Topologies on sets of polynomial knots and the homotopy types of the respective spaces
aa r X i v : . [ m a t h . GN ] J a n Topologies on sets of polynomial knots and thehomotopy types of the respective spaces
Hitesh Raundal
Harish-Chandra Research Institute, Chhatnag Road, Jhunsi, Prayagraj 211019, [email protected]
Abstract
A polynomial knot in R n is a smooth embedding R ֒ → R n such that the componentfunctions are real polynomials. In the earlier paper with Mishra, we have studied thespace P of polynomial knots in R with the inductive limit topology coming from thespaces O d for d ≥ , where O d is the space of polynomial knots in R with degree d and having some conditions on the degrees of the component polynomials. In thesame paper, we have proved that the space of polynomial knots in R has the samehomotopy type as S . The homotopy type of the space is the mere consequence of thetopology chosen. If we have another topology on P , the homotopy type may change.With this in mind, we consider in general the set L n of all polynomial knots in R n with various topologies on it and study the homotopy type of the respective spaces.Let L be the union of the sets L n for n ≥ . We also explore the homotopy type ofthe space L with some natural topologies on it. The space of knots as embeddings of S in S with some specific behavior at the northpole can be approximated by the spaces of embeddings of R into R n of the form t (cid:0) t d + a t d − + · · · + a d − t, t d + a t d − + · · · + a d − t, · · · , t d + a n t d − + · · · + a nd − t (cid:1) , for example see [1, 10, 11]. Let K nd be the space of maps from R to R n of the form aboveand let Σ nd be the discriminant of this space. In [11], Vassiliev discussed and proved thefollowing result: Theorem 1.1. [11, Theorem 1] For n ≥ the space K n \ Σ n is contractible and the space K n \ Σ n is homology equivalent to S n − . For n ≥ the space K n \ Σ n is also homotopyequivalent to S n − . The generator of the group H n − ( K n \ Σ n ) is equal to the linkingnumber in K n with the set of maps above with singular points. This result is about investigating the homology and homotopy of some specific kind ofspace of embeddings of R in R n whose component functions are real polynomials. Withthis motivation, Durfee and O’Shea [3] introduced the space M nd \ Σ nd , where M nd is the Key words and phrases:
Embedding, Polynomial knot, Topologies on a set, Homotopy type of a space,Homotopy equivalence. itesh Raundal space of maps from R to R n whose component functions are real polynomials, and Σ nd is the discriminant of the space M nd . If two members of the space M nd \ Σ nd are pathequivalent, then they are topologically equivalent (i.e. if two members belong to thesame path component of the space M nd \ Σ nd , then their extensions as embeddings of S in S are ambient isotopic), see [3]. A smooth embedding of R in R n given by t ( f ( t ) , f ( t ) , . . . , f n ( t )) , where f , f , . . . , f n are real polynomials, is referred as polynomialknot. Note that the spaces K nd \ Σ nd and M nd \ Σ nd are spaces of polynomial knots of degree d with some specific conditions on the component polynomials.Shatri [8] proved that every knot has polynomial representation. That is, for a givenknot as an embedding of S in S , there exists a polynomial knot φ : R ֒ → R n whoseextension is ambient isotopic to the original one. Furthermore, it has been proved that iftwo polynomial knots represent the same knot-type, then there exists a polynomial isotopy F : [0 , × R → R which connects them, see [9, Theorem 2.3]. If we have a long knot asan embedding of R in R , then it can be rolled to the infinity to make it into a straightline. In other words, for a long knot φ , there exists an isotopy H : [0 , × R → R of longknots that connects φ to an unknot (i.e. H s : R → R given by t H ( s, t ) is a long knotfor s ∈ [0 , H = φ and H is an unknot). Since a polynomial knot is also a long knot,there exists an isotopy of long knots (i.e. an isotopy of the type H above) connecting agiven polynomial knot to an unknot. Here, it is natural to ask that is there a polynomialisotopy that connects a given polynomial knot to a linear one (to an unknot)? We havethe affirmative answer to this question and this fact has been proved in my earlier paper[6] with Mishra. In the same paper, we have introduced the spaces O d , P d and Q d ofpolynomial knots as embeddings of R in R with the degrees of component polynomialsless than or equal to d and having some specific conditions on the degrees. We proved thatif two polynomial knots belong to the same path component of the space Q d , then theirextensions as embeddings of S in S are ambient isotopic. In other words, if we couldprove that if two knots as embeddings of S in S have polynomial representations in Q d belonging to different path components, then the knots are not ambient isotopic (are notequivalent as knots). In [5], we have found the minimal polynomial degree of all knots upto six crossings except the knots 5 , 6 , 6 and 6 that have representations in degree 7 (seeRolfsen’s table in [4, Appendix E] and [7] for the notations of the knots). If we could provethat all the path components of the space Q are occupied by the knots up to six crossings(each path component contains at least one member representing either of the knot up tosix crossings) except the knots 5 , 6 , 6 and 6 , then there will not be representations forthe knots 5 , 6 , 6 and 6 in degree 6, and hence their minimal polynomial degree willbe 7. With this regard, one can see that by studding spaces of polynomial knots wouldhelp in drawing some inferences in the point of view of knot theory. Reader may refer to[2, 4, 7] for basic knot theory.In [6], we studied the homotopy types of the spaces O d for d ≥ S . Using this result and some techniques in thedifferential topology, we proved that the space P of polynomial knots as embeddings of R in R with the inductive limit topology coming from the spaces O d for d ≥ S , see [6, Corollary 4.15]. The space P can be given various topologiesthat come in a natural way. According to the topology given, the homotopy type of thespace P may change. To look into this problem, we consider more generalized spaces L n ,for n ≥
1, of polynomial knots together with four types of various topologies on them.The space L n is the space of all polynomial knots as embeddings of R in R n . Note that2 opologies on sets of polynomial knots the space L is the space P discussed in [6]. We also consider the space L as union of thespaces L n for n ≥
1. By identifying a polynomial knot φ : R ֒ → R n with the embedding˜ φ : R ֒ → R ∞ via the standard embedding of R n in R ∞ and then eventually with a Λ-tuple( φ ij ) i,j , we can think of the sets L n and L as subsets of R Λ , where Λ = Z + × N and φ ij is the coefficient of t j in the i th component of ˜ φ . The sets L n and L can be given thesubspace topologies that inherit from the box and the product topologies of R Λ . We alsohave the topologies on L n and L induced by the metrics d r (for r ≥
1) and d ∞ given by d r ( φ, ψ ) = X i,j | φ ij − ψ ij | r /r and d ∞ ( φ, ψ ) = sup i,j | φ ij − ψ ij | for φ, ψ ∈ L . In this paper, we study the homotopy types of the spaces L n and L withrespect to the topologies discussed above. We see that the homotopy types of the spacesare independent on the topologies chosen and the results are compatible with the resultproved in [6] regarding the homotopy type of the space P .This paper is organized as follows: In Section 2, we define sets L n and L , and discusssome various natural topologies on them. In Section 3, we compare the topologies describedin the previous section. In Section 4, we prove the results (see Theorem 4.1 and Theorem4.2) regarding the homotopy types of the spaces L n and L with the topologies describedin Section 2. L n and L For a positive integer n , we define a polynomial knot in R n as follows: Definition 2.1.
A polynomial knot in R n is a smooth embedding R ֒ → R n such that thecomponent functions are real polynomials. In other words, a polynomial knot in R n is a map φ : R → R n given by t ( φ ( t ) , φ ( t ) , . . . , φ n ( t )) such that it satisfies the following conditions:(a) φ i ∈ R [ t ] for all i = 1 , , . . . , n ,(b) φ ( s ) = φ ( t ) for all s = t in R , and(c) φ ′ ( t ) = 0 for all t ∈ R . Notation 2.1.
For n ≥ , let L n denote the set of all polynomial knots in R n , and let L = S n ∈ Z + L n . Notation 2.2.
Let Λ denote the set Z + × N , where Z + is the set of all positive integersand N is the set of all nonnegative integers. Given a polynomial knot φ : R ֒ → R n , it can be thought of as an embedding ˜ φ of R in R ∞ via the standard embedding of R n in R ∞ . Let ˜ φ ( t ) = P j ( φ ij ) i t j for t ∈ R , i.e. φ ij (for ( i, j ) ∈ Λ) is the coefficient of t j in the i th component of ˜ φ , where Λ = Z + × N .We can identify the polynomial knot φ with the Λ-tuple ( φ ij ) i,j of real numbers. Withthis identification of a polynomial knot, we can think of the sets L n (for n ≥
1) and L as3 itesh Raundal subsets of R Λ (where R Λ is the set of all Λ-tuples of real numbers). The sets L n and L can be redefined as L = ( φ ij ) i,j ∈ R Λ | (i) φ ij = 0 for all but finitely many ( i, j ) ∈ Λ , and (ii) the map φ : R → R ∞ given by t X j ( φ ij ) i t j is a smooth embedding , L n = n ( φ ij ) i,j ∈ L | φ ij = 0 if i > n and j ≥ o . We have the subspace topologies on L n and L that inherit from the box and theproduct topologies of R Λ . The sets L n and L can be equipped with the metric topologies.Let r ≥ d r be the metric on L defined as d r ( φ, ψ ) = X i,j | φ ij − ψ ij | r /r for φ, ψ ∈ L , where φ ij and ψ ij (for ( i, j ) ∈ Λ) are the coefficients of t j in the i th compo-nents of φ and ψ respectively. Denote an open ball centered at φ ∈ L and of radius δ > B r ( φ, δ ), i.e. B r ( φ, δ ) = { ψ ∈ L : d r ( φ, ψ ) < δ } . This type of balls generate a topologyon L . We have another metric d ∞ on L defined as d ∞ ( φ, ψ ) = sup i,j | φ ij − ψ ij | for φ, ψ ∈ L . This metric induces a topology on L , i.e. the topology is generated bythe balls of the type B ∞ ( φ, δ ) = { ψ ∈ L : d ∞ ( φ, ψ ) < δ } , where φ ∈ L and δ >
0. Since L n is a subset of L , we have the topologies on L n induced by the metrics d r and d ∞ when restricted to L n . Note that these topologies are generated by the balls of the types B nr ( φ, δ ) = { ψ ∈ L n : d r ( φ, ψ ) < δ } and B n ∞ ( φ, δ ) = { ψ ∈ L n : d ∞ ( φ, ψ ) < δ } respectively,where φ ∈ L n and δ >
0. We denote the topologies on L n and L as in Table 1.Table 1: Topologies on L n and L Description Notations for thetopologies on L Notations for thetopologies on L n The subspace topology that inherit fromthe box topology of R Λ T b T nb The subspace topology that inherit fromthe product topology of R Λ T p T np The metric topology induced by themetric d r on L (respectively by d r | L n ) T r T nr The metric topology induced by themetric d ∞ on L (respectively by d ∞ | L n ) T ∞ T n ∞ opologies on sets of polynomial knots Remark 2.1.
It is trivial that the spaces R Λ and R ∞ each with the box topology (or withthe product topology) are homeomorphic. Therefore, the spaces ( L n , T nb ) and ( L , T b ) canbe considered as subspaces of the space R ∞ with the box topology, and the spaces (cid:0) L n , T np (cid:1) and ( L , T p ) can be regarded as subspaces of the space R ∞ with the product topology. L n and L In this section, we compare the topologies T b , T p , T r , T s (for r > s ≥
1) and T ∞ on L and the topologies T nb , T np , T nr , T ns and T n ∞ on L n . Lemma 3.1.
Let r and s be real numbers such that r ≥ s ≥ and suppose a , a , . . . , a k be nonnegative real numbers. Then (cid:16)P ki =1 a ir (cid:17) /r ≤ (cid:16)P ki =1 a is (cid:17) /s .Proof. Let u = rs and let l be the integer such that u = l + v for some 0 ≤ v <
1. Notethat l ≥
1. Let b i = a is for i = 1 , , . . . , k and b = P ki =1 b i . We have following: k X i =1 b iu ≤ k X i =1 b iv b il ≤ k X i =1 b v b il ≤ b v k X i =1 b il ≤ k X i =1 b i ! v k X i =1 b i ! l ≤ k X i =1 b i ! u . Thus, P ki =1 a ir ≤ (cid:16)P ki =1 a is (cid:17) r/s and hence the result. Proposition 3.2.
For r > s ≥ , we have the relation T p ⊆ T ∞ ⊆ T r ⊆ T s ⊆ T b betweenthe topologies on L .Proof. We prove the proposition in the following parts:(a) To prove T p ⊆ T ∞ :The topology T p has the basis containing the sets of the type L T Q i,j U ij , where U ij is open in R for all ( i, j ) ∈ Λ and U ij = R for all but finitely many ( i, j ) ∈ Λ. Let U = L T Q i,j U ij be an open set in this basis. We show that U is open in ( L , T ∞ ).Suppose an element φ ∈ U be given. It is sufficient to show that B ∞ ( φ, δ ) ⊆ U for some δ >
0. Let Λ U = { ( i, j ) ∈ Λ : U ij = R } . Note that Λ U is a finite set. Since φ ij ∈ U ij for( i, j ) ∈ Λ U , we can choose a δ > φ ij − δ, φ ij + δ ) ⊆ U ij for all ( i, j ) ∈ Λ U . Nowconsider an open ball B ∞ ( φ, δ ). Suppose ψ ∈ B ∞ ( φ, δ ). Then d ∞ ( φ, ψ ) < δ and hence ψ ij ∈ ( φ ij − δ, φ ij + δ ) for all ( i, j ) ∈ Λ U . This implies that ψ ij ∈ U ij for all ( i, j ) ∈ Λ U .Note that ψ ij ∈ U ij for all ( i, j ) ∈ Λ \ Λ U , since U ij = R for ( i, j ) ∈ Λ \ Λ U . Thus, ψ ∈ U .This shows that B ∞ ( φ, δ ) ⊆ U .(b) To prove T ∞ ⊆ T r :Let B ∞ ( φ, ǫ ), for φ ∈ L and ǫ >
0, be an open ball in ( L , d ∞ ). We show that B ∞ ( φ, ǫ )is open in ( L , d r ). Suppose an element ψ ∈ B ∞ ( φ, ǫ ) be given. It is enough to show that B r ( ψ, δ ) ⊆ B ∞ ( φ, ǫ ) for some δ >
0. Take δ = ǫ − d ∞ ( φ,ψ )2 . Since d ∞ ( ψ, ω ) ≤ d r ( ψ, ω ) for all ω ∈ L , we have d ∞ ( φ, ω ) ≤ d ∞ ( φ, ψ ) + d ∞ ( ψ, ω ) ≤ d ∞ ( φ, ψ ) + d r ( ψ, ω ) ≤ d ∞ ( φ, ψ ) + δ < ǫ whenever d r ( ψ, ω ) < δ . This shows that B r ( ψ, δ ) ⊆ B ∞ ( φ, ǫ ).(c) To prove T r ⊆ T s :By Lemma 3.1, we have d r ( φ, ψ ) ≤ d s ( φ, ψ ) for all φ, ψ ∈ L . The rest of the prooffollows in a similar way as in the previous part.5 itesh Raundal (d) To prove T s ⊆ T b :Let B s ( φ, ǫ ), for φ ∈ L and ǫ >
0, be an open ball in ( L , d s ). We show that B s ( φ, ǫ )is open in ( L , T b ). Suppose an element ψ ∈ B s ( φ, ǫ ) be given. The topology T b hasthe basis containing the sets of the type L T Q i,j V ij , where V ij is open in R for all( i, j ) ∈ Λ. It is sufficient to show that there is a basic open neighborhood V of ψ suchthat V ⊆ B s ( φ, ǫ ). Let δ = min { / , ǫ − d s ( φ, ψ ) } and V = L T Q i,j V ij , where V ij = (cid:0) ψ ij − δ i ( j +1) , ψ ij + δ i ( j +1) (cid:1) for ( i, j ) ∈ Λ. Note that ψ ∈ V . Suppose ω ∈ V . Then | ψ ij − ω ij | < δ i ( j +1) for all ( i, j ) ∈ Λ. Let Λ ω = { ( i, j ) ∈ Λ : ψ ij = 0 or ω ij = 0 } . Notethat Λ ω is a finite set. We have d ( ψ, ω ) ≤ P ( i,j ) ∈ Λ ω | ψ ij − ω ij | ≤ P ( i,j ) ∈ Λ ω δ i ( j +1) ≤ P ∞ i =1 P ∞ j =0 δ i ( j +1) ≤ P ∞ i =1 δ i − δ i ≤ P ∞ i =1 δ i ≤ δ − δ ≤ δ (since δ ≤ / d s ( ψ, ω ) ≤ d ( ψ, ω ); thus, d s ( φ, ω ) ≤ d s ( φ, ψ ) + d s ( ψ, ω ) ≤ d s ( φ, ψ ) + d ( ψ, ω ) ≤ d s ( φ, ψ ) + δ < ǫ (since δ ≤ ǫ − d s ( φ, ψ ) and δ ≤ / ω ∈ B s ( φ, ǫ ). Thisshows that V ⊆ B s ( φ, ǫ ).For n ∈ Z + and for real numbers r > s ≥
1, the topologies T np , T n ∞ , T nr , T ns and T nb canbe seen as the subspace topologies that inherit from the topologies T p , T ∞ , T r , T s and T b respectively. Thus, the next corollary follows trivially from Proposition 3.2. Corollary 3.2.1.
For n ∈ Z + and for r > s ≥ , we have the relation T np ⊆ T n ∞ ⊆ T nr ⊆ T ns ⊆ T nb between the topologies on L n . Proposition 3.3.
For n ∈ Z + and for r > s ≥ , we have the relation T p ( T ∞ ( T r ( T s ( T b between the topologies on L and the relation T np ( T n ∞ ( T nr ( T ns ( T nb betweenthe topologies on L n .Proof. Since the topologies T np , T n ∞ , T nr , T ns and T nb can be viewed as the subspace topolo-gies that inherit from the topologies T p , T ∞ , T r , T s and T b respectively, it is enough toshow that the set inclusions in Corollary 3.2.1 are proper. We do this in the followingparts:(a) To show the set inclusion T np ⊆ T n ∞ is proper:Let φ : R → R n be defined as t ( t, , . . . , φ ∈ L n and B n ∞ ( φ, / ∈ T n ∞ .We show that B n ∞ ( φ, / / ∈ T np . Suppose contrary that B n ∞ ( φ, / ∈ T np . Then there isa basic open neighborhood U = L n T Q i,j U ij of φ such that U ⊆ B n ∞ ( φ, / U ij is open in R for all ( i, j ) ∈ Λ and U ij = R for all but finitely many ( i, j ) ∈ Λ.Choose an odd positive integer k such that U k = R . Let ψ : R → R n be defined as t (cid:0) t k + t, , . . . , (cid:1) . One can see that ψ ∈ L n , ψ ij ∈ U ij for all ( i, j ) ∈ Λ \ { (1 , k ) } (since ψ ij = φ ij and φ ij ∈ U ij ), ψ k ∈ U k (since U k = R ) and d ∞ ( φ, ψ ) = 1. Thus, ψ ∈ U and ψ / ∈ B n ∞ ( φ, / T n ∞ ⊆ T nr is proper:Let φ be defined as in the first part. Note that B nr ( φ, / ∈ T nr . We show that B nr ( φ, / / ∈ T n ∞ . Suppose contrary that B nr ( φ, / ∈ T n ∞ . Then there is a δ > B n ∞ ( φ, δ ) ⊆ B nr ( φ, / k > δ − r . Let ω : R → R n be defined as t (cid:16) k − r (cid:0) t k +1 + t k − + · · · + t (cid:1) + t, , . . . , (cid:17) . We can see that ω ∈ L n , d ∞ ( φ, ω ) = k − r < δ and d r ( φ, ω ) = 1. Thus, ω ∈ B n ∞ ( φ, δ ) and ω / ∈ B nr ( φ, / opologies on sets of polynomial knots (c) To show the set inclusion T nr ⊆ T ns is proper:Let φ be defined as in the first part. Note that B ns ( φ, / ∈ T ns . We show that B ns ( φ, / / ∈ T nr . Suppose contrary that B ns ( φ, / ∈ T nr . Then there is a δ > B nr ( φ, δ ) ⊆ B ns ( φ, / k > δ rss − r . Let σ : R → R n be defined as t (cid:16) k − s (cid:0) t k +1 + t k − + · · · + t (cid:1) + t, , . . . , (cid:17) . One can see that σ ∈ L n , d r ( φ, σ ) = k s − rrs < δ and d s ( φ, σ ) = 1. Thus, σ ∈ B nr ( φ, δ ) and σ / ∈ B ns ( φ, / T ns ⊆ T nb is proper:Let φ be defined as in the first part. Let U = L n T Q i,j U ij , where U ij = (cid:16) − , i ( j +1) (cid:17) .Note that U ∈ T nb . Since φ ij = 0 ∈ U ij for all ( i, j ) ∈ Λ \ { (1 , } and φ = 1 ∈ U , φ ∈ U . We show that U / ∈ T ns . Suppose contrary that U ∈ T ns . Then there is a δ > B ns ( φ, δ ) ⊆ U . Choose an odd positive integer k > − δδ . Let ξ : R → R n bedefined as t (cid:16) k +1 t k + t, , . . . , (cid:17) . Since d s ( φ, ξ ) = k +1 < δ and ξ k ≥ k +1 > k +1 (i.e. ξ k / ∈ U k ), we have ξ ∈ B ns ( φ, δ ) and ξ / ∈ U . This is a contradiction. L n and L In this section, we prove the following theorems regarding the homotopy types of thespaces L n and L with the topologies as in Table 1. Theorem 4.1.
The space L n having any of the topologies T nb , T np , T nr (for r ≥ ) and T n ∞ has the same homotopy type as S n − . Theorem 4.2.
The space L having any of the topologies T b , T p , T r (for r ≥ ) and T ∞ is contractible. Let us go through some lemmas that will be used to prove Theorem 4.1 and Theorem4.2.
Lemma 4.3.
Let F : R Λ → R ∞ be defined as ( x ij ) i,j ( x i ) i , where ( x i ) i is a sequencehaving its i th coordinate x i for i ∈ Z + . Suppose G : R ∞ → R Λ be defined as ( y i ) i (cid:0) | j − | y i (cid:1) i,j , where = 1 and the tuple on the right has its ( i, j ) th coordinate | j − | y i for ( i, j ) ∈ Λ . Suppose R Λ and R ∞ have the box topologies or both have the product topologies.Then the maps F and G are continuous.Proof. (a) To prove F is continuous:Let U = Q i U i be a basic open set in the box topology of R ∞ (i.e. U i is open in R forall i ∈ Z + ). One can see that F − ( U ) = Q i,j U ij , where U i = U i for i ∈ Z + and U ij = R for ( i, j ) ∈ Z + × N \ { } . Thus, F − ( U ) is open in the box topology of R Λ . If U i = R forall but finitely many i ∈ Z + , i.e. if U is a basic open set in the product topology of R ∞ ,then F − ( U ) is open in the product topology of R Λ .(b) To prove G is continuous:Let V = Q i,j V ij be a basic open set in the box topology of R Λ (i.e. V ij is open in R for all ( i, j ) ∈ Λ). If there is an ( i, j ) ∈ Z + × N \ { } such that 0 / ∈ V ij , then G − ( V ) is an7 itesh Raundal empty set. If 0 ∈ V ij for all ( i, j ) ∈ Z + × N \ { } , then G − ( V ) = Q i V i , where V i = V i for i ∈ Z + . Thus, in either case, the set G − ( V ) is open in the box topology of R ∞ . If V ij = R for all but finitely many ( i, j ) ∈ Λ, i.e. if V is a basic open set in the producttopology of R Λ , then G − ( V ) is open in the product topology of R ∞ . Notation 4.1.
Let E = { ( x i ) i ∈ R ∞ \ { } | x i = 0 for all but finitely many i ∈ Z + } , andfor n ≥ , let E n = { ( x i ) i ∈ R ∞ \ { } | x i = 0 for all i > n } . We have the subspace topologies on E and E n that inherit from the box and the producttopologies of R ∞ . Let r ≥ ρ r be a metric on E defined as ρ r ( x, y ) = X i | x i − y i | r ! /r for x, y ∈ E . This metric induces a topology on E , i.e. the topology is generated by theballs of the type C r ( x, δ ) = { y ∈ E : ρ r ( x, y ) < δ } for x ∈ E and δ >
0. We have anothermetric ρ ∞ on E defined as ρ ∞ ( x, y ) = sup i | x i − y i | for x, y ∈ E . Let C ∞ ( x, δ )) be an open ball in the metric space ( E , ρ ∞ ) centered at x ∈ E and of radius δ >
0, i.e. C ∞ ( x, δ )) = { y ∈ E : ρ ∞ ( x, y ) < δ } . This type of balls generate atopology on E . Since E n (for n ≥
1) is a subset of E , the set E n has the topologies inducedby the metrics ρ r and ρ ∞ when restricted to it. We denote the topologies on E and E n asin Table 2. Table 2: Topologies on E and E n Description Notations for thetopologies on E Notations for thetopologies on E n The subspace topology that inherit fromthe box topology of R ∞ S b S nb The subspace topology that inherit fromthe product topology of R ∞ S p S np The metric topology induced by themetric ρ r on E (respectively by ρ r | E n ) S r S nr The metric topology induced by themetric ρ ∞ on E (respectively by ρ ∞ | E n ) S ∞ S n ∞ Remark 4.1.
We have F ( L ) ⊆ E and G ( E ) ⊆ L , where F and G be the maps defined inLemma 4.3. This can be checked as follows:(a) Let φ be an element in L . By the definition of L , φ i = 0 for all but finitely many i ∈ Z + . Furthermore, we have φ i = 0 for some i ∈ Z + (otherwise φ ′ (0) = 0 ). Thisshows that F ( φ ) = ( φ i ) i is contained in E .(b) Let x = ( x i ) i be an element in E . Choose an n ∈ Z + such that x i = 0 for all i > n .Note that x i = 0 for some i ≤ n . Let φ : R → R n be defined as t ( x t, x t, . . . , x n t ) ,i.e. φ ij = 0 | j − | x i for all ( i, j ) ∈ Λ . Note that G ( x ) = φ and φ ∈ L . Thus, G ( x ) ∈ L . opologies on sets of polynomial knots Lemma 4.4.
Let f : L → E and g : E → L be the restrictions of the maps F and G respectively (where F and G be the maps defined in Lemma 4.3). Suppose E and L havethe topologies(a) S b and T b respectively, or(b) S p and T p respectively, or(c) S r and T r (for r ≥ ) respectively, or(d) S ∞ and T ∞ respectively.Then f and g are continuous.Proof. We prove the lemma by considering each of the four cases separately.(a) If E and L have the topologies S b and T b respectively:Since S b and T b are the subspace topologies inherited by the box topologies of R ∞ and R Λ respectively, the maps f and g (being the restrictions of F and G respectively) areobviously continuous by Lemma 4.3.(b) If E and L have the topologies S p and T p respectively:Since S p and T p are the subspace topologies inherited by the product topologies of R ∞ and R Λ respectively, the maps f and g are obviously continuous by Lemma 4.3.(c) If E and L have the topologies S r and T r (for r ≥
1) respectively:Note that ρ r ( f ( φ ) , f ( ψ )) ≤ d r ( φ, ψ ) for φ, ψ ∈ L and d r ( g ( x ) , g ( y )) ≤ ρ r ( x, y ) for x, y ∈ E , since X i | φ i − ψ i | r ≤ X i,j | φ ij − ψ ij | r & X i,j | x ij − y ij | r ≤ X i | x i − y i | r ≤ X i | x i − y i | r , where x = ( x i ) i , y = ( y i ) i , g ( x ) = ( x ij ) i,j and g ( y ) = ( y ij ) i,j . Thus, f and g are Lipschitzcontinuous.(d) If E and L have the topologies S ∞ and T ∞ respectively:As in the previous case, it can be seen that f and g are Lipschitz continuous in thepresent case. Remark 4.2.
One can check that F ( L n ) ⊆ E n and G ( E n ) ⊆ L n , where F and G be themaps defined in Lemma 4.3. With the same type of topologies (i.e. the topologies having the same subscripts asin Table 1 and Table 2), the spaces L n and E n are subspaces of the spaces L and E respectively. The next lemma follows immediately from Lemma 4.4. Lemma 4.5.
Let f n : L n → E n and g n : E n → L n be the restrictions of the maps F and G respectively (where F and G be the maps defined in Lemma 4.3). Suppose E n and L n have the topologies(a) S nb and T nb respectively, or(b) S np and T np respectively, or itesh Raundal (c) S nr and T nr (for r ≥ ) respectively, or(d) S n ∞ and T n ∞ respectively.Then f n and g n are continuous. Lemma 4.6.
Let H : [0 , × R Λ → R Λ be defined as (cid:16) s, ( x ij ) i,j (cid:17) (cid:16) s | j − | x ij (cid:17) i,j , where = 1 and the tuple on the right has its ( i, j ) th coordinate s | j − | x ij for ( i, j ) ∈ Λ .Suppose R Λ has the product topology. Then H is continuous.Proof. Let P : [0 , × R Λ → [0 ,
1] and P : [0 , × R Λ → R Λ be the projections onto thefirst and second coordinates respectively. For ( i, j ) ∈ Λ, let P ij : R Λ → R be the projectiononto the ( i, j ) th coordinate. Note that the projections P , P and P ij , for ( i, j ) ∈ Λ, arecontinuous. For ( i, j ) ∈ Λ, let Q ij : [0 , × R Λ → [0 , × R be defined as Q ij ( s, x ) = ( P ( s, x ) , P ij ( P ( s, x )))for ( s, x ) ∈ [0 , × R Λ . This map is continuous, since its component functions P and P ij ◦ P are continuous. For j ∈ N , let M j : [0 , × R → R be defined as ( s, t ) s | j − | t .Note that this map is continuous. It can be checked that H ( s, x ) = ( M j ( Q ij ( s, x ))) i,j for ( s, x ) ∈ [0 , × R Λ , where the tuple on the right has its ( i, j ) th coordinate M j ( Q ij ( s, x ))for ( i, j ) ∈ Λ. This shows that H is continuous, since its component functions M j ◦ Q ij ,for ( i, j ) ∈ Λ, are continuous.
Lemma 4.7.
Let H be the map defined in Lemma 4.6. Then H ([0 , × L n ) ⊆ L n (for n ∈ Z + ) and H ([0 , × L ) ⊆ L .Proof. Since L = S n ∈ Z + L n , it is enough to show that H ([0 , × L n ) ⊆ L n for n ∈ Z + .Suppose a positive integer n be given. Let ( s, φ ) ∈ [0 , × L n . Define ψ : R → R ∞ by t X j ( φ ij ) i s | j − | t j , where φ ij is the coefficient of t j in the i th component of φ for ( i, j ) ∈ Λ. Since φ ij = 0for all but finitely many ( i, j ) ∈ Λ and φ ij = 0 whenever i > n and j ≥
0, we have ψ ij = s | j − | φ ij = 0 for all but finitely many ( i, j ) ∈ Λ and ψ ij = s | j − | φ ij = 0 when-ever i > n and j ≥
0. If s = 0, then ψ ( u ) − ψ ( v ) = P j ≥ ( φ ij ) i s | j − | ( u j − v j ) = s − P j ≥ ( φ ij ) i (cid:0) ( su ) j − ( sv ) j (cid:1) = s − ( φ ( su ) − φ ( sv )) = 0 for u = v in R and ψ ′ ( t ) = P j ≥ j ( φ ij ) i ( st ) j − = φ ′ ( st ) = 0 for t ∈ R (since φ is a polynomial knot). Thus, if s = 0,then ψ is a polynomial knot. If s = 0, then ψ ( t ) = ( φ i t ) i for t ∈ R , and hence ψ is alinear knot (since φ i = 0 for some i ≤ n ). In either case, ψ ∈ L n . One can see that H ( s, φ ) = ψ , and thus H ( s, φ ) ∈ L n . This shows that H ([0 , × L n ) ⊆ L n .By Lemma 4.7, the restriction of H to [0 , × L has its image contained in L . Withthis in mind, we have the following: 10 opologies on sets of polynomial knots Lemma 4.8.
Let h : [0 , × L → L be the restriction of the map H defined in Lemma 4.6.Suppose L has any of the topologies T b , T p , T r (for r ≥ ) and T ∞ . Then h is continuous.Proof. To prove the lemma, we consider the cases as follows:(a) If L has the topology T b :Let U = L T Q i,j U ij be a basic open set in ( L , T b ) (i.e. U ij is open in R for all( i, j ) ∈ Λ). We show that h − ( U ) is open in [0 , × L . If h − ( U ) is empty, then it isobviously an open set. Suppose h − ( U ) is not empty. Let an element ( s, φ ) ∈ h − ( U ) begiven. It is enough to show that ( s, φ ) ∈ V × W ⊆ h − ( U ) for some open set V in [0 ,
1] andan open set W in L . Since h ( s, φ ) = (cid:0) s | j − | φ ij (cid:1) i,j and h ( s, φ ) ∈ U , s | j − | φ ij ∈ U ij for all( i, j ) ∈ Λ. Let Λ φ = { ( i, j ) ∈ Λ : φ ij = 0 } . Note that this is a finite set. Since U ij is openin R for all ( i, j ) ∈ Λ φ , we can find an ǫ > (cid:0) s | j − | φ ij − ǫ, s | j − | φ ij + ǫ (cid:1) ⊆ U ij for all ( i, j ) ∈ Λ φ . Since φ ij = 0 (i.e. s | j − | φ ij = 0) for ( i, j ) ∈ Λ \ Λ φ , 0 ∈ U ij forall ( i, j ) ∈ Λ \ Λ φ . For ( i, j ) ∈ Λ \ Λ φ , let ǫ ij > − ǫ ij , ǫ ij ) ⊆ U ij . Let m = max { j ∈ N : ( i, j ) ∈ Λ φ for some i ∈ Z + } and M = max {| φ ij | : ( i, j ) ∈ Λ φ } . We canfind a δ > (cid:12)(cid:12)(cid:12) s | j − | − t | j − | (cid:12)(cid:12)(cid:12) < ǫ M whenever | s − t | < δ and j ≤ m. (1)Let V = ( s − δ, s + δ ) ∩ [0 ,
1] and W = L T Q i,j W ij , where W ij = ( φ ij − ǫ/ , φ ij + ǫ/ i, j ) ∈ Λ φ and W ij = ( − ǫ ij , ǫ ij ) for ( i, j ) ∈ Λ \ Λ φ . Note that ( s, φ ) ∈ V × W . Suppose( t, ψ ) ∈ V × W . By Expression (1), we have (cid:12)(cid:12)(cid:12) s | j − | φ ij − t | j − | ψ ij (cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12) s | j − | − t | j − | (cid:12)(cid:12)(cid:12) | φ ij | + t | j − | | φ ij − ψ ij |≤ ǫ M M + ǫ < ǫ for ( i, j ) ∈ Λ φ , since t ∈ V (i.e. | s − t | < δ and 0 ≤ t ≤ j ≤ m and ψ ij ∈ W ij (i.e. | φ ij − ψ ij | < ǫ/ t | j − | ψ ij ∈ (cid:0) s | j − | φ ij − ǫ, s | j − | φ ij + ǫ (cid:1) ⊆ U ij for all ( i, j ) ∈ Λ φ .Also, we can see that t | j − | ψ ij ∈ ( − ǫ ij , ǫ ij ) ⊆ U ij for all ( i, j ) ∈ Λ \ Λ φ , since ψ ij ∈ W ij = ( − ǫ ij , ǫ ij ) for ( i, j ) ∈ Λ \ Λ φ . Therefore, h ( t, ψ ) = (cid:0) t | j − | ψ ij (cid:1) i,j ∈ U , and hence( t, ψ ) ∈ h − ( U ). This shows that V × W ⊆ h − ( U ).(b) If L has the topology T p :Since T p is the subspace topology inherited by the product topology of R Λ , the map h (being the restriction of H ) is obviously continuous by Lemma 4.6.(c) If L has the topology T r (for r ≥ s, φ ) ∈ [0 , × L . We show that h is continuous at ( s, φ ). Let an ǫ > δ and δ ′ such that d r ( h ( s, φ ) , h ( t, ψ )) < ǫ whenever | s − t | < δ ′ , ≤ t ≤ d r ( φ, ψ ) < δ. Let δ = ǫ/ m = max { j ∈ N : φ ij = 0 for some i ∈ Z + } and R = (cid:16)P i ≥ P j ≤ m | φ ij | r (cid:17) /r .Choose a δ ′ > (cid:12)(cid:12)(cid:12) s | j − | − t | j − | (cid:12)(cid:12)(cid:12) < δR whenever | s − t | < δ ′ and j ≤ m. (2)11 itesh Raundal Since φ ij = 0 for i ∈ Z + and j > m , by Expression (2), we have d r ( h ( s, φ ) , h ( t, φ )) ≤ X i ≥ X j ≤ m (cid:12)(cid:12)(cid:12) s | j − | − t | j − | (cid:12)(cid:12)(cid:12) r | φ ij | r /r ≤ δR X i ≥ X j ≤ m | φ ij | r /r ≤ δ whenever | s − t | < δ ′ and 0 ≤ t ≤
1. Also, we note that d r ( h ( t, φ ) , h ( t, ψ )) ≤ X i,j t | j − | r | φ ij − ψ ij | r /r ≤ X i,j | φ ij − ψ ij | r /r ≤ δ whenever 0 ≤ t ≤ d r ( φ, ψ ) < δ . Therefore, d r ( h ( s, φ ) , h ( t, ψ )) ≤ d r ( h ( s, φ ) , h ( t, φ )) + d r ( h ( t, φ ) , h ( t, ψ )) ≤ δ + δ < ǫ whenever | s − t | < δ ′ , 0 ≤ t ≤ d r ( φ, ψ ) < δ .(d) If L has the topology T ∞ :By slight modifications in the previous case, it can be shown that h is continuous inthe present case.The restriction of H to [0 , × L n has its image contained in L n (see Lemma 4.7).Since T nb , T np , T nr (for r ≥
1) and T n ∞ are the subspace topologies that inherit from thetopologies T b , T p , T r and T ∞ respectively, we have an immediate consequence of Lemma4.8 as follows: Lemma 4.9.
Let h n : [0 , × L n → L n be the restriction of the map H defined in Lemma4.6. Suppose L n has any of the topologies T nb , T np , T nr (for r ≥ ) and T n ∞ . Then h n iscontinuous. Lemma 4.10.
The space E n having any of the topologies S nb , S np , S nr (for r ≥ ) and S n ∞ is homeomorphic to R n \ { } .Proof. Let α : E n → R n \ { } be a map defined as ( x , x , x , . . . ) ( x , x , . . . , x n ) andsuppose β : R n \ { } → E n be defined as ( y , y , . . . , y n ) ( y , y , . . . , y n , , , . . . ). Notethat the compositions α ◦ β and β ◦ α are the identity maps of R n \ { } and E n respectively.Thus, α is a bijective map and β is its inverse. Now, it is enough to show that α and β are continuous. To do this, we consider the cases as follows:(a) If E n has the topology S nb :The sets of the type E n T Q i U i , where U i is open in R for all i ∈ Z + , form a basisfor the topology S nb . Let U = E n T Q i U i be an open set in this basis. If 0 / ∈ U i for some i > n , then U is an empty set and hence so is the set β − ( U ). If 0 ∈ U i for all i > n , then β − ( U ) = ( U × U × · · · × U n ) \{ } . In either case, β − ( U ) is open in R n \{ } . This shows12 opologies on sets of polynomial knots that β is continuous. It can be seen that the sets of the type ( V × V × · · · × V n ) \ { } ,where V i is open in R for all i = 1 , , . . . , n , form a basis for the usual (standard) topologyof R n \ { } . Suppose V = ( V × V × · · · × V n ) \ { } be an open set in this basis. Then α − ( V ) = E n T Q i W i , where W i = V i for i ≤ n and W i = R for i > n . Thus, α − ( V ) isopen in ( E n , S nb ). This shows that α is continuous.(b) If E n has the topology S np :By slight modifications in the previous case, it can shown that α and β are continuousin the present case. (Note that the sets of the type E n T Q i U i , where U i is open in R forall i ∈ Z + and U i = R for all but finitely many i ∈ Z + , form a basis for the topology S np .)(c) If E n has the topology S nr (for r ≥ C nr ( x, δ ) = { z ∈ E n : ρ r ( x, z ) < δ } , for x ∈ E n and δ >
0, form a basis for the topology S nr , and the sets of the type U nr ( y, δ ) = w ∈ R n \ { } : n X i =1 | y i − w i | r ! /r < δ for y ∈ R n \ { } and δ >
0, form a basis for the usual topology of R n \ { } . It can bechecked that β − ( C nr ( x, δ )) = U nr ( α ( x ) , δ ) and α − ( U nr ( y, δ )) = C nr ( β ( y ) , δ ) for x ∈ E n , y ∈ R n \ { } and δ >
0. Thus, α and β are continuous.(d) If E n has the topology S n ∞ :By slight modifications in the previous case, it can be shown that α and β are contin-uous in the present case. Proof of Theorem 4.1.
The composition f n ◦ g n is the identity map of E n , where f n and g n be the maps defined in Lemma 4.5. Also, it can be checked that h n (0 , φ ) = g n ( f n ( φ ))and h n (1 , φ ) = φ for φ ∈ L n , where h n is the map defined in Lemma 4.9. Thus, the map g n ◦ f n is homotopy equivalent to the identity map of L n , and hence the spaces E n and L n , with the topologies as in Lemma 4.5, have the same homotopy type. Since R n \ { } and S n − have the same homotopy type, the result follows by Lemma 4.10. Lemma 4.11.
The space E having any of the topologies S b , S p , S r (for r ≥ ) and S ∞ iscontractible.Proof. Let S : [0 , × E → E and T : [0 , × E → E be maps defined as S ( s, ( x i ) i ) =((1 − s ) x i + sx i − ) i and T ( s, ( x i ) i ) = ((1 − s ) x i − + sa i ) i for ( s, ( x i ) i ) ∈ [0 , × E , where x = 0, a = 1 and a i = 0 for i ≥
2. Note that S (0 , x ) = Id E ( x ), S (1 , x ) = T (0 , x ) = S ( x )and T (1 , x ) = C a ( x ) for all x ∈ E , where Id E is the identity map of E , S is the shift map( x , x , x , . . . ) (0 , x , x , . . . ) and C a is the constant map ( x , x , x , . . . ) (1 , , , . . . ).If we can show that S and T are continuous, then the identity map of E will be null-homotopic and hence E will be contractible. To show that S and T are continuous, weconsider the cases as follows:(a) If E has the topology S b :To prove S is continuous: Let U = E T Q i U i be a basic open set in ( E , S b ) (i.e. U i isopen in R for all i ∈ Z + ). We show that S − ( U ) is open in [0 , × E . If S − ( U ) is empty,then it is obviously an open set. Suppose S − ( U ) is not empty. Let ( s, ( x i ) i ) ∈ S − ( U )be given. It is enough to show that ( s, ( x i ) i ) ∈ V × W ⊆ S − ( U ) for some open set13 itesh Raundal V in [0 ,
1] and an open set W in E . Let m ≥ x i = 0 for all i ≥ m and let M = max {| x i | : i < m } . Since S ( s, ( x i ) i ) ∈ U , (1 − s ) x i + sx i − ∈ U i for i ∈ Z + . Choose an ǫ > − s ) x i + sx i − − ǫ, (1 − s ) x i + sx i − + ǫ ) ⊆ U i for all i ≤ m . Since x i = x i − = 0 (i.e. (1 − s ) x i + sx i − = 0) for i > m , 0 ∈ U i for all i > m . Thus, for i > m , we can find an ǫ i > − ǫ i , ǫ i ) ⊆ U i . Let δ = min { ǫ , ǫ m +1 } and δ i = min { ǫ i , ǫ i +1 } for i > m . Take V = ( s − δ/M, s + δ/M ) ∩ [0 , W = E T Q i W i , where W i = ( x i − δ, x i + δ ) for i ≤ m and W i = ( − δ i , δ i ) for i > m .Note that ( s, ( x i ) i ) ∈ V × W . Suppose ( t, ( y i ) i ) ∈ V × W . Since t ∈ V (i.e. | s − t | < δ/M and 0 ≤ t ≤
1) and y i ∈ W i (i.e. | x i − y i | < δ ) for i ≤ m , we have | (1 − s ) x i + sx i − − (1 − t ) y i − ty i − | ≤ | (1 − s ) x i + sx i − − (1 − t ) x i − tx i − | + | (1 − t ) x i + tx i − − (1 − t ) y i − ty i − |≤ | (1 − s ) − (1 − t ) | | x i | + | s − t | | x i − | + (1 − t ) | x i − y i | + t | x i − − y i − |≤ δM M + δM M + (1 − t ) δ + tδ< ǫ for i ≤ m . Thus, (1 − t ) y i + ty i − ∈ ((1 − s ) x i + sx i − − ǫ, (1 − s ) x i + sx i − + ǫ ) ⊆ U i forall i ≤ m . Also, we note that | (1 − t ) y m +1 + ty m | ≤ | y m +1 | + | y m | ≤ | y m +1 | + | x m − y m | ≤ δ m +1 + δ < ǫ m +1 and | (1 − t ) y i + ty i − | ≤ | y i | + | y i − | ≤ δ i + δ i − < ǫ i for i ≥ m + 2,since 0 ≤ t ≤ x m = 0, y m ∈ W m (i.e. | x m − y m | < δ ) and y i ∈ W i (i.e. | y i | < δ i )for i > m . In other words (1 − t ) y i + ty i − ∈ ( − ǫ i , ǫ i ) ⊆ U i for all i > m . Thus, wehave S ( t, ( y i ) i ) = ((1 − t ) y i + ty i − ) i ∈ U and hence ( t, ( y i ) i ) ∈ S − ( U ). This shows that V × W ⊆ S − ( U ), since ( t, ( y i ) i ) was an arbitrary element in V × W .To prove T is continuous: Let U = E T Q i U i be a basic open set in ( E , S b ). We showthat T − ( U ) is open in [0 , × E . If T − ( U ) is empty, then there is nothing to prove.Suppose T − ( U ) is not empty. Let ( s, ( x i ) i ) ∈ T − ( U ) be given. It is enough to showthat ( s, ( x i ) i ) ∈ V × W ⊆ T − ( U ) for an open set V in [0 ,
1] and an open set W in E .Let m ≥ x i = 0 for all i ≥ m and let M = max {| x i | : i < m } .Since T ( s, ( x i ) i ) ∈ U , (1 − s ) x i − + sa i ∈ U i for all i ∈ Z + . Choose an ǫ > − s ) x i − + sa i − ǫ, (1 − s ) x i − + sa i + ǫ ) ⊆ U i for each i ≤ m . Since x i − = a i = 0(i.e. (1 − s ) x i − + sa i = 0) for i > m , 0 ∈ U i for all i > m . For i > m , choose an ǫ i > − ǫ i , ǫ i ) ⊆ U i . Let δ = min { ǫ , ǫ m +1 } and δ i = ǫ i +1 for i > m . Take V = (cid:16) s − δM +1 , s + δM +1 (cid:17) ∩ [0 ,
1] and W = E T Q i W i , where W i = ( x i − δ, x i + δ ) for i ≤ m and W i = ( − δ i , δ i ) for i > m . Note that ( s, ( x i ) i ) ∈ V × W . Suppose ( t, ( y i ) i ) ∈ V × W .Since t ∈ V (i.e. | s − t | < δM +1 and 0 ≤ t ≤
1) and y i ∈ W i (i.e. | x i − y i | < δ ) for i ≤ m ,we have | (1 − s ) x i − + sa i − (1 − t ) y i − − ta i | ≤ | (1 − s ) x i − − (1 − t ) y i − | + | sa i − ta i |≤ | (1 − s ) − (1 − t ) | | x i − | + (1 − t ) | x i − − y i − | + | s − t | | a i | opologies on sets of polynomial knots ≤ | s − t | | x i − | + | x i − − y i − | + | s − t |≤ δM + 1 M + δ + δM + 1 < ǫ for i ≤ m . Thus, (1 − t ) y i − + ta i ∈ ((1 − s ) x i − + sa i − ǫ, (1 − s ) x i − + sa i + ǫ ) ⊆ U i for all i ≤ m . Also, we have | (1 − t ) y m + ta m +1 | ≤ | y m | ≤ | x m − y m | ≤ δ < ǫ m +1 and | (1 − t ) y i − + ta i | ≤ | y i − | ≤ δ i − < ǫ i for i ≥ m + 2, since 0 ≤ t ≤ a i = 0for i > m , x m = 0, y m ∈ W m (i.e. | x m − y m | < δ ) and y i ∈ W i (i.e. | y i | < δ i ) for i > m . In other words (1 − t ) y i − + ta i ∈ ( − ǫ i , ǫ i ) ⊆ U i for all i > m . Therefore, T ( t, ( y i ) i ) = ((1 − t ) y i − + ta i ) i ∈ U and hence ( t, ( y i ) i ) ∈ T − ( U ). This shows that V × W ⊆ T − ( U ), since ( t, ( y i ) i ) was arbitrary.(b) If E has the topology S p :To prove S is continuous: Let U = E T Q i U i be a basic open set in ( E , S p ) (i.e. U i is open in R for all i ∈ Z + and U i = R for all but finitely many i ∈ Z + ). We show that S − ( U ) is open in [0 , × E with respect to the topology S p on E . If S − ( U ) is empty,then it is trivially an open set. Therefore, assume that it is not empty. Since U is also anopen set in ( E , S b ), by the previous case, S − ( U ) is open in [0 , × E with respect to thetopology S b on E . Thus, we have S − ( U ) = S j V j × W j , where j runs over some indexingset J , and for j ∈ J , V j is a basic open set in [0 ,
1] and W j = E T Q i W ji is a basic open setin ( E , S b ) (i.e. W ji is open in R for all i ∈ Z + ). Since we can remove a set V j × W j that isempty, we assume that the sets V j and W j are not empty for all j ∈ J . Thus, the set W ji is not empty for all ( i, j ) ∈ Z + × J , and for j ∈ J , 0 ∈ W ji for all but finitely many i ∈ Z + .Since U i = R for all but finitely many i ∈ Z + , we can find a positive integer m such that U i = R for all i > m . For j ∈ J , let W ′ j = E T Q i W ′ ji , where W ′ ji = W ji for i ≤ m and W ′ ji = R for i > m . Note that W ′ j is a basic open set in ( E , S p ) for all j ∈ J . Now, it isenough to prove that S − ( U ) = S j V j × W ′ j . We note that S − ( U ) ⊆ S j V j × W ′ j , since W j ⊆ W ′ j for all j ∈ J . Suppose ( s, ( x i ) i ) ∈ S j V j × W ′ j . Then ( s, ( x i ) i ) ∈ V j × W ′ j forsome j ∈ J and hence x i ∈ W ′ ji = W ji for all i ≤ m . Let y i = x i for i ≤ m . Since W ji is not an empty set for all i > m and 0 ∈ W ji for all but finitely many i > m , we canchoose y i ∈ W ji for each i > m such that at least one of the y i is nonzero and y i = 0 forall but finitely many i > m . It is clear that ( s, ( y i ) i ) ∈ V j × W j ⊆ S − ( U ). In other words S ( s, ( y i ) i ) ∈ U . Thus, (1 − s ) x i + sx i − = (1 − s ) y i + sy i − ∈ U i for all i ≤ m , since y i = x i for i ≤ m and S ( s, ( y i ) i ) = ((1 − s ) y i + sy i − ) i . Note that (1 − s ) x i + sx i − ∈ U i for all i > m (since U i = R for i > m ). Therefore, S ( s, ( x i ) i ) ∈ U and hence ( s, ( x i ) i ) ∈ S − ( U ).This shows that S j V j × W ′ j ⊆ S − ( U ).The continuity of T can be proved in a similar way as in the case of S .(c) If E has the topology S r (for r ≥ S is continuous: Let ( s, ( x i ) i ) ∈ [0 , × E . We show that S is continuous at( s, ( x i ) i ). Suppose an ǫ > δ and δ ′ such that ρ r ( S ( s, ( x i ) i ) , S ( t, ( y i ) i )) < ǫ whenever | s − t | < δ ′ , ≤ t ≤ ρ r (( x i ) i , ( y i ) i ) < δ. itesh Raundal Let R = ( P i | x i | r ) /r , δ = ǫ and δ ′ = ǫ R . We have the following: ρ r ( S ( s, ( x i ) i ) , S ( t, ( y i ) i )) ≤ ρ r ( S ( s, ( x i ) i ) , S ( t, ( x i ) i )) + ρ r ( S ( t, ( x i ) i ) , S ( t, ( y i ) i )) ≤ X i | (1 − s ) x i + sx i − − (1 − t ) x i − tx i − | r ! /r + X i | (1 − t ) x i + tx i − − (1 − t ) y i − ty i − | r ! /r ≤ X i | (1 − s ) − (1 − t ) | r | x i | r ! /r + X i | s − t | r | x i − | r ! /r + X i (1 − t ) r | x i − y i | r ! /r + X i t r | x i − − y i − | r ! /r ≤ | s − t | X i | x i | r ! /r + (1 − t + t ) X i | x i − y i | r ! /r ≤ δ ′ R + δ< ǫ whenever | s − t | < δ ′ , 0 ≤ t ≤ ρ r (( x i ) i , ( y i ) i ) < δ .To prove T is continuous: Let ( s, ( x i ) i ) ∈ [0 , × E . We show that T is continuous at( s, ( x i ) i ). Suppose an ǫ > δ = ǫ and δ ′ = ǫ R , where R = ( P i | x i | r ) /r .We have the following expression: ρ r ( T ( s, ( x i ) i ) , T ( t, ( y i ) i )) ≤ X i | (1 − s ) x i − + sa i − (1 − t ) y i − − ta i | r ! /r ≤ X i | sa i − ta i | r ! /r + X i | (1 − s ) x i − − (1 − t ) y i − | r ! /r ≤ X i | s − t | r | a i | r ! /r + X i | (1 − s ) − (1 − t ) | r | x i − | r ! /r + X i (1 − t ) r | x i − − y i − | r ! /r ≤ | s − t | X i | a i | r ! /r + | s − t | X i | x i | r ! /r opologies on sets of polynomial knots + (1 − t ) X i | x i − y i | r ! /r ≤ δ ′ + δ ′ R + δ< ǫ whenever | s − t | < δ ′ , 0 ≤ t ≤ ρ r (( x i ) i , ( y i ) i ) < δ . Therefore, for every ǫ >
0, thereexist positive real numbers δ and δ ′ such that ρ r ( T ( s, ( x i ) i ) , T ( t, ( y i ) i )) < ǫ whenever | s − t | < δ ′ , ≤ t ≤ ρ r (( x i ) i , ( y i ) i ) < δ. This shows that T is continuous at ( s, ( x i ) i ).(d) If E has the topology S ∞ :By sight modifications in the previous case, it can shown that S and T are continuousin the present case. Proof of Theorem 4.2.
The composition f ◦ g is the identity map of E , where f and g be themaps defined in Lemma 4.4. Also, it can be seen that h (0 , φ ) = g ( f ( φ )) and h (1 , φ ) = φ for φ ∈ L , where h is the map defined in Lemma 4.8. Thus, the map g ◦ f is homotopyequivalent to the identity map of L , and hence the spaces E and L , with the topologies asin Lemma 4.4, have the same homotopy type. Now, the result follows by Lemma 4.11. Acknowledgments
The author is thankful to the Harish-Chandra Research Institute, Prayagraj, India for thepostdoctoral fellowship during this research work.
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