Maximal lineability of the set of Continuous Surjections
aa r X i v : . [ m a t h . F A ] A p r MAXIMAL LINEABILITY OF THE SET OF CONTINUOUS SURJECTIONS
NACIB GURGEL ALBUQUERQUE
Abstract.
Let m, n be positive integers. In this short note we prove that the set of all continuousand surjective functions from R m to R n contains (excluding the 0 function) a c -dimensional vectorspace. This result is optimal in terms of dimension. Preliminaries
Lately the study of the linear structure of certain subsets of surjective functions in R R (suchas everywhere surjective functions, perfectly everywhere surjective functions, or Jones functions)has attracted the attention of several authors working on Real Analysis and Set Theory (see, e.g.[1, 2, 4, 6, 7]). The previously mentioned functions are, indeed, very “pathological”: for instance aneverywhere surjective function f in R R verifies that f ( I ) = R for every interval I ⊂ R and theother classes (perfectly everywhere surjective functions and Jones functions) are particular casesof everywhere surjective functions and, thus, with even “worse” behavior. It has been shown [5]that there exists a 2 c -dimensional vector space every non-zero element of which is a Jones functionand, thus, everywhere surjective (here, c stands for the cardinality of R ). Of course, this previousresult is optimal in terms of dimension since dim( R R )= 2 c . However, all the previous classes arenowhere continuous, thus, it is natural to ask about the set of continuous surjections. The aim ofthis short note is to prove, in a more general framework that of R R , that (for every m, n ∈ N ) theset of continuous surjections from R m onto R n is c -lineable [1] (that is, it contains a c -dimensionalvector space every non-zero element of which is a continuous surjective function from R m onto R n ).Since dim C ( R m , R n ) = c we have that this result would be the best possible in terms of dimension,that is, the set of continuous surjections from R m onto R n is maximal lineable [3].While there are many trivial examples of surjective continuous functions in R R , coming up witha concrete example of a continuous surjective function from R onto R is a totally different story.The existence of a continuous surjection from R onto R (a Peano type function) can be found in[8, p. 42] or [9, p. 274]. Both references use the existence of a continuous surjection from [0 , , (a Peano curve in [0 , or a space filling curve ). The existence of this curve is proved,for instance, in [8] invoking a result due to A. D. Alexandrov: there is a continuous surjectionfrom the Cantor space K onto any arbitrary nonempty compact metric space (see [8, p. 40]); in[9, section 44] the construction of the Peano curve is done geometrically, and is a consequence ofthe completeness of the space C ( X, M ) of all continuous functions from a topological space X to acomplete metric space M , considering C ( X, M ) with the uniform metric.
Mathematics Subject Classification.
Key words and phrases. lineability; spaceability; algebrability; Peano type function.The author is supported by Capes. The lineability of the set of continuous surjections from R m to R n Let m and n be positive integers. Throughout this note we shall denote S m,n = { f : R m −→ R n ; f is continuous and surjective } . The following result shows that S m,n = ∅ , and uses the fact that S , = ∅ ([8, p. 42]). Proposition 2.1.
Let m, n ∈ N . There exists a continuous surjection f : R m → R n .Proof. Let us take f ∈ S , . If f i := π i ◦ f, i = 1 , i -coordinates functions of f ( f = ( f , f )), then the map id R × f : R −→ R defined by id R × f ( t, s ) := ( t, f ( s ) , f ( s ))is a continuous surjection. Thus, ( id R × f ) ◦ f is in S , . Proceeding in an induction manner,we can assure the existence of a function g belonging to S ,n for every n ∈ N . Hence, defining F : R m −→ R n by F := g ◦ π , i.e., F ( x ) = F ( x , . . . , x m ) = g ( x ) , for all x = ( x , . . . , x m ) ∈ R m ( π : R m −→ R denotes the canonical projection over the first coordinate), we conclude that F ∈ S m,n ( F is composition of continuous surjective functions). (cid:3) Attempting maximal lineability of S m,n (that is, c -lineability) we make use of the followingremark (inspired in a result from [1]), which indicates a method to obtain our main result. Remark 2.2.
Given a continuous surjection f : R m −→ R n , suppose we have X ⊂ C ( R n ; R n ) asubset of c -many linearly independent functions such that every nonzero element of span ( X ) is acontinuous surjection. Then, we have that Y := { F ◦ f } F ∈X ⊂ C ( R m ; R n ) has cardinality c , is linearly independent and is formed just by continuous surjections. Moreover,span ( Y ) ⊂ S m,n ∪ { } , obtaining the c -lineability of S m,n . In order to continue we shall need two lemmas and some notation. First, let us consider (for r >
0) the homeomorphism φ r : R → R given by φ r ( t ) := e rt − e − rt . Lemma 2.3.
The subset A := { φ r } r ∈ R + of R R is linearly independent, has cardinality c , and everynonzero element of span ( A ) is continuous and surjective.Proof. First let us prove that every nonzero element φ = P ki =1 α i · φ r i ∈ span( A ) is surjective. Wemay suppose that r > r > · · · > r k and α = 0. Writing φ ( t ) = e r t · α + k X i =2 α i · e ( r i − r ) t ! − k X i =1 α i · e − r i t , we conclude that lim t → + ∞ φ ( t ) = sign( α ) · ∞ and lim t →−∞ φ ( t ) = − sign( α ) · ∞ . Thus, the continuityof φ assures its surjection. Now let us see that A is linearly independent: suppose that ψ = P ni =1 λ i · φ s i = 0. If there is some λ j = 0, we may suppose that s > · · · > s n and λ = 0.Repeating the argument above, we obtainlim t → + ∞ ψ ( t ) = sign( λ ) · ∞ and lim t →−∞ ψ ( t ) = − sign( λ ) · ∞ , AXIMAL LINEABILITY OF THE SET OF CONTINUOUS SURJECTIONS 3 which contradicts ψ = 0. This proves that A is linearly independent. The other assertions are easyto prove. (cid:3) For each r = ( r , . . . , r n ) ∈ ( R + ) n , let ϕ r be the homeomorphism from R n to R n defined by ϕ r = ( φ r , . . . , φ r n ), i.e. , ϕ r ( x ) := ( φ r ( x ) , . . . , φ r n ( x n )) , for all x = ( x , . . . , x n ) ∈ R n . Working on each coordinate, and using the previous lemma, we have the following.
Lemma 2.4.
The set B = { ϕ r } r ∈ ( R + ) n of C ( R n ; R n ) is linearly independent, has cardinality c , andevery nonzero element of span ( B ) is continuous and surjective. Now it is time to state and prove our main result.
Theorem 2.5. S m,n is c -lineable.Proof. Let f ∈ S m,n . Using the notation of the previous lemma and the ideas of the Remark 2.2,we now prove that the set C = { F ◦ f } F ∈ B is so that span( C ) is the space we are looking for.The surjectivity of f assures that G ◦ f = 0 implies G = 0, for every function G from R n to R n .Thus, if G i ∈ B , i = 1 , . . . , k and0 = k X i =1 α i · G i ◦ f = k X i =1 α i G i ! ◦ f, then P ki =1 α i · G i = 0; so since B is linearly independent, we conclude that α i = 0 , i = 1 , . . . , k and thus, C is linearly independent. Thus, clearly, it has cardinality c . Furthermore, any nonzerofunction l X i =1 λ i · F i ◦ f = l X i =1 λ i F i ! ◦ f of span( C ) is continuous and surjective, since it is the composition of continuous surjective functions(recall that, from Lemma 2.4, P li =1 λ i F i is a continuous surjective function). Therefore, span( C )only contains, except the zero function, continuous surjective functions. (cid:3) Remark 2.6.
As we mentioned in the Introduction, and since dim C ( R m , R n ) = c , this result isthe best possible in terms of dimension. The next step (in sense of trying a similar result in higherdimensions) could be related to the lineability of S m, N (the set of the continuous surjections from R onto R N with the product topology). However this is not possible, since S m, N = ∅ ( [9, p. 275] ). References [1] R. Aron, V. I. Gurariy, and J. B. Seoane-Sep´ulveda,
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Departamento de Matem´atica,Universidade Federal da Para´ıba,58.051-900 - Jo˜ao Pessoa, Brazil.
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