# Lineability and Spaceability: A New Approach

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## Abstract

The area of research called “Lineability” looks for linear structures inside exotic subsets of vector spaces. In the last decade lineability/spaceability has been investigated in rather general settings; for instance, Set Theory, Probability Theory, Functional Analysis, Measure Theory, etc. It is a common feeling that positive results on lineability/spaceability are quite natural (i.e., in general “large” subspaces can be found inside exotic subsets of vector spaces, in quite different settings) and more restrictive approaches have been attempted. In this paper we introduce and explore a new approach in this direction.

## Introduction and Preliminaries

The notions of lineability and spaceability were introduced by Gurariy and Quarta (2004) and by Aron et al. (2005). For a comprehensive background on lineability and spaceability we recommend the recent monograph (Aron et al. 2016). This line of research investigates the presence of “large” linear subspaces in certain mathematical objects with a priori no linear structure. The properties of lineability and spaceability are studied in several contexts with interesting applications in different fields as norm-attaining operators, multilinear forms, homogeneous polynomials, sequence spaces, holomorphic mappings, absolutely summing operators, Peano curves, fractals, among others. See, for instance, Albuquerque et al. (2014), Barroso et al. (2013), Bernal-González and Cabrera (2014), Botelho et al. (2011), Botelho et al. (2009), Botelho and Fávaro (2015), Cariello et al. (2017), Cariello and Seoane-Sepúlveda (2014), Pellegrino and Teixeira (2009), Seoane-Sepúlveda (2006), and the references therein.

Let E be a vector space and $$\alpha$$ be a cardinal number. A subset A of E is called $$\alpha \text {-lineable}$$ if $$A\cup \left\{ 0\right\}$$ contains an $$\alpha$$-dimensional linear subspace of E;  if E is a topological vector space, it is called $$\alpha$$-spaceable if $$A\cup \left\{ 0\right\}$$ contains a closed $$\alpha$$-dimensional linear subspace of E. It is well known that, in general, positive results of lineability are rather usual and the feeling that “everything is lineable” is somewhat common. So, a more restrictive notion of lineability is in order. Our paper investigates a stronger notion of lineability/spaceability, which is rather more restrictive.

Let us establish some notations that will be carried out along this work. From now on all vector spaces are considered over a fixed scalar field $${\mathbb {K}}$$ which can be either $${\mathbb {R}}$$ or $${\mathbb {C}}$$. For any set X we shall denote by card(X) the cardinality of X; we also define $${\mathfrak {c}}=card\left( {\mathbb {R}}\right)$$ and $$\aleph _{0}=card\left( {\mathbb {N}}\right)$$. The following concepts are more restrictive notions of lineability/spaceability inspired by some ideas from Nogueira and Pellegrino (2015):

### Definition 1.1

Let $$\alpha ,\beta ,\lambda$$ be cardinal numbers and V be a vector space, with $$\dim V=\lambda$$ and $$\alpha <\beta \le \lambda$$. A set $$A\subset V$$ is:

1. (i)

$$\left( \alpha ,\beta \right)$$-lineable if it is $$\alpha$$-lineable and for every subspace $$W_{\alpha }\subset V$$ with $$W_{\alpha }\subset A\cup \left\{ 0\right\}$$ and $$\dim W_{\alpha }=\alpha$$, there is a subspace $$W_{\beta }\subset V$$ with $$\dim W_{\beta }=\beta$$ and $$W_{\alpha }\subset W_{\beta }\subset A\cup \left\{ 0\right\}$$.

2. (ii)

$$\left( \alpha ,\beta \right)$$-spaceable if it is $$\alpha$$-lineable and for every subspace $$W_{\alpha }\subset V$$ with $$W_{\alpha }\subset A\cup \left\{ 0\right\}$$ and $$\dim W_{\alpha }=\alpha$$, there is a closed subspace $$W_{\beta }\subset V$$ with $$\dim W_{\beta }=\beta$$ and $$W_{\alpha }\subset W_{\beta }\subset A\cup \left\{ 0\right\}$$.

The original notion of lineability (or spaceability) is just the case $$\alpha =0$$. Let $$\alpha _{1},\alpha _{2}$$ be cardinal numbers with $$\alpha _{1}<\alpha _{2}\le \beta$$. We start by showing that $$\left( \alpha _1,\beta \right)$$-lineability does not imply $$\left( \alpha _2,\beta \right)$$-lineability and vice versa. To provide examples, let us recall that for $$p>0$$, $$\ell _{p}$$ denotes the Banach space (p-Banach space if $$p<1$$) of the sequences $$\left( x_{j}\right) _{j=1}^{\infty }$$ such that

\begin{aligned} \left\| \left( x_{j}\right) _{j=1}^{\infty }\right\| _{p}=\left( \sum \limits _{j=1}^{\infty }\left| x_{j}\right| ^{p}\right) ^{1/p}<\infty . \end{aligned}

For each $$j\in {\mathbb {N}}$$, we denote by $$e_{j}$$ the canonical vector $$\left( 0,\ldots ,0,1,0,0,\ldots \right)$$ with 1 in the j-th coordinate.

More generally, let $$p\in [1,\infty )$$ and $$\Gamma$$ be an abstract nonempty set. We denote by $$\ell _{p}(\Gamma )$$ the vector space of all functions $$f:\Gamma \longrightarrow {\mathbb {K}}$$ such that $$\sum _{\gamma \in \Gamma }\vert f(\gamma )\vert ^{p}<\infty ,$$ which becomes a Banach space with the norm

\begin{aligned} \Vert f\Vert _{p}= \left( \sum _{\gamma \in \Gamma }\vert f(\gamma )\vert ^{p}\right) ^{1/p}, \end{aligned}

where the sum is defined by

\begin{aligned} \sum _{\gamma \in \Gamma }\vert f(\gamma )\vert ^{p}=\sup \left\{ \sum _{\gamma \in F}\vert f(\gamma )\vert ^{p}: F \text { is a finite subset of } \Gamma \right\} . \end{aligned}

We also denote the elements of the canonical generalized Schauder basis of $$\ell _{p}(\Gamma )$$ by $$e_{i}$$, $$i\in \Gamma .$$

It is clear that, when $$\Gamma ={\mathbb {N}}$$, we have $$\ell _{p}({\mathbb {N}} )=\ell _{p}.$$ For details about the space $$\ell _{p}(\Gamma )$$ and related results we refer to Fabian et al. (2011).

### Example 1.2

Let $$n\in {\mathbb {N}}$$ and consider the following subset of $$\ell _{p}$$:

\begin{aligned} A={{span}}\left\{ e_{1},e_{2},\ldots ,e_{n}\right\} \cup \left\{ \left( x_{j}\right) _{j=1}^{\infty }\in \ell _{p}:x_{1}=x_{2}=\cdots =x_{n}=0\right\} . \end{aligned}

Thus A is $$\left( n+1,{\mathfrak {c}}\right)$$-lineable, but it is not $$\left( n,{\mathfrak {c}}\right)$$-lineable.

### Example 1.3

Let $${\mathcal {B}}=\{e_{i}:i\in \Gamma \}$$ be the canonical generalized Schauder basis in $$\ell _{p}\left( \Gamma \right)$$ with $$card\left( \Gamma \right) =2^{\aleph _{1}}$$. Choose $$i_{0},i_{1}\in \Gamma ,$$ with $$i_{0}\ne i_{1}$$. We can easily write

\begin{aligned} {\mathcal {B}}-\{e_{i_{0}},e_{i_{1}}\}=\bigcup \limits _{\left( \lambda ,\mu \right) \in {\mathbb {R}}^{2}}{\mathcal {A}}_{\left( \lambda ,\mu \right) } \end{aligned}

as a pairwise disjoint union, with

\begin{aligned} card\left( {\mathcal {A}}_{\left( \lambda ,\mu \right) }\right) =2^{\aleph _{1}} \end{aligned}

for all $$\left( \lambda ,\mu \right) \in {\mathbb {R}}^{2}$$. Let

\begin{aligned} A=\bigcup \limits _{\left( \lambda ,\mu \right) \in {\mathbb {R}}^{2}}span\left( \left\{ \lambda e_{i_{0}}+\mu e_{i_{1}}\right\} \cup {\mathcal {A}}_{\left( \lambda ,\mu \right) }\right) . \end{aligned}

Note that there is no vector space $$W\subset \ell _{p}\left( \Gamma \right)$$ with $$\dim W=2^{\aleph _{1}}$$ and

\begin{aligned} span\{e_{i_{0}},e_{i_{1}}\}\subset W\subset A. \end{aligned}

In fact, if such W exists, since

\begin{aligned} e_{i_{0}}\in W\subset A, \end{aligned}

by the very definition of A we conclude that

\begin{aligned} W\subset \bigcup \limits _{\overset{\lambda \in {\mathbb {R}}}{\lambda \ne 0} }span\left( \left\{ e_{i_{0}}\right\} \cup {\mathcal {A}}_{\left( \lambda ,0\right) }\right) \bigcup span\{e_{i_{0}},e_{i_{1}}\}. \end{aligned}
(1.1)

Let us prove it: If the inclusion above was false, then there would exist $$w\in W$$ with

\begin{aligned} w\notin \bigcup \limits _{\overset{\lambda \in {\mathbb {R}}}{\lambda \ne 0} }span\left( \left\{ e_{i_{0}}\right\} \cup {\mathcal {A}}_{\left( \lambda ,0\right) }\right) \bigcup span\{e_{i_{0}},e_{i_{1}}\}, \end{aligned}

i.e.,

\begin{aligned} w\in \!\!\bigcup \limits _{\underset{\mu \ne 0}{\left( \lambda ,\mu \right) \in {\mathbb {R}}^{2}}}span\left( \left\{ \lambda e_{i_{0}}\!+\mu e_{i_{1} }\right\} \cup {\mathcal {A}}_{\left( \lambda ,\mu \right) }\right) \bigcup span\left( {\mathcal {A}}_{\left( 0,0\right) }\right) \! \text { and }w\notin span\{e_{i_{0}},e_{i_{1}}\}. \end{aligned}

Thus

\begin{aligned} w=r\left( \lambda e_{i_{0}}+\mu e_{i_{1}}\right) +v \end{aligned}

with $$v\in span\left( {\mathcal {A}}_{\left( \lambda ,\mu \right) }\right) ,v\ne 0,$$ and $$r\in {\mathbb {R}}$$, where $$\mu \ne 0$$ or $$\left( \lambda ,\mu \right) =(0,0).$$ First suppose that $$\mu \ne 0$$. Since W is a vector space containing $$e_{i_{0}}$$ and $$e_{i_{1}},$$ we have

\begin{aligned} w+\left( -r\lambda +\lambda \right) e_{i_{0}}+\left( -r\mu +\mu +s\right) e_{i_{1}}\in W, \end{aligned}

i.e.,

\begin{aligned} \lambda e_{i_{0}}+\left( \mu +s\right) e_{i_{1}}+v\in W \end{aligned}

for any choice of $$s\in {\mathbb {R}}.$$ Let us choose $$s\ne -\mu$$ such that

\begin{aligned} \left( \lambda ,\mu \right) \text { and }\left( \lambda ,\mu +s\right) \text { are linearly independent.} \end{aligned}

Then,

\begin{aligned} \lambda e_{i_{0}}+\left( \mu +s\right) e_{i_{1}}\ne 0, \end{aligned}

and since $$v\in span\left( {\mathcal {A}}_{\left( \lambda ,\mu \right) }\right)$$ we have that

\begin{aligned} v\notin \bigcup \limits _{\left( \alpha ,\beta \right) \ne \left( \lambda ,\mu \right) }span\left( {\mathcal {A}}_{\left( \alpha ,\beta \right) }\right) . \end{aligned}

Now it follows from the definition of A that

\begin{aligned} \lambda e_{i_{0}}+\left( \mu +s\right) e_{i_{1}}+v\notin A, \end{aligned}

Now, let us consider the case $$\left( \lambda ,\mu \right) =(0,0).$$ In this case

\begin{aligned} w\in span\left( {\mathcal {A}}_{\left( 0,0\right) }\right) \text { and so } w\notin \bigcup \limits _{\left( \alpha ,\beta \right) \ne \left( 0,0\right) }span\left( {\mathcal {A}}_{\left( \alpha ,\beta \right) }\right) . \end{aligned}

Since $$e_{i_{0}},e_{i_{1}}\in W$$, for any choice of $$\left( \alpha ,\beta \right) \ne (0,0)$$ we obtain

\begin{aligned} \alpha e_{i_{0}}+\beta e_{i_{1}}+w\in W. \end{aligned}

By the definition of A we have

\begin{aligned} \alpha e_{i_{0}}+\beta e_{i_{1}}+w\notin A, \end{aligned}

Analogously, we conclude that

\begin{aligned} W\subset \bigcup \limits _{\mu \in {\mathbb {R}}{\setminus }\{0\}}span\left( \left\{ e_{i_{1}}\right\} \cup {\mathcal {A}}_{\left( 0,\mu \right) }\right) \bigcup span\{e_{i_{0}},e_{i_{1}}\}. \end{aligned}
(1.2)

Thus, from (1.1) and (1.2),

\begin{aligned}&W \subset \left( \bigcup \limits _{\lambda \in {\mathbb {R}}{\setminus } \{0\}}span\left( \left\{ e_{i_{0}}\right\} \cup {\mathcal {A}}_{\left( \lambda ,0\right) }\right) \right) \bigcap \left( \bigcup \limits _{\mu \in {\mathbb {R}}{\setminus }\{0\}}span\left( \left\{ e_{i_{1}}\right\} \cup {\mathcal {A}}_{\left( 0,\mu \right) }\right) \right) \\&\qquad \bigcup span\{e_{i_{0}},e_{i_{1}}\}=span\{e_{i_{0}},e_{i_{1}}\}, \end{aligned}

a contradiction. Thus A is not $$\left( 2,2^{\aleph _{1}}\right)$$-lineable, but it is clear that A is $$\left( 1,2^{\aleph _{1}}\right)$$-lineable.

Obviously, every known lineability/spaceability result can be investigated in this more restrictive setting. In this paper we develop a technique to characterize the $$\left( \alpha ,{\mathfrak {c}}\right)$$-spaceability of $$\ell _{p} {\setminus } {\textstyle \bigcup _{0<q<p}} \ell _{q}$$ for all $$\alpha <{\mathfrak {c}}$$, with $$p>0$$, that illustrates the technicalities arisen by this new approach. We also propose open problems in the final section.

## Characterization of $$\left( \alpha ,{\mathfrak {c}}\right)$$-Spaceability in Sequence Spaces

For $$p>0$$, it is well known that $$\ell _{p}{\setminus } {\textstyle \bigcup _{0<q<p}} \ell _{q}$$ is $${\mathfrak {c}}$$-spaceable (see Botelho et al. 2011). In this section we characterize $$\left( \alpha ,{\mathfrak {c}}\right)$$-spaceability in this setting. For the sake of completeness, we begin by presenting a simple explicit example of a vector inside $$\ell _{p}{\setminus } {\textstyle \bigcup _{0<q<p}} \ell _{q}$$ that we were not able to find in the literature. Let $$\left\lceil {x}\right\rceil$$ denote the smaller integer greater than or equal to x. Let

\begin{aligned} \mathbb {N=}\bigcup \limits _{j=\left\lceil 1/p\right\rceil }^{\infty } {\mathbb {N}}_{j}, \end{aligned}
(2.1)

with $${\mathbb {N}}_{i}\cap {\mathbb {N}}_{j}=\emptyset$$ whenever $$i\ne j$$ and $$card\left( {\mathbb {N}}_{k}\right) =\aleph _{0}$$ for all $$k\in [\left\lceil 1/p\right\rceil ,\infty )$$. Denote, for all $$k\in [\left\lceil 1/p\right\rceil ,\infty ),$$

\begin{aligned} {\mathbb {N}}_{k}=\{k_{1},k_{2},...\} \end{aligned}

with $$k_{r}<k_{s}$$ whenever $$r<s$$, and define

\begin{aligned} x_{j}^{(k)}=\left\{ \begin{array} [c]{l} 0, \quad \text {if }j\notin {\mathbb {N}}_{k}\\ r^{-1/\left( p-k^{-1}\right) ,} \quad \text {if }j=k_{r}\in {\mathbb {N}}_{k}. \end{array} \right. \end{aligned}

Note that $$\left( x_{j}^{(k)}\right) _{j=1}^{\infty }\in \ell _{p} {\setminus }\ell _{p-\frac{1}{k}}$$ for all of those values of k. For each $$j\in {\mathbb {N}}$$, there is k such that $$j\in {\mathbb {N}}_{k}$$, so define

\begin{aligned} y_{j}=\frac{x_{j}^{(k)}}{2^{k}\left\| \left( x_{l}^{(k)}\right) _{l=1}^{\infty }\right\| _{p}}. \end{aligned}

Then, the sequence $$\left( y_{j}\right) _{j=1}^{\infty }$$ belongs to $$\ell _{p}{\setminus } {\textstyle \bigcup _{0<q<p}} \ell _{q}.$$ In fact,

\begin{aligned} \sum \limits _{j=1}^{\infty }\left| y_{j}\right| ^{p}&=\sum \limits _{k=\left\lceil 1/p\right\rceil }^{\infty }\sum \limits _{j\in {\mathbb {N}}_{k}}\left| y_{j}\right| ^{p}\\&=\sum \limits _{k=\left\lceil 1/p\right\rceil }^{\infty }\sum \limits _{j\in {\mathbb {N}}_{k}}\left( \frac{\left| x_{j}^{(k)}\right| }{2^{k}\left\| \left( x_{l}^{(k)}\right) _{l=1}^{\infty }\right\| _{p} }\right) ^{p}\\&=\sum \limits _{k=\left\lceil 1/p\right\rceil }^{\infty }\frac{1}{2^{kp}\left\| \left( x_{l}^{(k)}\right) _{l=1}^{\infty }\right\| _{p}^{p}}\sum \limits _{j\in {\mathbb {N}}_{k}}\left| x_{j}^{(k)}\right| ^{p}\\&=\sum \limits _{k=\left\lceil 1/p\right\rceil }^{\infty }\frac{1}{2^{kp} }<\infty , \end{aligned}

and a similar estimate shows that

\begin{aligned} \sum \limits _{j=1}^{\infty }\left| y_{j}\right| ^{q}=\infty \end{aligned}

for all $$0<q<p.$$

### Lemma 2.1

Let $$p>0$$ and $$x_{1},\ldots ,x_{n}$$ be linearly independent vectors of $$\ell _{p}$$. There exists $$n_{0}\in {\mathbb {N}}$$ such that the first $$n_{0}$$ coordinates of $$x_{1},\ldots ,x_{n}$$ form a linearly independent set of $${\mathbb {K}}^{n_{0}}$$.

### Proof

Suppose that, for each $$j\in {\mathbb {N}}$$, there exist $$a_{1j},\ldots a_{nj} \in {\mathbb {K}}$$, not all equal to zero, such that

\begin{aligned} a_{1j}x_{1}+\cdots +a_{nj}x_{n}=(0,\ldots ,0,\lambda _{j+1},\lambda _{j+2},\ldots ) \end{aligned}
(2.2)

It is plain that we may suppose $$\Vert \alpha _{j}\Vert _{1}=1,$$ where $$\alpha _{j}=(a_{1j},\ldots a_{nj})\in {\mathbb {K}}^{n}$$, for all $$j\in {\mathbb {N}}.$$ Since $$(\alpha _{j})_{j=1}^{\infty }$$ is bounded, there is a convergent subsequence $$\alpha _{j_{k}}\rightarrow \alpha =(a_{1},\ldots ,a_{n})\in {\mathbb {K}}^{n}$$. Since $$\Vert \alpha _{j}\Vert _{1}=1,$$ it follows that $$\alpha \ne 0.$$ Then

\begin{aligned} a_{1j_{k}}x_{1}+\cdots +a_{nj_{k}}x_{n}\rightarrow a_{1}x_{1}+\cdots +a_{n} x_{n},\quad \text { when } \quad k\rightarrow \infty . \end{aligned}
(2.3)

On the other hand, if $$\pi _{m}:\ell _{p}\rightarrow {\mathbb {K}}$$ denotes the m-th canonical projection, then it follows from (2.2) that

\begin{aligned} \pi _{m}(a_{1j_{k}}x_{1}+\cdots +a_{nj_{k}}x_{n})\rightarrow 0,\quad \text { when} \quad k\rightarrow \infty , \quad \text { for every } \quad m\in {\mathbb {N}}. \end{aligned}

Since convergence in (2.3) implies coordinatewise convergence, it follows that $$a_{1}x_{1}+\cdots +a_{n}x_{n}=(0,0,\ldots )$$, which is a contradiction because $$x_{1},\ldots ,x_{n}$$ are linearly independent. $$\square$$

Considering the Continuum Hypothesis we have the following result:

### Theorem 2.2

For all $$p>0$$ the set $$\ell _{p}{\setminus } {\textstyle \bigcup _{0<q<p}} \ell _{q}$$ is $$\left( \alpha ,{\mathfrak {c}}\right)$$-spaceable in $$\ell _{p}$$ if, and only if, $$\alpha <\aleph _{0}$$.

### Proof

If $$\alpha =\aleph _{0}$$, then the following example shows that $$\ell _{p}{\setminus }{\textstyle \bigcup _{0<q<p}} \ell _{q}$$ is not $$\left( \aleph _{0},{\mathfrak {c}}\right)$$-spaceable:

Split

\begin{aligned} \mathbb {N=}\bigcup \limits _{j=1}^{\infty }{\mathbb {N}}_{j} \end{aligned}

as in (2.1). Let

\begin{aligned} W:=span\left\{ x^{(j)}:j\in {\mathbb {N}}\right\} \end{aligned}

with

• $$x^{(1)}=\left( 1,\left( x_{j}^{(1)}\right) _{j>1}\right)$$ where

\begin{aligned} \left\{ \begin{array} [c]{l} x_{j}^{(1)}=0, \quad \text { if }j\notin {\mathbb {N}}_{1}\\ \left( x_{j}^{(1)}\right) _{j\in {\mathbb {N}}_{1}(j>1)}\in \ell _{p}{\setminus } {\textstyle \bigcup _{0<q<p}} \ell _{q} \end{array} \right. \end{aligned}

and

\begin{aligned} \left\| \left( x_{j}^{(1)}\right) _{j\in {\mathbb {N}}_{1}(j>1)}\right\| _{p}=2^{-1}. \end{aligned}
• $$x^{(2)}=\left( 1,\left( x_{j}^{(2)}\right) _{j>1}\right)$$ where

\begin{aligned} \left\{ \begin{array} [c]{l} x_{j}^{(2)}=0,\text { if }j\notin {\mathbb {N}}_{2}\\ \left( x_{j}^{(2)}\right) _{j\in {\mathbb {N}}_{2}(j>1)}\in \ell _{p} {\setminus } {\textstyle \bigcup _{0<q<p}} \ell _{q} \end{array} \right. \end{aligned}

and

\begin{aligned} \left\| \left( x_{j}^{(2)}\right) _{j\in {\mathbb {N}}_{2}(j>1)}\right\| _{p}=2^{-2}, \end{aligned}

and so on. It is obvious that $$\left( W{\setminus }\{0\}\right) \subset \ell _{p}{\setminus } {\textstyle \bigcup _{0<q<p}} \ell _{q}$$ and $$\dim \left( W\right) =\aleph _{0}$$ and there is no closed subspace $$W_{1}$$ of $$\ell _{p}$$ such that $$W\subset W_{1}$$ and $$\left( W_{1}{\setminus }\{0\}\right) \subset \ell _{p} {\setminus } {\textstyle \bigcup _{0<q<p}} \ell _{q}$$, because

\begin{aligned} \lim _{k\rightarrow \infty }\left\| \left( x_{j}^{(k)}\right) _{j\in {\mathbb {N}}}-e_{1}\right\| _{p}=\lim _{k\rightarrow \infty }\left\| \left( 0,x_{j}^{(k)}\right) _{j\in {\mathbb {N}}_{k}(j>1)}\right\| _{p} =\lim _{k\rightarrow \infty }\frac{1}{2^{k}}=0. \end{aligned}

Now we prove the theorem for $$\alpha =3.$$ The general case $$\alpha \in {\mathbb {N}}$$ is easily adapted from this case.

Let $$x=\left( x_{j}\right) _{j=1}^{\infty }$$, $$y=\left( y_{j}\right) _{j=1}^{\infty }$$ and $$z=(z_{j})_{j=1}^{\infty }$$ be linearly independent and

\begin{aligned} W:=span\{x,y,z\}\subset \left( \ell _{p}{\setminus }{\bigcup _{0<q<p}} \ell _{q}\right) \cup \{0\} \end{aligned}

be a 3-dimensional subspace of $$\ell _{p}.$$ Since $$\left( x_{j}\right) _{j=1}^{\infty },\left( y_{j}\right) _{j=1}^{\infty }$$ and $$(z_{j} )_{j=1}^{\infty }$$ are linearly independent, Lemma 2.1 assures that there is a natural number $$n_{0}$$ such that the first $$n_{0}$$ coordinates of $$\left( x_{j}\right) _{j=1}^{\infty },\left( y_{j}\right) _{j=1}^{\infty }$$ and $$(z_{j})_{j=1}^{\infty }$$ form a linearly independent set of $${\mathbb {K}} ^{n_{0}}$$.

It is easy to pick an infinite set $${\mathbb {N}}_{1}:=\{\alpha _{1}<\alpha _{2}<\cdots \}$$ of positive integers such that

\begin{aligned} \left\{ \begin{array} [c]{c} {\displaystyle \sum \nolimits _{j\notin {\mathbb {N}}_{1}}} x_{j}e_{j}\in \ell _{p}{\setminus } {\textstyle \bigcup _{0<q<p}} \ell _{q},\\ {\displaystyle \sum \nolimits _{j\notin {\mathbb {N}}_{1}}} y_{j}e_{j}\in \ell _{p}{\setminus } {\textstyle \bigcup _{0<q<p}} \ell _{q},\\ {\displaystyle \sum \nolimits _{j\notin {\mathbb {N}}_{1}}} z_{j}e_{j}\in \ell _{p}{\setminus } {\textstyle \bigcup _{0<q<p}} \ell _{q} \end{array} \right. \end{aligned}

and such that

\begin{aligned} \max \{\left| x_{\alpha _{l}}\right| ,\left| y_{\alpha _{l} }\right| ,\left| z_{\alpha _{l}}\right| \}<\frac{1}{2^{l}} \end{aligned}
(2.4)

for all $$l\in {\mathbb {N}}$$. Of course, $${\mathbb {N}}{\setminus }{\mathbb {N}}_{1}$$ is also infinite. Let

\begin{aligned} {\mathbb {O}}:={\mathbb {N}}_{1}{\setminus }\{1,\ldots ,n_{0}\}, \end{aligned}
(2.5)

and note that

\begin{aligned} \left\{ \begin{array} [c]{c} {\displaystyle \sum \nolimits _{j\notin {\mathbb {O}}}} x_{j}e_{j}\in \ell _{p}{\setminus } {\textstyle \bigcup _{0<q<p}} \ell _{q},\\ {\displaystyle \sum \nolimits _{j\notin {\mathbb {O}}}} y_{j}e_{j}\in \ell _{p}{\setminus } {\textstyle \bigcup _{0<q<p}} \ell _{q},\\ {\displaystyle \sum \nolimits _{j\notin {\mathbb {O}}}} z_{j}e_{j}\in \ell _{p}{\setminus } {\textstyle \bigcup _{0<q<p}} \ell _{q}. \end{array} \right. \end{aligned}
(2.6)

Split

\begin{aligned} \mathbb {O=} {\displaystyle \bigcup \limits _{i=1}^{\infty }} {\mathbb {O}}_{i} \end{aligned}

with $${\mathbb {O}}_{i}:=\{i_{1}<i_{2}<\cdots \}$$ and $${\mathbb {O}}_{i} \cap {\mathbb {O}}_{j}=\emptyset$$ for all $$i\ne j.$$ It is plain that $$\mathbb {N{\setminus } O}$$ is infinite and let $$f:{\mathbb {N}}\rightarrow {\mathbb {N}}$$ be injective such that

\begin{aligned} \mathbb {N{\setminus } O}:=\{f(1),f(2),\ldots \} \end{aligned}

and define, for all i, the vector

\begin{aligned} \varepsilon _{i}=\sum _{j=1}^{\infty }x_{f(j)}e_{i_{j}}\in {\mathbb {K}}^{{\mathbb {N}} }. \end{aligned}

It is obvious that $$\varepsilon _{i}\in \ell _{p}$$ for all i. Define $${\widetilde{p}}=1$$ if $$p\ge 1$$ and $${\widetilde{p}}=p$$ if $$0<p<1$$. For $$(a_{i})_{i=1}^{\infty }\in \ell _{{\widetilde{p}}}$$,

\begin{aligned} \sum _{i=1}^{\infty }\Vert a_{i}\varepsilon _{i}\Vert _{p}^{{\widetilde{p}}} =\sum _{i=1}^{\infty }|a_{i}|^{{\widetilde{p}}}\Vert \varepsilon _{i}\Vert _{p}^{{\widetilde{p}}}\le \Vert x\Vert _{p}^{{\widetilde{p}}}\sum _{i=1}^{\infty }\left| a_{i}\right| ^{{\widetilde{p}}}=\Vert x\Vert _{p}^{{\widetilde{p}} }\left\| (a_{i})_{i=1}^{\infty }\right\| _{{\widetilde{p}}}^{{\widetilde{p}} }<\infty . \end{aligned}

Thus $$\sum _{i=1}^{\infty }\Vert a_{i}\varepsilon _{i}\Vert _{p}^{{\widetilde{p}} }<\infty$$ and thus the series $$\sum _{i=1}^{\infty }a_{i}\varepsilon _{i}$$ converges in $$\ell _{p}$$. Hence, the operator

\begin{aligned} \left\{ \begin{array} [c]{l} T:\ell _{{\widetilde{p}}}\longrightarrow \ell _{p},\\ ~T\left( \left( a_{i}\right) _{i=0}^{\infty }\right) =a_{0}x+a_{1} y+a_{2}z+\sum \nolimits _{i=3}^{\infty }a_{i}\varepsilon _{i} \end{array} \right. \end{aligned}

is well defined. It is easy to see that T is linear and injective. In fact, if

\begin{aligned} T\left( \left( a_{i}\right) _{i=0}^{\infty }\right) =0 \end{aligned}

then

\begin{aligned} a_{0}x+a_{1}y+a_{2}z+\sum \limits _{i=3}^{\infty }a_{i}\varepsilon _{i}=0. \end{aligned}

In particular, since the first $$n_{0}$$ coordinates of each $$\varepsilon _{i}$$ are all zero, we have

\begin{aligned} a_{0}\left( x_{j}\right) _{j=1}^{n_{0}}+a_{1}\left( y_{j}\right) _{j=1}^{n_{0}}+a_{2}\left( z_{j}\right) _{j=1}^{n_{0}}=0 \end{aligned}

and then

\begin{aligned} a_{0}=a_{1}=a_{2}=0. \end{aligned}
(2.7)

From (2.7) and since the supports of the vectors $$\varepsilon _{i}$$ are disjoint, we have $$a_{i}=0$$ for all i.

Note that $$T\left( \left( a_{i}\right) _{i=0}^{\infty }\right) \notin \ell _{q}$$ for all $$0<q<p$$ and $$\left( a_{i}\right) _{i=0}^{\infty }\ne 0$$. In fact, let $$i_{0}$$ be such that $$a_{i_{0}}\ne 0.$$ If $$a_{0}=a_{1}=a_{2}=0$$ we have

\begin{aligned} \left\| T\left( \left( a_{i}\right) _{i=0}^{\infty }\right) \right\| _{q}=\left\| \sum \limits _{i=3}^{\infty }a_{i}\varepsilon _{i}\right\| _{q}\ge \left| a_{i_{0}}\right| \left\| \varepsilon _{i_{0} }\right\| _{q}=\infty . \end{aligned}

If $$a_{i}\ne 0$$ for some $$i=0,1,2,$$ since the coordinates of $$\varepsilon _{i}$$ belonging to $$\mathbb {N{\setminus } O}$$ are zero, we have

\begin{aligned} \left\| T\left( \left( a_{i}\right) _{i=0}^{\infty }\right) \right\| _{q}=\left\| a_{0}x+a_{1}y+a_{2}z+\sum \limits _{i=3}^{\infty }a_{i} \varepsilon _{i}\right\| _{q}\ge \left\| \left( a_{0}x_{j}+a_{1} y_{j}+a_{2}z_{j}\right) _{j\notin {\mathbb {O}}}\right\| _{q}. \end{aligned}

Since

\begin{aligned} \infty= & {} \left\| \left( a_{0}x_{j}+a_{1}y_{j}+a_{2}z_{j}\right) _{j\in {\mathbb {N}}}\right\| _{q}^{q}=\left\| \left( a_{0}x_{j}+a_{1} y_{j}+a_{2}z_{j}\right) _{j\in {\mathbb {O}}}\right\| _{q}^{q}\\&+\left\| \left( a_{0}x_{j}+a_{1}y_{j}+a_{2}z_{j}\right) _{j\notin {\mathbb {O}} }\right\| _{q}^{q} \end{aligned}

and by (2.4) and (2.5) we have

\begin{aligned} \left\| \left( a_{0}x_{j}+a_{1}y_{j}+a_{2}z_{j}\right) _{j\in {\mathbb {O}} }\right\| _{q}<\infty , \end{aligned}

then

\begin{aligned} \left\| \left( a_{0}x_{j}+a_{1}y_{j}+a_{2}z_{j}\right) _{j\notin {\mathbb {O}}}\right\| _{q}=\infty . \end{aligned}
(2.8)

Thus $$T\left( \left( a_{i}\right) _{i=0}^{\infty }\right) \notin \ell _{q}$$ for all $$0<q<p.$$ We have thus just proved the $$\left( 3,{\mathfrak {c}}\right)$$-lineability (and of course the $$\left( n,{\mathfrak {c}}\right)$$-lineability for $$n\in {\mathbb {N}}$$ is analogous). Now let us prove the spaceability.

Of course, $$\overline{T\left( \ell _{{\widetilde{p}}}\right) }$$ is a closed infinite-dimensional subspace of $$\ell _{p}$$. We just have to show that

\begin{aligned} \overline{T\left( \ell _{{\widetilde{p}}}\right) }\mathbb {{\setminus }}\left\{ 0\right\} \subset \ell _{p}{\setminus } {\textstyle \bigcup _{0<q<p}} \ell _{q}. \end{aligned}

Let $$w=\left( w_{n}\right) _{n=1}^{\infty }\in \overline{T\left( \ell _{{\widetilde{p}}}\right) }$$, $$w\ne 0$$. There are sequences $$\left( a_{i}^{(k)}\right) _{i=0}^{\infty }\in \ell _{{\widetilde{p}}}$$, $$k\in {\mathbb {N}}$$, such that $$w=\lim _{k\rightarrow \infty }T\left( \left( a_{i}^{(k)}\right) _{i=0}^{\infty }\right)$$ in $$\ell _{p}.$$ Note that, for each $$k\in {\mathbb {N}}$$,

\begin{aligned} T\left( \left( a_{i}^{(k)}\right) _{i=0}^{\infty }\right)&=a_{0} ^{(k)}x+a_{1}^{(k)}y+a_{2}^{(k)}z+\sum \limits _{i=3}^{\infty }a_{i} ^{(k)}\varepsilon _{i}\\&=a_{0}^{(k)}x+a_{1}^{(k)}y+a_{2}^{(k)}z+\sum \limits _{i=3}^{\infty } a_{i}^{(k)}\sum _{j=1}^{\infty }x_{f(j)}e_{i_{j}}\\&=a_{0}^{(k)}x+a_{1}^{(k)}y+a_{2}^{(k)}z+\sum \limits _{i=3}^{\infty } \sum \limits _{j=1}^{\infty }a_{i}^{(k)}x_{f(j)}e_{i_{j}}. \end{aligned}

Since the coordinates of $$\varepsilon _{i}$$ belonging to $$\mathbb {N{\setminus } O}$$ are zero, then the coordinates of $$\sum \nolimits _{i=3}^{\infty }a_{i} ^{(k)}\varepsilon _{i}$$ belonging to $$\mathbb {N{\setminus } O}$$ are also zero. Thus, the coordinates of $$T\left( \left( a_{i}^{(k)}\right) _{i=0}^{\infty }\right)$$ belonging to $$\mathbb {N{\setminus } O}$$ are the respective coordinates of $$a_{0}^{(k)}x+a_{1}^{(k)}y+a_{2}^{(k)}z$$ for all k. Hence the limit

\begin{aligned} \lim _{k\rightarrow \infty }\left( a_{0}^{(k)}x_{j}+a_{1}^{(k)}y_{j}+a_{2} ^{(k)}z_{j}\right) _{j\in \mathbb {N{\setminus } O}} \end{aligned}

exists in $$\ell _{p}$$. Since $$\{\left( x_{j}\right) _{j\in \mathbb {N{\setminus } O}},\left( y_{j}\right) _{j\in \mathbb {N{\setminus } O} },\left( z_{j}\right) _{j\in \mathbb {N{\setminus } O}}\}$$ is linearly independent (because $$\{1,...,n_{0}\}\subset \mathbb {N{\setminus } O}$$), it is obvious that this limit can be written in an unique form as

\begin{aligned} \left( ax_{j}+by_{j}+cz_{j}\right) _{j\in \mathbb {N{\setminus } O}}. \end{aligned}

Applying a linear functional that sends $$x_{j}$$ in 1 and $$y_{j}$$ and $$z_{j}$$ in 0 we obtain that

\begin{aligned} \lim _{k\rightarrow \infty }\left( a_{0}^{(k)}\right) =a. \end{aligned}

Analogously we obtain

\begin{aligned} \lim _{k\rightarrow \infty }\left( a_{1}^{(k)}\right) =b\text { and } \lim _{k\rightarrow \infty }\left( a_{2}^{(k)}\right) =c. \end{aligned}

Thus

\begin{aligned} w_{j}=\left( \lim _{k\rightarrow \infty }a_{0}^{(k)}\right) x_{j}+\left( \lim _{k\rightarrow \infty }a_{1}^{(k)}\right) y_{j}+\left( \lim _{k\rightarrow \infty }a_{2}^{(k)}\right) z_{j} \end{aligned}

for all positive integers j in $$\mathbb {N{\setminus } O}.$$ To finish the proof consider the following cases:

• First case: $$\lim _{k\rightarrow \infty }a_{i}^{(k)}\ne 0,$$ for some $$i=0,1,2.$$

This case is simple because

\begin{aligned} \left\| w\right\| _{q}^{q}\ge \sum \limits _{j\in \mathbb {N{\setminus } O} }\left| a_{0}x_{j}+a_{1}y_{j}+a_{2}z_{j}\right| ^{q}\overset{(2.8)}{=}\infty \end{aligned}

and the proof is done.

• Second case: $$\lim _{k\rightarrow \infty }a_{i}^{(k)}=0$$ for all $$i=0,1,2.$$

In this case $$w_{j}=0$$ for all j in $$\mathbb {N{\setminus } O}.$$ Since $$w\ne 0,$$ for all there is $$r\in {\mathbb {O}}$$ such that $$w_{r}\ne 0.$$

Since $${\mathbb {O}}=\bigcup _{j=1}^{\infty }{\mathbb {O}}_{j}$$, there are (unique) $$m,t\in {\mathbb {N}}$$ such that $$e_{m_{t}}=e_{r}$$. Thus, for each $$k\in {\mathbb {N}}$$, the r-th coordinate of $$T\left( \left( a_{i}^{(k)}\right) _{i=0}^{\infty }\right)$$ is the number $$a_{0}^{(k)}x_{r}+a_{1}^{(k)} y_{r}+a_{2}^{(k)}z_{r}+a_{m}^{(k)}x_{f(t)}.$$ So

\begin{aligned}&0\ne w_{r}=\lim _{k\rightarrow \infty }\left( a_{0}^{(k)}x_{r}+a_{1}^{(k)} y_{r}+a_{2}^{(k)}z_{r}+a_{m}^{(k)}x_{f(t)}\right) \\&\quad =\lim _{k\rightarrow \infty }a_{m}^{(k)}x_{f(t)}=x_{f(t)}\cdot \lim _{k\rightarrow \infty }a_{m}^{(k)}. \end{aligned}

It follows that $$x_{f(t)}\ne 0.$$ Hence $$\lim _{k\rightarrow \infty }|a_{m} ^{(k)}|=\frac{\left| w_{r}\right| }{\left| x_{f(t)}\right| }\ne 0$$. For $$j,k\in {\mathbb {N}}$$, the $$m_{j}$$-th coordinate of $$T\left( \left( a_{i}^{(k)}\right) _{i=0}^{\infty }\right)$$ is

\begin{aligned} a_{0}^{(k)}x_{m_{j}}+a_{1}^{(k)}y_{m_{j}}+a_{2}^{(k)}z_{m_{j}}+a_{m} ^{(k)}x_{f(j)}. \end{aligned}

Defining $$\alpha _{m}=\frac{\left| w_{r}\right| }{\left| x_{f(t)}\right| }$$ we have

\begin{aligned}&\lim _{k\rightarrow \infty }\left| a_{0}^{(k)}x_{m_{j}}+a_{1}^{(k)}y_{m_{j} }+a_{2}^{(k)}z_{m_{j}}+a_{m}^{(k)}x_{f(j)}\right| =\lim _{k\rightarrow \infty }\left| a_{m}^{(k)}x_{f(j)}\right| \\&\quad =\left| x_{f(j)} \right| \cdot \lim _{k\rightarrow \infty }|a_{m}^{(k)}|=\alpha _{m}\left| x_{f(j)}\right| \end{aligned}

for every $$j\in {\mathbb {N}}$$. On the other hand, coordinatewise convergence gives us

\begin{aligned} \lim _{k\rightarrow \infty }\left| a_{0}^{(k)}x_{m_{j}}+a_{1}^{(k)}y_{m_{j} }+a_{2}^{(k)}z_{m_{j}}+a_{m}^{(k)}x_{f(j)}\right| =\left| w_{m_{j} }\right| , \end{aligned}

so $$\left| w_{m_{j}}\right| =\alpha _{m}\left| x_{f(j)}\right|$$ for each $$j\in {\mathbb {N}}$$. Hence

\begin{aligned}&\left\| w\right\| _{q}^{q}=\sum \limits _{n=1}^{\infty }\left| w_{n}\right| ^{q}\ge \sum \limits _{j=1}^{\infty }\left| w_{m_{j} }\right| ^{q}=\sum \limits _{j=1}^{\infty }\alpha _{m}^{q}\cdot \left| x_{f(j)}\right| ^{q}=\alpha _{m}^{q}\cdot \left\| \left( x_{f(j)} \right) _{j=1}^{\infty }\right\| _{q}^{q}\\&\quad =\alpha _{m}^{q}\cdot \left\| \left( x_{j}\right) _{j\in \mathbb {N{\setminus } O}}\right\| _{q}^{q} \overset{{(2.6)}}{=}\infty . \end{aligned}

Therefore $$w\notin {\textstyle \bigcup _{0<q<p}} \ell _{q}$$, so $$\overline{T\left( \ell _{{\widetilde{p}}}\right) }{\setminus }\left\{ 0\right\} \subseteq \ell _{p}{\setminus } {\textstyle \bigcup _{0<q<p}} \ell _{q}.$$$$\square$$

## Some Open Problems

Of course, any positive result of lineability and/or spaceability is a potential problem to be investigated in our more general framework (for instance, the results of Albuquerque et al. 2014; Bernal-González and Cabrera 2014; Botelho et al. 2009, 2011, 2012a, b, 2014; Pellegrino and Teixeira 2009). Our feeling is that, in general, new techniques and tools are needed to deal with these new problems. We finish this paper by illustrating situations in which we were able just to prove $$\left( 1,{\mathfrak {c}}\right)$$-lineability or $$\left( 1,{\mathfrak {c}} \right)$$-spaceability.

### Non Injective Linear Operators

Lineability and spaceability are investigated in the framework of (non) injective and (non) surjective functions/operators in several papers (see, for instance, Albuquerque 2014; Aron et al. 2018; Bernal-González 2017; Gámez-Merino 2011; Jiménez-Rodríguez et al. 2016 and the references therein). In this section we prove a related result in the new context initiated in the present paper; our result provides only $$\left( 1,{\mathfrak {c}}\right)$$-lineability; the general case seems to be an interesting open problem, as well as the case of (non) surjective operators.

### Theorem 3.1

Let $$p,q\ge 1$$ and $$A:=\left\{ T\in {\mathcal {L}}\left( \ell _{p};\ell _{q}\right) :\text { }T\text { is non injective}\right\}$$. Then, A is $$\left( 1,{\mathfrak {c}} \right)$$-lineable.

### Proof

In fact, let $$T\in A{\setminus }\{0\}$$ and consider $$W_{1}$$ the subspace generated by T . By hypothesis, there exist xy in $$\ell _{p}$$ with $$x\ne y$$ such that

\begin{aligned} T\left( x\right) =T\left( y\right) . \end{aligned}
(3.1)

Since $$T\ne 0$$, there is a $$z\in \ell _{p}$$ such that $$T\left( z\right) \ne 0$$. Let $$j_{0}$$ be such that $$\left( Tz\right) _{j_{0}}\ne 0,$$ where $$\left( Tz\right) _{j_{0}}$$ denotes the $$j_{0}$$-th coordinate of the sequence T(z). Let us choose $$\left( {\mathbb {N}}_{k}\right) _{k=1}^{\infty }$$ a sequence of pairwise disjoint subsets of $${\mathbb {N}}$$ with $$card\left( {\mathbb {N}}_{k}\right) =\aleph _{0}$$ for all k, and such that $$j_{0}\notin {\textstyle \bigcup \nolimits _{k=1}^{\infty }} {\mathbb {N}}_{k}$$.

Define, for each $$k\in {\mathbb {N}}$$, a sequence of linear operators $$T_{k} :\ell _{p}\longrightarrow \ell _{q}$$ by

\begin{aligned} \left\{ \begin{array} [c]{l} \left( T_{k}(x)\right) _{j}=0,\text { if }j\notin {\mathbb {N}}_{k}\\ \left( T_{k}(x)\right) _{j}=\left( T(x)\right) _{i}\text { if }j=k_{i} \in {\mathbb {N}}_{k}. \end{array} \right. \end{aligned}

Note that $$T\left( x\right) =T\left( y\right) \Rightarrow T_{k}\left( x\right) =T_{k}\left( y\right)$$. We conclude that $$T_{k}\in A$$ for all $$k\in {\mathbb {N}}$$. In addition, the set $$\left\{ T,T_{k}:k\in {\mathbb {N}} \right\}$$ is linearly independent. In fact, let $$a,a_{1},\ldots ,a_{k}$$ be scalars and suppose that

\begin{aligned} aT+a_{1}T_{1}+\cdots +a_{k}T_{k}=0. \end{aligned}

We have

\begin{aligned} aT\left( z\right) +a_{1}T_{1}\left( z\right) +\cdots +a_{k}T_{k}\left( z\right) =0. \end{aligned}

In particular

\begin{aligned} a\left( Tz\right) _{j_{0}}+a_{1}\left( T_{1}z\right) _{j_{0}}+\cdots +a_{k}\left( T_{k}z\right) _{j_{0}}=0. \end{aligned}

Since $$j_{0}\notin {\textstyle \bigcup \nolimits _{k=1}^{\infty }} {\mathbb {N}}_{k}$$ , it follows that $$\left( T_{1}z\right) _{j_{0}}=\cdots =\left( T_{k}z\right) _{j_{0}}=0.$$ So, $$a\left( Tz\right) _{j_{0}}=0$$ and thus $$a=0$$. Consequently

\begin{aligned} a_{1}T_{1}+\cdots +a_{k}T_{k}=0. \end{aligned}

Since $$\{T_{1},\ldots ,T_{k}\}$$ is linearly independent, we conclude that $$a_{1}=\cdots =a_{k}=0$$.

Now consider the linear operator

\begin{aligned} \Psi :\ell _{1}\longrightarrow {\mathcal {L}}\left( \ell _{p};\ell _{q}\right) \end{aligned}

given by

\begin{aligned} \Psi \left( \left( a_{k}\right) _{k=1}^{\infty }\right) =a_{1}T+ {\textstyle \sum \limits _{j=2}^{\infty }} a_{j}T_{j-1}. \end{aligned}

Note that $$\Psi$$ is well defined, because

\begin{aligned} \left\| \Psi \left( \left( a_{k}\right) _{k=1}^{\infty }\right) \right\| _{{\mathcal {L}}\left( \ell _{p};\ell _{q}\right) }&=\left\| a_{1}T+ {\textstyle \sum \limits _{j=2}^{\infty }} a_{j}T_{j-1}\right\| _{{\mathcal {L}}\left( \ell _{p};\ell _{q}\right) }\\&\le \left\| a_{1}T\right\| _{{\mathcal {L}}\left( \ell _{p};\ell _{q}\right) }+ {\textstyle \sum \limits _{j=2}^{\infty }} \left\| a_{j}T_{j-1}\right\| _{{\mathcal {L}}\left( \ell _{p};\ell _{q}\right) }\\&\le \left| a_{1}\right| \left\| T\right\| _{{\mathcal {L}} \left( \ell _{p};\ell _{q}\right) }+ {\textstyle \sum \limits _{j=2}^{\infty }} \left| a_{j}\right| .\left\| T\right\| _{{\mathcal {L}}\left( \ell _{p};\ell _{q}\right) }\\&\le {\textstyle \sum \limits _{j=2}^{\infty }} \left| a_{j}\right| .\left\| T\right\| _{{\mathcal {L}}\left( \ell _{p};\ell _{q}\right) }<\infty . \end{aligned}

Moreover, it is not difficult to prove that $$\Psi$$ is injective. Let $$a=\left( a_{k}\right) _{k=1}^{\infty }\in \ell _{1},$$ with $$a_{k}\ne 0$$ for some $$k\in {\mathbb {N}}$$ . So, from (3.1),

\begin{aligned} \Psi \left( \left( a_{k}\right) _{k=1}^{\infty }\right) \left( x\right)&=a_{1}T\left( x\right) +a_{2}T_{1}\left( x\right) +a_{3}T_{2}\left( x\right) +\cdots \\&=a_{1}T\left( y\right) +a_{2}T_{1}\left( y\right) +a_{3}T_{2}\left( y\right) +\cdots \\&=\Psi \left( \left( a_{k}\right) _{k=1}^{\infty }\right) \left( y\right) , \end{aligned}

and $$\Psi \left( \left( a_{k}\right) _{k=1}^{\infty }\right)$$ is non injective. Hence, $$\Psi \left( \ell _{1}{\setminus }\left\{ 0\right\} \right) \subseteq A.$$ Moreover,

\begin{aligned} T\in W_{1}\subseteq \Psi \left( \ell _{1}{\setminus }\left\{ 0\right\} \right) \subseteq A\cup \left\{ 0\right\} . \end{aligned}

Therefore, A is $$\left( 1,{\mathfrak {c}}\right)$$-lineable. $$\square$$

### $$\left( \alpha ,{\mathfrak {c}}\right)$$-Spaceability in $$L_{p}$$ Spaces

Recall that, for $$p>0$$, $$L_{p}\left[ 0,1\right]$$ denotes the classical space of the (class of equivalence of) measurable functions $$f:\left[ 0,1\right] \longrightarrow {\mathbb {K}}$$ equipped with the norm (p-norm if $$0<p<1$$) defined by

\begin{aligned} \left\| f\right\| _{p}=\left( \int \nolimits _{0}^{1}\left| f(t)\right| ^{p}{d}t\right) ^{\frac{1}{p}}. \end{aligned}

By mimicking our constructive example of the beginning of the previous section, we can easily provide a simple construction of a function in $$L_{p}\left[ 0,1\right] \diagdown \bigcup \nolimits _{q>p}L_{q}\left[ 0,1\right] .$$

In Botelho et al. (2012a) it was proved that $$L_{p}\left[ 0,1\right] \diagdown \bigcup \nolimits _{q>p}L_{q}\left[ 0,1\right]$$ is spaceable, but the proof does not assure that $$L_{p}\left[ 0,1\right] \diagdown \bigcup \nolimits _{q>p} L_{q}\left[ 0,1\right]$$ is $$\left( \alpha ,{\mathfrak {c}}\right)$$-spaceable for some cardinal $$\alpha >0.$$ The next result shows that this is true for $$\alpha =1.$$ The question for a cardinal $$1<\alpha <{\mathfrak {c}}$$ remain unanswered. In the next proof, for any $$X\subset [0,1]$$, the characteristic function of X on [0, 1] is denoted by $$\chi _{X}$$.

### Theorem 3.2

$$L_{p}\left[ 0,1\right] \diagdown {\textstyle \bigcup \nolimits _{q>p}} L_{q}\left[ 0,1\right]$$ is $$\left( 1,{\mathfrak {c}}\right)$$-spaceable in $$L_{p}\left[ 0,1\right] ,$$ for every $$p>0$$.

### Proof

Let $$f\in L_{p}\left[ 0,1\right] \diagdown {\textstyle \bigcup _{q>p}} L_{q}\left[ 0,1\right] .$$ It is obvious that

\begin{aligned} {\widetilde{f}}=f\chi _{[0,1/2]}\text { or }\widetilde{{\widetilde{f}}}=\text { }f\chi _{[1/2,1]} \end{aligned}

belongs to $$L_{p}\left[ 0,1\right] \diagdown {\textstyle \bigcup _{q>p}} L_{q}\left[ 0,1\right] .$$ Without loss of generality let us assume that

\begin{aligned} {\widetilde{f}}\in L_{p}\left[ 0,1\right] \diagdown {\textstyle \bigcup _{q>p}} L_{q}\left[ 0,1\right] . \end{aligned}

Split [1 / 2, 1) as an infinite sequence of disjoint intervals $$I_{n} =[c_{n},d_{n})$$. Notice that, for every $$n\in {\mathbb {N}}$$ and every $$x\in I_{n}$$, there is a unique $$t_{x,n}\in \left[ 0,1\right)$$ such that

\begin{aligned} x=(1-t_{x,n})c_{n}+t_{x,n}d_{n}. \end{aligned}

Define

\begin{aligned} f_{n}(x)=\left\{ \begin{array}[c]{ll} {\widetilde{f}}(t_{x,n}) &{} \quad \quad \mathrm {if~}x\in I_{n},\\ 0 &{} \quad \quad \mathrm {if~}x\notin I_{n}. \end{array} \right. \end{aligned}

It is simple to verify that our construction provides

\begin{aligned} \left\| f_{n}\right\| _{p}<\left\| {\widetilde{f}}\right\| _{p}, \end{aligned}

for every $$n\in {\mathbb {N}}$$ and so $$f_{n}\in L_{p}\left[ 0,1\right]$$.

Define $${\widetilde{p}}=1$$ if $$p\ge 1$$ and $${\widetilde{p}}=p$$ if $$0<p<1$$. For $$(a_{i})_{i=1}^{\infty }\in \ell _{{\widetilde{p}}}$$,

\begin{aligned} \sum _{i=1}^{\infty }\Vert a_{i}f_{i}\Vert _{p}^{{\widetilde{p}}}=\sum _{i=1}^{\infty }|a_{i}|^{{\widetilde{p}}}\Vert f_{i}\Vert _{p}^{{\widetilde{p}}} \le \Vert {\widetilde{f}}\Vert _{p}^{{\widetilde{p}}}\sum _{i=1}^{\infty }\left| a_{i}\right| ^{{\widetilde{p}}}=\Vert {\widetilde{f}}\Vert _{p}^{{\widetilde{p}} }\left\| (a_{i})_{i=1}^{\infty }\right\| _{{\widetilde{p}}}^{{\widetilde{p}} }<\infty . \end{aligned}

Thus $$\sum _{i=1}^{\infty }\Vert a_{i}f_{i}\Vert _{p}^{{\widetilde{p}}}<\infty$$ and since $$L_{p}\left[ 0,1\right]$$ is Banach (p-Banach if $$0<p<1$$), then the series $$\sum _{i=1}^{\infty }a_{i}f_{i}$$ converges in $$L_{p}\left[ 0,1\right]$$. Hence, the operator

\begin{aligned} T:\ell _{{\widetilde{p}}}\longrightarrow L_{p}\left[ 0,1\right] ,\quad T\left( \left( a_{i}\right) _{i=0}^{\infty }\right) =a_{0} f+\sum \limits _{i=1}^{\infty }a_{i}f_{i} \end{aligned}

is well defined. It is easy to see that T is linear and injective. In fact, if

\begin{aligned} T\left( \left( a_{i}\right) _{i=0}^{\infty }\right) =0 \end{aligned}

then

\begin{aligned} a_{0}f+\sum \limits _{i=1}^{\infty }a_{i}f_{i}=0 \end{aligned}

and choosing $$x\in [0,1/2]$$ we have

\begin{aligned} a_{0}f(x)=a_{0}f(x)+\sum \limits _{i=1}^{\infty }a_{i}f_{i}(x)=0. \end{aligned}

Since f is non null on [0, 1 / 2] we conclude that $$a_{0}=0$$. Since $$\{f_{i}:i\in {\mathbb {N}}\}$$ is linearly independent (they have disjoint supports) we obtain $$a_{i}=0$$ for all i.

Thus $$\overline{T\left( \ell _{{\widetilde{p}}}\right) }$$ is a closed infinite-dimensional subspace of $$L_{p}\left[ 0,1\right]$$. We just have to show that

\begin{aligned} \overline{T\left( \ell _{{\widetilde{p}}}\right) }-\left\{ 0\right\} \subseteq L_{p}\left[ 0,1\right] \diagdown {\textstyle \bigcup _{q>p}} L_{q}\left[ 0,1\right] . \end{aligned}

Indeed, let $$g\in \overline{T\left( \ell _{{\widetilde{p}}}\right) } {\setminus }\{0\}$$. Thus $$g\ne 0$$ a.e., that is, the set $$[0,1]-A$$ has null measure, where

\begin{aligned} A=\{x\in [0,1]:g(x)\ne 0\}. \end{aligned}

Let us consider sequences $$\left( a_{i}^{(k)}\right) _{i=0}^{\infty }\in \ell _{{\widetilde{p}}}$$ ($$k\in {\mathbb {N}}$$) such that $$g=\lim _{k\rightarrow \infty }T\left( \left( a_{i}^{(k)}\right) _{i=0}^{\infty }\right)$$ in $$L_{p}[0,1].$$ By the definition of T we have

\begin{aligned} T\left( \left( a_{i}^{(k)}\right) _{i=0}^{\infty }\right) =a_{0}^{\left( k\right) }f+\sum \limits _{i=1}^{\infty }a_{i}^{\left( k\right) }f_{i} \overset{k\rightarrow \infty }{\longrightarrow }g\text { in }L_{p}\left[ 0,1\right] . \end{aligned}

In particular

\begin{aligned} T\left( \left( a_{i}^{(k)}\right) _{i=0}^{\infty }\right) \chi _{[0,1/2]}\overset{k\rightarrow \infty }{\longrightarrow }g\chi _{[0,1/2]}\text { in }L_{p}\left[ 0,1\right] . \end{aligned}

Since $$f_{i}$$ is null on the interval [0, 1 / 2] for all i,  we have

\begin{aligned} a_{0}^{\left( k\right) }{\widetilde{f}}=T\left( \left( a_{i}^{(k)}\right) _{i=0}^{\infty }\right) \chi _{[0,1/2]}\overset{k\rightarrow \infty }{\longrightarrow }g\chi _{[0,1/2]}\text { in }L_{p}\left[ 0,1\right] . \end{aligned}
(3.2)

On the other hand, the convergence in $$L_{p}\left[ 0,1\right]$$ given in (3.2) ensures that there is a subsequence $$\left( a_{0}^{\left( k_j\right) }{\widetilde{f}}\right) _{j=1}^{\infty }$$ such that

\begin{aligned} a_{0}^{\left( k_j\right) }{\widetilde{f}}(x)\longrightarrow g\chi _{[0,1/2]}(x) \text { a.e, when } j\rightarrow \infty . \end{aligned}

Since $${\widetilde{f}}$$ is non-null a.e., there is $$x_0\in [0,1/2]$$ such that $${\widetilde{f}}(x_0)\ne 0$$ and

\begin{aligned} a_{0}^{\left( k_j\right) }{\widetilde{f}}(x_0)\longrightarrow g\chi _{[0,1/2]}(x_0) \text { when } j\rightarrow \infty . \end{aligned}

So, $$\lim _{j\rightarrow \infty }a_{0}^{\left( k_j\right) }$$ exists and clearly,

\begin{aligned} a_{0}^{\left( k_j\right) }{\widetilde{f}}\overset{j\rightarrow \infty }{\longrightarrow }\alpha {\widetilde{f}}\quad \text { in } \quad L_{p}\left[ 0,1\right] , \end{aligned}
(3.3)

where $$\alpha =\lim _{j\rightarrow \infty }a_{0}^{\left( k_j\right) }$$. By (3.2) and (3.3) we have

\begin{aligned} g\chi _{[0,1/2]}=\alpha {\widetilde{f}}\quad \text { a.e.}\quad \end{aligned}

So, if $$\alpha \ne 0$$

\begin{aligned} \left\| g\right\| _{q}^{q}=\int \nolimits _{0}^{1}\left| g\left( t\right) \right| ^{q}{d}t\ge \int \nolimits _{0}^{\frac{1}{2}}\left| g\left( t\right) \right| ^{q}{d}t=\alpha ^{q}\left\| f\chi _{[0,1/2]}\right\| _{q}^{q}=\infty , \end{aligned}

proving that $$g\notin L_{q}\left[ 0,1\right]$$.

If $$\alpha =0,$$ then $$g\chi _{[0,1/2]}=0$$ a.e. Define

\begin{aligned} {\widetilde{A}}=\{x\in [1/2,1]:g(x)\ne 0\}. \end{aligned}

Since A has positive measure and $$g\chi _{[0,1/2]}=0$$ a.e., then $${\widetilde{A}}$$ has positive measure. Since

\begin{aligned} \left\| \left( a_{0}^{(k)}f+\sum _{n=1}^{\infty }a_{n}^{(k)}f_{n}-g\right) \chi _{[1/2,1]}\right\| _{p}\overset{k\rightarrow \infty }{\longrightarrow }0, \end{aligned}

there is a subsequence

\begin{aligned} \left( \left( a_{0}^{(k_{j})}f+\sum _{n=1}^{\infty }a_{n}^{(k_{j})} f_{n}\right) \chi _{[1/2,1]}\right) _{j=1}^{\infty } \end{aligned}

such that

\begin{aligned} \left( a_{0}^{(k_{j})}f(x)+\sum _{n=1}^{\infty }a_{n}^{(k_{j})}f_{n}(x)\right) \chi _{[1/2,1]}(x)\overset{j\rightarrow \infty }{\longrightarrow } g(x)\chi _{[1/2,1]}(x)\text { a.e.} \end{aligned}

Hence the set $$[1/2,1]-B$$, where $$B=\left\{ x\in [1/2,1]:\text { the limit above holds}\right\}$$, has measure zero. Since

\begin{aligned} {\widetilde{A}}=(B\cap {\widetilde{A}})\cup (([1/2,1]-B)\cap {\widetilde{A}}), \end{aligned}

it follows that $$B\cap {\widetilde{A}}$$ has positive measure. Let $$C=\{x\in [0,1/2]:f(x)=0\}$$. Since $${\widetilde{f}}=f\chi _{[0,1/2]}\in L_{p}\left[ 0,1\right] \diagdown {\textstyle \bigcup _{q>p}} L_{q}\left[ 0,1\right]$$ it follows that C has measure zero. By the fact that each $$f_{n}$$ is the reproduction of f on the interval $$I_{n}$$, it follows that the set $$C_{n}=\{x\in I_{n}:f_{n}(x)=0\}$$ has measure zero, for all $$n\in {\mathbb {N}}$$. Since

\begin{aligned} B\cap {\widetilde{A}}=(B\cap {\widetilde{A}}\cap C_{n})\cup (B\cap {\widetilde{A}} \cap (I_{n}-C_{n})), \end{aligned}

and $$B\cap {\widetilde{A}}\cap C_{n}$$ has measure zero, then $$B\cap \widetilde{A}\cap (I_{n}-C_{n})$$ has positive measure, for each $$n\in {\mathbb {N}}.$$ Fixing $$r\in {\mathbb {N}}$$ and choosing $$x_{0}\in B\cap {\widetilde{A}}\cap (I_{r}-C_{r})$$, with $$x_{0}\ne 1$$, we have that $$x_{0}\in I_{r}$$, $$f_{r}(x_{0})\ne 0$$, $$g(x_{0})\ne 0$$ and

\begin{aligned} a_{0}^{(k_{j})}f(x_{0})+a_{r}^{(k_{j})}f_{r}(x_{0})=a_{0}^{(k_{j})} f(x_{0})+\sum _{n=1}^{\infty }a_{n}^{(k_{j})}f_{n}(x_{0})\longrightarrow g(x_{0})\text { when }j\rightarrow \infty . \end{aligned}

Since $$\lim _{j\rightarrow \infty }a_{0}^{\left( k_j\right) }=\alpha =0$$ we obtain

\begin{aligned} \lim _{j\rightarrow \infty }a_{r}^{(k_{j})}=\frac{g(x_{0})}{f_{r}(x_{0})} =\eta \ne 0. \end{aligned}

Since

\begin{aligned} f_{r}\chi _{I_{r}}(x)a_{r}^{(k_{j})}\longrightarrow g\chi _{I_{r}}(x)\text { a.e.},\text { when }j\rightarrow \infty , \end{aligned}

by the uniqueness of the limit we have

\begin{aligned} g\chi _{I_{r}}=\eta f_{r}\chi _{I_{r}}\text { a.e., } \end{aligned}

which implies that $$g\chi _{I_{r}}\notin L_{q}[0,1]$$ and consequently $$g\notin L_{q}[0,1]$$ (regardless of the $$q>p$$) finishing the proof. $$\square$$

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## Acknowledgements

The authors thank Daniel Cariello, Fernando Vieira Costa Junior for important comments/suggestions, and also thank the anonymous referee for his/her careful reading and suggestions that helped to improve the final version of this paper.

## Author information

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Correspondence to D. Pellegrino.

### Publisher's Note

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Daniel Tomaz is supported by Capes.

Daniel Pellegrino is supported by CNPq and Pronex-Fapesq.

Vinicius Fávaro is supported by FAPEMIG Grants APQ-03181-16, PPM-00217-18; and CNPq Grant 310500/2017-6.

Fávaro, V.V., Pellegrino, D. & Tomaz, D. Lineability and Spaceability: A New Approach. Bull Braz Math Soc, New Series 51, 27–46 (2020). https://doi.org/10.1007/s00574-019-00142-3

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### Keywords

• Cardinal numbers
• Lineability
• Spaceability

• 15A03
• 46A16
• 46A45