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 in the Department of MATHEMATICS We accept this thesis as conforming to the required standard THE UNIVERSITY OF BRITISH COLUMBIA June, 1973 In presenting t h i s thesis i n p a r t i a l f ulfilment of the requirements for an advanced degree at the University of B r i t i s h Columbia, I agree that the Library s h a l l make i t freely available for reference and study. I further agree that permission for extensive copying of t h i s thesis for scholarly purposes may be granted by the Head of my Department or by his representatives. I t i s understood that copying or publication of t h i s thesis for f i n a n c i a l gain s h a l l not be allowed without my written permission. Department of MAV^M/\JlCS The University of B r i t i s h Columbia Vancouver 8, Canada Date Supervisor: Professor J , V. Whittaker i i ABSTRACT 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 r e s u l t of this thesis i s to characterize the property that (0,0) and (1,1) are in the same connected component of G(f,g) = {(x,y)jf(x)=g ( y ) } . In Chapter I, we study conditions implying that (0,0) and (1,1) are i n 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 I I , we give a simple proof of a theorem by Sikorski and Zarankiewicz. In Chapter I I I , we obtain our main r e s u l t . In Chapter IV, 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 question of s l i d i n g a chord of given length along a path. An example i s given to show that t h i s i s not always p o s s i b l e . i i i TABLE OF CONTENTS Page INTRODUCTION . . . . . . . . . . . . 1 CHAPTER I : CHARACTERIZATIONS OF [0,1] . . . . . . 3 CHAPTER II : CONNECTEDNESS ; . . . . 17 CHAPTER III : COMPATIBILITIES OF FUNCTIONS 27 CHAPTER IV : SOME PROPERTIES OF MOUNTAIN CLIMBING AND APPLICATIONS . . . . . 49 CHAPTER V : CHORD SLIDING ALONG A PATH 59 BIBLIOGRAPHY 70 ACKNOWLEDGEMENT I am deeply indebted to Professor J , V. Whittaker for his valuable assistance throughout the research and prepa-ration of thi s t h e s i s . The f i n a n c i a l support of the University of B r i t i s h Columbia and the National Research Council of Canada i s g r a t e f u l l y acknowledged. 1 INTRODUCTION Let £ and g be two real continuous functions defined on [0,1] with f(0) = g(0) and f ( l ) = g ( l ) . The question of whether there e x i s t continuous functions h and j from [0,1] to [0,1] with h(0) = j(0) and h(l) = j(1) such that fh(x) = gj(x) for a l l x e [0,1] has been studied by Sikorski and Zarankiewicz [1], Homma [7], Huneke [6] and Whittaker [2], If we consider the graphs of f and g to be mountains, then the existence of h and j t e l l s us that climbers can always climb the mountains maintaining a common elevation ( h and j denote the horizontal progress of the climbers). I f we consider the set G(f,g) defined by {(x,y)| f(x) = g(y)}, then the existence of h and j i s equivalent to the existence of a path i n G(f,g) joining (0,0) and (1,1). The main purpose of this thesis i s to study the connectedness of G(f,g) between (0,0) and (1,1). This i s important when we want to know i f the functions h and j e x i s t . 2 The s t a r t i n g point w i l l be a theorem of Sikorski and Zarankiewicz which states that i f f and g are continuous functions from [0,1] to [0,1] with f(0) = g CO) = 0 and f ( l ) » g(l') = 1, then (0,0) and (1,1) are i n the same connected component of G(f,g). In Chapter I, we show that the above nice result i s only true for functions from [0,1] [0,1], It turns out we can characterize [0,1] as well as the points 0 and 1 in terms of connectedness properties for G(f,g). In Chapter I I , we get a theorem which generalizes the above theorem proved by Sikorski and Zarankiewicz. Actually the r e s u l t we get i s known, but the o r i g i n a l proof requires a l o t of advanced topological means. The proof we have i s a very simple d i r e c t proof. In Chapter I I I , we study the connectedness i n G(f,g) between (0,0) and (1,1) i n more d e t a i l . We can give a necessary and s u f f i c i e n t condition that (0,0) and (1,1) are i n the same connected component of G(f,g) i n the most general case. In Chapter IV, we study some properties of pathwise connectedness i n G(f,g) and give some ap p l i c a t i o n s . In Chapter V, we discuss the question studied by Fenn [3] of s l i d i n g a chord of fixed length along a path from our point of view. An i n t e r e s t i n g counter-example i s included. After I had the example, Professor R. 0. Davies showed me a better example i n terms of a simple curve. CHAPTER I 3 CHARACTERIZATIONS OF [0,1] Let a' and-.-, b be two. points of a topological space Y. Throughout this t h e s i s , we s h a l l denote by F(a,b;Y) the class of a l l continuous functions from the closed unit i n t e r v a l I into Y with f(0) = a and f ( l ) = b. For s i m p l i c i t y , we s h a l l write F(a,b) and F instead of F(a,b;R), (where R i s the set of a l l r e a l numbers), and F(0,1;I) respectively. I f f , f^ ,«•«, f n belong to F(a,b;Y), then G(f ,f , . . , , f ) i s the set: •{(* ,x x ) e I n I f.U.) = f.(x.) for a l l i , j } . 1 2 n ' x i J j Now we state two theorems which R. Sikorski and K. Zarankiewicz proved i n [1]. Theorem 1.1: Let f, g belong to F. Then the points (0,0) and (1,1) are i n the same connected component of G(f,g). 4 Theorem 1.2: Let f ^ , £ 2 » • « « , . £ n belong to F with each f^ consisting of a f i n i t e number of monotone pieces. Then there e x i s t ^ , y ,... , ^ e F such that for a l l x i n [ 0 , l ] o In this chapter, we would l i k e to consider Theorem 1.1 i n the case where we do not assume £ and g are r e a l functions. F i n a l l y , we are able to give theorems to characterize [0,1], Remark 1.3: A space Y i s c a l l e d pathwise connected i f for each p a i r of d i s t i n c t points a, b i n Y, there i s a con-tinuous function £ from [0,1] to Y such that £(0) = a and £(1) = b. I f we can always choose £ to be a homeomorphism, then we c a l l Y arcwise connected. Lemma 1.4: Let a, b be two d i s t i n c t points of a pathwise connected space Y. I f for every £, g i n F(a,b;Y), 5 (0,0) and (1,1) are in the same connected component of G(f,g), then every h i n F(a,b;Y) i s s u r j e c t i v e . Proof: I f h(I) i s not equal to Y, then there exists a point c e Y - h ( I ) . By pathwise connectedness, there exists j e F(a,b;Y) with j(1/2) = c. It i s clear that h(x) i s not equal to j (1/2) for any x i n [0,1], There-fore the set {(x,y) e I2 | x = 1/2} and G(j,h) are d i s j o i n t . Hence (0,0) and (1,1) are not i n the same connected component of G(j,h). This contradicts our assumption that (0,0) and (1,1) are i n the same connected component of G(j,h). Hence every h £ F(a,b;Y) i s s u r j e c t i v e . Lemma 1.5. Let Y be a Hausdorff space and a, b be two d i s t i n c t points i n the same path-connected component of Y. Suppose that every f i n F(a,b;Y) i s s u r j e c t i v e . Then Y i s homeomorphic to the closed unit i n t e r v a l I. 6 Proof: Since a, b are i n the same path-connected com-ponent of Y, there exist g e F(a,b;Y). C l e a r l y , g(I) = Y. By the Hahn-Mazurkiewiez Theorem [5, p. 129], Y i s a compact, l o c a l l y connected, connected metric space, i . e . , Y i s a Peano space, and so Y i s arcwise connected. Hence there exists a homeomorphism fl from I into Y with fl(0) = a and 11(1) = b. Since tl e F(a,b;Y), so 11(1) = Y. Hence Y i s homeomorphic to the closed unit i n -t e r v a l I. Using the same argument as above, we immediately get a simple proof of the following known r e s u l t . Theorem 1.6: I f Y i s a Hausdorff space, then Y i s arcwise connected i f and only i f Y i s pathwise connected. Example 1.7: The following example shows pathwise connectedness does not imply arcwise connectedness i n general. Let Y consist of the two points {0,1} with the topology T = {4>f{0},Y}« We have only one pai r of d i s t i n c t p o i n t s , i . e . , 0, and 1 to consider. Define f from [0,1] to Y with f ( [ 0 , l / 2 ) ) - 0 and f ( [ l / 2 , l ] ) = 1. Then f i s continuous. Hence Y. i s pathwise connected space. Since Y has only two p o i n t s , Y i s not arcwise connected. Theorem 1.8: Let Y be a pathwise connected, Hausdorff space. Given two d i s t i n c t points a, b of Y, the following con-ditions are equivalent: (1) If f , • g are i n F(a,b;Y), then (0,0) and (1,1) are i n the same connected component of G(f,g). (2) Y i s homeomorphic to [0,1] with a = 0 and b = 1 (or a =1, b = 0), Proof: (2) implies (1). This i s just Theorem 1.1. Conversely i f we assume (1), then by Lemma 1.4, every h e F(a,b;Y) i s s u r j e c t i v e . By Lemma 1.5 Y i s ho-meomorphic to [0,1]. Now we claim a ~ 0 and b - 1 (or a - 1, b = 0). I f this i s not true, then a continuous fun-c t i o n f can be constructed such that f i n F(a,b;Y) and f i s not s u r j e c t i v e . This i s a co n t r a d i c t i o n , since we •8 have proved that a l l such £ are s u r j e c t i v e . Hence a = 0 and b = 1 (or a = 1, b = 0). Lemma 1,9: Let f , g be two homeomorphisms from I to Y with f(0) = g(0) = a, f ( l ) = g(l) = b. I f f ( I ) i g ( I ) , then f ( l ) - g(l) i <j> and g(l) - f(I) i 4>. Proof: Assume f(I) - g(I) = <{>, i . e . , f ( I) i s a proper subset of g ( I ) . Then there exists a point t i n (0,1) such that g(t) £ f ( I ) . It i s clear g(t) i s a cut point of g ( I ) , i . e . , g(I) - g(t) = U U V, where U and V are separated. Note we have a e U, b e V. Since f ( I ) i s contained i n g(I) - g(t) and i s a connected s e t , so f(I ) i s contained i n one of U, V. and not the other. But f ( l ) = g(l) = b e V and F(0) = g(0) = a e U, which i s im-pos s i b l e . Hence f(I) - g(I) 1 <t>. S i m i l a r l y , g(I) - f ( I ) f <}» Lemma 1,10: Let Y be an arcwise connected space and p be a given point of Y, Suppose that for every f e F(p,p;Y), 9 the points (0,1) and (1,0) are in the same connected com-ponent of G(f,f) = {(x,y).-j f(x) = £ ( y ) } . Then (a) Y lias a unique arc (Arc i s a homeomorphic image of the closed unit i n t e r v a l I.) be-tween any point x and p which we denote by px. (b) For any x^, x 2 i n Y, we have px-^ Q px"2 or px 2 C P X J . Proof: (a) I f we have two homeomorphisms f , g from I to Y with f(0) = g(0) = p, f ( l ) = g(l) = x and £(I) f g(I)> then by Lemma 1.9, there e x i s t t ^ , t 2 i n [0,1] with f ^ g C ^ ) = (j), g"1fCt 2) = <fr. Let f f(2t) for t i n [0,1/2] gC2-2t) for t i n [1/2,1] j ( t ) Then j e F(p,p;Y). By the choice of t 2 , j ( t ) f j ( t 2 / 2 ) for any t i n [1/2,1], Furthermore, j i s one to one .o.n [0,1/2],. It follows that j ( t ) t J(t 2/2) i f t f t 2/2 S i m i l a r l y , j ( t ) j j((2-t 1)/2) i f t f (2-^/2 C l e a r l y , G(j,j) is d i s j o i n t from S1 \j S 2 , where S 1 = {(x,y) | x = t 2/2, t 2/2 < y < 1} and S 2 = {(x,y) | y = (2-t 1)/2, 0 < x < (2-^/2} Hence (0,1) and (1,0) are not i n the same connected com-ponent of G(j,j) and this i s a contr a d i c t i o n . Therefore for any two homeomorphisms f , g with f(0) = g(0) = p and f ( l ) = g(l) = x, we-have that f(I) = g ( D . (b) Let f , g be two homeomorphisms from I to Y with f(0) = g(0) = p and f ( l ) = x-^ g ( l ) = x 2 < Define t1 - sup {t | f ( t ) e g(I)> Let f(t}) = x 3 = g ( t 2 ) . Then f(.[0tt^]) i s an arc from p to X3 and g([0,t 2]) i s an arc from p to x 3. By part (a) f C [ o , t i ] ) ' = g ( [ o , t 2 ] ) - Jx"3. We claim that x = x-^ or x = x 2. If this i s not tru e , 11 then t-^ f 1 and t 2 f 1. Let h(x) = f g(4x) g( 4 ( t 2 l ) x + 2 f(4(1 - t 2 ) x + 3 t 1 - 2) ^ £(4 - 4x) for x i n [0,1/4]. for x i n [1/4,1/2], for x i n [1/2,3/4], for x i n [3/4,1], C l e a r l y , h is i n F(p,p;Y). By the choice of t ^ , f ( 4 ( l - t i ) x + 3 t i - 2) i g ( I ) for any x in (1/2,3/4], Therefore h(y) i h(x) for any (x.,y) e S1 = [0,1/2] x (1/2,3/4], S i m i l a r l y , h(x) i h(y) for any (x,y) e S 2 = [1/4,1/2) x [1/2,1] Hence Si U S2 and G(h,h) are d i s j o i n t . That means (0,1) and (1,0) are not in the same connected component of G(h,h) and this contradicts our assumption. Therefore x 3 = x 1 or x 3 = x 2, i.e., px 1 C P x 2 or px 2 C Theorem 1.11: 12 Let p be a given point of a Peano space Y. Sup-pose that for every f e F(p,p;Y), the points (0,1) and (1,0) are in the same connected component of G £ f , f ) . Then Y i s homeomorphic to [0,1], Proof: In order to prove that Y i s homeomorphic to [0,1], we would l i k e to introduce an ordering on Y f i r s t . Define x < y i f px C py for x, y of Y. By part (b) of Lemma 1.10, (Y,<) becomes a l i n e a r l y ordered s e t . Remark: < i s also a dense ordering i n the follow-ing sense: Given any a and b of Y with a < b, there ex i s t an element c such that a < c < b. Since a < b means pa C pb", so there i s a homeomorphism f with. £([0,1]) = pF, f(0) = p, f ( l ) = b. f ( [ 0 , t ] ) = pa for some t i n (0,1). Take any s e ( t , l ) and l e t f(s) = c. It i s c l e a r that paC pc C pb", i . e . , there i s an element c with a < c < b. This proves our remark. To prove the theorem, define A(x) = {y e Y | x y}, we want to prove that A(x) i s a closed set for any x i n Y. Take any e Y - A(x). Note that Y - A(x) ="px - {x}. Hence e px and there i s an open set U of px such that z, e U, x f. U, U = px n V, where V i s an open set 13 of Y. It i s clear that x : i s not i n V. Since Y i s a Peano space, in p a r t i c u l a r Y i s l o c a l l y connected, so there i s an open connected set K containing z-^ and contained i n V. We know that a connected open subset of a Peano space i s arcwise connected [5, p. 118].. Hence K i s arcwise con-nected. I f there i s an element z 2 e K Pi A(x) , then z^, z 2 c a n be J o in e <i by an arc T i n K. It i s clear that T u pT-^ i s a pathwise connected, Hausdorff space. Therefore T U pz" i s arcwise connected. Hence pz"? C "T y p z . 1 ^ 1 Recall x 4 T \j pzT^. - Since z 2 e A(x) , we have x < z 2 , i . e . , px C P2*2* This i s impossible because x f. pz"2. Hence K n A(x) = <p. This proves that the complement of A(x) i s open and so A(x) i s closed. Now the class of a l l A(x) has the f i n i t e i n t e r -section property. The compactness of Y implies that n{A(x) | x e Y} i <j>. Now we claim that n (A(x) | x eY} has exactly one element. If this i s not true , say we have x^, x 2 of ^(A(x) | x e Y}, then we have either x-^ < x 2 or x 2 < x-^ . Let us say x^ < x 2. By d e f i n i t i o n of A ( x 2 ) , we know x--, t A(x 2) and so x-^ t H {A(x) | x e Y}. This contradicts our choice of x-^ . Hence <^ (A(x) j x e Y} has only one point q. Now we want to show pq = y» Take any r c Y. I f q < r , then q £ A(r) and that implies q f. H{A(x) | x e Y}. This contradicts our choice of q. Therefore r <, q, i . e . pr C pq, and r £ pq". Hence pq = Y, so Y . i s homeomor phic to [0,1]. Lemma 1.12: Suppose f i s a continuous function from [0,1] to [0,1] with f(0) = f ( l ) = 0. Then the points (0,1) and (1,0) are i n the same connected component of G(f,f) = {(x,y) | f(x) = f ( y ) } . Proof: Let c be a point where f attains i t s maximum. Define: h-^x) = f ( c x ) / f ( c ) for x i n [0,1], h 2(x) = f ( ( c - l ) x + l ) / f ( c ) for x i n [0,1]. Then h , h belong to F. By Theorem 1.1, the points (0,0), (1,1) are i n the same connected component of G(h h 2 ) . Let 15 "G = {(x.y) e G(f,f) | 0 < x < c, c < y < 1} The mapping: (x,y) —— * (ex, ( c - l ) y + 1) defines a homeomorphism from GCh^,}^) onto G-^ . From the fact that the points (0,0) and (1,1) are i n the same connected component of Gfh-pl^), we i n f e r that the points (0,1) and (c,c) are i n the same connected component of G^ C G ( f , f ) . Since G(f,f) i s symmetric with respect to to the diagonal set {(x,x) | x e [0,1]}, we get that (1,0) and (c,c) are i n the same connected component of G ( f , f ) . Therefore (1,0) and (0,1) are i n the same connected com-ponent of G ( f , f ) . Theorem 1.15: Let p be a given point of a Peano space Y. Then the following conditions are equivalent: (1) Y = [0,1] and p = 0 (or p = 1) ; (2) For every f i n F(p:^p;Y), the points (0,1) and (1,0) are in the same connected compo-nent of G ( f , f ) . Proof: (1) implies ( 2 ) . This i s just Lemma 1.12. Let (2) hold. By Theorem 1.11, Y i s homeo-morphic to [0,1], We claim that p i s equal to either or 0. If p f 0 and p f 1, then take 0 < x 2 < p < x-^ so that p3r2 i s not a subset of p x , and px"-^ i s not subset of px*2. This contradicts part (b) of Lemma 1.10. Hence p = 0 or p = 1. This completes the proof of Theorem 1.13, 17 CHAPTER II CONNECTEDNESS In this chapter we s h a l l prove the following theo-rem which i s a generalization of Theorem 1.1. Theorem 2.1; I f f ^ , f 2 , . •« , f n (0,0,...,0) and (1,1,...,1) component of G ( f , f 2 ,. .. , f n ) . The proof of Theorem lemmas. Lemma 2.2: belong to F, then the points are i n the same connected 2.1 i s based on the following Suppose f e F and e > 0. Then there exists h e F, where h consists of a f i n i t e number of monotone pie c e s , such that |f(x) - h(x)| < e for any x i n [0,1], Proof: Because f is a continuous function defined on a compact set [0,1], i t i s uniformly continuous on [0,1], For every e > 0, there exists a 6 > 0 such that |f(x) - £ ( y ) | < e/2 whenever |x - yj < 6. Choose 0 = a „ < a. < . . . < a =1 such that I a. -.a. ., I < x 0 1 n 1 l l - l1 ^ 0 for each . i . Define h(a i) = f ^ ) and l e t h be li n e a r in [a. ,,a.]. Then h e F and h consists of a f i n i t e number l - l I of monotone pieces. For any x i n [0,1], there exists a j such that x e [a. ...a.]. Therefore |f(x) - 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 This completes the proof of Lemma 2.2. Lemma 2.3: -Suppose f , g belong to F and e > 0. Then there e x i s t 41, £ of F such that |£^(x) - g £ ( x ) | < e for every x i n [0,1]. Proof: By Lemma 2.2, there exi s t h, j of F with each consisting of a f i n i t e number of monotone pieces, such that |f(x) - h(x)| < e/2 and |g(x) - j ( x ) | < e/2 for any x i n [0,1]. By Theorem 1.2, there exi s t £ of F such that h^(x) = jS(x) for every x i n [0,1]. Because |4'Xx) - h\|i(x)| < e/2 and |g£(x)- j£(x)| < 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 f ,, f„,..., f e F and e > 0. Then there 1 2 n exist ij; , <j>2,..., ^ e F such that |f.¥. (x) - £.*. (x) | < e for any x i n [0,1] and i , j = 1,2,..., n. Proof: We s h a l l prove i t by induction. It i s true for n = 2 (Lemma 2.3). Assume i t i s true for n = k, i . e . , there ex i s t £^, f ^ , . . . , £jc of F such that I f.tyx) - f.£.(x)| < E/2 for any x i n [0,1] and i , j = 1,2,..., k. Since £\%>\* ^k+1 e ^* t n e n by Lemma 2.3, there e x i s t h h 2 e F such that |£ 1€ 1h 1(x) - f k + 1 h 2 ( x ) | < e/2. Let KjJ^i ~ £ ° r 1 = 1» 2»«*«» k « a n d n2 = ^k+1 Then (1) l £j*j.W " fi¥i0O| < e/2 In p a r t i c u l a r , l f 1 ^ 1 ( x ) - f.*.W\ < e/2 Since 21 I V l ( x ) " fk*l^k+lOO I < e/2, we have (2) [fjI'jCx) - f k + i ^ k + i ( x ) | < e From (1) and (2), j f ^ C x ) - £ ^ j ( x ) | < e. for any x e [0,1] and i , j = 1, 2,..., k+1. Hence the proof of Lemma 2.4 i s completed. Lemma 2.5: Let X be a compact Hausdorff space and G be a compact subset of X. I f a and .b are not in the same component of G, then there exist d i s j o i n t open sets U, V of X such that G C V u U and a e U, b e V. Proof: Since G i s a compact Hausdorff space, the com-ponents and quasi-components are i d e n t i c a l . That a and b are not i n the same connected component of G implies that a and b are not in the same quasi-component of G. By d e f i n i t i o n of quasi-component, there ex i s t d i s j o i n t closed sets H, K of G with G = HuK, a e H and b e K. Since G i s 22 closed in X, so H and K are closed i n X. Every compact Hausdorff space is normal. Therefore there exist d i s j o i n t open sets U and V of X such that H C U and K C V. It i s clear that G C U ( j V and a e U, b e V. This completes the proof of Lemma 2.5. Now we have established enough lemmas to prove • Theorem 2.1. Proof of Theorem 2.1: F i r s t note that G ( f ^ , f 2 , . . . ,f ) i s a compact subset of the compact Hausdorff space In. Assume (0, 0, 0) and (1, 1, .... 1) are not in the same connected component of G = G ( f , f 2 , . . . ,f ) . Then by Lemma 2.5, there e x i s t d i s j o i n t open sets U, V, of In with G C U U V and (0, 0, 0) e U, (1, 1, ...,1) e V. For every p o s i t i v e integer k, there ex i s t ^2k» ••• » ^ nk e P such that for i , j « 1, 2, •. •, n, and the mapping t ~~* ^ l k( t )» ^ 2 k( t )» ••*» *nk( t ) ) 23 defines a path joining (0, 0, . .., 0) and (1, 1, ...„ 1), Let Ak = { ( * l k ( t ) , ^(t), . * n k ( t ) ) | t e [0,1] }. Since Ak i s connected and contains (0, 0, 0) and (1, 1, 1), by our choice of U, V, we have Ak - CUUV) f <j). Let ( x l k , x 2 k , ..., x n k) e A k - (UUV), where x^k = ^ i k ^ ) £ o r s o m e t« The sequence (y k) where y k = C-^ik* x21-' •••» xn]<) contains a convergent subsequence. So without loss of g e n e r a l i t y , we may assume {yk} converges to y = (x-p x 2, x n ) , Given e > 0, we can choose k large enough so that < If.Cx.) - £.(x i k)| • |£.(x.k) - f . ( x . k ) | • |£.(x j k) - f j C o < e/3 + l £ i * i k ( t ) " V j k ( t ) ' + e / 3 < e/3 + 1/k + e/3 < e. This means that IfiCxi) - f j C x ^ l = o for every i , j - 1, 2, . ,.,'n. Hence (1) y = ( x x , x 2, x n) e G C U u V. 24 On the other hand, y k e A k - (U U V) C i n - (U U V) . Since U and V are open, In — (U U V) i s closed. Hence (2) y e In - (U U V) . From (1) (2), we get a con t r a d i c t i o n . This means (0, 0, 0) and (1, 1, 1) are i n the same connected component of G (f , f 2 ,... , f n ) . This completes the proof of Theorem 2.1. Corollary 2.6: Suppose f i s a real-valued continuous function defined on [0,1], f(x) 2. 0 for every x i n [0,1] and fCa^) = f ( a 2 ) - ... = f ( a n ^ = 0 f o r 0 = a-^ < a 2 <••••< a n =1. Then the n11 points of the form (b^, b 2 , b ) , where every b^ = a., f o r some j (b s may equal bj_ i f s f k ) , are i n the same connected component of G ( f , f , . . . , f ) . Proof: I f f = 0, then the res u l t i s true. Let f f 0. Assume that f(x) attains i t s maximum at point c. Note c f a-p a 2 , a n . Let £(c) = M. For every p o s i t i v e integer k, define g^(x) = ~ Dk )x + b^J/M, where 25 (b^, b^, ... f b R) is a point of the required form. The g^'s s a t i s f y the conditions of Theorem 2.1. Hence (0, 0, ..., 0) and (1. 1, 1) are i n the same connected component of G(g^ p g^» • g n ) . Let G' be the set: {(x,,x7,...,x ) e G ( f , f , . . . , f ) | x. i s between b. and c ) . The mapping (x , x2, ... , x ) •*• ((c - b 1 ) x 1 + b l f (c - b 2 ) x 2 + b 2 (c - b n ) x n + b p defines a homeomorphism from G (g^ ,g2,.. . ,g ) onto G*. Because (0, 0, 0) (b^. b 2 , b ) and (1, 1, 1) -*• (c, c, c ) , we have ( b ^ b 2 , b R) and (c, c, ..., c) i n the same connected component of G' C G ( f , f , . . . , f) . This i s true for any (b^, b 2 , b n ) , where every b^ = a^ for some j . Hence the n n points of this form are i n the same connected component of G ( f , f , . . . , f ) . We can use the same method as i n Corollary 2.6 and use Theorem 1.2 to prove the following: 26 Corollary 2.7: Suppose £ i s a continuous function defined on [0,1], f(x) > 0 for every x i n [0,1], f consists of a f i n i t e number of monotone pieces and f C a ^ = f ( a 2 ) = = f ( a n ) = 0 for 0 = < &2 < ... < a n = 1, Then the n n points of the form V°\t'^>2r *•• » b n) > where every b^ = aj for some j ( b_ may equal b. i f s f k ) , are i n the same pathwise k _ connected component of G ( f , f , . . . , f ) . 27 CHAPTER III COMPATIBILITIES OF FUNCTIONS From Theorem 1.1, we know that i f we have two continuous functions f , g, from [0,1] to [0,1] with f(0) = g(0) = a = 0 and f ( l ) = g(l) = b = 1, then (0,0) and (1,1) are i n the same connected component of G(f,g). If we do not assume a = 0 or b = 1, then the points (0,0) and (1,1) need not be i n the same -connected component of G(f,g), as the following example shows. Let f 2 x for x e [0,-j] f(x) = , 3 • 2 x for x e [£,l] f 4 x for x e [0,1] But i n thi s case we are s t i l l able to give a necessary and s u f f i c i e n t condition that (0,0) .and (1,1) w i l l l i e i n the same connected component of G(f,g). 28 D e f i n i t i o n 3.1: We say that a closed i n t e r v a l [p>q] i s an a-interval of f i f f [p,q] f [ f ( p ) , f ( q ) ] i f f(p) < f(q) I [f(q),f(p)3 i f f(q) < f(p). D e f i n i t i o n 3.2: |[s,t] i f condition (A) or (B) holds. i s c a l l e d right compatible ivith g |[c,d] Condition (A): there e x i s t points s = x o < x^ < ... < x m = t , c = y < y-^ < .. . < y m = d such that 1. fCx^ = gCyj) i = 0, 1, ... , m. 2. Every closed i n t e r v a l [x^,x^ +^] i s an a-interval of f . 3. Every closed i n t e r v a l [ y ^ , y + i s an a-interval of g. Condition (B): there e x i s t s t r i c t l y monotone increasing sequences {x^}, (y^) such that 29 1. X q = s, y Q = c and x^ < t , y^ < d for a l l i . 2. I f lim x i = xOT and lira y n = y^, then f([x r o,tD = g([y w,cl]) = f ( t ) . 3. fCx^ •= g(x i) for a l l i . 4. Every closed i n t e r v a l [x^,x^ +^] i s an a-interval of f . 5. Every closed i n t e r v a l [y^ »y^ + ^] is a n a - i n t e r v a l of g. D e f i n i t i o n 3.3 f . r , i s compatible with g l r i f condition |[s,t] |[c,d] (A) holds. D e f i n i t i o n 3.4; f j j-g t-j i s l e f t compatible with S | [ c d] i f condition (A') or (B1) holds: Condition (A'): there e x i s t points s = x_ n < x _ n + i < ••• x Q = t , c = y_ n < y . n + 1 < ... < y 0 = d such that 1. f ( x _ i ) = gCy.j) for i « 0, 1, ... , n. 2. Every closed i n t e r v a l [x .,x . ] i s an a - i n t e r v a l - l - i + lJ of f . 3. Every closed i n t e r v a l [y ^,y 1 S an a-interval of g. 30 Condition (B'): there exist s t r i c t l y monotone decreasing sequences {x_^} and {y_i> s a t i s f y i n g the following r e l a t i o n s : 1. x = t , y = d and x . > s, y . > c for a l l i . o ' o -1 •-1 2. If lim x_^ = x_ro and lim y.^ = y-co, then f C f s . x . J ) - g ( [ c , y _ J ) = f ( s ) . 3. f(xmi) = gCy . i ) for a l l i . 4. Every closed i n t e r v a l [x- i »x- i + i l *s a n a-interval of f . 5. Every closed i n t e r v a l [y-_i-,y_-j_ + ^ ]) i s an a-int e r v a l of g. Theorem 3.5: Suppose that f , g, belong to F ( 0 , b ; [ 0 , » ) ) . Then the points (0,0) and (1,1) are i n the same connected component of G(f,g) i f and only i f f|[o l ] is r ignt compatible with gj[o l ] ' Proof: First, l e t us suppose that f | [ 0 , l ] *s ri gn t compatible with g|[Q i ] * Suppose condition (A) holds. Then by Theorem 1.1, 31 (xi + l » y i + l) (xiiYO are in the same connected component of G i - G C f , , , , ) . where t ± - 1, [ x . and 8 - 8 , ^ . ^ ] Since (0,0) = (x 0,y 0) and (x 1,y 1) are i n the same connected component of G Q and ( x j ^ - ^ ) , ( x2»y2^ a r e * n t I i e s a m e connected component of G^, hence (0,0), Cx2»^2^ a r e * n the same connected component of G Q U G-^ . Using the same argument, we can show that (0,0) and (1,1) = (xm,y ) are m-1 i n the same connected component of u G.. Because G . C G ( f , g ) i = o 1 1 m 1 for each i , we have u Gi C G(f,g). Hence (0,0) and i=o (1,1) are i n the same connected component of G ( f , g ) . Suppose condition (B) holds. Then i t i s c l e a r that (xoo,yoo) and (1,1) are i n the same connected component of G(f,g) because f ( [ x 0 0 , l ] ) = g ( [ y 0 o , l ] ) . For each i , (x i,y i) and (xi + i » y i + i ) a r e 1 1 1 t n e same connected component of Gi = G(f.,g.), where f. = f | f , and g. = g, 1 1 1 1 Uxi »xi + l J 1 | [ y i , y i + 1 ] I f we define H. = ?t G-, then we have that (0,0) and 3 i=o i ^xj + l ' ^ j + l ^ a r e *n s a m e connected component of H... Let 00 H = .U^ H_. , then H has a component C containing (x^,y^) for each i . By d e f i n i t i o n of x^, y w , we have that ( xco.y«>) belongs to the closure of C r e l a t i v e to the usual topology of R 2 . Since C i s contained i n G(f,g) and 32 G(f,g) i s a closed set of R2, the closure of C i s a subset of G(f,g). Therefore the closure of C i s a connected set containing both (0,0) and ( x t t , y j . We have proved that' ( x ^ y o o ) and (1,1) are i n the same connected component of G(f,g). Hence the points (0,0) and (1»1) are i n the same connected component of G(f,g). Now assume that (0,0) and (1,1) are i n the same connected component of G(f,g). It i s cle a r that £(I) = g(l)» for otherwise there exists q such that •f" lg(oJ = $'• Then (0,0) and (1,1) are not i n the same connected component of G(f,g). Now we want to show that condition (A) or (B) holds. Define X Q = 0 , YQ ~ °» and x = sup (x c [0,1]| f(x) attains i t s absolute maximum} . y 1 = sup (y e [0,1] | g (y) attains i t s absolute maximum}. Now f and g are continuous functions defined on a compact set [0,1], so x^ and y^ are well-defined. Since f(I) ' = g ( I ) , we have f ( x 1 ) = g ( y 1 ) . Define sup {x e [x^,l]( f(x) attains i t s minimum in [x^,l]} sup {x e [ y ^ , l ] | g(y) attains i t s minimum in [ y ^ l ] } . It i s c l e a r that y^ ^ < y2 and x^ < x^. We claim that 33 f ( * 2 ) = S ( y 2 ) . Suppose f(x ) > g ( y 2 ) . Then f(x) f g(y ) for x t x^ and fCx-^) f g(y) for y > y^. Therefore the set G(f,g) i s d i s j o i n t from U L 2 , where L-L =: {(x,y) | x = x p y > y1> L 2 85 -C(x.y) | x * x ^ y = y 2>. This contradicts our assumption that ( 0 , 0 ) and ( 1 , 1 ) are in the same connected component of G(f,g). S i m i l a r l y , we cannot assume that f (x2 ) < g ( y 2 ) » Hence f (x2 ^ = S(y 2)* Suppose that x. and y have been constructed. J 3 Assume x_. $ 1 and j i s even, i . e . , £(x^) < f ( l ) = b. Define: x j + l = sup {x e [ x . , l ] | f(x) attains i t s maximum i n [ X j , l ] } yi + l ~ S UP ty e [y •, 1]I g(y) attains i t s maximum i n [y.,l]> We claim that f ( X j + 1 ) = g(y\. + 1 ) . Assume f ( X j + 1 ) < g(yj + 1) Then f(x) f g ( y j + 1 ) for x i n [x^,l] and f ( x j ) f g(y) for y > y.. Therefore the set G(f,g) i s d i s j o i n t from L' U L*, where L» = {(x,y) I y = y , x £ x^} L* = ((x,y) | x = X J , y > yj>. 34 This contradicts our assumption that (0,0) and (1,1) are in the same connected component of G(f,g). S i m i l a r l y , we cannot assume that f(x.. + 1) > g(y\j + 1 ) . Hence f ( X j + 1 ) = g(yj + Suppose that x.. i 1 and j i s odd, i . e . , f ( x j ) > f ( l ) = b Define: sup {x e [x.,1]| f(x) attains i t s minimum i n [x.,1]} 3 3 sup {y e [ y j , l ] | g(y) attains i t s minimum i n [ y j , l ] } . Using the same method as when we proved that f ( x 2 ^ = g ( y 2 ) » we can prove that f ( X j + -^ ) = g(yj + ] L). This completes the induction step to construct (x^}, ( y ^ ) . I f for some m, x^ = 1, then from f (xm ) = S(y m) and f (xm ) = f ( l ) = D» we have g(y m) = b. By d e f i n i t i o n of y m , we get y^ = 1. In thi s case, we get condition (A). If for any i , x^ ^ 1, then y^ f 1 for any i . Because {x^} and {y^} are bounded monotone increasing sequences, we have lim x^ = x^ and lim y^ = y^,. By our construction cf {x^} and {y^.}, we have f(x^) > b i f i i s odd and f(x^) < b i f i i s even. Hence f(x ) = f (x ) - g(xO T) = b. In order to get condition (B), the only 35 t h i n g t o prove i s fCtx^,,!]) = g([yco»l]) = b • Assume t h a t Xo, ^ 1 and £(x) j b f o r some x i n [ x M , l ] ; say f(x) < b. S i n c e ^( x2i J * s m o n o t o n e i n c r e a s i n g t o b, t h e r e i s an x 2 j such t h a t f(x) < f ( x 2 j ) < b. By c o n t i n u i t y o f f , t h e r e i s a p o i n t c such t h a t x«> < c < x and f ( c ) = f(x ). The f a c t s x < c and f ( c ) = f(x ) <*3 £3 2 j c o n t r a d i c t our c h o i c e o f x„ .. That means f(x) cannot be l e s s than b whenever x i s i n [x^,!]. For the same r e a s o n , f(x) cannot be g r e a t e r than b whenever x i s i n [ X o o , l ] , Hence f([-x»,l]) = b. S i m i l a r l y , g ( [ y o o , l ] ) = b. T h i s completes t h e p r o o f o f Theorem 3.5. I t i s c l e a r t h a t the f o l l o w i n g theorems are t r u e . Theorem 3.6: L e t f, g e F ( a , b ) , where f(0) = g(0) = a i s the maximum v a l u e o f f ( x ) and g ( x ) . Then (0,0) and (1,1) are i n the same c o n n e c t e d component o f G ( f , g ) i f and o n l y i f f | [ o l ] * s r i g n t c o m p a t i b l e w i t h S J [0 1 ] * Theorem 3 . 7 : L e t f , g e F ( a , b ) , where f ( l ) = g ( l ) = b i s the maximum (minimum) v a l u e o f £(x) and g ( x ) . Then (0,0) and 36 (1,1) are i n the same con n e c t e d component o f G ( f , g ) i f and o n l y i f £ | [ 0 1] ^ s ^e^t c o m p a t i b l e w i t h g|[o ]_] • Now we want t o d i s c u s s more g e n e r a l c a s e , i . e . , w i t h o u t assuming a n y t h i n g about a o r b. D e f i n i t i o n 5,8: D e f i n e : a l = i n f {x e [ 0 . 1 ] | f ( x ) a t t a i n s i t s maximum} a 2 = sup {x e [ o , i ] | f ( x ) a t t a i n s i t s maximum} a 3 = i n f {x e [0,1] | f ( x ) a t t a i n s i t s minimum} a 4 = sup {x e [0,1] | f ( x ) a t t a i n s i t s minimum} b l = i n f {x e [0,1] | g(x) a t t a i n s i t s maximum} b 2 = sup {x e [0,1] I g(x) a t t a i n s i t s maximum} b 3 = i n f {x e [0,1] I g(x) a t t a i n s i t s minimum} b 4 = sup (x e [0,1] 1 g(x) a t t a i n s i t s minimum}. I f f and g are not c o n s t a n t f u n c t i o n s , t h e n t h e r e a re s i x p o s s i b l e ways t o o r d e r a^, a 2 , a^, a^ as w e l l as b j , b 2 , b 3 , b ^ 1. a^ £ a 2 < a 3 £ a^ 3. a j < a 3 <L a^ < a 2 5. • < < < a 2 2. a^ < a^ < a 2 < a^ 6. a 3 a-^ £ a 2 < a^« 37 Lemma 3.9: Suppose that f , g e F(a,b) and (0,0), (1,1) are in the same connected component of G(f,g). Then the orderings of the a^'s and b^'s have the same i n i t i a l and terminal s u b s c r i p t s . Remark: As before, that (0,0) and (1,1) are i n the same connected component of G(f,g) implies that f(I) = g ( I ) . Hence f ( a x ) = £ ( a 2 ) = g O ^ ) = g(b 2) and £ ( a 3 ) = f ( a 4 ) = g(b 3) = g ( b 4 ) . Proof of Lemma 3.9: Assume for the given f , we have a^ £ a 2 < a 3 ^ a^. We want to show that for g we have b^ ^ b 2 < b 3 $ b 4 or b^ < b 3 < b 2 < b^. Since g cannot be a constant function, we have b 3 < b-^ or b-^ < b^. Assume b 3 < b^. By d e f i n i t i o n of b^ and the fact that f(a^) = g(b^), we have: f C a ^ ¥ g(y) for any y e [ 0 ^ ) . By d e f i n i t i o n of b 3 and the fact that g(b 3) = £ ( a 3 ) , we have: 38 T g ( b 3 ) f ° r a n y x e [ 0 , a 3 ) . T h i s i m p l i e s t h a t (0,0) and (1,1) are not i n the same co n n e c t e d component o f G ( f , g ) . Hence < b^. The s e t { ( x , y ) | x = a 4 , y e ( b 4 , 1 ] } u { (x ,y) | x e ( a 2 , l ] , y = b 2> i s always d i s j o i n t from G ( f , g ) . I f we assume b^ < b,2, th e n the above s e t s e p a r a t e s (0,0) and ( 1 , 1 ) . T h i s i s i m p o s s i b l e . Hence b 2 < b^. T h e r e f o r e i f we assume a l - a 2 < a 3 ^ a4» w e n a v e b^ < b 2 < b 3 < b 4 or b^ < b^ < b 2 < b^. U s i n g the same k i n d o f argument, we can check the o t h e r f i v e c a s e s . T h i s completes the p r o o f o f Lemma 3.9. Theorem 3.10: Suppose t h a t f , g e F ( a , b ) . Then (0,0) and (1,1) are i n the same co n n e c t e d component o f G ( f , g ) i f a n c j o n l y i f t h e r e e x i s t X Q , y Q such t h a t f|[o x ] ^ s i e f ' t c o m p a t i b l e w i t h g ) r n v i and f , r ,, i s r i g h t c o m p a t i b l e I 1 » yo J I L xo» J w i t h , g | [ y 0 , i ] ' P r o o f : Assume t h a t (0,0) and (1,1) are i n the same c o n n e c t e d component o f G ( f , g ) . If we have (1) a i ' — a 2 < a 3 - a4» 39 or (2) a-^ < a 3 < a 2 < a 4 , or (3) a-j, < a 3 < a 4 < a 2 , then we choose: x = a, and y = b -o 1 o 1 If we have (4) a 3 <_. a 4 < a-^ < a 2 , or (5) a 3 < a^ < a 4 < a 2 , or (6) a 3 < a-^ < a 2 < a 4 , then we choose: x = a 7 and y = b T . o ••> o 5 Now we s h a l l discuss case (1). The other f i v e cases can b.e treated s i m i l a r l y . By d e f i n i t i o n of a^ and b^, we have f f a ^ = g O ^ ) i f(x) for any x i n [ 0 , 3 ^ and g(y) 1 f ( for any y i n [ 0 ^ ) . Suppose that ( a ^ b p and (0,0) are not i n the same connected component of G(f,g). Then i t is c l e a r that (0,0) and (1,1) are not i n the same connected component of G(f,g). Therefore (a-^b^) and (0,0) are i n the same connected component of G(f,g). By the above observation, we know that (0,0) and (a^,b^) are i n the same connected component of: G* {(x,y) e G(f,g)| 0 < a 1 ( O i y i b ^ By Theorem 3.7, f . r „ . , i s l e f t compatible with g, | £ ° » a l l lEO.bj] S i m i l a r l y , we can show that f ( r i s right compatible I Ia<, iJ 40 with -g | [b ^ 1]* Since we assume a-^ < a 2 < a 3 <_ a 4 , we have b-^ < b 4 by Lemma 3.9. Because [a^,a 4] and [b-pb4] are a-intervals of f and g re s p e c t i v e l y , £| [ a 1] *s right compatible with 8 | [ b ^ , l ] « Now suppose for some X q , y Q , f | [o x ] * s l e £ t compatible with 2|[0,y ] a n c* £|[x ,1] ^s right compatible with S|[y i y Then we can show that (0,0) and ( x 0 » y D ) are i n the same connected component of G(f,g) and that (x 0,y 0) and (1,1) are i n the same connected component of G(f,g) by using the same method as we used i n Theorem 3.5. Hence (0,0) and (1,1) are i n the same connected component of G(f,g). D e f i n i t i o n 3.11: For f , g e F(a,b), define f - g i f (0,0) and (1,1) are i n the same connected component of G(f,g). Theorem 3.12: = i s an equivalence r e l a t i o n . Proof: f - f because the set {(x,x)| x e [0,1]} belongs 41 to G ( f , g ) and connects (0,0) and (1,1). I f f - g, then (0,0) and (1,1) are i n the same connected component of G(£,g). Also G(f,g) i s homeomorphic to G(g,f) under the mapping (x,y) -> (y,x), and (0,0), (1,1) are mapped to (0,0), (1,1). Hence g - f« Now we prove that - i s a t r a n s i t i v e r e l a t i o n . From Theorem 3.5 and Theorem 3.10, we know that each f has a unique sequence associated with i t . Assume that f - g. By Theorem 3.10, f i r n , i s l e f t compatible with g i r n n IL u »xoJ I l u »yoJ and f|[x ,1] i 5 r i g h t compatible with g|[y f i ] * And i f g = h, then S|[0 y'] ^ s compatible with h.j j-Q Z J and o o and gj[y» i ] i s r i g h t compatible with h | [ z iy Because the sequences for f, g, h are unique as chosen i n Theorem 3.10, we must have y^ = y[ and f(x-) = g(y^) = h(z for a l l i . Therefore f|[o x ] * s l e r > t compatible with k|[0» z o] a n c * £ | [ x 0 , l ] ^ s r i 2 n t compatible with h | [ z f i ] . Hence f = h. Therefore - i s an equivalence r e l a t i o n i n F(a,b). D e f i n i t i o n 3.13: Let f, g e F(p,p). Then define: f £(2x) for x 1) for x Lemma 3.14: I f £ 0 - f i and g 0 = g 1 # then f 0 ° g 0 - £i°gi. Proof: = f 0 ( x ) = f i ( y ) = fi°gi(2")' Since fQ - flt we have that ( 0 , 0 ) , (—,—) are i n the same connected component of 2 2 G C f 0 o g 0 , f 1 o g 1 ) . For every (x,y) e G ( g o , g 1 ) , we have ^1T»^§^ e G ( £ o o g o ' £ l o g l ) * s i n c e S D ~ g l » w e h a v e t h a t (i,—) and (1,1) are i n the same connected component of 2 2 G ( f ( ) o g 0 , f 1 o g 2 ) . Hence (0,0) and (1,1) are i n the same connected component of G ( f Q o g 0 , f ^ o g ^ ) , i . e . , fo°go s fl ° g l -Lemma 3.15: The equivalence classes of F(p,p) form a semi-group. 43 Remark 3.16: Let f , g e F(a,b). Define f «v, g i f (0,0) and (1,1) are i n the same path-connected component of G(f,g). Then the following example shows that ^ i s not an equivalence r e l a t i o n . Example 3.17: Let f , g e F. Then f ^ g i f and only i f there exist h, j e F such'that fh(x) = gj (x) for a l l x i n [0,1]. There exist f , g e F such that £ | g ( c f . [ l ] ) . It i s clear that i f i(x) = x for a l l x e [0,1], then f ^ i and i ^ g for any f , g. If we choose f , g such that f 4' g, then we cannot have the t r a n s i t i v e r e l a t i o n . We know from [1] that i f f , g e F(a,b), and f , g consist of a f i n i t e number of monotone p i e c e s , then . (0,0) and (1,1) are i n the same path-connected component of G(f,g) i f and only i f (0,0) and (1,1) are in the same connected component of G(f,g). So we immediately have the following theorem as a c o r o l l a r y of Theorem 3.10. Theorem 3.18: Suppose that f , g e F(a,b), and f , g consist 44 o£ a f i n i t e number of monotone pi e c e s . Then (0,0) and (1,1) l i e i n the same path-connected component of G(f,g). i f and only i f f|[o 1] *s c o mPa t l Dl e with g| j-Q . Theorem 3.19: Suppose that f e F(p,p) and f consists of a f i n i t e number of monotone pieces. Then (0,1) and (1,0) are in the same path-connected component of G(f,f) i f and only i f there e x i s t 0 = x_ n < x _ n + i < < x c < x i < < x n = 1 such that every i n t e r v a l [x^,x^ +^] i s an a-int e r v a l of f and f ( x i ) = f ( x , i ) for a l l i . Proof: F i r s t we assume that the conditions hold. We want to prove that (0,1) and (1,0) are pathwise connected i n G ( f , f ) . Define f x ( x ) = f(x Qx) and f 2 ( x ) = f ( ( x Q - l ) x + l ) . By Theorem 3.18, (0,0) and (1,1) are pathwise connected i n G ( f l f . f 2 ) . Let G' be the set: {(x,y) e G ( f , f ) | x e [0, x Q] , y e [ x 0 , l ] } . 45 The mapping (x,y) -> (x Qx, ( x Q - l ) y + 1) defines a homeomo-rphism from G(f-L,f 2) onto G'. Hence (0,1) and (x 0,x Q) are pathwise connected in G.' C G ( f , f ) . Now G(f,f) i s symmetric with respect to the diagonal. Hence (1,0) and (x Q,x 0) are also pathwise connected in G ( f , f ) . Therefore (1,0) and (0,1) are pathwise connected i n G ( f , f ) . Usually we have six possible ways to order a^, &3 and , but i f we assume (0,1) and (1,0) are pathwise connected i n G ( f , f ) , then we claim that either al < a3 — a 4 K a2 o r a3 < a l a2 < a 4 * By d e f i n i t i o n of a l t a 2 and a 3 , f(x) i f(aj^) for x e ( a 2 , l ] and f(x) 5* f ( a 3 ) for x £ [0, a 3 ) . I f al - a2 < a3» t h e n i*- i s d e a r that (0,1) and (1,0) are not i n the same connected component of G ( f , f ) . Hence we cannot have a^ .< a 2 < a 3 < a^. S i m i l a r l y we cannot have the other three cases. Now suppose (0,1) and (1,0) are i n the same path-connected component of G ( f , f ) . I f we have < a j i < a 2 , then we want to show that (0,1) and (a ,a ) are i n the same path-connected component of G ( f , f ) . X *» We have f(x) ? f(a^) for x > a 2 and f ( a 2 ) i f(x) for x < a^. Hence the set {(x,y)| y = a 2 , x < a-^ } and the set {(x,y)| x = a l P y > a2> are d i s j o i n t from G ( f , f ) . Therefore the points (0,1) and (a-^,a2) are i n the same 46 p a t h - c o n n e c t e d component o f the s e t : - { ( x , y ) e G ( f , f ) | 0 £ x <. a-^ a 2 < y ± 1 } . Define: 'flOO = f C a j X ) f o r x e [0,1] f 2 ( x ) = f ( ( a 2 - l ) x + 1 ) f o r x e [ 0 , 1 ] . Then by Theorem 3.18, £l | [ o 1] i s c o m p a t i b l e w i t h £ 2 | [ 0 , 1 ] * Hence we have 0 = x . n < x . n + 1 < .... < x . i - a 2 a 2 = x l < x 2 < ••. < x n = 1 such t h a t each i n t e r v a l [ x i , x ^ + 1 ] i s an a - i n t e r v a l o f f and f ( x ^ ) = f ( x _ ^ ) . I f we l e t x Q = a 3 , the theorem i s p r o v e d f o r the case a-j_ < a 3 <. a 4 < a 2 . S i m i l a r l y , we can prove the second case i n w h i c h a 3 < a-£ a 2 < a 4 . T h i s c o m p l e t e s the p r o o f o f Theorem 3-. 19. Theroem 3.20: L e t f , g be c o n t i n u o u s f u n c t i o n s from [0,1] t o SX »"{x e R 2| |x| = 1} w i t h f ( 0 ) = g(0) = f ( l ) = g ( l ) . 47 I f (0,0) and (1,1) are in the same connected component of" G(f,g), then f i s homotopic to g. Proof: For any f and g from [0,1] to S-^ , there exist f and g from [0,1] to R such that f(0) = g(0), f = ef and g = eg, where e(t) = (cos t , s i n t ) . Let H(x,y) = {f(x) - g(y)}/2TT. I f (x,y) e G(f,g), then f(x) = fi-(y) ,. i.e.. , ef(x) = eg(y). Hence f(x) - g(y) = 2nir, where n i s an integer. Therefore H i s a continuous function from the connected component G' of G(f,g) containing (0,0), (1,1) to the set of integers. Since H(0,0) = 0 and G' i s connected, we must have H(x,y) = 0 for any (x,y) e G'. Hence f ( l ) - g(l) = 0. Therefore f , g have the same degree, i . e . , f i s homotopic to g. Theorem 3.21: Let f , g e F(a,b;Y), and l e t (0,0) and (1,1) be i n the same path-connected component of G(f,g). Then f i s homotopic to g. Proof: Since (0,0) and (1,1) are pathwise connected i n 48 G ( f , g } , we have i|>:[0.1] + [0,1] x [0,1] such that ^ ( x ) = ( * 1 ( x ) ^ 2 C x ) } , vpx COD = 0 = ^ ( 0 ) , = ^ ( 1 ) = 1 and ( ^ ( x ) ,T|> 2(X)) e G(f,g) for any x, i . e . , f ^ O O = g«J>2(x). Define H(x,t) = f ( t ^ 1 ( x ) + ( l - t ) x ) . Then H i s a continuous function from [0,1] x [0,1] to Y with H(x,0) = f(x) and H(x,l) = f i ^ C x ) . But we have fi/i^Cx) = g'[ J 2(x)j hence f i s homotopic to g^ 2. Define E(x,t) = g(ti | J 2(x) + ( l - t ) x ) . Then E i s a continuous function from [0,1] x [0,1] to Y with E(x,0) = g(x) and E ( x , l ) = g(^ 2C x))« Hence g i s homotopic to g i r ) 2 . It follows that f i s homotopic to g. 49 CHAPTER IV SOME PROPERTIES OF MOUNTAIN CLIMBING AND APPLICATIONS If we have four continuous functions f , g, h and j from [0,1] into [0,1] with fh = gj and these functions also s a t i s f y some other conditions, then we s h a l l discuss more f u l l y the relatio n s h i p among f , g, h and j in this chapter. Lemma 4.1: Suppose that f i s a continuous function defined on [0,1] that contains no constant piece and that there exists a continuous function h from [0,1] to [0,1] with h(0) = 0 and h(l) = 1 such that f(x) = fh(x) for any x i n [0,1], Then h(x) = x for any x i n [0,1]. Proof: If there i s an x„ such that x_ < hfx ) , then o o o we w i l l construct a sequence i n d u c t i v e l y . Using the inequality 0 - h(0) < x Q < h.(xQ) , we know from the continuity of h that there i s an x^ i n (0,x o) such that h(x^) = X Q . Using t n e inequality 0 =» h(0) < x^ < X q = h(x^), we know that there i s an x 2 i n (0,x^) such that h(x 2) = x-^ . 50 Assume that x n has been constructed, and h(x n) = xn - i a n <i x n < x n_^. Using the inequality 0 = h(0) < x n < x n_^ = h ( x n ) , and by continuity of h, there i s an x n + j i n (0,x n) with h(x n +^) = x n. This completes the induction step to construct our sequence. Hence there i s a decreasing sequence {x n) with the property h(x n) = xn_-^ for any n. Since we have h(xi) = x Q and by assumption we have fh(x^) = f (x^) , hence f (x0 ) = f In general, f ( x n ) = fh (x n) B ' f ( x n _ 1 ) » f h ( x n - l ^ = f C x n - 2 ^ = ' .. = f ( x 0 ) . Since {xn} i s a bounded decreasing sequence, i t has a l i m i t . Say {xn} converges to x. By continuity of f , we have f ( x n ) converges to f ( x ) . Since f ( x n ) = f ( x 0 ) for any n, so we have: (1) £(x n) = f(x) = £(x c) for any n. Since f has no constant piece there i s a t^ in ( X j , x D ) such that (2) f ( x 0 ) i f C t ^ . Using the i n e q u a l i t y h(x 2) = x± < t± < xQ= h ^ ) , there i s a t 2 i n (x2,x1) such that h ( t 2 ) = t 1 # We can s t a r t from here to construct a sequence i n d u c t i v e l y . Hence there i s a decreasing sequence {tn> with t n e (xn,x n_ 1) •-• and h ( t n ) = t n - 1 . Since x n converges to x and 51 and t e (xn »xn - i ) > s o must converge to x. So we also have f ( t n ) converges to f ( x ) . By (1) we have (3) £Ctn5 converges to f ( x 0 ) . On the other hand, f ( t n ) = f h ( t n ) = f C t ^ ) = f l i C t ^ ) = .. .. = fCt - ^ ) . Therefore we have (4) f ( t n ) = f ( t x ) for any n. By (3) and (4), we have (5) £ ( t 1 ) = £ ( x Q ) . But this contradicts (2). Hence there i s no x with x < h(x). Using exactly the same method, we can show that there i s no x with x > h(x). Hence h(x) = x for any x in [0,1], i . e . , h i s the i d e n t i t y function. Theorem 4.2; Suppose that f i s a continuous function defined on [0,1] which contains no constant piece and that there exists a continuous function h from [0,1] to [0,1] with h(0) = 1, h ( l ) = 0 and f(x) = fh(x) for any x i n [0,1]. Then hh(x) = x, i . e . , h i s an i n v o l u t i o n , hence h i s a 52 one to one function and so h i s a homeomorphism, and h is also unique. Proof: Consider hh. We have hh(0) = 0 and hh(l) = 1, and we also have fhh(x) = fh(x) = f (x) for any x in [0,1]• Hence by Lemma 4.1-, hh(x) = x. Now we want to show that h is one-to-one. Let h(x) = h( y ) . Then we have x = hh(x) = hh(y) = y. Hence x = y. Now .we want to show that h i s unique. Assume there i s another continuous function j from [0,1] to [0,1] with f j ( x ) = f(x) for any x i n [0,1] and j(0) = 1 , j(1) = 0. Since £h(j(x)) = f j ( x ) = f(x) for any x i n [0,1], and hj (0) = 0, hj(1) = 1 , so by Lemma 4.1, we have hj(x) = x. We have already proved that hh(x) = x and h i s one-to-one, hence h(x) = hhj (x) = j (x), i . e . h = j . Therefore h i s unique. Theorem 4.2 has the following simple i n t e r p r e t a t i o n i n terms of mountain-climbing. Two men stand at opposite sides of a mountain range (graph of f(x) ) . The condition f(x) = fh(x) with h(0) = 1 and f ( l ) = 0 means that they can climb the mountain range from one side to the other i n such a way that t h e i r elevations remain equal a l l the time, and one man always moves forward along the way. The conclusion of Theorem 4.2 t e l l s us that the other man must also move 53 f o r w a r d a l l the t i m e . I t i s s u r p r i s i n g t h a t even i f we know t h a t two men can c l i m b the mountain range from one s i d e t o the o t h e r , m a i n t a i n i n g a common e l e v a t i o n , t h e r e i s no guarantee t h a t they can c l i m b p r o p e r l y such t h a t a t any i n s t a n t t h e r e i s at l e a s t one man moving f o r w a r d . The f o l l o w i n g example shows t h a t we w i l l have two men b o t h g o i n g backwards a t some i n s t a n t . D e f i n e : L e t f ( 0 ) = f ( l ) - 0, £(i) - | , £(|) - i £(~) = 1, f(|o = j and f ( ~ ) = -|, and d e f i n e f t o be l i n e a r i n a l l the r e m a i n i n g i n t e r v a l s . I f we l o o k a t the graph o f G ( f , f ) (See the f i g u r e ) , t h e n t h e r e i s e s s e n t i a l l y one p a t h c o n n e c t i n g (0,1) and ( 1 , 0 ) , and t h i s p a t h shows t h a t at.some i n s t a n t b o t h men go backwards. ( O . D R - . 71 (1,0) 54 Corollary 4.5: Suppose that f i s a continuous function defined on [0,1] which contains no constant piece and that there e x i s t continuous functions h and g from [0,1] to [0,1] with h(0) = 1, h ( l ) = 0, g(0) = 0 and g ( l ) = 1 such that fg(x) = fh(x) for a l l x i n [0,1], I f g i s a one-to-one function, then h i s a homeomorphism and i s unique. Proof: Since g i s on-to-one, i t has an inverse function g"1. From fgCg" 1 x) = fh (g~ 1 (x)) , we have f(x) = £ h g " 1 ( x ) and hg"1(0) = 1, h g '1( l ) = 0. By Theorem 4.2, we have hg"*hg~*(x) = x, i . e . , hg'^hCx) = g(x). Assume we have h(x) = h ( y ) , then hg _ 1 h(x) = hg _ 1h(y) and hence g(x) = g(y). But g i s one-to-one, so we get x = y, i . e . , h i s one-to-one. Now we want to show'that h i s unique. Suppose j i s another function that has a l l the required p r o p e r t i e s . Then fg(x) = fh(x) and fg(x) = f j (x) give us f j (x) = f h ( x ) . Now h has inverse h- 1, so f j h_ 1( x ) = f(x) for a l l x, and j h "1^ ) = 0, j h_ 1( l ) = 1. By Lemma 4.1, jh"^"(x) = x. Hence j (x) = h(x) for a l l x. That means h = j and h i s unique. 55 Lemma 4.4: Suppose that £ i s a continuous function defined on R, f(x) =0 for x e R-(0,1) and £(x) > 0 for a l l x i n (0,1). Then given any d > 0, there ex i s t a, b such that (a + b)/2 e (0,1), [a - bj = d and f(a) = f (b). Proof: I f d > 1, i t i s clear that we can choose a = (1 - d)/2, b = (l'+ d)/2. Suppose d < 1. Let {£ n> be a sequence of continuous functions which converges to f and .f consists of a f i n i t e number of monotone pieces. We may choose £ n C x ) = 0 for x e R- (0,1) and f n ( x ) £. 0 for x i n [0,1], Now from Whittaker's r e s u l t [ 2 ] , for each n there e x i s t continuous functions j n , h n from [0,1] to [0,1] such that (1) j n ( 0 ) = 0, j n ( l ) = 1, h n(0) = 1, h n ( l ) = 0, and f n j n ( x ) = f n h n ( x ) for a l l x i n [0,1], From (1) and the continuity of j and h n , i t follows that t h e i r graphs must cross at some point t e [0,1], Then (2) j (t ) = h (t ). Define H R(x) = |j n ( x ) - h n ( x ) | . From (1) and (2), we 56 have H n(0) « 1 and. H ( t Q ) = 0, From the continuity of H n, i t follows that there exists a point t n e (0,t Q) such that { In ( t n ) = d. Let J n ^ n ^ = an a n c* nn ( t n ) = b n. Then |a n - b n| = d and f n ( a n ) = •£Ii:iri(.tTl) = f n h n ( t n ) = f n ( b n ) . Without loss of g e n e r a l i t y , we may assume that the sequences {an> and {bn> converge to a and b re s p e c t i v e l y . Then f n ( a n ) converges to f(a) and f n ( b n ) converges to f ( b ) . Hence f(a) = f ( b ) . It i s clear that |a - b| - d. Since a n , b n e [0,1], we have (a + b)/2 e (0,1). Lemma 4.6: Let D be a unit disc i n the plane R2. Let f be a continuous function defined on R2 such that f(x) > 0 i f x e D and f(x) =0 i f x | D, Let the number d > 0 be given, then there are three points i n R2 forming a tria n g l e with area (or perimeter) equal to d and with the center of the tr i a n g l e i n D such that f takes the same value on the vertices of the t r i a n g l e . 57 P roof: We s h a l l get a tria n g l e with area d. The case of a triangle with given perimeter i s treated i n the same way. Actually we can use this kind of method to get other s i m i l a r r e s u l t s . 3/3/4 i s the maximum area we can get from a tr i a n g l e inside D. I f d > 3/3/4, i t i s clear that we can choose three points outside D forming a triangle with area d. and center in D. As a matter of fact 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 center of the unit d i s c . I f d < 3/3/4, then take any three points p, q and r from the boundary of the disc such that the tr i a n g l e pqr i s e q u i l a t e r a l , so i t s area i s 3/3/4. Let m be a point where f(x) attains i t s maximum value. Take three arcs inside D joining p, q and r to m. Denote these three arcs, by pm, qm and rm. Let h-^, h 2 , and I13 be homeomorphisrns from [0,1] onto pm, qm and rm respectively such that h-^ CO) = p, h 2(0) = q, h 3(0) = r and h l ( l ) - h 2 ( l ) = h 3 ( l ) - m. Let ^ = f , _ , £ 2 = f | — and f 3 = fj- j ^ - . Then f]h]_» ^2^2* a n c* £3n3 a r e c o n t i n u o u s functions defined on [0,1], Let ^ f ^ n ) » ^ 2 n ^ » ^£3n^ °e sequences of continuous functions which converge to f ini » £2b2 ' ^3^3 respectively such that those functions ^ f ^ n ^ » ^£2n^> ^£3n-^ consist of a f i n i t e number of monotone pieces. 58 By Theorem 1.2, there exist continuous functions g]_ n, §2n» g 3 n i n F such that f l n g l n = f 2 n g 2 n - f 3 n g 3 n . Define H (t) to be the area of the tr i a n g l e witli vertices h ^ g ^ C t ) , h 2 g 2 n ( t ) , h 3 g 3 n ( t ) . We have H n(0) = 3/3/4 and H n ( l ) = 0. By continuity of H n, there i s a t i n [0,1] such that H n ( t n ) - d, i . e . , the tria n g l e with vertices ^ i S i n (tn ^ » n2&?n^tn^' ^2^3n^tn^ ^a s a r e a ^* Without loss of generality, we may assume that g i n ( t n ) , gZn^n) » g3nUn) converge to h ^ C a ) , h ^ O ) , h ^ f c ) . r e s p e c t i v e l y , where a e pm, b e qm, c e rm. The area of tr i a n g l e abc i s equal to d by the above conditions. We have f i n S i n ^ n ) = f2nS2nC tn) = = £ 3 n g 3 n C t n ) for a l l n. Therefore f ^ h i V ) = f ^ h ^ C b ) = f j h s h^Cc), i . e . , f x ( a ) = f 2 ( b ) = = f 3 ( c ) . Hence f(a) = f(b) = f ( c ) . This completes the proof of Lemma 4.7. 5 9 CHAPTER V CHORD SLIDING ALONG A PATH R. Ferm [3] c o n s i d e r e d the q u e s t i o n o f s l i d i n g a chor d o f l e n g t h d from one end o f a p a t h t o the o t h e r , i . e . , i f I = [0,1] i s the u n i t i n t e r v a l and h: I ->- X a p a t h i n a m e t r i c space ( X , p ) , then do t h e r e e x i s t maps p and q from I t o I such t h a t p (hp (s) , h q ( s ) ) = d f o r a l l s e I , p(0) = 0, and q ( l ) = 1? He p r o v e d t h a t the answer i s yes i f the p a t h i s r e a s o n a b l e , say p i e c e w i s e a n a l y t i c , and d i s not g r e a t e r than the d i s t a n c e between the end p o i n t s , p (h (0),h(1)). We w i s h t o d i s c u s s a s i m i l a r q u e s t i o n 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 we a l s o w i s h t o g i v e an example t o show t h a t t h e r e are paths f o r w h i c h s l i d i n g cannot take p l a c e f o r c e r t a i n v a l u e s o f d. The q u e s t i o n we asked i s c o m p l e t e l y s i m i l a r t o the mountain c l i m b i n g p r o b l e m s . A l l we asked i s whether t h e r e e x i s t x and y such t h a t (x,0) and ( l , y ) are p a t h w i s e c o n n e c t e d i n the s e t E d = { ( s , t ) e I 2 | p ( h ( s ) , h ( t ) = d}, where 0 < d £ p ( h ( 0 ) , h ( l ) ) . We are a b l e t o show t h a t t h e r e 60 are x and y such that (x ,0) and (l,y) are in the same component C of E^ by means of elementary point set topology. And i f the path s a t i s f i e s property (*) which we w i l l define l a t e r , then we are able to show that C i s l o c a l l y connected. It i s clear that C i s a compact metric space. Then by the Arcwise Connectedness Theorem [ 4 ,p.3 6 ] , (x ,0) and (l,y) can be joined i n C by a simple arc. That means we w i l l have the maps p and q. Before we do anything, we wish to quote a theorem from [4 ,p.109] , Z o r e t t i Theorem: I f K i s a component of a compact set M i n the plane, and e i s any pos i t i v e number, then there exists a simple closed curve J which encloses K and i s such that JHM = <!>, and every point of J i s at a distance less than e from some point of K. Lemma 5 . 1 : Let h be a continuous function from I to a metric space (X,p). Then there e x i s t x and y in I such that (x,o) and (l,y) are i n the same connected component C of E . 61 Proof: Our r e s u l t i s t r i v i a l i f d = p(h(l),h(0)) or d = 0. Assume that 0 < d < p(h(1),h(0)). Since i s a compact set which does not meet the diagonal nor contain the points (0,1) and (1,0), then there exists a set L x = {(x,0)| 0 < a < x <_b < 1} for some a and b such that E d n {(x,0) | 0 < x < 1} C L r Now we want to show that any component of E^ which does not i n t e r s e c t w i l l be a component of F^ d u Lj,. Let E be a component of E^ such that E H L1 = o>. Let L = {(x,0)| -1 < x < 2}. Since E e l x I, and E n = <f>, then by the Z o r e t t i Theorem, there exists a simple closed curve J which encloses E and . J f*V = (J>, and every point of J i s at a distance less than e from some point of E. I f e i s small, say e i s less than the distance between L 2 and E, then J cannot be far away from I x I because E C I x I. This means that L 2 cannot be completely inside J . But i f some of L 2 i s i n the i n t e r i o r of J , then by the connectedness of L 2 , we must have some z e J H h^. From z e J , we i n f e r that the distance between z and E i s less than e, from z e L ? , we i n f e r that the distance between z and E i s 62 i s g r e a t e r than E. T h i s i s i m p o s s i b l e . .Hence i s c o m p l e t e l y i n the e x t e r i o r o f J . L e t E' be .the...component o f E d u L x such t h a t E C E 1 . I f E' n h = i . e . , E' C E^ s then s i n c e E i s a component o f E^, we have E = E* . I f E' H f <J>, Then E' cannot be c o n n e c t e d because E l . c o n t a i n s E w h i c h i s i n the i n t e r i o r o f J and i s i n the e x t e r i o r o f J and J ^ ( E ^ u L^) = <J>. Hence E = E' and E i s a component o f E^ \j L^. Now l e t A be the u n i o n o f and a l l the components o f E^ w h i c h i n t e r s e c t L^. We s h a l l use the above r e s u l t t o prove t h a t A i s a component o f E^ U L^. Suppose A i s n o t a component o f E^ u L]_. Then t h e r e e x i s t s a p o i n t r ^ A, b u t r and A are i n the same component o f E j U L^. By d e f i n i t i o n o f A, i t i s c l e a r t h a t r i s i n a component B o f E^, w h i c h does n o t i n t e r s e c t . L j_ . Hence B i s a component o f E^ U L^. The f a c t t h a t B n L^ = 4> g i v e s us a c o n t r a d i c t i o n s i n c e we assume t h a t r i s i n a component o f E ^ u L^ w h i c h c o n t a i n s L^. T h e r e f o r e A i s a component o f E ^ U L^. Assume t h a t A n L3 = (j>, where L 3 ~ ' { ( l , y ) | 0 < y < 1}. By the Z o r e t t i Theorem, t h e r e e x i s t s a s i m p l e c l o s e d curve J w h i c h e n c l o s e s A and J n (Ed u L^) = <f>. I f e i s s m a l l enough, th e n J i s i n the i n t e r i o r o f [ 0 , 1 ] x [-1,1] and J H A, f <j>, 63 {(x,y) | y = 0, O i x <a) and {(x,y)| y = 0, b < x < 1}, We want to show that there i s an arc N C J and N C I2 such that N i s not d i s j o i n t from A-^ and A 2. It i s c l e a r that each component of J n I2 i s not d i s j o i n t from Let be the closed region bounded by and the l i n e segment between c| and c|. If A i s d i s j o i n t from C^t then we can deform the arc to a line segment (the l i n e between and c?) without touching A. I f we can do th i s for any C^t then we can shrink J to a single point without touching A. This contradicts our choice of J which encloses A. Hence A C\ j <j> for some i . Since A i s connected and does not meet C^, we have that A C C| for some i . This i s the arc N which we want. Let (x,,0) e N n A-, and (x ?,0) e N n A?t Let ^1 u A2* i n f {(x,0)| (x,0) e C i H (A U } c* i sup {(x,0)| (x,0) e C± H (Ax U A 2)}. R = N U { ( x , 0 ) | 0 ^ x < x 1 ) U { ( x , 0 ) | x 2 < x < l } . 64 D e f i n e H(x,y) = p (h (x) ,h ( y ) ) . Then H(0,0) = 0 and H(1,0) > d. S i n c e R i s a co n n e c t e d s e t and H i s a c o n t i n u o u s f u n c t i o n , hence t h e r e i s a p o i n t ( s , t ) e R such t h a t H ( s , t ) = d. T h i s means t h a t ( s , t ) e E^, b u t R H E^ = o). T h i s i s a c o n t r a d i c t i o n . Hence we must have A H f a). T h e r e f o r e t h e r e i s a component C o f E^ which c o n t a i n s (x,0) and ( l , y ) f o r some x and y. P r o p e r t y (*) : There i s a dense s u b s e t D o f I such t h a t f o r any e e D, the l i n e segment { ( x , e ) | 0 <, x < 1} i n t e r s e c t s C i n a f i n i t e number o f c o n n e c t e d components, and the l i n e segment { ( e , y ) | 0 < y < 1} i n t e r s e c t s C i n a f i n i t e number o f c o n n e c t e d components. Suppose h i s a f u n c t i o n from I i n t o ( X , p ) . Then we draw a c i r c l e o f r a d i u s d, c e n t e r a t h ( t ) f o r t e l . I f each such c i r c l e i n t e r s e c t s h ( I ) i n a f i n i t e number o f c o n n e c t e d components, t h e n we w i l l have P r o p e r t y (*) p r o v i d e d the p a t h i s s i m p l e . Theorem 5.2: L e t h be a c o n t i n u o u s f u n c t i o n from I t o a 65 m e t r i c space (X,p) such t h a t P r o p e r t y (*) h o l d s . Then t h e r e i s a c h o r d o f l e n g t h d wh i c h can be s l i d from one end o f the p a t h t o the o t h e r p r o v i d e d t h a t 0 < d s p ( h ( 0 ) , h ( l ) ) . P r o o f : We have p r o v e d t h a t t h e r e e x i s t x and y such t h a t ( x , 0 ) and ( l , y ) are i n the same c o n n e c t e d component C o f E^. Now we want t o use P r o p e r t y (*) t o show t h a t 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 , a l l we have to prove i s t h a t f o r any p o i n t ( s 0 > t 0 ) i n C. ev e r y n e i g h b o r h o o d o f ( s Q , t 0 ) i n C c o n t a i n s a c o n n e c t e d , n e i g h b o r h o o d o f ( s 0 , t 0 ) i n C. For any n e i g h b o r h o o d o f ( s 0 , t o ) , we can f i n d a s m a l l e r n e i g h b o r h o o d V = C O G', where G«= { ( s , t ) | sx-5 < s < s 2 + 6 , t 1 - 6 < t < t 2 + 6 } and ( s o , t 0 ) i s i n T x = { ( s , t ) | S 2 < s < s 2 , , < t < t 2 > f o r s l f s 2 , t-,^, t 2 i n D. L e t T 2 be the c l o s u r e o f T x. Then T £ n C C V. ( T 2 ~ T 1 ) n C has a f i n i t e number o f components by P r o p e r t y (*), Now we want t o show t h a t T 2 n C has o n l y a f i n i t e number o f components. Assuming t h a t t h i s i s not t r u e , t h e n t h e r e e x i s t s at l e a s t one 66 component K with K n (T 2 - T^) = <j>. Since K i s a component o£ the compact set T 2 n C, we know that i£ we choose e small enough, then by the Zorette Theorem, we can have a simple closed curve J enclosing K such that j n T 2 n c = <j> and J i s inside T j . Evidently (x,0) and (l,y) are i n the exterior of J , Hence C i s separated by J . This is impossible. So we have that T 2 n C has only a f i n i t e number of components. Hence every component of T 2 n C i s both closed and open i n T 2 n C. Let G be the component of T 2 n C such that (s ,t ) e G. Since G i s open i n T_ n C, we have o' o 2 ' G = (T 2 n C) C\ U where U i s open i n C, C s 0» t 0^ i s i n (T-L H C) n ' U C G and T 1 n C n U i s open i n C. Hence G i s a connected neighborhood of ( s 0 , t 0 ) i n C, and G i s contained i n V. Hence C i s l o c a l l y connected. This implies that C i s arcwise connected and proves the theorem. The following example shows that i f the curve i s not n i c e , then for some d, 0 < d < p(h(0),h(1)), s l i d i n g cannot take place. This example i s a curve (Figure I) i n the plane. Let h be the function from [0,1] to the plane with h(0) = T, h ( l ) = W, where h(x) goes from T to R, then returns to R via smaller and smaller c i r c l e s . In fact h(x) goes through R i n f i n i t e l y many times, and those c i r c l e s converge to the point R. Any two points i n the 67 W T ( F i g u r e I ) ar c PQ w i l l have the same d i s t a n c e from R. We c a l l t h i s d i s t a n c e d. T h i s example shows t h a t we can not s l i d e any c h o r d o f l e n g t h d from one end o f the p a t h to the o t h e r . Suppose t h e r e e x i s t mappings p and q w i t h the r e q u i r e d p r o p e r t i e s . I f h p ( s ) i s n e a r R and has not y e t p a s s e d R, t h e n we w i l l have h q ( s ) n e a r Q and i n the a r c SQ. I f h p ( s ) i s n e a r R and has j u s t p a s s e d R, t h e n we w i l l have h q ( s ) n e a r P and i n the a r c PT. Hence t h e r e e x i s t s an i n c r e a s i n g sequences {s n> such t h a t h p ( s n ) converges t o R, but hq(s2n) converges to P and hq(s2n+l) converges t o Q. T h i s means t h a t q cannot be c o n t i n u o u s . 68 In the above example, i f we take any chord of length b i- d, then we are s t i l l able to s l i d e the chord from one end of the path to the other. Now we want to extend this example by adjoining other copies of h to get a path H on which s l i d i n g cannot take place for any chord of length b e Y, where Y i s a countable sequence converging to 0 and Y C [0,p(H(0),H'(l)]. The construction of H i s made by f i r s t choosing an a r b i t r a r y p o s i t i v e number a. Then i t i s cl e a r that there i s a sequence {s^} converging to 0, which l i e s i n (0,a). We make a copy of h i n which s-^ i s the forbidden chord length. Then we extend the arc t 0 0^ such that the distance between T-^ and 0-^ i s a, and connect W^ with T (Figure I I ) . We c a l l this new path g which we 1 s 1 consider defined on t^,^] with gs-^C0) = 0^ and g s i(~) = T-p For each s. ( i i 1 ) , we can get a copy of h in which s^ i s the forbidden chord length. Then we connect with T^ to get a new closed path g s^ which we consider defined on [ ( i - 1 ) / i , i / ( i + 1 ) ] with g s . ( i^0 = T. = g s. ( r i - ) . b l i 1 si l + l We j o i n the g s^ for a l l i i n such a way that the points T- a l l c o i n c i d e , and the new copies g Q. with i f 1 are in any po s i t i o n as long as they do not meet the i n t e r i o r of the c i r c l e with center at 0-^ and radius a (Figure I I I ) . Let H(x) = g s.(x) i f x e [ ( i - l ) / i , i / ( i + l ) ] . Then H(.O) = and H(l) = T 1 # Since {s^} converges to 0, H must be continuous at 1. For any s^, there i s a unique chord of length s^ to s t a r t s l i d i n g . It i s cle a r that we cannot s l i d e the chord from one end of the path H to the other. 70 BIBLIOGRAPHY R. Sikorski and K, Zarankiewicz, On Uniformization of Functions (I) ( I I ) , Fund. Math. 41 (1954), 339-350. J . V. Whittaker, A Mountain-Climbing Problem, Canadian Journal of Mathematics, 18 (1966), 873-882. R. Fenn, S l i d i n g a Chord and the Width and Breadth of a Closed Curve, J . London Math. Soc. (2), 5 (1972), 39-47. G. T. Whyburn, Analytic Topology, A. M. S. Colloquium Publ. XXVII 1963. J . Hocking and G. Young, Topology, Addison-Wesley Publishing Company, Inc. 1961. J . Huneke, Mountain Climbing, Amer, Math. Soc Trans., 139 (1963), 383-391. T. I-Ioinma, A Theorem on Continuous Functions, Kodai Math. Sem. Reports, 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 |
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UBCV |
Scholarly Level | Graduate |
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