Equivalent metrics and compactifications

Equiv alen t metrics and compactifications Y oung Deuk Kim Sc ho ol of Mathematical Sciences Seoul National Unive rsity Seoul, 151-74 7, Korea (ydkim us@yahoo.com) No v em b er 20, 2018 Abstract Let ( X, d ) be a metric space and m ∈ X . Supp ose that φ : X × X → R is a nonnegative symmetric funct ion. W e d efine a metric d φ,m on X whic h is equ ival ent to d . If d φ,m is totally b ounded , its completion is a compact- ification of ( X, d ). As examples, we construct tw o compactifications of ( R s , d E ), where d E is th e Euclidean metric and s ≥ 2. ke y w ords. equiv alen t metric; completion; compactification Mathematics Sub ject C lassifications (2000 ). 54E35, 54D35 1 The metric d φ,m Let ( X, d ) b e a metr ic s pa ce and m ∈ X . Suppo se tha t φ : X × X → R is a nonnegative symmetric function. As usual, t wo metrics d 1 and d 2 on a set X are called equiv alent if ( X, d 1 ) and ( X , d 2 ) are homeomo rphic. In this section, we will define a metric d φ,m on X which is equiv alent to d . F or each x, y ∈ X , let δ φ,m ( x, y ) = min  d ( x, y ) , 1 1 + d ( m, x ) + φ ( x, y ) + 1 1 + d ( m, y )  . And for each x, y ∈ X and n ∈ N , let Γ n x,y = { ( x 0 , · · · , x n ) | x 0 = x, x n = y and x i ∈ X fo r all i } and Γ x,y = [ n ∈ N Γ n x,y . Notice that Γ x,y 6 = ∅ for all x, y ∈ X . In the following definition, the infimum runs over all elemen ts of Γ x,y . 1 Definition 1.1 S u pp ose that x, y ∈ X . L et d φ,m ( x, y ) = inf Γ x,y n X i =1 δ φ,m ( x i − 1 , x i ) . (1) F or the sak e o f s implicity , we w ill simply write d φ , δ φ to denote d φ,m , δ φ,m resp ectively . In particular, we write eq. (1) as d φ ( x, y ) = inf Γ x,y n X i =1 δ φ ( x i − 1 , x i ) . Notice that ( x, y ) ∈ Γ x,y , and therefor e d φ ( x, y ) = inf Γ x,y n X i =1 δ φ ( x i − 1 , x i ) ≤ δ φ ( x, y ) ≤ d ( x, y ) . (2) Notice als o that d φ is nonneg ativ e. Therefore fr om eq. (2), we hav e d φ ( x, x ) = 0 for a ll x ∈ X . (3) The follo wing subset ∆ x,y of Γ x,y is useful in the pro of of Lemma 1 .1. ∆ x,y = { ( x 0 , · · · , x n ) ∈ Γ x,y | δ φ ( x i − 1 , x i ) 6 = d ( x i − 1 , x i ) for some 1 ≤ i ≤ n } . Lemma 1 .1 Su pp ose that d φ ( x, y ) 6 = d ( x, y ) . Then d φ ( x, y ) ≥ 1 2(1 + d ( m, x )) . Pro of. Suppo s e that d φ ( x, y ) 6 = d ( x, y ). By eq. (3) w e hav e x 6 = y , and by eq. (2) we have d φ ( x, y ) < d ( x, y ) . (4) If ( x 0 , · · · , x n ) ∈ Γ x,y − ∆ x,y , then n X i =1 δ φ ( x i − 1 , x i ) = n X i =1 d ( x i − 1 , x i ) ≥ d ( x, y ) . Therefore fro m eq. (4), we hav e ∆ x,y 6 = ∅ and d φ ( x, y ) = inf ∆ x,y n X i =1 δ φ ( x i − 1 , x i ) . (5) Suppo se tha t ( x 0 , · · · , x n ) ∈ ∆ x,y . Let k b e the smallest integer such that δ φ ( x k , x k +1 ) 6 = d ( x k , x k +1 ). Notice that if k ≥ 1 then δ φ ( x i − 1 , x i ) = d ( x i − 1 , x i ) for all 1 ≤ i ≤ k . 2 If d ( x 0 , x k ) ≥ 1 + d ( m, x 0 ) then we have k ≥ 1, and therefor e n X i =1 δ φ ( x i − 1 , x i ) ≥ k X i =1 δ φ ( x i − 1 , x i ) = k X i =1 d ( x i − 1 , x i ) ≥ d ( x 0 , x k ) ≥ 1 + d ( m, x 0 ) = 1 + d ( m, x ) . (6) If d ( x 0 , x k ) < 1 + d ( m, x 0 ) then 1 + d ( m, x k ) ≤ 1 + d ( m, x 0 ) + d ( x 0 , x k ) < 2 + 2 d ( m, x 0 ) . Therefore n X i =1 δ φ ( x i − 1 , x i ) ≥ δ φ ( x k , x k +1 ) = 1 1 + d ( m, x k ) + φ ( x k , x k +1 ) + 1 1 + d ( m, x k +1 ) > 1 1 + d ( m, x k ) > 1 2(1 + d ( m, x 0 )) = 1 2(1 + d ( m, x )) . (7) Hence from eq. (5), (6) and (7), w e ha ve d φ ( x, y ) ≥ min  1 + d ( m, x ) , 1 2(1 + d ( m, x ))  = 1 2(1 + d ( m, x )) . Now we s ho w that d φ is a metric o n X . Theorem 1.1 d φ is a metric on X . Pro of. F ro m eq. (1) and (3), reca ll that d φ is nonnegative a nd d φ ( x, x ) = 0 for all x ∈ X . Supp ose that d φ ( x, y ) = 0. By Lemma 1.1, we have d ( x, y ) = d φ ( x, y ) = 0. Th us x = y . Suppo se that x, y ∈ X . Notice that ( x 0 , x 1 , · · · , x n ) ∈ Γ x,y if and only if ( x n , x n − 1 , · · · , x 0 ) ∈ Γ y ,x . Since φ is symmetric, so is δ φ . Therefor e n X i =1 δ φ ( x i − 1 , x i ) = n X i =1 δ φ ( x n +1 − i , x n − i ) for all ( x 0 , x 1 , · · · , x n ) ∈ Γ x,y . 3 Hence d φ ( x, y ) = d φ ( y , x ). Suppo se that x, y , z ∈ X and ǫ > 0. There exist ( x 0 , x 1 , · · · , x n ) ∈ Γ x,y and ( y 0 , y 1 , · · · , y m ) ∈ Γ y ,z such that n X i =1 δ φ ( x i − 1 , x i ) < d φ ( x, y ) + ǫ 2 and m X j =1 δ φ ( y j − 1 , y j ) < d φ ( y , z ) + ǫ 2 . Notice that ( x 0 , · · · , x n = y = y 0 , · · · , y m ) ∈ Γ x,z . Ther efore d φ ( x, z ) ≤ n X i =1 δ φ ( x i − 1 , x i ) + m X j =1 δ φ ( y j − 1 , y j ) < d φ ( x, y ) + ǫ 2 + d φ ( y , z ) + ǫ 2 = d φ ( x, y ) + d φ ( y , z ) + ǫ. Since ǫ is ar bitrary , we have d φ ( x, z ) ≤ d φ ( x, y ) + d φ ( y , z ). By the following le mma, the identit y map from ( X , d φ ) to ( X , d ) is contin u- ous. Lemma 1 .2 F or al l x ∈ X , ther e exists an op en b al l B x in ( X , d φ ) , with c enter x , such that d φ ( y , z ) = d ( y , z ) for al l y , z ∈ B x . Pro of. F or e a c h x ∈ X , let B x =  y ∈ X | d φ ( y , x ) < 1 8(1 + d ( m, x ))  . Suppo se that y ∈ B x . B y Lemma 1.1, we hav e d φ ( x, y ) = d ( x, y ), and therefore d ( m, y ) ≤ d ( m, x ) + d ( x , y ) = d ( m, x ) + d φ ( x, y ) < d ( m, x ) + 1 8(1 + d ( m, x )) < d ( m, x ) + 1 + d ( m, x ) = 1 + 2 d ( m, x ) . (8) Suppo se that y, z ∈ B x . F rom eq. (8 ), we hav e 1 + d ( m, y ) < 2 + 2 d ( m, x ). Therefore d φ ( y , z ) ≤ d φ ( y , x ) + d φ ( x, z ) < 1 8(1 + d ( m, x )) + 1 8(1 + d ( m, x )) = 1 4(1 + d ( m, x )) < 1 2(1 + d ( m, y )) . 4 Hence by Lemma 1.1, we hav e d φ ( y , z ) = d ( y , z ). By the following corolla ry , d φ is equiv alent to d for all φ and m . Corollary 1 .1 The identity map fr om ( X , d φ ) to ( X , d ) is a home omorphism. Pro of. By eq. (2) and Lemma 1.2, it is trivial. 2 The compactification A compactification of a top ological space X is a compact Hausdorff space Y containing X a s a s ubspace such that X = Y . It is known that every metric space has a c o mpactification (se e [6 ], § 38). With the equiv alent metric in the previous se c tio n, we a re able to construct v arious c ompactifications of a metric space. Let ( X , d ) b e a metric s pace. Supp ose that m ∈ X and φ : X × X → R is a nonnegative s ymmetric function. T o get a co mpactification, we assume that ( X, d φ ) = ( X , d φ,m ) is totally b ounded , ie. there is a finite cov ering by ǫ balls for every ǫ > 0. Then o ur compactifica tio n of ( X , d ) is the completion ( X , ρ ) o f the totally b ounded metric spa ce ( X , d φ ). Notice that X is a dense subset o f X and ( X , ρ ) is a compa ct metric space (see [6], § 45 and [3], § XIV.3 for details). X can b e considere d as the set of equiv- alence clas ses of all Cauch y sequences in ( X , d φ ) with the equiv alence r elation (see [4], § V.7) x i ∼ y i if a nd only if lim i →∞ d φ ( x i , y i ) = 0 , where a point x in X is iden tified to the eq uiv alence cla ss of constant Cauch y sequence { x } . Suppo se that { x i } , { y i } ∈ X . The metric ρ is given by ρ ( { x i } , { y i } ) = lim i →∞ d φ ( x i , y i ) . In pa rticular, we hav e ρ ( { x } , { y } ) = d φ ( x, y ) for all x, y ∈ X . In 2 002, the a uthor had tried to a pply this compa ctification to the r esearch on the tameness conjecture of Mar den([5]) whic h was proved by Agol([1]) a nd Calegar i-Gabai([2 ]) in 20 04, indep endently . The a uthor think that the com- pactification could b e useful in the study of T e ic hm ¨ uller space. In the next t wo sections, w e apply the c o mpactification to the Euclidean metric space R s with s ≥ 2. 5 3 The standard compactification of ( R s , d E ) Let O = (0 , · · · , 0) ∈ R s . W e write d E to denote the E uclidean metr ic on R s . In this section, as an example of the co mpa ctification in Section 2, we constr uct a compactification of ( R s , d E ), which will be called the standar d c omp actific ation , which is homeomorphic to the Euclidean clo sed unit ball B s = { x ∈ R s | d E ( O, x ) ≤ 1 } . Notice that w e need to define a nonnegative symmetric f unction φ : R s × R s → R such that ( R s , d φ ) = ( R s , d φ,O E ) is totally b ounded, where we wrote d φ to denote d φ,O E for the sake of simplicity . F or all m ∈ N , let a m = 1 + 1 2 + · · · + 1 m and S m = { x ∈ R s | d E ( O, x ) = a m } . Note that a m is an increasing sequence and lim m →∞ a m = ∞ . F or all p, q ∈ N , le t h p,q : S p → S q be the homeomorphism defined by h p,q ( x ) = a q a p x for all x ∈ S p . Notice that if h p,q ( x ) = y then h q,p ( y ) = x . W e define the nonnega tiv e sym- metric function φ as follows. Definition 3.1 φ ( x, y ) =      0 if h p,q ( x ) = y for some p, q ∈ N 1 a m d E ( x, y ) = d E  x a m , y a m  if x, y ∈ S m for some m ∈ N d E ( x, y ) otherwise Suppo se tha t x ∈ R s and r > 0. W e write B r ( x ) to deno te the Euclidean op en ball with cen ter x and radius r , and B φ r ( x ) to denote the op en ball in ( R s , d φ ). Now we show that ( R s , d φ ) is totally bounded. Lemma 3 .1 ( R s , d φ ) is total ly b ounde d. Pro of. Let ǫ > 0. W e ma y a ssume tha t ǫ < 1. Cho ose k ∈ N suc h that 1 1 + k < ǫ 4 and 1 1 + a k < ǫ 4 , (9) and let B k +1 = { x ∈ R s | d E ( O, x ) ≤ a k +1 } . 6 Since B k +1 is compact in ( R s , d E ), so is in ( R s , d φ ) by Corollar y 1 .1. Therefore we can cover B k +1 with finite num b er of ǫ -balls in ( R s , d φ ). Notice that S k ⊂ B k +1 . Since S k is a lso co mpact in ( R s , d E ), we can c over S k with finite nu mber of Euclidea n ǫ 4 -balls with centers x 1 , x 2 , · · · , x N ∈ S k . F rom eq. (2), we have S k ⊂ N [ i =1 B ǫ 4 ( x i ) ⊂ N [ i =1 B ǫ ( x i ) ⊂ N [ i =1 B φ ǫ ( x i ) . Note that if z ∈ S k then ther e exists x i ∈ { x 1 , x 2 , · · · , x N } ⊂ S k such that d E ( z , x i ) < ǫ 4 . T o show that ( R s , d φ ) is totally bo unded, it is enough to show that if x / ∈ B k +1 then there exists x i ∈ { x 1 , x 2 , · · · , x N } such that d φ ( x, x i ) < ǫ . Suppos e that x / ∈ B k +1 . Ther e exists m ∈ N suc h that a m ≤ d E ( O, x ) < a m +1 . Since x / ∈ B k +1 , w e have k < m . Let y = a m d E ( O, x ) x ∈ S m . F rom eq. (9), w e hav e d E ( x, y ) < 1 1 + m < 1 1 + k < ǫ 4 . (10) Let z be the p oint in S k such that h k,m ( z ) = y . Cho ose x i ∈ { x 1 , x 2 , · · · , x N } such that d E ( z , x i ) < ǫ 4 . (11) F rom eq. (2), (9), (10 ) and (11), we hav e d φ ( x, x i ) ≤ d φ ( x, y ) + d φ ( y , z ) + d φ ( z , x i ) ≤ d E ( x, y ) + δ φ ( y , z ) + d E ( z , x i ) < ǫ 4 + 1 1 + a m + 1 1 + a k + ǫ 4 < ǫ. Since ( R s , d φ ) is totally b o unded, its co mpletion ( R s , ρ ) = ( R s , ρ φ ) is a compactification of ( R s , d E ), where we wrote simply ρ to denote ρ φ for the sake of simplicit y . Recall that a n element of ( R s , ρ ) is an equiv alence clas s of Cauch y sequence in ( R s , d φ ), where t wo Cauch y seq uences { x i } and { y i } a re equiv a le n t if a nd only if lim i →∞ d φ ( x i , y i ) = 0 . 7 Notice that if { x i } is a Ca uc hy seq uence in ( R s , d φ ) which conv erges to x , then { x i } and the constant Cauch y sequence { x } are equiv alent. Notice also that if { y i } is a subsequence of a Cauch y seque nc e { x i } , then they are equiv alent. Since for all x ∈ S 1 , we hav e d φ ( a i x, a j x ) ≤ δ φ ( a i x, a j x ) ≤ 1 1 + a i + 1 1 + a j , it is clea r that { a i x } is a Cauch y seq ue nc e in ( R s , d φ ). By Lemma 1.1, we can show that { a i x } is not equiv alent to any constant Cauch y sequence (see the pro of of Lemma 3.4). F ur thermore, we hav e Lemma 3 .2 If { x i } is a Cauchy se quenc e in ( R s , d φ ) which is not e quivalent to a c onstant Cauchy se quenc e, then it is e quivalent to { a i x } for some x ∈ S 1 . Pro of. Suppo se tha t { x i } is a Cauch y sequence in ( R s , d φ ) which is not equiv - alent to a cons ta n t Cauch y sequence. If { x i } is b ounded in ( R s , d E ), then it has a conv ergent subseq uenc e { y i } , whic h co n verges to a p oint y in ( R s , d E ). Notice that { y i } co n verges to y in ( R s , d φ ), to o. Therefore { x i } is equiv a len t to { y i } , a nd hence to the co nstan t Ca uc hy sequence { y } . This is a con tradictio n. Since { x i } is unbounded in ( R s , d E ), we can choos e a subsequence of x i , which we will call x i again, suc h that 0 < d E ( O, x i ) < d E ( O, x i +1 ) for all i ∈ N and there exists at most one x i such that a m ≤ d E ( O, x i ) < a m +1 for ea c h m ∈ N . Notice tha t m → ∞ a s i → ∞ . Since 1 d E ( O, x i ) x i ∈ S 1 for all i ∈ N and ( S 1 , d E ) is co mpact, x i has a s ubsequence, whic h w e will call x i again, such that x i d E ( O, x i ) conv erges to x for some x ∈ S 1 . Suppo se that a m ≤ d E ( O, x i ) < a m +1 . Let y i = a m x . Notice that { y i } is a subsequence of { a i x } . L e t z i = a m d E ( O, x i ) x i . Since d E ( x i , z i ) ≤ 1 m +1 , w e have lim i →∞ d φ ( x i , y i ) ≤ lim i →∞  d φ ( x i , z i ) + d φ ( z i , y i )  8 ≤ lim i →∞  d E ( x i , z i ) + δ φ ( z i , y i )  ≤ lim i →∞  1 1 + m + 1 1 + a m + d E  x i d E ( O, x i ) , x  + 1 1 + a m  = 0 . Therefore { x i } a nd { y i } a r e eq uiv alent, a nd th us { x i } is e q uiv alent to { a i x } . T o show that ( R s , ρ ) is homeomorphic to ( B s , d E ), we define a function h : ( B s , d E ) → ( R s , ρ ) as follows. h ( x ) =  1 1 − d E ( O,x ) x (the constant Cauch y sequence) if d E ( O, x ) < 1 { a i x } if d E ( O, x ) = 1 Notice that h  1 1 + d E ( O, y ) y  = y for all y ∈ R s . Ther efore from Lemma 3 .2, it is clear that h is sur jectiv e. W e will need the following le mma to sho w that h is injective. Lemma 3 .3 Su pp ose t hat d E ( O, x ) ≥ 1 and d E ( O, y ) ≥ 1 . L et ( x 0 , x 1 , · · · , x m ) ∈ Γ x,y with d E ( O, x i ) < 1 for al l 1 ≤ i ≤ m − 1 . Then m X i =1 δ φ ( x i − 1 , x i ) ≥ d E  x d E ( O, x ) , y d E ( O, y )  . Pro of. Notice that w e may ass ume x d E ( O, x ) 6 = y d E ( O, y ) . If m = 1 then m X i =1 δ φ ( x i − 1 , x i ) = δ φ ( x, y ) = min  d E ( x, y ) , 1 1 + d E ( O, x ) + φ ( x, y ) + 1 1 + d E ( O, y )  ≥ min  d E ( x, y ) , 1 1 + d E ( O, x ) + d E  x d E ( O, x ) , y d E ( O, y )  + 1 1 + d E ( O, y )  ≥ d E  x d E ( O, x ) , y d E ( O, y )  . 9 Suppo se that m 6 = 1. Notice that δ φ ( x i − 1 , x i ) = d E ( x i − 1 , x i ) for all 1 ≤ i ≤ m and therefo r e m X i =1 δ φ ( x i − 1 , x i ) ≥ m X i =1 d E ( x i − 1 , x i ) ≥ d E ( x, y ) ≥ d E  x d E ( O, x ) , y d E ( O, y )  . Now we s ho w that h is injectiv e. Lemma 3 .4 h is inje ctive. Pro of. Supp ose tha t h ( x ) = h ( y ). W e will show that x = y . If d E ( O, x ) < 1 and d E ( O, y ) < 1, then 1 1 − d E ( O, x ) x = 1 1 − d E ( O, y ) y (12) and therefo r e 1 1 − d E ( O, x ) d E ( O, x ) = 1 1 − d E ( O, y ) d E ( O, y ) . Hence d E ( O, x ) = d E ( O, y ). Thus from eq. (12 ), we hav e x = y . If d E ( O, x ) = 1 and d E ( O, y ) = 1, then the Cauch y sequences { a i x } and { a i y } a re equiv alent. Supp ose that x 6 = y . W e will get a co n tradiction. Let ( x 0 , x 1 , · · · , x m ) ∈ Γ a i x,a i y . Using Lemma 3.3, we c a n show that m X i =1 δ φ ( x i − 1 , x i ) ≥ d E ( x, y ) . and therefo r e d φ ( a i x, a i y ) ≥ d E ( x, y ) > 0 for all i. (13) Hence lim i →∞ d φ ( a i x, a i y ) 6 = 0. This is a cont ra diction. Suppo se that d E ( O, x ) < 1, d E ( O, y ) = 1 and lim i →∞ d φ  1 1 − d E ( O, x ) x, a i y  = 0 . W e will g et a co n tradiction. Notice that if i is large enough, then d φ  1 1 − d E ( O, x ) x, a i y  6 = d E  1 1 − d E ( O, x ) x, a i y  . 10 Therefore by Lemma 1.1, for large enough i , w e have d φ  1 1 − d E ( O, x ) x, a i y  ≥ 1 2  1 + d E  O, 1 1 − d E ( O,x ) x  > 0 . Hence lim i →∞ d φ  1 1 − d E ( O, x ) x, a i y  6 = 0 . This is a co n tradiction. Since h is bijective, we can consider its inv erse function. Recall Lemma 3 .2 and let k : ( R s , ρ ) → ( B s , d E ) be the function defined b y k ( { x i } ) =  1 1+ d E ( O,x ) x if { x i } = { x } is a constant Cauch y sequence x if x i = a i x for some x ∈ S 1 . It is easy to s ho w that k is the in verse function o f h . In the following t wo lemmas, we will show tha t h a nd k ar e contin uous. Therefore ( R s , ρ ) is homeomorphic to ( B s , d E ). Lemma 3 .5 h is c ontinuous. Pro of. Supp ose that x n → x in ( B s , d E ). W e will s how that h ( x n ) → h ( x ) in ( R s , ρ ). If d E ( O, x ) < 1, then it is trivial to show that h ( x n ) → h ( x ) in ( R s , d E ). Ther efore from eq. (2), we hav e h ( x n ) → h ( x ) in ( R s , d φ ), and hence in ( R s , ρ ). Suppo se that d E ( O, x ) = 1. Notice that it is enough to consider only the following tw o cases, (a) d E ( O, x n ) = 1 for all n (b) d E ( O, x n ) < 1 for all n . F or the ca se (a), w e have ρ ( h ( x n ) , h ( x )) = lim i →∞ d φ ( a i x n , a i x ) ≤ lim i →∞  1 1 + a i + d E ( x n , x ) + 1 1 + a i  = d E ( x n , x ) . Therefore if x n → x in ( B s , d E ), then h ( x n ) → h ( x ) in ( R s , ρ ). F or the ca se (b), if a m ≤ d E ( O, h ( x n )) = d E  O, 1 1 − d E ( O, x n ) x n  < a m +1 , 11 let z n = a m d E ( O, h ( x n )) h ( x n ) = a m d E ( O, x n ) x n . Notice that z n ∈ S m , and m → ∞ as n → ∞ . Therefor e from eq. (2), w e hav e lim n →∞ ρ ( h ( x n ) , h ( x )) = lim n →∞ lim i →∞ d φ ( h ( x n ) , a i x ) ≤ lim n →∞ lim i →∞  d φ ( h ( x n ) , z n ) + d φ ( z n , a m x ) + d φ ( a m x, a i x )  ≤ lim n →∞ lim i →∞  d E ( h ( x n ) , z n ) + δ φ ( z n , a m x ) + δ φ ( a m x, a i x )  ≤ lim n →∞ lim i →∞  1 1 + m + 1 1 + a m + d E  h ( x n ) d E ( O, h ( x n )) , x  + 1 1 + a m + 1 1 + a m + 1 1 + a i  ≤ lim n →∞ d E  x n d E ( O, x n ) , x  = 0 . Therefore h ( x n ) → h ( x ) as n → ∞ . Lemma 3 .6 k is c ontinuous. Pro of. Supp ose that x n = { x n,i } conv erg es to x = { x i } in ( R s , ρ ). W e will show that k ( x n ) co n verges to k ( x ) in ( B s , d E ). Suppo se that x is e q uiv alent to a co nstan t Cauchy sequence { x } in ( R s , d φ ). If x n is equiv alent to { a i x n } with x n ∈ S 1 for infinitely many n, then choo se a subsequence o f x n , which we will call x n again, s uc h that x n = { a i x n } with x n ∈ S 1 . Notice that there exists I > 0, which do es not depend on n , such that d E ( a i x n , x ) ≥ 1 2(1 + d E ( O, x )) for all i > I . Therefore by Lemma 1.1, w e have d φ ( a i x n , x ) ≥ 1 2(1 + d E ( O, x )) for all i > I . Hence x n do es not conv erges to x in ( R s , ρ ). This is a contradiction. Therefore x n = { x n } is a constant Cauch y sequence in ( R s , d φ ) for la rge eno ug h n . Since x n conv erges to x in ( R s , d φ ), b y Corollar y 1.1, x n conv erges to x in ( R s , d E ). Therefore k ( x n ) = 1 1 + d E ( O, x n ) x n conv erges to k ( x ) = 1 1 + d E ( O, x ) x. 12 If x = { x i } is no t equiv a len t to a constant Cauchy sequence in ( R s , d φ ), then b y Lemma 3 .2, we ma y assume x i = a i x for s ome x ∈ S 1 . Notice that we may c o nsider only the following tw o ca ses. (a) F or all n , x n,i = a i x n for s o me x n ∈ S 1 . (b) F or all n , x n = { x n } is a consta n t Cauch y sequence. F or the ca se (a), from eq. (13) w e have 0 = lim n →∞ ρ ( x n , x ) = lim n →∞ lim i →∞ d φ ( a i x n , a i x ) ≥ lim n →∞ d E ( x n , x ) = lim n →∞ d E ( k ( x n ) , k ( x )) . F or the ca se (b), suppose that lim n →∞ d E  1 1 + d E ( O, x n ) x n , x  6 = 0 . W e will g et a co n tradiction. Cho ose a subsequence { y n } o f { x n } such that 1 1 + d E ( O, y n ) y n → y 6 = x in ( B s , d E ) . Since h is con tinuous and injective, we ha ve y n = h  1 1 + d E ( O, y n ) y n  → h ( y ) 6 = h ( x ) = x in ( R s , ρ ) . Therefore lim n →∞ ρ ( y n , x ) 6 = 0. This is a cont ra diction. 4 A compactification of ( R s , d E ) whic h is not equiv- alen t to the standard compactification Two compactifications Y 1 and Y 2 of a top ological space X ar e called e quiv alent if ther e exists a homeomorphism h : Y 1 → Y 2 such that h ( x ) = x for all x ∈ X . Recall that s ≥ 2. In this se c tion, w e constr uc t a compactification of ( R s , d E ) which is homeomor phic to the clo sed unit ball ( B s , d E ), but not equiv alent to the standard compactification ( R s , ρ φ ) in Section 3. W e define a nonnegative symmetric function ψ : R s × R s → R as follows. Cho ose 0 < δ < π 4 and let A + = { x ∈ S 1 | ∠ xO a 1 ≤ δ } , A − = { x ∈ S 1 | ∠ xO ( − a 1 ) ≤ δ } , where a 1 = (1 , 0 , · · · , 0) and − a 1 = ( − 1 , 0 , · · · , 0) ∈ R s . F or each x ∈ S 1 , let P x = { t a 1 + t ′ x ∈ R s | t, t ′ ∈ R } . 13 W e define an infinite ray L x ⊂ P x starting from x a s follo ws. See Figure 1, where θ = π π − 2 δ ( ∠ xO a 1 − δ ) . P S f r a g r e p la c e m e n t s O L x L x x x y a 1 θ δ Figure 1 : L x L x =    { x + t a 1 | t ≥ 0 } if x ∈ A + { x + t a 1 | t ≤ 0 } if x ∈ A − { x, y ∈ P x | ∠ ( y − x ) O a 1 = π π − 2 δ ( ∠ xO a 1 − δ ) } if x ∈ S 1 \ ( A + ∪ A − ) and let L = { L x | x ∈ S 1 } . Notice that (i) If ∠ xO a 1 = π 2 , then L x = { tx | t ≥ 1 } . (ii) F or all x ∈ S 1 , the a ngle b et ween t wo rays L x and { tx | t ≥ 1 } is not greater than δ . (iii) F or all y ∈ R s with d E ( O, y ) ≥ 1, ther e ex ists unique ray in L which is through y . F or all p, q ∈ N , le t h p,q : S p → S q be the homeomorphism defined by h p,q ( x ) = the intersection of S q and the ray in L which is through x. In particular , we hav e h p,p ( x ) = x , and if h p,q ( x ) = y then h q,p ( y ) = x . The nonnegative s ymmetric function ψ is defined a s follows. Definition 4.1 ψ ( x, y ) =    0 if h p,q ( x ) = y for some p, q ∈ N d E ( h m, 1 ( x ) , h m, 1 ( y )) if x, y ∈ S m for some m ∈ N d E ( x, y ) otherwise. 14 Similarly as in Section 3, we can show that ( R s , d ψ ) = ( R s , d ψ ,O E ) is to- tally b ounded, and its completion ( R s , ρ ψ ) is homeo morphic to ( B s , d E ) b y the following homeo morphism h : ( B s , d E ) → ( R s , ρ ψ ), h ( x ) =          1 1 − d E ( O,x ) x if d E ( O, x ) < 1 2 y ∈ L x d E ( O,x ) such that d E  x d E ( O,x ) , y  = d E ( O,x ) − 1 2 1 − d E ( O,x ) if 1 2 ≤ d E ( O, x ) < 1 { h 1 ,i ( x ) } if d E ( O, x ) = 1 . Suppo se that A, B ⊂ R s . Le t d E ( A, B ) = inf { d E ( x, y ) | x ∈ A, y ∈ B } . In spherica l co ordinate sy s tem the distance b etw een ( ρ 1 , φ 1 , θ 1 ) and ( ρ 2 , φ 2 , θ 2 ) is q ρ 2 1 + ρ 2 2 − 2 ρ 1 ρ 2 { sin φ 1 sin φ 2 cos( θ 1 − θ 2 ) + cos φ 1 cos φ 2 } . (14) The following tw o Lemmas are us e ful to show that h is a ho meomorphism. Lemma 4 .1 Su pp ose that x, y ∈ S 1 . Then d E ( L x , L y ) ≥ 1 2 √ 2 d E ( x, y ) . Pro of. P S f r a g r e p la c e m e n t s a 1 α β O L x L ∗ x L w x w φ 2 Figure 2 : Since 0 ≤ α ≤ δ < π 4 , w e have π 4 < β ≤ 3 π 4 . W e may as sume that x 6 = y . Since there ex is ts a 3-dimensional subspa ce which contains O , a 1 , − a 1 , x and y , we may a ssume that R s = R 3 . In spherical co or- dinates ( ρ, φ, θ ), let O = (0 , 0 , 0), a 1 = (1 , 0 , 0), − a 1 = (1 , π , 0), x = (1 , φ 1 , θ 1 ) 15 and y = (1 , φ 2 , θ 2 ). By exchanging x and y if necessar y , we may a ssume that 0 ≤ φ 1 ≤ π 2 and φ 1 ≤ φ 2 . Suppo se that φ 2 ≤ π 2 . Let z = (1 , φ 1 , θ 2 ) and w = (1 , φ 2 , θ 1 ). Since d E ( x, y ) ≤ d E ( x, z ) + d E ( z , y ) and d E ( x, w ) = d E ( z , y ), we hav e d E ( x, z ) ≥ 1 2 d E ( x, y ) or d E ( x, w ) ≥ 1 2 d E ( x, y ) . If d E ( x, z ) ≥ 1 2 d E ( x, y ), let P b e the plane containing x and z which is p er- pendicula r to a 1 . Let L ′ x be the pro jection of L x to the plane P and so is L ′ y . Notice that we have d E ( L x , L y ) ≥ d E ( L ′ x , L ′ y ) ≥ d E ( x, z ) ≥ 1 2 d E ( x, y ) . Suppo se that d E ( x, w ) ≥ 1 2 d E ( x, y ). Let L ∗ x be the ray starting from x with the same direction as L w . Since L w = { ( ρ, φ, θ 1 ) | ( ρ, φ, θ 2 ) ∈ L y } , from eq. (14) and Figur e 2, we ha ve d E ( L x , L y ) ≥ d E ( L x , L w ) ≥ d E ( L ∗ x , L w ) ≥ 1 √ 2 d E ( x, w ) ≥ 1 2 √ 2 d E ( x, y ) . P S f r a g r e p la c e m e n t s α α O L ′′ x L ′′ y L ∗∗ x L ∗∗ y L ∗ z ′ x y z ′ Figure 3 : π 4 ≤ α ≤ π 2 Suppo se that φ 2 > π 2 . Let P ′ be the plane which c o n tains the g reatest circle in S 1 through the p oints x a nd y . Let z ′ be the p oint on the greatest circle such 16 that ∠ xOz ′ = ∠ y Oz ′ ≤ π 2 . Let L ′′ x be the pro jection of L x to the plane P ′ and so is L ′′ y . Le t L ∗ z ′ = { tz ′ | t ≥ 1 } . Let L ∗∗ x be the ray from x to the direction of L ∗ z ′ and so is L ∗∗ y . F rom Figur e 3, we have d E ( L x , L y ) ≥ d E ( L ′′ x , L ′′ y ) ≥ d E ( L ∗∗ , L ∗∗ y ) = d E ( L ∗∗ x , L ∗ z ′ ) + d E ( L ∗ z ′ , L ∗∗ y ) ≥ 1 √ 2 d E ( x, z ′ ) + 1 √ 2 d E ( z ′ , y ) ≥ 1 √ 2 d E ( x, y ) . Lemma 4 .2 Su pp ose that a m ≤ d E ( O, x ) < a m +1 . L et y b e the interse ction of S m and the r ay in L which is t hr ough x . Then we have d E ( y , x ) ≤ 1 cos δ 1 m + 1 ≤ √ 2 m + 1 . Pro of. Recall that 0 < δ < π 4 . F rom Figure 4, the pro of is trivia l. P S f r a g r e p la c e m e n t s α β O L z x y z S 1 S m S m +1 δ Figure 4 : β < α ≤ δ < π 4 The following Lemma is also us e ful to show that h is a homeomorphism. Similarly as Lemma 3 .3, we can pr o ve this lemma. 17 Lemma 4 .3 Su pp ose that d E ( O, x ) ≥ 1 and d E ( O, y ) ≥ 1 . S upp ose also that x ∈ L x ′ and y ∈ L y ′ with x ′ , y ′ ∈ S 1 . L et ( x 0 , x 1 , · · · , x m ) ∈ Γ x,y with d E ( O, x i ) < 1 for al l 1 ≤ i ≤ m − 1 . Then m X i =1 δ ψ ( x i − 1 , x i ) ≥ d E ( L x ′ , L y ′ ) . Now w e show that the compactifica tion ( R s , ρ ψ ) of ( R s , d E ) is not equiv a len t to the standard compactification ( R s , ρ φ ) in Section 3 . Prop osition 4 .1 ( R s , ρ ψ ) and ( R s , ρ φ ) ar e not e qu ival ent c omp actific ations. Pro of. Supp ose that they are equiv alent. There exists a homeomorphism h : ( R s , ρ ψ ) → ( R s , ρ φ ) such that h ( x ) = x for all x ∈ R s . Cho ose a p oint b 1 ∈ S 1 such that ∠ b 1 O a 1 = δ 2 . Let a = { a i } a nd b = { b i } , wher e a i = h 1 ,i ( a 1 ) = ( a i , 0 , · · · , 0) and b i = h 1 ,i ( b 1 ) for a ll i ∈ N . Notice that sin δ 2 ≤ d E ( a i , b i ) ≤ δ 2 for a ll i. Suppo se that ( x 0 , x 1 , · · · , x m ) ∈ Γ a i , b i . Using Lemma 4.1 a nd 4 .3, we can show that m X i =1 δ ψ ( x i − 1 , x i ) ≥ 1 2 √ 2 d E ( a 1 , b 1 ) ≥ 1 2 √ 2 sin δ 2 > 0 . Therefore d ψ ( a i , b i ) ≥ 1 2 √ 2 sin δ 2 for a ll i, and hence ρ ψ ( a , b ) 6 = 0. Thus a 6 = b in ( R s , ρ ψ ). But we have ρ φ ( h ( a ) , h ( b )) = lim i →∞ ρ φ ( h ( a i ) , h ( b i )) = lim i →∞ ρ φ ( a i , b i ) = lim i →∞ d φ ( a i , b i ) ≤ lim i →∞ δ φ ( a i , b i ) ≤ lim i →∞  1 1 + a i + 1 a i d E ( a i , b i ) + 1 1 + a i  ≤ lim i →∞  2 1 + a i + δ 2 a i  = 0 . 18 Therefore h ( a ) = h ( b ) in ( R s , ρ φ ). This is a contradiction. References [1] I. Agol, T ameness of hyp erb olic 3-manifolds , preprint, arXiv:math. GT/0405 568. [2] D. Calega ri a nd D. Gabai, S hrinkwr apping and taming of hyp erb olic 3- manifolds , J. Amer. Math. So c. 1 9 (2006), no.2, 385-44 6. [3] J. Dugundji, T op olo gy , Allyn and Bacon, Boston, 1966. [4] S. A. Gaal, Point set toplo gy , Academic Press, New Y ork and London, 196 4. [5] A. Marden, The ge ometr y of finitely gener ate d Kleinian gr oups , Ann. of Math. (2) 99 (19 74), 383-4 62. [6] J. R. Munkres, T op olo gy at fi rst c ourse , Prentice-Hall, New Jersey , 197 5. 19