Olivier’s theorem: ideal convergence, algebrability and Borel classification

The classical Olivier’s theorem says that for any nonincreasing summable sequence (a(n)) the sequence (na(n)) tends to zero. This result was generalized by many authors. We propose its further generalization which implies known results. Next we consider the subset AOS\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathcal {AOS}}$$\end{document} of ℓ1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$ \ell _{1} $$\end{document} consisting of sequences for which the assertion of Olivier’s theorem is false. We study how large and good algebraic structures are contained in AOS\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathcal {AOS}}$$\end{document} and its subsets; this kind of study is known as lineability. Finally we show that AOS\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathcal {AOS}}$$\end{document} is a residual Gδσ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$ \mathcal {G}_{\delta \sigma } $$\end{document} but not an Fσδ-set\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$ {\mathcal {F}}_{\sigma \delta } \text {-set} $$\end{document}.


Introduction
Throughout the paper by a = (a(n)), b = (b(n)), etc. we denote sequences of reals. The sequences of positive integer numbers are denoted by (n k ), (m k ) and so one.
It is a very basic convergence test, that for convergent series a(n) the limit of its summands lim n→∞ a(n) is zero. However, the following classical Olivier's theorem seems a bit surprising for many mathematicians. Theorem 1.1 [13] For any nonincreasing summable sequence a = (a(n)) the sequence (na(n)) tends to zero. B Szymon Głab szymon.glab@p.lodz.pl Artur Bartoszewicz artur.bartoszewicz@wmii.uni.lodz.pl Agnieszka Widz agnieszkawidzenfp@gmail.com 1 Many authors have tried to omit the assumption of monotonicity in different ways. For example the authors of [7] have found the full characterization of sequences (a(n)) for which the assertion of the Olivier's theorem holds. The second way is to consider the ideal convergence instead of the usual convergence of sequences. Recall that for an admissible ideal I on N, where admissible means containing finite sets and non-equal to P(N), we have Ilim a(n) = 0 or a(n) I → 0 if and only if for any ε > 0 the set A ε = {n : |a(n)| ≥ } belongs to I, see for example [9]. Clearly for I = Fin (the ideal of finite sets) the ideal convergence reduces to the usual convergence. The main result in [14] is the following.

Theorem 1.3 Let a = (a(n)) be an arbitrary sequence of nonnegative numbers. Then the convergence of the series a(n) implies the I-convergence to zero of the sequence (na(n))
if and only if an ideal I contains the summable ideal I ( 1 n ) . Recall that an ideal I (x) consists of the sets A ⊂ N for which n∈A x(n) < ∞ for divergent series x(n) = ∞ with x(n) ≥ 0. Note that for a summable sequence x = x(n) the ideal I (x) = P(N), so it is not admissible. If there exists c > 0 such that we have x(n) > c for all n ∈ N, then I (x) = Fin. The result of Toma and Šalát was further generalized in [8] where the authors characterized ideals I for which the implication holds true for a fixed positive p.
Recently Mišík and Tóth in [12] have proved the following theorem that generalizes those of [8,14]. Theorem 1.4 Let p, q be fixed positive numbers and α, β be fixed nonnegative ones. Moreover, let a = (a(n)) be an arbitrary sequence with positive terms. Then the convergence of the series n α a p (n) implies the I-convergence to zero of the sequence (n β a q (n)) if and only if an ideal I contains the summable ideal I (n α−β p/q ) .
The paper is organized as follows. The main result of Sect. 2 further generalizes Theorem 1.4. We show that it implies several known results and it cannot be inferred from them. Next we consider the subset of 1 consisting of sequences for which the assertion of Olivier's theorem is false. We shall call such sequences Anti-Olivier and the set of all them denote by AOS. In fact we consider the algebraic and topological properties of some subsets of AOS. In Sect. 3 we study how large and good algebraic structures are contained in AOS and its subsets; this kind of study is known as lineability, see [1,5]. Section 4 is devoted to the Borel classification of considered sets in 1 . The main result here is that AOS is a residual G δσ but not an F σ δ -set. (1) For any sequence a = (a(n)) of positive numbers the convergence of series f (n)a p (n)

Olivier's theorem and the ideal convergence
implies that g(n)a q (n) Let us assume that I ( f g − p/q ) ⊂ I and f (n)a p (n) < ∞. We shall show that g(n)a q (n) I → 0. Namely, we need to prove that Let us fix ε > 0. Since p, q, f and g are positive, we have We have Hence A ε ∈ I ( f g − p/q ) . Since I ( f g − p/q ) ⊂ I, then A ε ∈ I and g(n)a q (n) Let us assume that the implication f (n)a p (n) < ∞ ⇒ g(n)a q (n) holds for any sequence a of positive numbers. Let us put A a,ε := {n ∈ N : g(n)a q (n) ≥ ε}. Now, our assumption can be written as follows We will show that I ( f g − p/q ) ⊂ I. To do this, we will show that any set B belonging to the ideal I ( f g − p/q ) is a subset of A a,ε for some ε > 0 and some sequence a of positive numbers, in symbols Let us take a set B for which n∈B f (n)/g p q (n) < ∞. Let us define a as follows.
Hence, an arbitrary set B ∈ I ( f g − p/q ) is contained in A a,ε for ε = 1 and for some sequence Let us consider some particular cases of the Theorem 2.1.

Algebraic structures in AOS and its subsets
Having an algebraic structure A and its subset E ⊂ A one can ask if E ∪ {0} contains a large substructure of A. Roughly speaking, if the answer is affirmative, then we say that E is structurable. Let κ be a cardinal number. Assume that V is a linear space (resp. an algebra).
contains an κ-dimensional subspace (resp. κ-generated algebra). We say that an algebra is κ-generated if the minimal cardinality of sets of its generators is equal to κ. Moreover, we say that a subset E of an algebra V is strongly κ-algebrable [4] if there exists algebra A ⊂ E ∪ {0} which is isomorphic with the κ-generated free algebra. Let us recall that X = {x α : α < κ} ⊂ E is a set of free generators of a free algebra A ⊂ E if the set X of elements of the form x k 1 α 1 . . . x k n α n is linearly independent and all linear combinations of elements from X are in E ∪ {0}. It means that for any nonzero polynomial P in n variables without a constant term and any distinct We have the following obvious implications: κ-strong algebrability ⇒ κ-lineability spaceabilit y ⇒ c-lineability.
More information and more examples on these notions can be found in survey article [5] and book [1].
Let us define the following subsets of 1 strictly connected with Olivier's Theorem.
• Set of AntiOlivier sequences AOS = {a ∈ 1 : na(n) 0} • Set of sequences with finite number of terms equal to zero:

Theorem 3.3 The sets AO S b and AO S ub are spaceable.
Proof of Theorem 3. 1 We say that a family F of subsets of N (or of another countable set) is almost disjoint if it consists of infinite sets and the intersection of any two distinct members of F is finite. Probably, Sierpiński was the first who observed that there exists an almost disjoint family F with the cardinality 2 ω = c.
Let A ∈ I (1/n) be infinite, ⊆ (2, ∞) be linearly independent over the field of rational numbers Q with | | = c and F = {A α : α ∈ } be an almost disjoint family of subsets of A such that F = A. Let us define sequences a α , α ∈ , by the formula: Let us consider a linear combination b = (c 1 a α 1 +· · ·+c k a α k ) for pairwise distinct α 1 , . . . , α k and c i ∈ R\{0}, for i ≤ k. Clearly b ∈ 1 as a linear combination of 1 -sequences. Since F consists of almost disjoint sets, the set N is a disjoint union of k + 1 infinite sets: We will show that the sequence (nb(n)) tends on A α i \ j =i A α j to c i = 0. This shows that the sequence (nb(n)) is bounded and at most finitely many b(n)'s equal zero there. We will also show that |b(n)| > 0 for almost the reasoning is the same). Then n(b(n)) = n c 1 1 n + c 2 1 n α 2 + · · · + c k 1 n α k = c 1 + c 2 n 1−α 2 + · · · + c k n 1−α k .
Thus the sequence (n(b(n))) tends to c 1 = 0 along indices belonging to A α 1 \ j≥2 A α j . Let n / ∈ m i=1 A α i . We may assume that α 1 < α 2 < · · · < α k . Hence which is greater than zero for sufficiently large numbers n. Finally note that the set AOS b is not 1-algebrable. Indeed, assume that for all n we have |na(n)| < M. It means that a(n) < M/n and hence a 2 (n) < M 2 /n 2 . Therefore na 2 (n) → 0, which means that a 2 / ∈ AOS.

Proof of Theorem 3.2 Let us define an increasing sequence of positive integers
It is enough to take m k := k k+1 + 1. Let ⊂ (2, ∞) be a linearly independent set over rational with | | = c. Denote the set {m 1 , m 2 , . . .} by M. Let us define sequences a α for α ∈ by the formula Let P(x 1 , . . . , x q ) be a polynomial in q variables without a constant term. Hence P is of is a matrix with nonnegative integer terms and with nonzero, pairwise distinct rows. Then, for m k ∈ M we have where r i = α 1 β i 1 + · · · + α q β i q . Let us observe that by the linear independence of the numbers r 1 , r 2 , . . . , r p are nonzero and pairwise distinct. We may assume that r 1 = min(r 1 , r 2 , . . . , r p ). Hence, Since r 1 − r j < 0, then for sufficiently large k Hence P(a α 1 , . . . , a α q ) ∈ AOS ub and P(a α 1 , . . . , a α q )(n) = 0 for almost all n ∈ M. Then the same argument shows P(a α 1 , . . . , a α q )(n) = 0 for all but finitely many numbers n ∈ N\M. Consequently P(a α 1 , . . . , a α q ) ∈ FZ.

Proof of Theorem 3.3
For x ∈ 1 the set {n ∈ N : x(n) = 0} is called the support of x. It is well known that if we take a sequence (x (i) ) i∈N in 1 with pairwise disjoint supports and x (i) = 1, then the closed subspace V generated by ( To prove the spaceability of AOS ub let us divide N into infinite number of infinite sets N i . So = (b(n)) be an arbitrary fixed sequence belonging to AOS ub with b = 1. Let us define a family (b (i) ) i∈N by Let us observe, that the subspace V b generated by (b (i) ) i∈N is contained in AOS ub . Indeed, we have lim sup nb(n) = ∞. Hence, for N i = {n Let A 1 , A 2 , . . . be a sequence of subsets of N such that n∈A i 1 2n−1 = 2 i−1 . Let us define the sequences (a (i) ) i∈N by Let us observe that a (i) = 1 their supports are pariwise disjoint (the supports of a (i) is contained in N i ) and the subspace V a generated by (a (i) ) i∈N is contained in For similar constructions as the above one, see f.e. [2,6]. The next theorem shows, that set AOS ∩ FZ is not spaceable.

Theorem 3.4
Let Y ⊂ 1 be a closed, infinitely dimensional subspace. Then there exists z ∈ Y \{0} and infinite set N ⊂ N such that z(n) = 0 for n ∈ N . In other words, FZ is not spaceable.
By the construction a has no zeros. For any n ∈ N one can find k i > n, and Finally, we show that AOS ∩ FZ is not F σ δ . Put This is a G δσ subset of N N considered with the product topology which is not F σ δ , see [11,Section 23A]. Our aim is to construct a continuous function f : That would show that AOS ∩ FZ is not F σ δ as D is not F σ δ and continuous preimages preserve Borel pointclasses. Let us fix a strictly increasing sequence Let us define f : N N → 1 as follows To prove that f is continuous let us assume that x n → x in N N . Then ∀m ∃k ∀n ≥ k ∀i ≤ m (x n (i) = x(i)), which means that for any m sequences x n eventually equal to x up to m-th coordinate. Let ε > 0. Find i 0 such that i≥i 0 1 k i < ε. Let m := k i 0 − 1. find k with ∀n ≥ k ∀i ≤ m (x n (i) = x(i)).
Let us observe that n f (x)(n) → 0 ⇐⇒ k n f (x)(k n ) → 0. It follows from the fact that n/2 n → 0.
Assume that x ∈ D. Then lim inf n→∞ x(n) = M < ∞. Then there is a strictly increasing sequence of indices (n l ) with x(n l ) = M for every l. Then f (x)(k n l ) = Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/.