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Strong Algebrability of Sets of Sequences and Functions

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Strong Algebrability of Sets of Sequences and Functions PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 141, Number 3, March 2013, Pages 827–835 S 0002-9939(2012)11377-2 Article electronically published on July 20, 2012 STRONG ALGEBRABILITY OF SETS OF SEQUENCES AND FUNCTIONS ARTUR BARTOSZEWICZ AND SZYMON GLA¸B (Communicated by Thomas Schlumprecht) Abstract. We introduce a notion of strong algebrability of subsets of linear algebras. Our main results are the following. The set of all sequences from c0 which are not summable with any power is densely strongly c–algebrable. The set of all sequences in l∞ whose sets of limit points are homeomorphic to the Cantor set is comeager and strongly c-algebrable. The set of all non- measurable functions from RR is 2c–algebrable. These results complete several by other authors, within the modern context of lineability. 1. Introduction Assume that B is a linear algebra, that is, a linear space that is also an algebra. According to [4], a subset E ⊂ B is called algebrable whenever E ∪{0} contains an infinitely generated algebra. If κ is a cardinal, E is said to be κ–algebrable if E∪{0} contains an algebra generated by an algebraically independent set with cardinality κ. Hence algebrability equals ω-algebrability, where ω denotes the cardinality of the set N of positive integers. In addition to algebrability, there exist other related properties such as lineability, dense-lineability and spaceability; see [3], [8], [9] and [14]. In this note we will focus on algebrability. This notion was considered by many authors. For instance Aron, P´erez-Garc´ıa and Seoane-Sep´ulveda proved in [4] that the set of non-convergent Fourier series is algebrable. Aron and Seoane-Sep´ulveda established in [5] algebrability of everywhere surjective functions on C,whichwas strengthened in [2]. Other algebrability results can be found in [11] and [12]. Here we will discuss some strengthening of the notion of algebrability. For a cardinal κ,wesaythatA is a κ-generated free algebra if there exists a subset X = {xα : α<κ} of A such that any function f from X to some algebra A can be uniquely extended to a homomorphism from A into A.ThenX is called a set of free generators of the algebra A. A subset X = {xα : α<κ} of a commutative algebra B generates a free subalgebra A if and only if for each polynomial P and any xα1 ,xα2 , ..., xαn we have P (xα1 ,xα2 , ..., xαn ) = 0 if and only if P =0. This leads us to the following definition. We say that a subset E of a commutative linear algebra B is strongly κ–algebrable if there exists a κ-generated free algebra A contained in E ∪{0}. If additionally, B is equipped with some topological structure and A is dense in B,thenwesaythatE is densely strongly κ–algebrable. Received by the editors May 24, 2011 and, in revised form, July 19, 2011 and July 23, 2011. 2010 Mathematics Subject Classification. Primary 15A03; Secondary 28A20, 46J10. Key words and phrases. Algebrability, non-summable sequences, non-measurable functions, Banach-Mazur game, accumulation points, Cantor sets. c 2012 American Mathematical Society Reverts to public domain 28 years from publication 827 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 828 ARTUR BARTOSZEWICZ AND SZYMON GLA¸B Note that X = {xα : α<κ}⊂E is a set of free generators of a free algebra ⊂ ˜ k1 k2 ··· kn A E if and only if the set X of elements of the form xα1 xα2 xαn is linearly independent and all linear combinations of elements from X˜ are in E ∪{0}.(In particular, it follows that sets considered in [7] were strongly algebrable.) It is easy to see that free algebras have no divisors of zero. In general, there are subsets of linear algebras which are algebrable but not strongly algebrable. Let c00 be a subset of c0 consisting of all sequences with real terms equal to zero from some place. Having members x1, ..., xn of some linear algebra and a polynomial P ,anelementP (x1, ..., xn) will be called an algebraic combination of x1, ..., xn. Proposition 1. The set c00 is algebrable in c0 but is not strongly 1–algebrable. Proof. Note that c00 is a linear subalgebra of c0 which can be generated by a linearly independent set of cardinality ω,namely,by{e1,e2, ...},whereei is a sequence consisting of 1 on the i-th coordinate and zeros on the remaining coordinates. Note also that c00 cannot be generated by finitely many elements, since any linear (or algebraic) combination of finitely many elements from c00 eventually equals zero. Suppose that c00 is strongly 1–algebrable; i.e., there is a sequence a ∈ c00 which is algebraically independent. In particular, this means that the collection 2 3 {a, a ,a , ...} is linearly independent. Since a ∈ c00,thereisn ∈ N such that a(k)=0fork ≥ n.Moreoveram(k) = 0 for all k ≥ n and m ∈ N. Note that the linear space generated by a, a2,a3, ... is isomorphic to a linear subspace of Rn and 2 3 Rn the restriction vectors an,an,an, ... are linearly independent in , a contradic- tion. The aim of this note is to prove strong c–algebrability of some sets of sequences and functions, where c denotes cardinality of the continuum. These are: the set of all sequences tending to zero which are not summable with any power, the set of non-measurable functions, and the set of bounded real sequences whose set of accumulation points is of measure zero and can be decomposed into a finite set and a set homeomorphic to the Cantor set. This last set happens to be co-meager. ∞ 2. Algebrability in c0 and l To prove Theorem 2 we will use the idea from the paper [7]. Our theorem is p related to [10, Theorem 1.3], from which the spaceability of c0 \ {l : p ≥ 1} is extracted as a consequence. Recall that a subset E of a topological vector space X is called spaceable if E ∪{0} contains a closed infinite-dimensional subspace of X. p Theorem 2. The set c0 \ {l : p ≥ 1} is densely strongly c–algebrable in c0. Proof. Let {rα : α<c} be a linearly independent (with respect to the field Q) subset of positive reals containing 1. Let A be a linear algebra generated by the set ∞ 1 S = rα : α<c . ln (n +1) n=1 We will show that any non-trivial algebraic combination of elements from S is either a null sequence or it is not summable with any power. To prove this, we need to show that for every matrix (kil : i ≤ m, l ≤ j) of non-negative integers with non- zero and distinct rows and any β1, ..., βm ∈ R which do not vanish simultaneously, License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use STRONG ALGEBRABILITY OF SOME SETS 829 the series ∞ 1 1 (1) β1 + ... + βm k11rα1 +...+k1j rαj km1rα1 +...+kmj rαj n=1 ln (n +1) ln (n +1) is not summable with any power. Note that the sequence 1 (2) k r +...+k r ln 11 α1 1j αj (n +1) n∈N tends to zero; i.e. it is in c0. We need to prove that the series (1) is divergent. Since { } the set rα : α<c is linearly independent, the numbers k11rα1 + ... + k1j rαj ,..., km1rα1 + ... + kmjrαj are distinct. To simplify the notation, put ki := ki1rα1 + ... + kijrαj for every i =1, ..., m. We may assume that k1 <k2 < ... < km and β1 =0. k Since ln i (n+1) →∞(n →∞) for all i =2, ..., m, we obtain that there is an N lnk1 (n+1) such that | | 1 1 1 β1 (3) β1 + β2 + ... + βm ≥ lnk1 (n +1) lnk2 (n +1) lnkm (n +1) 2lnk1 (n +1) ∞ ≥ 1 ∈ p ≥ for all n N.Since lnq (n+1) / l for any p 1andq>0, the result follows. n=1 Now, we prove that A is dense in c0. Consider the subalgebra A1 of space c (of all ∞ 1 convergent sequences with the sup-norm), generated by two elements: ln(n+1) ∞ n=1 1 and1=(1, 1, 1, ...). Let A2 be a subalgebra of A generated by ln(n+1) .Note n=1 that A1 = {x +(a, a, a, ...):x ∈ A2,a ∈ R}. Equip the set of positive integers N with the discrete topology and let αN be a one-point compactification of N.Note that c is isometrically isomorphic to the space C(αN) of all continuous real functions ∞ N 1 on α with the sup-norm. Since ln(n+1) is a strictly decreasing sequence of n=1 positive numbers tending to zero, then it separates the points of αN. Therefore by the Stone–Weierstrass theorem, the algebra A1 is dense in c. Take any x ∈ c0 and ε>0. There is y ∈ A1 with ||x − y|| <ε/2. Note that | limn y(n)|≤ε/2. Puty ˆ =(y(1)−limn y(n),y(2)−limn y(n),y(3)−limn y(n), ...) ∈ c0.Then||x − yˆ|| <ε. Sincey ˆ ∈ A2 ⊂ A, we obtain that A2, and therefore A,is dense in c0. In the sequel we will need the following observation. Let S be a set defined in the proof of Theorem 2. Let S = {aα : α<c} and let (nk) be any increasing sequence of natural numbers. Then S := {(aα(nk))k∈N : α<c} is also a set of free generators in c0. This easily follows from (3). In [1] it was proved that the set E of all sequences from l∞ which are not convergent is c-lineable; i.e.
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