , f n ) . p r o o f o f Theorem 2.1 i s based on the f o l l o w i n g lemmas. Lemma 2.2: Suppose h e F, where h f e F and e > 0. Then there e x i s t s c o n s i s t s o f a f i n i t e number o f monotone p i e c e s , such that |f(x) - h ( x ) | < e f o r any x in [0,1], Proof: Because compact s e t [0,1], f i s a continuous f u n c t i o n d e f i n e d i t i s u n i f o r m l y continuous on on a [0,1], For every |f(x) e > 0, there e x i s t s a - £ ( y ) | < e/2 whenever 0 = a „ < a. < . . . < a = 1 0 1 n f o r each . i . [a. , , a . ] . l-l I Define Then j such t h a t |f(x) x e such that and h For any [ a . ...a.]. such that |x - y j < 6. h(ai) = f ^ ) he F of monotone p i e c e s . 6 > 0 x 1 I a. l Choose - . a . .,1 I 0 < x l - l ^ and l e t h be l i n e a r i n c o n s i s t s o f a f i n i t e number in [0,1], there e x i s t s a Therefore - h ( x ) | < |f(x) - f ( a . ) | + |f(a ) - h ( x ) | < e/2 + \hCa.) - h(x) | < e/2 + |h( a j ) - h C a ^ ^ i < e/2 + e/2 = e T h i s completes the p r o o f o f Lemma 2.2. Lemma 2.3: -Suppose 41, there e x i s t f o r every Proof: x in f, g £ of [0,1]. belong F to such that F and e > 0. |£^(x) - g £ ( x ) | Then < e By Lemma 2.2, there e x i s t each c o n s i s t i n g o f a f i n i t e h, j of F with number o f monotone p i e c e s , such that |f(x) - h ( x ) | < e/2 and |g(x) f o r any x in - j ( x ) | < e/2 [0,1]. By Theorem 1.2, there e x i s t h^(x) = j S ( x ) £ of f o r every x F such in that [0,1]. Because |4'Xx) - h\|i(x)| < e/2 |g£(x)- j£(x)| and < e/2, therefore" |£^(x) - g £ ( x ) | < mix) - h^(x)| + jh^(x) - gg(x)| < e/2 + IjSCx) - g£(x) | < e/2 + e/2 = e Lemma 2.4: Let exist f , , f„,..., f 1 2 n ij; , <j> ,..., ^ 2 e F e such F and e > 0. Then there that | f . ¥ . (x) - £ . * . (x) | < e f o r any x in [0,1] and i , j = 1,2,..., n. Proof: We s h a l l prove i t by i n d u c t i o n . I t i s true f o r n = 2 (Lemma 2.3). Assume i t i s true f o r n = k, i . e . , there e x i s t £^, f ^ , . . . , £j of c F such t h a t I f . t y x ) - f.£.(x)| < E/2 f o r any x [0,1] in Since £\%>\* h such t h a t e 2 F ^k+1 ^* e i , j = 1,2,..., k. and t n e n by Lemma 2.3, there e x i s t h |£ € h (x) - f 1 Let KjJ^i ~ 1 1 r1 £ ° = k + 1 h ( x ) | < e/2. 1» 2 » « * « » 2 k « a n d n 2 Then (1) l j*j.W " fi¥i0O| < e/ £ In p a r t i c u l a r , lf ^ (x) 1 Since 1 - f.*.W\ < e/2 2 = ^k+1 21 IV l ( x e "k*l^k+lOO I < / 2 , ) f we have (2) [fjI'jCx) - f From (1) k + i^ k + i(x)| < e and ( 2 ) , j f ^ C x ) - £ ^ j ( x ) | < e. f o r any x e [0,1] and i , j = 1, 2,..., k+1. Hence the p r o o f o f Lemma 2.4 i s completed. Lemma 2.5: Let X compact subset component o f V of X be a compact Hausdorff of G, X. If a and .b then there e x i s t such t h a t G C V u U space and G be a are n o t i n the same d i s j o i n t open s e t s and a e U, U, b e V. Proof: Since G i s a compact H a u s d o r f f s p a c e , ponents and quasi-components are i d e n t i c a l . are not i n the same connected component o f a and b That G the coma and b implies that are not i n the same quasi-component o f G. By d e f i n i t i o n o f quasi-component, there e x i s t d i s j o i n t c l o s e d s e t s H, K of G with G = HuK, a e H and b e K. Since G is 22 c l o s e d i n X, Hausdorff sets U so and H K are c l o s e d i n X. space i s normal. Therefore and c l e a r that V of X G CU(jV such t h a t and there e x i s t d i s j o i n t open H C U a e U, Every compact and b e V. K C V. It i s T h i s completes the p r o o f o f Lemma 2.5. Now we have e s t a b l i s h e d enough lemmas to prove • Theorem 2.1. Proof o f Theorem 2.1: F i r s t note that subset G ( f ^ , f 2 , . . . ,f ) o f the compact Hausdorff Assume ( 0 , 0, space 0) and not i n the same connected component o f i s a compact In. ( 1 , 1, . . . . 1) are G = G(f,f 2 Then by Lemma 2.5, there e x i s t d i s j o i n t open s e t s In with (1, 1, ...,1) e V. G C U U V ( 0 , 0, i , j « 1, 2, t •. •,n, ~~* ^ l k ( t ) » such t h a t and the mapping ^2k(t)» U, V, o f 0) e U, For every p o s i t i v e i n t e g e r ^2k» ••• » ^nk e P exist for and ,... ,f ) . ••*» * n k ( t ) ) k, there 23 d e f i n e s a path j o i n i n g ( 0 , 0, . .., 0) and ( 1 , 1, ...„ 1), Let Ak Since (1, Ak 1), *nk(t))| by our choice of CUUV) f <j). Let where . i s connected and contains 1, Ak - yk = { ( * l k ( t ) , ^(t), x^ k £ o r = ^ik^) = C-^ik* x21-' (xlk, x2k, s o m e •••» ]<) x n ( 0 , 0, U, V, t« The y = (x-p x 2 , xn), }. 0) we ..., xnk) and have e A k sequence (UUV), (yk) where contains a convergent subsequence. So without l o s s of g e n e r a l i t y , we may to t e [0,1] Given assume e > 0, we {y k } converges can choose k l a r g e enough so that < If.Cx.) - £ . ( x ) | • |£.(x. ) - f . ( x . ) | • |£.(x ) - f j C o < e/3 + l£i*ik < e/3 + 1/k i k + k ( t ) " V j k ( t ) k ' + e / jk 3 e/3 < e. T h i s means that IfiCxi) - f j C x ^ l for (1) every i,j- 1, 2, . ,.,'n. y = (xx, x2, xn) = o Hence e G CU u V. 24 On the other hand, Since U and V yk e A - (U U k are open, y e In - (2) From (1) V) C i I n — (U U (U U V) n - (U U V) . i s c l o s e d . Hence V) . ( 2 ) , we get a c o n t r a d i c t i o n . T h i s means (0, 0, 0) component o f and ( 1 , 1, 1) G ( f , f2 ,... , f n ) . are i n the same connected T h i s completes the p r o o f of Theorem 2.1. Corollary 2.6: Suppose d e f i n e d on [0,1], every n 11 i s a r e a l - v a l u e d continuous f u n c t i o n f ( x ) 2. 0 f o r every . . . = f ( a n ^= fCa^) = f ( a 2 ) Then the f 0 p o i n t s o f the form b^ = a., f o r some j (bs f o r x in [0,1] 0 = a-^ < a (b^, b 2 , may equal are i n the same connected component o f 2 and <••••< a n b ), =1. where bj_ i f s f k ) , G(f,f,...,f). Proof: If Assume t h a t f = 0, f ( x ) a t t a i n s i t s maximum at p o i n t c f a-p a 2 , k, define then the r e s u l t i s t r u e . L e t an. g^(x) = Let £ ( c ) = M. ~Dk) x+ c. f f Note For every p o s i t i v e b^J/M, where 0. integer 25 ( b ^ , b^, . . . f b R ) g^'s satisfy the c o n d i t i o n s o f Theorem 2.1. Hence (0, 0, . . . , 0) component o f and ( 1 . 1, G ( g ^ p g^» • {(x,,x 7 ,...,x The mapping i s a p o i n t o f the r e q u i r e d form. The (x , x2, l f ( 1 , 1, and 1) -*• ( c , c , G' C G ( f , f , . . . , f) . where every x. i s between b. and c ) . from (c - b n ) x n G (g^ ,g 2 ,.. . ,g ) (b^.b 2 , b ) c ) , we have onto + b p G*. and ( b ^ b2, b ) R i n the same connected component o f T h i s i s t r u e f o r any b^ = a^ be the s e t : (c - b 2 ) x 2 + b 2 0) ( c , c , . . . , c) G' . . . , x ) •*• d e f i n e s a homeomorphism ( 0 , 0, are i n the same connected g n ) . Let ) e G(f,f,...,f)| ((c - b 1 ) x 1 + b Because 1) f o r some (b^, b 2 , j . Hence the nn t h i s form are i n the same connected component o f bn), points of G(f,f,...,f). We can use the same method as i n C o r o l l a r y 2.6 use Theorem 1.2 t o prove the f o l l o w i n g : and 26 C o r o l l a r y 2.7: Suppose [0,1], f(x) > 0 £ i s a continuous f u n c t i o n d e f i n e d on f o r every x in [0,1], f c o n s i s t s of a f i n i t e number o f monotone p i e c e s and fCa^ for form = f(a2)= < &2 < . . . < a n = 1, 0 = V°\t'^ 2r > *•• » b n ) > = f(an) = 0 Then the where every nn b^ = aj points o f the f o r some j b. i f s f k ) , are i n the same pathwise k _ connected component o f G(f,f,...,f). ( b_ may equal 27 CHAPTER I I I COMPATIBILITIES OF FUNCTIONS From Theorem 1.1, we know t h a t i f we have two continuous f u n c t i o n s f ( 0 ) = g(0) = a = 0 and (1,1) of and G(f,g), and from [0,1] to [0,1] w i t h f ( l ) = g ( l ) = b = 1, then are i n the same connected component o f I f we do not assume (0,0) f , g, a = 0 or b = 1, (0,0) G(f,g). then the p o i n t s (1,1) need not be i n the same -connected component as the f o l l o w i n g example shows. f2x for x e [0,-j] for x e for x e [0,1] Let f(x) = , 3 •2 x f4x But i n t h i s case we are s t i l l s u f f i c i e n t c o n d i t i o n that same connected component o f able [£,l] to give (0,0) .and G(f,g). a n e c e s s a r y and (1,1) w i l l l i e i n the 28 Definition 3.1: We a-interval say that a c l o s e d i n t e r v a l of f [p>q] i s an if f [f(p),f(q)] i f f(p) < f(q) I [f(q),f(p)3 i f f(q) < f ( p ) . f [p,q] Definition 3.2: is |[s,t] if condition Condition (A) (A): c = y c a l l e d r i g h t compatible ivith or (B) < y-^ < . . . < y m Condition = d such < x^ < . . . < x m = t, that [x^,x^+^] i s an a-interval [y^,y+ i s an a-interval f. 3. Every c l o s e d i n t e r v a l of s = xo i = 0, 1, ... , m. 2. Every c l o s e d i n t e r v a l of |[c,d] holds. there e x i s t p o i n t s 1. f C x ^ = gCyj) g g. (B): there e x i s t s t r i c t l y monotone i n c r e a s i n g sequences {x^}, (y^) such that 29 1. X q = s, 2. I f yQ lim x = x i f([x ,tD ro 3. f C x ^ •= = c OT and x^ and lira y n g(x ) i for all = y^, for a l l i . then i . [x^,x^ ^] i s an [y^ »y^ + ^] i a-interval f. s a n a-interval g. of 3.3 f.r , |[s,t] (A) < d + 5. Every c l o s e d i n t e r v a l Definition y^ = g([yw,cl]) = f ( t ) . 4. Every c l o s e d i n t e r v a l of < t, i s compatible with g i f condition l r |[c,d] holds. Definition 3.4; f j j- g t - j condition (A') Condition (A'): xQ i s l e f t compatible w i t h or (B 1 ) c = y_n 1. f ( x _ i ) = gCy.j) < y.n+1 for 2. Every c l o s e d i n t e r v a l of 3. if s = x_ n < x_n+i < ... < y0 = d < ••• such t h a t i « 0, 1, ... , n. [x .,x . ] -l -i+lJ i s an a-interval f. Every c l o s e d i n t e r v a l of d] holds: there e x i s t p o i n t s = t, S|[c g. [y ^,y 1 S an a-interval 30 Condition (B'): there e x i s t s t r i c t l y monotone sequences {x_^} following and {y_i> decreasing s a t i s f y i n g the relations: 1. x = t , y = d and x . > s , y . > c f o r a l l i . o ' o -1 •-1 2. I f l i m x_^ = x_ro and l i m y.^ = y-co, then fCfs.x.J) 3. f(x ) - g([c,y_J) = gCy.i) mi for a l l i . x 4. Every c l o s e d i n t e r v a l of Theorem x s a n [ -i» -i+il * a-interval [y-_i-,y_-j_ + ^]) i s an a - i n t e r v a l f. 5. Every c l o s e d i n t e r v a l of =f(s). g. 3.5: Suppose that the p o i n t s (0,0) component o f compatible and G(f,g) with f , g, belong (1,1) to F(0,b;[0,»)). Then are i n the same connected i f and only i f f|[o l ] i s r i gnt gj[o l ] ' Proof: F i r s t , l e t us suppose t h a t compatible with f|[0,l] *s r ig n t g|[Q i ] * Suppose c o n d i t i o n (A) h o l d s . Then by Theorem 1.1, 31 ( iiYO x ( x i + l » y i + l) of G i - G C f , , , , ) . where Since (0,0) = ( x 0 , y 0 ) component o f G ± - 1, [ x . and (x1,y1) G^, hence same connected component o f G a r e (0,0), Q 8 - 8 , ^ . ^ ] are i n the same connected (x2»y2^ (xj^-^), connected component o f the t and and Q are in the same connected component U tIie s a m e C x 2»^2^ a r e * n * n G-^. Using the same and (1,1) = ( x m , y ) are m-1 i n the same connected component o f u G.. Because G.CG(f,g) 1 i =o 1 m 1 f o r each i , we have u G i C G ( f , g ) . Hence (0,0) and i=o argument, we can show that (1,1) are i n the same connected component o f Suppose c o n d i t i o n and (xoo,yoo) G(f,g) and (0,0) (1,1) because Gi = G ( f . , g . ) , 1 a r e 1 1 1 t n e 3 ^xj+l'^j+l^ a r e * For each f. = f , 1 Uxi»xi+lJ | n s a m e i , (xi,yi) same connected component o f and f g. = g, |[yi,yi+1] 1 ?t G - , then we have t h a t i=o i H. = that are i n the same connected component o f where 1 I f we d e f i n e h o l d s . Then i t i s c l e a r f([x00,l]) = g([y0o,l]). (xi+i»yi+i) 1 (B) G(f,g). (0,0) connected component o f and H... Let 00 H = .U^ H_. , f o r each ( co.y«>) x then H has a component i . By d e f i n i t i o n o f x^, belongs t o the c l o s u r e o f topology of R 2 . Since C C yw, C containing (x^,y^) we have t h a t relative i s contained i n t o the u s u a l G(f,g) and 32 G(f,g) i s a closed set of subset o f G(f,g). R, Therefore the c l o s u r e o f (0,0) connected s e t c o n t a i n i n g both have proved that' (x^yoo) connected component o f (1»1) and G(f,g). g(l)» •f" g(oJ l $'• = (x t t is a is a ,yj. We are i n the same (0,0) Hence the p o i n t s (0,0) (1,1) and G(f,g). (0,0) Then (A) or (1,1) and G(f,g). (B) holds. G(f,g). are i n the I t i s c l e a r that f o r otherwise there e x i s t s connected component o f condition C C are i n the same connected component o f same connected component o f = and (1,1) and Now assume that £(I) the c l o s u r e o f 2 q such t h a t are not i n the same Now we want t o show that Define X Q YQ =0, ~ °» and x .y 1 = sup (x c [0,1]| f ( x ) a t t a i n s i t s absolute maximum} = sup (y e [0,1] | g (y) a t t a i n s i t s absolute maximum}. Now f set [0,1], f(I) ' = and g so are continuous f u n c t i o n s x^ and g ( I ) , we have y^ d e f i n e d on a compact are w e l l - d e f i n e d . f(x ) = g(y ). 1 1 Since Define sup {x e [ x ^ , l ] ( sup {x e [ y ^ , l ] | g ( y ) a t t a i n s i t s minimum i n It i s c l e a r that y^^ < y2 f ( x ) a t t a i n s i t s minimum i n and x^ < x^. We c l a i m that [x^,l]} [y^l]}. 33 f(*2) = S(y2). Suppose f(x ) > g(y2). for x t x^ and fCx-^) f g(y) set G(f,g) i s disjoint This 85 from U cannot assume that f( Suppose that x 2 )< g ( y 2 ) » x. and J Assume x_. $ 1 and j L 2 , where y = y2>. c o n t r a d i c t s our assumption that of ( 0 , 0 ) and ( 1 , 1 ) are G ( f , g ) . S i m i l a r l y , we Hence y Therefore the y > y1> p -C(x.y) | x * x ^ i n the same connected component f ( x ) f g(y ) f o r y > y^. L-L =: {(x,y) | x = x L2 Then x f( 2 have been ^= S(y2)* constructed. 3 i s even, i . e . , £ ( x ^ ) < f ( l ) = b. Define: x j l sup {x e [ x . , l ] | f ( x ) a t t a i n s i t s maximum i n [ X j , l ] } = + y i +l ~ SU P ty e We c l a i m t h a t Then for L' [y •, 1]I f(Xj + 1 f(x) f g ( y j + 1 ) y > y.. g(y) a t t a i n s i t s maximum i n [ y . , l ] > ) = g(y\.+ 1 ) . for x in Assume f ( X j ) < g ( y j+ 1 ) [x^,l] and T h e r e f o r e the s e t G(f,g) +1 i s disjoint U L * , where L» = {(x,y) I y = y f ( x j ) f g(y) , x £ x^} from 34 L* = ((x,y) | x = X J , This in y > yj>. c o n t r a d i c t s our assumption that the same connected component o f cannot assume that f(x.. + 1 ) x.. i 1 Suppose that (0,0) G(f,g). > g(y\j ) . j (1,1) are S i m i l a r l y , we Hence + 1 and and f(Xj i s odd, i . e . , + 1 ) = g(yj + f(xj) > f ( l ) = b Define: sup {x e [x.,1]| f ( x ) a t t a i n s i t s minimum i n [x.,1]} 3 sup 3 {y e [ y j , l ] | g(y) a t t a i n s i t s minimum i n [ y j , l ] } . f ( 2 ^= Using the same method as when we proved t h a t we can prove t h a t i n d u c t i o n step If and of f(xm) ym, Because = f ( X j -^) = g ( y j+ ] L ). f o r some f ( l )= we get f(x » f o r any {x^} and i s odd cf and ) - g(xOT) = m, (x^}, x^ = 1, we have y^ = 1. sequences, we have i D If construction T h i s completes the + to c o n s t r u c t f(xm) = S(ym) By d e f i n i t i o n then y^ f 1 (A). f o r any i . are bounded monotone i n c r e a s i n g and f(x^) < b then from In t h i s c a s e , we get c o n d i t i o n l i m x^ = x^ {x^} (y^). g ( y m ) = b. i , x^ ^ 1, {y^} g(y2)» x and {y^.}, i f i l i m y^ = y^,. we have is even. b . In order to get c o n d i t i o n By our f(x^) > b i f Hence f(x ) = ( B ) , the only 35 thing Xo, ^ 1 f(x) is and < b. an f, fCtx^,,!]) t o prove i s £(x) less b than of whenever c a n n o t be g r e a t e r Hence f([-x»,l]) = b . completes t h e proof It Theorem o n o t o n x [x ,l]; x„ .. x than and f(x) 2j ) c a n n o t be F o r t h e same whenever x is in g([yoo,l]) = b. Similarly, there and f ( c ) = f(x [x^,!]. b b, By c o n t i n u i t y o f T h a t means is in say M x«> < c < x < c £3 Assume t h a t increasing to e such t h a t The f a c t s our choice m in 2 c contradict s x b• = f(x) < f ( x j ) < b. i s a point f ( c ) = f ( x ). <*3 f(x) * such t h a t 2 there f o r some ^( x2i J Since x j j b = g([yco»l]) reason, [Xoo,l], This 3.5. o f Theorem i s c l e a r t h a t the f o l l o w i n g theorems a r e t r u e . 3.6: f, g e F ( a , b ) , Let maximum v a l u e of where f ( x ) and f ( 0 ) = g(0) = a g ( x ) . Then are i n t h e same c o n n e c t e d component o f if f|[o l ] Theorem 3.7: Let * s r ig n t compatible f , g e F(a,b), maximum (minimum) v a l u e of with where £(x) (0,0) G(f,g) and i sthe (1,1) i f and o n l y S J [0 1 ] * f(l) = g(l)= b and g ( x ) . Then i s the (0,0) and 36 (1,1) a r e i n t h e same c o n n e c t e d c o m p o n e n t o f only i f £ | [ 0 1] ^ ^^ s e compatible t with assuming anything about a or i f and g | [ o ]_] • Now we w a n t t o d i s c u s s more g e n e r a l without G(f,g) case, i . e . , b. D e f i n i t i o n 5,8: Define: l a a a a 2 3 4 b l b 2 b 3 b 4 = i n f {x e [ 0 . 1 ] | f(x) attains i t s maximum} = s u p {x e [ o , i ] | f(x) attains i t s maximum} e [0,1] | f ( x ) attains i t s minimum} = s u p {x e [0,1] | f ( x ) attains i t s minimum} = i n f{x e [ 0 , 1 ] | g ( x ) attains i t s maximum} = i n f{x g(x) attains i t s maximum} = i n f{x e g(x) attains i t s minimum} g(x) attains i t s minimum}. = sup (x e If there well I [0,1] I [0,1] 1 = s u p {x e [0,1] f and g are not constant a r e s i x p o s s i b l e ways t o o r d e r as b j , b , 1. a^ £ a 2 < a 3. aj < a 3 <L a^ < a 2 < 2 5. • b , 2 < 3 3 a^, f u n c t i o n s , then a , a^, 2 b ^ £ a^ < a 2. a^ < a ^ < a 2 < a^ 6. a 2 < a^« 3 a-^ £ a a^ as 37 Lemma 3.9: Suppose that f , g e F(a,b) and are i n the same connected component o f o r d e r i n g s o f the a^'s and b^'s (0,0), G(f,g). (1,1) Then the have the same i n i t i a l and terminal s u b s c r i p t s . Remark: As b e f o r e , that connected component o f Hence (0,0) and G(f,g) (1,1) are i n the same implies that f ( a x ) = £ ( a 2 ) = g O ^ ) = g(b2) f(I) = g(I). and £ ( a 3 ) = f ( a 4 ) = g(b3) = g ( b 4 ) . Proof o f Lemma 3.9: Assume f o r the g i v e n We want to show t h a t f o r g b^ < b 3 < b 2 < b^. we have of b^ b 3 < b-^ or By d e f i n i t i o n o f ¥ g(y) b3 a^ £ a 2 < a 3 ^ a ^ . we have b^ ^ b 2 < b 3 $ b 4 g or cannot be a c o n s t a n t f u n c t i o n , b-^ < b^. Assume and the f a c t that fCa^ have: Since f , we have b 3 < b^. f(a^) = g(b^), By d e f i n i t i o n we have: f o r any y e [ 0 ^ ) . and the f a c t t h a t g(b3) = £ ( a 3 ) , we 38 T This implies g( 3) f° b that (0,0) and c o n n e c t e d component o f r a n y e x [0,a ). 3 ( 1 , 1 ) a r e n o t i n t h e same G(f,g). Hence < b^. { ( x , y ) | x = a , y e ( b , 1 ] } u { (x ,y) | x e ( a , l ] , 4 is 4 G(f,g). t h e n t h e above s e t s e p a r a t e s impossible. a l - 2 a < a Hence 3 ^ 4» a b^ < b ^ < b 2 w b 2 n a e < b^. check the other < b^. v y = b > 2 always d i s j o i n t from (0,0) and Using < b 2 < b 3 t h e same k i n d f i v e cases. This 2 b ^ < b, , 2 (1,1). T h e r e f o r e i f we b^ < b e I f we assume 4 The s e t This i s assume or o f a r g u m e n t , we c a n completes the p r o o f o f Lemma 3.9. Theorem 3.10: Suppose t h a t (1,1) i fthere XQ , exist compatible with g ) r I 1 i t h Then a r e i n t h e same c o n n e c t e d component o f only w f , g e F(a,b). y i n v » o y such that Q and f, ,, I o» r J Lx (0,0) and G(f,g) i f ie ^s f't i s right compatible J g 0 Proof: (0,0) and c o n n e c t e d component o f If we h a v e ( 1 , 1 ) a r e i n t h e same G(f,g). (1) a i ' —2 a < a 3 - 4» a j f|[o x ] , |[y ,i]' Assume t h a t a n c 39 or (2) a-^ < a 3 < a 2 < a 4 , or (3) a-j, < a 3 < a 4 < a 2 , then we choose: x or (5) o = a, 1 I f we have (4) and y = b 1 o a 3 <_. a 4 < a-^ < a 2 , a 3 < a^ < a 4 < a 2 , or (6) a 3 < a-^ < a 2 < a 4 , then we choose: x o = a7 ••> and y o = bT . 5 Now we s h a l l d i s c u s s case ( 1 ) . The other five cases can b.e t r e a t e d s i m i l a r l y . By d e f i n i t i o n o f ffa^ = g O ^ ) i f(x) f o r any y in a^ f o r any and x b^, in we have [ 0 ^ ) . Suppose t h a t (a^bp are n o t i n the same connected component o f is clear that (0,0) and connected component o f (0,0) and [0,3^ and (0,0) G(f,g). Then i t (1,1) are not i n the same G(f,g). Therefore (a-^b^) are i n the same connected component o f the above g(y) 1 f ( o b s e r v a t i o n , we know that and G(f,g). (0,0) and By (a^,b^) are i n the same connected component o f : G* {(x,y) e G ( f , g ) | 0 By Theorem 3.7, f. r „ ., |£°» ll < a1( O i y i b ^ i s l e f t compatible with a S i m i l a r l y , we can show t h a t f ( r I Ia<, iJ i s right g, lEO.bj] compatible 40 with -g | [b ^ 1]* have b-^ < b 4 by Lemma 3.9. Because are Since we assume a-intervals of r i g h t compatible with f and g with S|[y X q , 2|[0,y ] G(f,g) Hence of anc * £ yQ, | [ x ,1] component o f f ^ i y Then we can show t h a t and and £ we [b-pb 4 ] *s | [ a 1] 8|[b^,l]« are i n the same connected (x0,y0) [a^,a 4 ] respectively, Now suppose f o r some compatible with a-^ < a 2 < a 3 <_ a 4 , *s | [o x ] s right (0,0) G(f,g) (1,1) are i n the same connected l e £ t compatible and (x0»yD) and t h a t component o f by using the same method as we used i n Theorem 3.5. (0,0) and (1,1) are i n the same connected component G(f,g). D e f i n i t i o n 3.11: For (1,1) f , g e F(a,b), define are i n the same connected f - g component o f i f (0,0) and G(f,g). Theorem 3.12: = i s an e q u i v a l e n c e relation. Proof: f - f because the s e t { ( x , x ) | x e [0,1]} belongs 41 to G(f,g) and connects If f - g, same connected g (0,0), then G(g,f) (1,1) and (0,0) component of homeomorphic to and (0,0) (1,1). and G(£,g). (1,1) Also G(f,g) under the mapping are mapped to (0,0), are i n the is (x,y) -> ( y , x ) , (1,1). Hence - f« Now we From Theorem 3.5 a unique prove and f | [ x ,1] then gj[y» i i] Theorem 3.10, r know t h a t each i z o a n c * 5 f = h. r n u right s f, ^ x 0 ,l] Therefore f, J and Z o compatible w i t h g, h are unique y^ = y[ f|[o x ] ^ s r - h|[ i2 n t * and s l e r > z iy Because as chosen i n f ( x - ) = g(y^) = h ( z t compatible compatible w i t h h|[ with z f i ] . i s an e q u i v a l e n c e r e l a t i o n i n 3.13: Let h.j j-Q compatible w i t h s F(a,b). Definition f - g. x we must have |[ has f for £ f n i . Therefore k|[0» ] Hence we relation. , i s l e f t compatible w i t h gi n I L » oJ I l »yoJ r i g h t compatible w i t h g|[y i ] * And i f f i S|[0 y'] o the sequences for a l l is a transitive sequence a s s o c i a t e d w i t h i t . Assume t h a t u g = h, - and Theorem 3.10, By Theorem 3.10, and that g e F(p,p). Then d e f i n e : f £(2x) 1) for x for x Lemma 3.14: If £ 0 - f i and g0 = g1# then f0°g 0 - £i°gi. Proof: = f 0 ( x ) = f i ( y ) = fi°gi(2")' GCf og ,f og ). 0 0 1 ^1T»^§^ (i,—) 2 2 e G ( £ and ( ) For every 1 o - f we have t h a t lt o g o'£l o g 1 Hence connected component o f s f l ) (x,y) e G ( g , g ) , o s i n c e * we have 1 SD ~ g l » w e h a v e t h a t ( 1 , 1 ) are i n the same connected component o f G(f og0,f og2). o°go Q (—,—) are i n the same connected component o f 2 2 (0,0), f f Since ( 0 , 0 ) and ( 1 , 1 ) are i n the same G(f og ,f^og^), i.e., Q 0 l°gl- Lemma 3 . 1 5 : The e q u i v a l e n c e c l a s s e s semi-group. of F(p,p) form a 43 Remark 3.16: Let (1,1) f, g e F(a,b). Define f «v, g i f (0,0) and are i n the same path-connected component o f Then the f o l l o w i n g example shows that equivalence Example 3.17: h, [0,1]. f, j e F g e F. f, I t i s c l e a r that i f and such that i ^ g f 4' g, Then such'that There e x i s t f ^ i i s not an relation. Let exist ^ G(f,g). i f and only i f there f h ( x ) = gj (x) g e F i(x) = x f o r any f ^ g such t h a t for a l l x £ | g (cf.[l]). f o r a l l x e [0,1], f, g. in I f we choose then f, g then we cannot have the t r a n s i t i v e relation. We know from f, g and G(f,g) (1,1) f , g e F(a,b), and i f and only i f (0,0) and G(f,g). (1,1) are i n the So we immediately have f o l l o w i n g theorem as a c o r o l l a r y o f Theorem Theorem . are i n the same path-connected component same connected component o f the that i f c o n s i s t o f a f i n i t e number of monotone p i e c e s , then (0,0) of [1] 3.10. 3.18: Suppose t h a t f , g e F(a,b), and f, g consist 44 o£ a finite (1,1) if number o f monotone p i e c e s . Then (0,0) l i e i n the same path-connected component o f *s and only i f f|[o 1] c o m P a t l D le with g| j- Q and G(f,g). . Theorem 3.19: Suppose that finite f e F(p,p) and number o f monotone p i e c e s . Then f (0,1) are i n the same path-connected component o f only i f there < < xc < x i < such t h a t every i n t e r v a l and and (1,0) G(f,f) i f and exist 0 = x_n < x _ n + i f consists of a f(xi)= f(x,i) [x^,x^+^] < xn = 1 i s an a - i n t e r v a l of for a l l i . Proof: F i r s t we assume t h a t the c o n d i t i o n s h o l d . to prove t h a t G(f,f). (0,1) Define Let (1,0) f x ( x ) = f(xQx) By Theorem 3.18, G(flf.f2). and G' (0,0) and We want are pathwise connected i n and (1,1) f2(x)= f ((xQ-l)x + l ) . are pathwise connected i n be the s e t : {(x,y) e G ( f , f ) | x e [ 0 , x Q ] , y e [x0,l]}. 45 The mapping (x,y) -> ( x Q x , ( x Q - l ) y homeomo-rphism (x0,xQ) from onto (xQ,x0) Therefore defines a G'. are pathwise connected i n i s symmetric with and G(f-L,f 2 ) + 1) Hence G.' C G ( f , f ) . r e s p e c t to the d i a g o n a l . and (0,1) Now Hence are a l s o pathwise connected i n (1,0) (0,1) and (1,0) G(f,f). are pathwise connected i n U s u a l l y we have s i x p o s s i b l e ways to order &3 and , but i f we assume pathwise connected i n a l < a 3 — 4 a K a o r 2 a 3 G(f,f), < a By d e f i n i t i o n for a x e (a2,l] - a2 l < a 3» of a 2 (0,1) i*- i s < a a2 and the o t h e r that a^ .< a 2 < a 3 < a^. three (0,1) X f ( x ) i f(aj^) and If (1,0) are G(f,f). Similarly (0,1) and p a t h - c o n n e c t e d component o f (a ,a ) a3, Hence we we cannot have cases. Now suppose < aj i (1,0) are for x £ [0,a3). not i n the same connected component o f cannot have a^, 4* alt dear and G(f,f). then we c l a i m t h a t e i t h e r f ( x ) 5* f ( a 3 ) and t h e n l G(f,f) < a2, (1,0) G(f,f). are i n the same I f we have then we want to show that (0,1) and are i n the same path-connected component of G(f,f). *» We have f(x) ? f(a^) x < a^. Hence the s e t {(x,y)| x = a Therefore l P for x > a2 y > a2> the p o i n t s and {(x,y)| y = a 2 , are d i s j o i n t (0,1) and f ( a 2 ) i f(x) for x < a-^} from (a-^,a2) and the s e t G(f,f). are i n the same 46 path-connected component o f t h e s e t : - { ( x , y ) e G ( f , f ) | 0 £ x <. a-^ a < y 2 ±1}. Define: ' f lOO = f ( ) x fCajX) f((a -l)x = 2 £ 2|[0,1]* a and £ 2 = x < x. n l < x f(x^) = f(x_^). < a-£ Theroem a S X [0,1]. compatible with i s < ••. < x [x ,x^ i a-j_ < a < a . 2 3 + 1 ] <. a 4 can prove Q a 2 = 1 n is let x This completes 4 an = a , 3 a-interval of f the theorem i s < a . 2 the second case i n which t h e p r o o f o f T h e o r e m 3-. 1 9 . 3.20: Let to x e < .... < x . i - I f we S i m i l a r l y , we 3 1] n + 1 2 each i n t e r v a l p r o v e d f o r the case a l|[o for Hence we h a v e 0 = x. such t h a t +1) 2 T h e n by Theorem 3.18, x e [0,1] for »"{x f, e R| 2 g be c o n t i n u o u s f u n c t i o n s |x| = 1} with from [0,1] f ( 0 ) = g(0) = f ( l ) =g ( l ) . 47 If (0,0) and of" G(f,g), (1,1) are i n the same connected component then f i s homotopic to For any f and and g from g. Proof: exist f f = ef and g g = e g , where = fi-(y) ,. i.e.. , where n function H(0,0) = 0 and g G' (x,y) e G'. S-^, such that there f(0) = g(0), e ( t ) = (cos t , s i n t ) . Let If (x,y) e G ( f , g ) , ef(x) = eg(y). Hence H from the connected component (0,0), f, R i s an i n t e g e r . T h e r e f o r e containing f o r any [ 0 , 1 ] to [ 0 , 1 ] to H(x,y) = { f ( x ) - g(y)}/2TT. f(x) from then f ( x ) - g(y) = 2nir, i s a continuous G' of G(f,g) (1,1) to the s e t o f i n t e g e r s . Since i s c o n n e c t e d , we must have Hence H(x,y) = 0 f ( l ) - g ( l ) = 0. T h e r e f o r e have the same d e g r e e , i . e . , f i s homotopic to g. Theorem 3.21: Let f , g e F(a,b;Y), and l e t be i n the same path-connected component o f f i s homotopic t o (0,0) and ( 1 , 1 ) G(f,g). Then g. Proof: Since (0,0) and (1,1) are pathwise connected i n 48 G(f,g}, i|>:[0.1] + we have and = H with g«J> (x). Define 2 i s a continuous H(x,0) = f ( x ) fi/i^Cx) = g ' [ ( x ) j f o r any 2 H(x,t) = f ( t ^ ( x ) and [0,1] from E ( x , l ) = g(^ C ))« x 2 f i s homotopic to follows that f x [0,1] Hence [0,1] x [0,1] H(x,l) = f i ^ C x ) . 2 g Then to E Y g. to But we have g^ . Define 2 i s a continuous with E ( x , 0 ) = g(x) i s homotopic to i s homotopic to = 1 + (l-t)x) . 1 E ( x , t ) = g(ti|J (x) + ( l - t ) x ) . function = ^(1) x, i . e . , f u n c t i o n from hence J such t h a t x 2 Then Y 2 ( ^ ( x ) ,T|> (X)) e G ( f , g ) f^OO x [0,1] vp COD = 0 = ^ ( 0 ) , ^(x) = ( * ( x ) ^ C x ) } , 1 [0,1] gi ) r 2 . It and 49 CHAPTER IV SOME PROPERTIES OF MOUNTAIN CLIMBING AND APPLICATIONS I f we have four continuous f u n c t i o n s and j from [0,1] i n t o [0,1] with f , g, f h = gj and these f u n c t i o n s a l s o s a t i s f y some other c o n d i t i o n s , then we d i s c u s s more f u l l y the r e l a t i o n s h i p among j i n this h f , g, h shall and chapter. Lemma 4.1: Suppose that on f i s a continuous f u n c t i o n d e f i n e d [0,1] t h a t c o n t a i n s no constant p i e c e and t h a t exists with any a continuous f u n c t i o n h(0) = 0 x in and h from [0,1] to [0,1] h ( l ) = 1 such that [0,1], Then h(x) = x there i s an there f ( x ) = fh(x) f o r f o r any x in [0,1]. Proof: If x„ o such that x_ < h f x ) , o o then we w i l l c o n s t r u c t a sequence i n d u c t i v e l y . Using the i n e q u a l i t y 0 - h(0) < x Q < h.(xQ) , that there Using t n e that there i s an x^ inequality i s an x2 we know from the c o n t i n u i t y o f in (0,x o ) such t h a t 0 =» h(0) < x^ < X in (0,x^) q h(x^) = X = h(x^), such that h Q . we know h ( x 2 ) = x-^. 50 Assume that h(xn) x = a n < - i n 0 = h(0) < xn i xn xn+j has been c o n s t r u c t e d , and < x n _^. < x n _^ there i s an xn Using = h(xn), in (0,x n ) the inequality and by c o n t i n u i t y o f with h, h(xn+^) = xn. This completes the i n d u c t i o n step to c o n s t r u c t our sequence. Hence there i s a d e c r e a s i n g sequence h(xn) by = xn_-^ f o r any assumption we n. have Since we f ( x ) = fh ( x ) .. = f ( x 0 ) . Since of f, f(xn) n n {xn} we have = f(x0) f(xn) n hence » 'f(x _ ) 1 h(xi) = xQ f h x ( n f( = -l^ x 0 f converges to n, so we x. f(x). By C and ) = x -2^ n f = ' sequence, continuity Since have: £(x ) = f(x) = £(x ) n for any in (Xj,xD) n. Since such f c has no constant p i e c e there i s a t^ that (2) f(x0) i f C t ^ . Using the i n e q u a l i t y a B converges to f o r any (1) have property i s a bounded d e c r e a s i n g {xn} a l i m i t . Say with the fh(x^) = f (x^) , In g e n e r a l , i t has {x n ) t 2 in (x ,x ) 2 1 h(x2) = x ± such t h a t < t ± < x= Q h(t2) = t 1 # h^) We , can there i s start from here to c o n s t r u c t a sequence i n d u c t i v e l y . Hence there i s a d e c r e a s i n g sequence h(tn) = t n - 1 . Since xn {t n > with t n e ( x n , x n _ 1 ) •-• and converges to x and 51 and t e (xn»xn-i)> a l s o have must converge to f ( t n ) converges to (3) £ Ctn5 On the other hand, .. = fCt-^). converges to (1) and f o r any ( 4 ) , we have But t h i s c o n t r a d i c t s ( 2 ) . Hence there i s no Using there i s no x with h x > h(x). Then with Hence h(x) = x f o r any x 4.2; [0,1] = 1, f i s a continuous function defined which c o n t a i n s no constant p i e c e and t h a t there e x i s t s a continuous h(0) x i s the i d e n t i t y f u n c t i o n . Suppose t h a t on £(xQ). e x a c t l y the same method, we can show that [0,1], i . e . , Theorem = .. n. £(t1) = in we have f(x0). (5) x < h(x). So we f ( t n ) = fh(tn) = f C t ^ ) = f l i C t ^ ) f(tn) = f(tx) (3) f ( x ) . By x. Therefore we have (4) By s o h(l) = 0 function and hh(x) = x, i . e . , h from f(x) = fh(x) h [0,1] to f o r any x i s an i n v o l u t i o n , hence [0,1] with in [0,1]. h is a 52 one t o one f u n c t i o n and so is also h i s a homeomorphism, and h unique. Proof: Consider h h . We have hh(0) = 0 and we a l s o have fhh(x) = fh(x) = f (x) Hence by Lemma 4.1-, hh(x) = x. i s one-to-one. that h x j in from [0,1] to [0,1] and x = y. hj(1) = 1 , that f o r any hh(x) = x h(x) = hhj (x) = j ( x ) , i . e . x in Since [0,1], h f ( x ) = fh(x) with Therefore hj (0) = 0, We have h i s unique. interpretation Two men s t a n d a t o p p o s i t e (graph o f h(0) = 1 and i s one-to-one, hence Theorem 4.2 has the f o l l o w i n g simple s i d e s o f a mountain range h f j ( x ) = f(x) f o r h j ( x ) = x. and h = j . i n terms o f m o u n t a i n - c l i m b i n g . [0,1]• continuous j ( 1 ) = 0. so by Lemma 4.1, we have a l r e a d y proved in Now .we want to show [0,1] w i t h j(0)= 1 , £h(j(x)) = f j ( x ) = f(x) x Then we have i s unique. Assume there i s another function any Hence f o r any h h ( l ) = 1, Now we want t o show that L e t h(x) = h ( y ) . x = hh(x) = hh(y) = y . and and f ( x ) ) . The c o n d i t i o n f(l)= 0 means t h a t they can climb the mountain range from one s i d e t o the o t h e r i n such a way t h a t t h e i r e l e v a t i o n s remain equal a l l the t i m e , and one man always moves forward along the way. The c o n c l u s i o n o f Theorem 4.2 t e l l s us t h a t the other man must a l s o move 53 forward a l l the time. It can i s surprising t h a t e v e n i f we know t h a t two men c l i m b t h e m o u n t a i n r a n g e f r o m one s i d e t o t h e o t h e r , maintaining a common e l e v a t i o n , t h e r e i s no g u a r a n t e e they can c l i m b p r o p e r l y such that t h a t a t any i n s t a n t t h e r e i s a t l e a s t one man m o v i n g f o r w a r d . The f o l l o w i n g e x a m p l e shows t h a t we w i l l instant. Define: Let £(~) = 1, linear f ( 0 ) = f ( l ) - 0, f(|o = j and G(f,f) £(i) - | , f ( ~ ) = -|, i n a l l the remaining graph o f one h a v e two men b o t h g o i n g b a c k w a r d s a t some path connecting (0,1) and d e f i n e intervals. (See t h e f i g u r e ) , and £(|) - i t o be I f we l o o k a t t h e then there i s e s s e n t i a l l y (1,0), and t h i s p a t h t h a t a t . s o m e i n s t a n t b o t h men go b a c k w a r d s . (O.DR-. f 71 (1,0) shows 54 Corollary 4.5: Suppose t h a t on [0,1] f i s a continuous f u n c t i o n d e f i n e d which contains no constant p i e c e and e x i s t continuous f u n c t i o n s h with h(0) g(0) that f g ( x ) = fh(x) = 1, h ( l ) = 0, h g from = 0 x for a l l to-one f u n c t i o n , then and in and that [0,1] g(l) = 1 [0,1], If g i s a homeomorphism and there to [0,1] such is a one- i s unique. Proof: Since g"1. g is f g C g " x ) = fh ( g ~ 1 (x)) , From 1 hg" 1 (0) = 1, and h g ' 1 ( l ) = 0. hg"*hg~*(x) = x, i . e . , h(x) = h ( y ) , But g is t o - o n e . Now j on-to-one, i t has one-to-one, so we get we want to show'that i s another f u n c t i o n that has Then fg(x) = fh(x) f j (x) = f h ( x ) . for a l l x, jh"^"(x) = x. h = j and Now and and h has h j (x) = h(x) h i s unique. and hence h have have g(x) is = g(y). one- i s u n i q u e . Suppose a l l the r e q u i r e d p r o p e r t i e s . inverse = 0, Assume we x = y, i . e . , fg(x) = f j (x) jh"1^) Hence By Theorem 4.2, we = hg_1h(y) _1 f(x) = £ h g " 1 ( x ) have hg'^hCx) = g ( x ) . hg h(x) then we an i n v e r s e function h-1, give so j h _ 1 ( l ) = 1. f o r a l l x. us f j h _ 1 ( x ) = f(x) By Lemma 4.1, That means 55 Lemma 4 . 4 : Suppose t h a t on x f(x) =0 R, (0,1). in £ i s a continuous f u n c t i o n f o r x e R-(0,1) Then given d > 0, any £(x) > 0 and for a l l there e x i s t (a + b)/2 e ( 0 , 1 ) , [a - b j = d such t h a t and defined a, b f(a) = f (b). Proof: d > 1, If i t i s c l e a r t h a t we can choose a = (1 - d ) / 2 , b = (l'+ d ) / 2 . Suppose be a sequence and may .f x each in n £ n Cx) = 0 [0,1], f o r x e R- (0,1) [0,1] (1) j n ( 0 ) = 0, f j (x) = f n j n [2], for , h from n such t h a t j ( l ) = 1, h n h n ( 0 ) = 1, n n (x) for a l l x j and graphs must cross at some p o i n t j (t ) (2) = H (x) = |j (x) - h (x)|. R x n Now from Whittaker's r e s u l t and the c o n t i n u i t y o f Define f ( ) £. 0 and there e x i s t continuous f u n c t i o n s to n n o f continuous f u n c t i o n s which converges t o f [0,1] and L e t {£ > c o n s i s t s o f a f i n i t e number of monotone p i e c e s . We choose for d < 1. n n in h , n t h ( l ) = 0, n [0,1], From i t follows that e [0,1], (1) their Then h (t ). From (1) and ( 2 ) , we 56 have H n (0) « 1 Hn, and. H ( t Q ) = 0, From the c o n t i n u i t y o f i t f o l l o w s t h a t there e x i s t s a p o i n t { I that n ( t n ) = d. |an - bn| = d Let Jn^n^ = a n and fn(an) Hence an, {bn> converge to a converges to f(a) = f ( b ) . bn e [0,1], * f n ( a n ) = •£Ii:iri(.tTl) and Without l o s s o f g e n e r a l i t y , we may {an> anc f(a) and n n (tn) n e (0,tQ) = bn. such Then = fnhn(tn) = fn(bn). assume t h a t the sequences and b fn(bn) It i s c l e a r that we have t r e s p e c t i v e l y . Then converges to f ( b ) . |a - b| - d. Since (a + b)/2 e ( 0 , 1 ) . Lemma 4.6: Let D be a u n i t d i s c i n the plane be a continuous f u n c t i o n d e f i n e d on R2 such f(x) > 0 if x e D f(x) =0 i f x | D, R2. Let f that and Let the number in d R2 d > 0 be g i v e n , then there are three forming a t r i a n g l e with and w i t h area (or p e r i m e t e r ) equal to the center of the t r i a n g l e i n takes the same value points D such that on the v e r t i c e s o f the t r i a n g l e . f 57 P roof: We s h a l l get a t r i a n g l e with area a t r i a n g l e w i t h g i v e n perimeter d. The case o f i s t r e a t e d i n the same way. A c t u a l l y we can use t h i s k i n d o f method to get other similar results. 3/3/4 i s the maximum area we can get from a triangle inside D. If d > 3/3/4, choose three p o i n t s o u t s i d e d. and c e n t e r i n D. D i t i s c l e a r that we can forming a t r i a n g l e with area As a matter o f f a c t we can even choose t h i s t r i a n g l e to be e q u i l a t e r a l with center at the c e n t e r o f the u n i t d i s c . p, q and triangle m r If d < 3/3/4, pqr i s e q u i l a t e r a l , so i t s area i s three arcs i n s i d e I13 D joining p, pm, and qm be homeomorphisrns from r e s p e c t i v e l y such that [0,1] h-^CO) = p , and h l ( l ) - h 2 ( l ) = h 3 ( l ) - m. and f = f j - j ^ - . Then f]h]_» f u n c t i o n s d e f i n e d on sequences o f continuous £ 2b2' ^£2n^> ^3^3 [0,1], q and Let Let to m. h-^, h 2 , onto pm, qm h2(0) = q, anc ^ * and = f , _ , £ ^f^n)» 3n3 and rm h3(0) = r a r e £2 = f | — c o n t i n u o u s ^2n^» ^£3n^ ° e f u n c t i o n s which converge to consist of a f i n i t e Denote Let Let ^2^2* r rm. r e s p e c t i v e l y such t h a t those ^£3n-^ 3/3/4. f ( x ) a t t a i n s i t s maximum v a l u e . Take three arcs, by 3 any three p o i n t s from the boundary o f the d i s c such that the be a p o i n t where these then take functions fini» ^f^n^» number o f monotone p i e c e s . 58 g ] _ n , §2n» By Theorem 1.2, there e x i s t continuous f u n c t i o n s g3n in H (t) F f such t h a t l n g l n = f 2 n g 2 n - f 3 n g 3 n h3g3n(t). By c o n t i n u i t y of We have Hn, H n (0) = 3/3/4 there i s a in t and [0,1] 2&?n^ t n ^' ^2^3n^ t n^ we may assume t h a t ^ a s a r e a gin(tn), ^* gZn^n) » g3nUn) h ^ O ) , h ^ f c ) . r e s p e c t i v e l y , where b e qm, c e rm. The area of t r i a n g l e = £3ng3 Ct ) n n f ^ h i V ) = f3(c). for a l l n. We have H n ( l ) = 0. such that Without l o s s o f g e n e r a l i t y , h^Ca), the above c o n d i t i o n s . h^g^Ct) , ^iSin(tn^» H n ( t n ) - d, i . e . , the t r i a n g l e w i t h v e r t i c e s by Define to be the area of the t r i a n g l e witli v e r t i c e s h2g2n(t), n . abc finSin^n) converge t o a e pm, i s e q u a l to = f2nS2nC t n) d = Therefore = f ^ h ^ C b ) = fjhsh^Cc), i . e . , fx(a) = f ( b ) = Hence 2 f(a) = f(b) = f ( c ) . p r o o f o f Lemma 4.7. This completes the 59 CHAPTER V CHORD SLIDING ALONG A PATH R. Ferm [3] c o n s i d e r e d chord of length i.e., i f I = [0,1] path p i n a metric and q for a l l the d from s e I, f r o m one e n d o f a p a t h i s the unit i n t e r v a l space (X,p), I I to and d then p(0) = 0, and a to the other, and h : I ->- X a do t h e r e e x i s t maps such t h a t answer i s yes i f t h e p a t h analytic, the q u e s t i o n o f s l i d i n g p (hp ( s ) , h q ( s ) ) q ( l ) = 1? i s reasonable, i s not g r e a t e r than He p r o v e d that say piecewise the d i s t a n c e between the end p o i n t s , p (h ( 0 ) , h ( 1 ) ) . similar question from the p o i n t o f view o f p o i n t s e t t o p o l o g y . And t o g i v e an e x a m p l e t o show t h a t t h e r e a r e we a l s o w i s h paths of f o r which s l i d i n g We w i s h = d to discuss a cannot take p l a c e for certain values d. The q u e s t i o n we a s k e d i s c o m p l e t e l y s i m i l a r t o the m o u n t a i n c l i m b i n g p r o b l e m s . A l l we a s k e d i s w h e t h e r there exist pathwise x and y such t h a t (x,0) and ( l , y ) are connected i n the s e t E where d = {(s,t) e I 2 | p(h(s),h(t) = d}, 0 < d £ p ( h ( 0 ) , h ( l ) ) . We a r e a b l e t o show t h a t there 60 are x and y same component set such that (x,0) C by means o f elementary p o i n t of E^ and ( l , y ) are i n the t o p o l o g y . And i f the path s a t i s f i e s p r o p e r t y (*) we w i l l d e f i n e l a t e r , then we are able to show that l o c a l l y c o n n e c t e d . I t i s c l e a r that C and ( l , y ) can be j o i n e d i n That means we w i l l have the maps p C C is i s a compact m e t r i c [4,p.36], space. Then by the Arcwise Connectedness Theorem (x,0) which by a simple a r c . and a n y t h i n g , we wish to quote a theorem from q. Before we do [4,p.109], Z o r e t t i Theorem: If plane, simple JHM e and K e i s a component o f a compact s e t M i s any p o s i t i v e number, then there e x i s t s a c l o s e d curve = <!>, i n the J which e n c l o s e s and every p o i n t o f from some p o i n t o f J K and i s such t h a t i s at a d i s t a n c e l e s s than K. Lemma 5 . 1 : Let m e t r i c space h be a continuous (X,p). such that (x,o) and component C of E . f u n c t i o n from Then there e x i s t x and I y to a in I ( l , y ) are i n the same connected 61 Proof: Our d = 0. result is t r i v i a l i f Assume that d = p(h(l),h(0)) 0 < d < p(h(1),h(0)). Since a compact s e t which does not meet the d i a g o n a l the p o i n t s (0,1) and Lx f o r some a and Now we which does not F^d u Lj,. E H L1 and E n such want to show that any E Let be a component o f L = {(x,0)| a simple c l o s e d curve J = (J>, f*V than less be L2, J -1 E. than the d i s t a n c e between L2 I x I cannot be z e J that the d i s t a n c e between z e L?, we J, z E^ that < x < 2}. Since Eel If e L2 and E CI then by H and E and . h^. E less i s s m a l l , say then E, x I. J. e J is cannot T h i s means But i f some o f the connectedness From z e J, i s l e s s than i n f e r that the d i s t a n c e between x exists i s at a d i s t a n c e completely i n s i d e we must have some such E^ J because i s i n the i n t e r i o r of r a component o f which encloses from some p o i n t of L Z o r e t t i Theorem, there and every p o i n t o f f a r away from that L2 e a set component of w i l l be then by the = <f>, contain < 1} {(x,0) | 0 < x < 1} C n d nor then there e x i s t s 0 < a < x <_b is that intersect Let = o>. = {(x,0)| b E (1,0), or z and we of infer e, from E is I, 62 is greater than completely i n the e x t e r i o r o f of E L u d E' C E ^ E. This such that x then s i n c e s E C E . f because E and i s i n the e x t e r i o r o f .contains E = E' and Now components o f result E^ exists A component o f that r E Then E' J J i s a component o f A but L^. r L^. E^ and B of E^ \j We L]_. u A L^. of E^, A, E^u L^ A E^U L^. L 3 ~'{(l,y)| exists J n the a simple (Ed u interior L^) = <f>. of L3 = (j>, [0,1] If e U L^. J which Theorem, which encloses and we contains there A i s s m a l l enough, then x [-1,1] The where By t h e Z o r e t t i c l o s e d curve E^ us a c o n t r a d i c t i o n s i n c e i s a component o f 0 < y < 1}. i t i s clear w h i c h does n o t i s i n a component o f n there a r e i n t h e same r A U Then L^ = 4> Assume t h a t L^. E^ n Therefore L^) = <J>. u s e t h e above fact that L^. J shall i s a component o f gives of (E^ u Hence B have connected ^ i n t e r s e c t . Lj_. assume t h a t we and a l l t h e By d e f i n i t i o n i s i n a component B i.e., E^, i s a component o f r ^ A, = c a n n o t be and which i n t e r s e c t Ej U h n which i s i n the i n t e r i o r i s n o t a component o f a point E' is be .the...component be t h e u n i o n o f to prove that Suppose <J>, E let A E' i s a component o f If l Let If 1 E H J. E = E* . Hence E' i s i m p o s s i b l e . .Hence J H A, f <j>, and J isin 63 {(x,y) | y = 0, O i x <a) and { ( x , y ) | y = 0, b < x < 1}, We want to show t h a t there i s an a r c such t h a t N i s not d i s j o i n t that each component ^1 u A of from J n A-^ I2 and and N C I A2. 2 It i s clear i s not d i s j o i n t from 2* c* i Let inf {(x,0)| (x,0) e C i H (A U sup {(x,0)| (x,0) e C± H (A x U A )}. be the c l o s e d r e g i o n bounded by segment between c| and c|. If then we can deform the arc between f o r any and C^t A. A. Hence i . This (x,,0) e N n A-, R = N A. A C\ to a s i n g l e and (x ? ,0) e N U{(x,0)| point of J i . Since f o r some C^, we have that N 1 A C C| which we want. L e t A?t n 0^x<x (the l i n e j <j> i s the arc C^t from I f we can do t h i s T h i s c o n t r a d i c t s our choice i s connected and does not meet f o r some and the l i n e is disjoint J } 2 to a l i n e segment then we can s h r i n k which e n c l o s e s A c?) without touching without touching A N C J Let )U{(x,0)|x 2 <x<l}. 64 Define H ( x , y ) = p (h (x) ,h ( y ) ) . H ( 1 , 0 ) > d. continuous that Since R H E^ = o). A H f a). = d. (*) C e e D, (x,0) there and the l i n e Suppose ( l , y ) f o r some h D of I t e l . I f each such c i r c l e intersects C and y. such t h a t f o r provided h intersects and t h e in a of radius d, I center intersects line finite the path into at h(I) number o f c o n n e c t e d c o m p o n e n t s , t h e n we w i l l Let E^ { ( x , e ) | 0 <, x < 1} i s a f u n c t i o n from draw a c i r c l e 5.2: x of components. Then we Theorem such but C number o f c o n n e c t e d c o m p o n e n t s , number o f c o n n e c t e d (*) (s,t)e R i s a component segment { ( e , y ) | 0 < y < 1} Property is a : in a finite segment H ( s , t ) e E^, There i s a dense s u b s e t any and i s a c o n t r a d i c t i o n . Hence we m u s t h a v e Therefore which contains and i s a point T h i s means t h a t This H(0,0) = 0 i s a connected s e t f u n c t i o n , hence t h e r e H(s,t) Property R Then (X,p). h(t) in a for finite have i s simple. be a c o n t i n u o u s f u n c t i o n from I to a 65 metric space there end (X,p) such that Property i s a chord of length d which (*) holds. c a n be s l i d o f the path t o the other provided Then f r o m one that 0 < d s p(h(0) , h ( l ) ) . Proof: We h a v e p r o v e d (x,0) that E^. and that there exist x and y ( l , y ) a r e i n t h e same c o n n e c t e d Now we w a n t t o u s e of C i s l o c a l l y c o n n e c t e d . By d e f i n i t i o n o f l o c a l l y c o n n e c t e d n e s s , we h a v e t o p r o v e every neighborhood neighborhood of we c a n f i n d a o Then l T f £ x 2 n that C C. 0 (s C contains 0 > t ) V = C O < s < s2+6, that i n C. 0 a connected, F o r any n e i g h b o r h o o d G', of (s ,t ), 0 where t 1 - 6 < t < t2+6} = {(s,t)| S 2 < s < s , , 2 2 C C V. components by T in smaller neighborhood s , t-,^, t n 0 t o show is in 0 T s Q (s ,t )in 0 f o r any p o i n t (s ,t ) x (s ,t ) for of { ( s , t ) | s -5 G«= and i s that (*) component C all Property such 2 i n Let (T ~T ) 2 Property has only D. 1 n C T 2 < t < t > 2 be t h e c l o s u r e o f has a f i n i t e number o f ( * ) , Now we w a n t t o show a finite number o f c o m p o n e n t s . t h i s i s not t r u e , then there exists T . that Assuming a t l e a s t one x o 66 component K with K n ( T 2 - T^) = <j>. component o £ the compact s e t choose e T2 and n T2 c = < j > and n C by J. n 2 (s C. G = (T2 n G we know that i £ we enclosing i s inside K such T j . Evidently J , Hence that (x,0) C is has only a f i n i t e number o f components. Hence Let ,t ) e G. o' o G T2 n C i s both c l o s e d and open i n be the component o f Since G C) C\ U i s open i n where H C) n ' U C G (T-L isa This i s i m p o s s i b l e . So we have that every component o f T J J ( l , y ) are i n the e x t e r i o r o f separated C, K s m a l l enough, then by the Z o r e t t e Theorem, we can have a simple c l o s e d curve j n T2 n Since and U T1 in i m p l i e s that C V. n C n U Hence C n C T_ n C, 2 ' i s open i n i s a connected neighborhood of i s contained T2 such we have Cs0»t0^ C, i s open (s0,t0) i s locally that in i s in C. C, and i n Hence G connected. T h i s i s arcwise connected and proves the theorem. The f o l l o w i n g example shows that i f the curve i s not n i c e , then f o r some cannot take p l a c e . the p l a n e . L e t h w i t h h(0) = T , then r e t u r n s to h(x) the f u n c t i o n from h ( l ) = W, R 0 < d < p(h(0),h(1)), This example i s a curve be goes through circles d, where h(x) sliding (Figure I) i n [0,1] to the plane goes from T to R, v i a s m a l l e r and s m a l l e r c i r c l e s . In f a c t R i n f i n i t e l y many t i m e s , and those converge to the p o i n t R. Any two p o i n t s i n the 67 W T (Figure arc PQ will distance d. h a v e t h e same d i s t a n c e T h i s e x a m p l e shows chord of length d If R, t h e n we w i l l If hp(s) have hp(s) near hq(s) R R, but converges to P hq(s2n) Q. R. We call can not s l i d e and R q this any w i t h the r e q u i r e d and has n o t y e t p a s s e d near Q and i n t h e a r c and h a s j u s t p a s s e d and i n t h e a r c an i n c r e a s i n g s e q u e n c e s to t h a t we p i s near have i s near hq(s) from f r o m one e n d o f t h e p a t h t o t h e o t h e r . Suppose t h e r e e x i s t mappings properties. I) {s > n converges PT. such t h a t to T h i s means t h a t P q and SQ. R, t h e n we will Hence there exists hp(s ) n converges hq(s2n+l) c a n n o t be c o n t i n u o u s . 68 In the above example, i f we take length b i- d , then we are s t i l l from one end o f the path able to s l i d e h to get a path example by a d j o i n i n g H on which cannot take p l a c e f o r any chord o f l e n g t h Y i s a countable the chord to the o t h e r . Now we want to extend t h i s other c o p i e s o f any chord o f sequence converging sliding b e Y, to 0 where and Y C [0,p(H(0),H'(l)]. The construction of an a r b i t r a r y p o s i t i v e number there i s a sequence (0,a). chord {s^} H i s made by f i r s t choosing a. Then i t i s c l e a r that converging We make a copy o f h to i n which 0, s-^ i s the f o r b i d d e n t 0 l e n g t h . Then we extend the a r c the d i s t a n c e between T-^ with We c a l l t h i s new path T (Figure I I ) . 1 c o n s i d e r d e f i n e d on and 0-^ t^,^] with which l i e s i n is a, gs-^C0) = 0^ such that and connect g 0^ W^ which we s1 and g s i ( ~ ) = T-p For each s. ( i i 1 ) , we can get a copy o f h i n which s^ i s the f o r b i d d e n chord l e n g t h . Then we connect with t o get a new c l o s e d path T^ d e f i n e d on [(i-1)/i,i/(i+1)] g .(i^0 s b l We j o i n T- the gs^ i for a l l gs^ which we c o n s i d e r with = T. = g s . ( r i - ) . 1 s i l+l i i n such a way that the p o i n t s a l l c o i n c i d e , and the new c o p i e s gQ. with i f 1 are in any p o s i t i o n as long as they do not meet the i n t e r i o r o f 0-^ the c i r c l e w i t h c e n t e r at and radius a (Figure I I I ) . Let H(x) = g . ( x ) i f xe s Then 0, H(.O) = H and H(l) = T must be continuous unique chord of l e n g t h that we cannot s l i d e to the o t h e r . s^ at [(i-l)/i,i/(i+l)]. 1 # Since 1. For any to s t a r t {s^} s^, sliding. converges to there i s a It i s clear the chord from one end o f the path H 70 BIBLIOGRAPHY R. S i k o r s k i and K, Z a r a n k i e w i c z , On U n i f o r m i z a t i o n of Functions (I) ( I I ) , Fund. Math. 41 (1954), 339-350. J . V. W h i t t a k e r , A Mountain-Climbing Problem, Canadian J o u r n a l o f Mathematics, 18 (1966), 873-882. R. Fenn, S l i d i n g a Chord and the Width and Breadth o f a C l o s e d C u r v e , J . London Math. Soc. ( 2 ) , 5 (1972), 39-47. G. T. Whyburn, A n a l y t i c Topology, A. M. S. Colloquium P u b l . XXVII 1963. J . Hocking and G. Young, Topology, Addison-Wesley P u b l i s h i n g Company, I n c . 1961. J . Huneke, Mountain C l i m b i n g , Amer, Math. Soc T r a n s . , 139 (1963), 383-391. T. I-Ioinma, A Theorem on Continuous F u n c t i o n s , Kodai Math. Sem. R e p o r t s , 1 (1952), 13-16.
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Some problems on mountain climbing Hung, Patrick Chia-Ling 1973
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Title | Some problems on mountain climbing |
Creator |
Hung, Patrick Chia-Ling |
Publisher | University of British Columbia |
Date Issued | 1973 |
Description | Let f and g be two continuous, real-valued functions defined on [0,1] with f(0) = g(0) and f(l) = g(l). The main result of this thesis is to characterize the property that (0,0) and (1,1) are in the same connected component of G(f,g) = {(x,y)|f(x)=g(y)}. In Chapter I, we study conditions implying that (0,0) and (1,1) are in the same connected component of G(f,g), where f and. g are not necessarily real-valued functions. We obtain theorems to characterize [0,1], In Chapter II, we give a simple proof of a theorem by Sikorski and Zarankiewicz. In Chapter III, we obtain our main result. In Chapter IV, we study pathwise connectedness in G(f,g) and give some applications. In Chapter V, we study the question of sliding a chord of given length along a path. An example is given to show that this is not always possible. |
Genre |
Thesis/Dissertation |
Type |
Text |
Language | eng |
Date Available | 2011-03-04 |
Provider | Vancouver : University of British Columbia Library |
Rights | For non-commercial purposes only, such as research, private study and education. Additional conditions apply, see Terms of Use https://open.library.ubc.ca/terms_of_use. |
DOI | 10.14288/1.0080475 |
URI | http://hdl.handle.net/2429/32040 |
Degree |
Doctor of Philosophy - PhD |
Program |
Mathematics |
Affiliation |
Science, Faculty of Mathematics, Department of |
Degree Grantor | University of British Columbia |
Campus |
UBCV |
Scholarly Level | Graduate |
Aggregated Source Repository | DSpace |
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