Ideal boundedness of subseries and rearrangements in Banach spaces vs Banach spaces possessing a copy of $c_0$

TL;DR

This paper characterizes Banach spaces that do not contain a copy of c_0 by the meagerness of sets coding bounded subseries and rearrangements, using Bessaga-Pe{ }lczy{\'n}ski's theorem and Baire ideal concepts.

Contribution

It establishes a new equivalence between the absence of c_0 in Banach spaces and the meagerness of certain sets related to series rearrangements.

Findings

Spaces without c_0 have meager sets for bounded subseries and rearrangements.

Uses Bessaga-Pe{ }lczy{\'n}ski c_0-Theorem and Baire ideal theory.

Provides a characterization involving ideal boundedness and Talagrand's theorem.

Abstract

Suppose that $X$ is a Banach space. We will show that $X$ does not contain a copy of $c_0$ if and only if for each series which is not unconditionally convergent in $X$ respective sets coding all bounded subseries and rearrangements are meager. We use Bessaga-Pe{\l}czy{\'n}ski $c_0-$Theorem and concept of uniformly unconditionally bounded series. Moreover we prove similar result for the ideal boundedness for a class of Baire ideals using Talagrand's characterisation.

Equations

$$E(\mathcal{I},(x_n))=\left\{s\in S:{\left(\sum _{i=1}^n{x}_{s(i)}\right)}_n\text{ is }\mathcal{I}-\text{bounded}\right\},$$

$$F(\mathcal{I},(x_n))=\left\{p\in P:{\left(\sum _{i=1}^n{x}_{p(i)}\right)}_n\text{ is }\mathcal{I}-\text{bounded}\right\},$$

$$U:=\{s\in S:s\text{mbox}extends(s_1,\dots ,s_k)\}$$

$$V:=\{p\in P:p\text{mbox}extends(p_1,\dots ,p_k)\}$$

$$B_m:=\left\{s\in S:{\exists }_{k>m}{\forall }_{l\in I_k}\Vert \sum _{i=1}^l{x}_{s(i)}\Vert >m\right\}.$$

$$U=\left\{w\in S:w\text{ extends }(s_1,\dots ,s_r)\right\}$$

$$\Vert \sum _{i=1}^r{x}_{s_i}+\sum _{i=r+1}^{l_r}{x}_{u(i)}\Vert >m+1.$$

$$\Vert \sum _{i=1}^r{x}_{s_i}+\sum _{i=r+1}^{l_r}{x}_{u(i)}+\sum _{i=l_r+1}^j{x}_{v_i}\Vert \ge \Vert \sum _{i=1}^r{x}_{s_i}+\sum _{i=r+1}^{l_r}{x}_{u(i)}\Vert -\Vert \sum _{i=l_r+1}^j{x}_{v_i}\Vert >m+1-1=m$$

$$V_U:=\left\{w\in S:w\text{ extends }(s_1,\dots ,s_r,u(r+1),\dots ,u(l_r),v_{l_r+1},\dots ,v_{n_{k+1}-1})\right\}\subset U\cap B_m$$

$$C_m:=\left\{p\in P:{\exists }_{k>m}{\forall }_{l\in I_k}\Vert \sum _{i=1}^l{x}_{p(i)}\Vert >m\right\}$$

$$U=\left\{w\in P:w\text{ extends }(p_1,\dots ,p_r)\right\}$$

$$\Vert \sum _{i=1}^r{x}_{p_i}+\sum _{i=z+1}^{m_r}{x}_{t(i)}+\sum _{i=m_r+1}^{j-r+z}{x}_{v_i}\Vert >m$$

$$V_U:=\left\{w\in P:w\text{ extends }(p_1,\dots ,p_r,t(z+1),\dots ,t(m_r),v_{m_r+1},\dots ,v_{n_{k+1}-1-r-z})\right\}\subset U\cap C_m.$$

Full text

1. Introduction

Denote by $Fin$ the family of all finite subsets of $\mathbb{N}$. Recall that an ideal $\mathcal{I}$ on $\mathbb{N}$ is the nonempty subfamily of the family $P(\mathbb{N})$ such that

1. (1)

$\emptyset\in\mathcal{I},$ 2. (2)

$A,B\in\mathcal{I}\Rightarrow A\cup B\in\mathcal{I},$ 3. (3)

$A\subset B\wedge B\in\mathcal{I}\Rightarrow A\in\mathcal{I},$ 4. (4)

$\mathbb{N}\notin\mathcal{I},$ 5. (5)

$Fin\subset\mathcal{I},$

where the last two conditions are not always included in the definition of ideal, but they are often considered as useful properties.

Since $\mathcal{I}\subset P(\mathbb{N})$ and $P(\mathbb{N})=\{0,1\}^{\mathbb{N}}$ has natural topological structure as product space of discrete spaces $\{0,1\}$, then we may investigate topological properties of ideals. Especially, recall that a subset $A$ of a metric space $X$ has the Baire property iff $A$ is symmetric difference of a meager set $N$ and an open set $U$. Family of these sets constitutes a $\sigma-$algebra on $X$. The following result (due to Jalali-Naini and Talagrand; see [4], [9]) gives a useful characterization of ideals with the Baire property.

Lemma 1.1**.**

An ideal $\mathcal{I}$ on $\mathbb{N}$ has the Baire property if and only if there is an infinite sequence $n_{1}<n_{2}<\dots$ in $\mathbb{N}$ such that no member of $\mathcal{I}$ contains infinitely many intervals $[n_{i},n_{i+1})\cap\mathbb{N}$.

An important example of an ideal distinct than $Fin$ and having the Baire property is the density ideal $\mathcal{I}_{d}:=\left\{A\subset\mathbb{N}\colon\frac{\operatorname{card}(A\cap\{1,\ldots,n\})}{n}\to 0\right\}.$

We say that a sequence $(x_{n})_{n}$ in a normed space is $\mathcal{I}-$bounded if there is $M>0$ with $\{n\in\mathbb{N}\colon\|x_{n}\|>M\}\in\mathcal{I}$. Evidently if we set $\mathcal{I}=Fin$ then we get usual notion of boundedness of a sequence. Moreover it can be shown that $\mathcal{I}-$convergent sequence is $\mathcal{I}-$bounded, too ([7], Theorem 2.2, by modification).

Put $S:=\left\{s\in\mathbb{N}^{\mathbb{N}}\colon s\textrm{ is increasing }\right\},P:=\left\{p\in\mathbb{N}^{\mathbb{N}}\colon p\textrm{ is a bijection }\right\}.$ It is easy to see that both sets are polish spaces, alike $\{0,1\}^{\mathbb{N}}.$ If $\sum x_{n}$ is a series in a normed space, than we may think that subseries of $\sum x_{n}$ is generated by a sequence $t\in\{0,1\}^{\mathbb{N}}$ or by a sequence $s\in S$ in such fashions: $\sum t(n)x_{n},\sum x_{s(n)}$, likewise $\sum x_{p(n)}$ is rearrangement if $p\in P.$

In this paper we are interested in ideal boundedness, hence we will only mention that following theorem has its equivalents in a case of ideal convergence in [1, 2].

Theorem 1.2**.**

([1], Theorem 4.1, Corollary 4.2) Suppose that $\sum x_{n}$ is a series which is not unconditionally convergent in a finite-dimensional Banach space $X$. Let $\mathcal{I}$ be an ideal with the Baire property. Then the sets

[TABLE]

[TABLE]

are meager in $S$ and $P$, respectively.

It is easy to see (analyzing the proof) that the above theorem holds changing coding of subseries from $S$ to $\{0,1\}^{\mathbb{N}}.$

In ([1], Example 3) the authors constructed a series $\sum x_{n}$ in $c_{0}$ which is not unconditionally convergent, but $E(Fin,(x_{n}))=S,F(Fin,(x_{n}))=P$, then both sets are evidently comeager. (Note that this example works for any Banach space possessing copy of $c_{0}$ and for any ideal $\mathcal{I}$, since boundedness implies ideal boundedness.) The above facts lead us to the question ([1], Question 4.3): for which (infinite-dimensional) Banach spaces for each series which is not unconditionally convergent both sets $E(\mathcal{I},(x_{n})),F(\mathcal{I},(x_{n}))$ are meager (in a case of Baire ideal $\mathcal{I}$)?

2. Main results

We will say that a series $\sum x_{n}$ in a Banach space is uniformly unconditionally bounded if there is $M>0$ such that for each $p\in P$ and $n\in\mathbb{N}$ we have $\left\|\sum_{i=1}^{n}x_{p(i)}\right\|\leq M.$ The following result should, in all likelihood, be known.

Lemma 2.1**.**

Suppose $\sum x_{n}$ is a series in a Banach space. The following conditions are equivalent:

1. (1)

$E(Fin,(x_{n}))=S,$** 2. (2)

there is $M^{\prime}>0$ s.t. for all $s\in S$ and $n\in\mathbb{N}$ we have $\left\|\sum_{i=1}^{n}x_{s(i)}\right\|\leq M^{\prime},$ 3. (3)

$\sum x_{n}$* is uniformly unconditionally bounded,* 4. (4)

$F(Fin,(x_{n}))=P.$**

Proof.

”(1) $\Rightarrow$ (2) ”

Suppose (1) holds but there is no such $M^{\prime}$ like in (2). Take an increasing sequence $(s_{1},\ldots,s_{k_{1}})$ such that $\|\sum_{i=1}^{k_{1}}x_{s_{i}}\|>0$. By the assumption, a series $\sum_{j=s_{k_{1}}+1}^{\infty}x_{j}$ does not satisfy (2), too. Then we can find indices $j_{1}<j_{2}<\ldots<j_{k_{2}}$ with $j_{1}>s_{k_{1}}$ such that $\|\sum_{i=1}^{k_{2}}x_{j_{i}}\|\geq 3\|\sum_{i=1}^{k_{1}}x_{s_{i}}\|.$ Since $\|x+y\|\geq|\|x\|-\|y\||,$ then if we put $s_{k_{1}+1}=j_{1},s_{k_{1}+2}=j_{1}+1,\ldots,s_{j_{k_{2}}-k_{1}+1}=j_{k_{2}}$ we get $\|\sum_{i=1}^{j_{k_{2}}-k_{1}+1}x_{s_{i}}\|\geq 2\|\sum_{i=1}^{k_{1}}x_{s_{i}}\|.$

Proceeding this way we can define $s\in S$ such that there is $u\in S$ with $\|\sum_{i=1}^{u(n)}x_{s(i)}\|\geq 2^{n}\|\sum_{i=1}^{k_{1}}x_{s_{i}}\|$ for each $n\in\mathbb{N}$ which shows that $s\notin E(Fin,(x_{n})).$

”(2) $\Rightarrow$ (3) ”

Suppose that all subseries have common bound $M^{\prime}>0$ and fix $p\in P$. Take a sequence $(s_{k})_{k}$ of elements of $S$ with property $\{s_{k}(1),\ldots,s_{k}(k)\}=\{p(1),\ldots,p(k)\}$ for each $k\in\mathbb{N}$. Then for all $k\in\mathbb{N}$ we obtain $\|\sum_{i=1}^{k}x_{p(i)}\|=\|\sum_{i=1}^{k}x_{s_{k}(i)}\|\leq M^{\prime}.$

”(3) $\Rightarrow$ (4) ” Obvious.

”(4) $\Rightarrow$ (1) ” Suppose (1) does not hold and fix $s\in S$ with unbounded $\sum x_{s(n)}.$ We will define an unbounded rearrangement of $\sum x_{n}.$ Take $n_{1}\in\mathbb{N}$ with $\left\|\sum_{i=1}^{n_{1}}x_{s(i)}\right\|\geq 1.$ Find $k_{1}>n_{1}$ and bijection $p_{1}\colon\{1,\ldots,k_{1}\}\to\{1,\ldots,k_{1}\}$ extending a sequence $(s(1),\ldots,s(n_{1})).$ By the assumption a series $\sum_{i=k_{1}+1}^{\infty}x_{s(i)}$ is unbounded, too. Then we can pick $n_{2}>k_{1}$ with $\left\|\sum_{i=1}^{k_{1}}x_{p_{1}(i)}+\sum_{i=k_{1}+1}^{n_{2}}x_{s(i)}\right\|\geq 2$. Find $k_{2}>n_{2}$ and a bijection $p_{2}\colon\{1,\ldots,k_{2}\}\to\{1,\ldots,k_{2}\}$ extending $p_{1}$ and $(s(1),\ldots,s(n_{2}))$ simultaneusly. Proceeding this concept ad infinitum we infer that $p:=\bigcup_{n\in\mathbb{N}}p_{n}\in P$ satisfy inequality $\left\|\sum_{i=1}^{n_{j}}x_{p(i)}\right\|\geq j$ for each $j\in\mathbb{N},$ thereby $p\notin F(Fin,(x_{n})).$

∎

Theorem 2.2**.**

Suppose $\sum x_{n}$ is a series in a Banach space. If $E(Fin,(x_{n}))\neq S$ (equivalently if $F(Fin,(x_{n}))\neq P$), then $E(Fin,(x_{n})),F(Fin,(x_{n}))$ are meager in $S$ and $P$, respectively.

Proof.

Suppose there is $s^{\prime}\in S\setminus E(Fin,(x_{n})).$ Write $E(Fin,(x_{n}))=\bigcup_{m\in\mathbb{N}}A_{m},$ where $A_{m}=\{s\in S\colon\left\|\sum_{i=1}^{n}x_{s(i)}\right\|\leq m\mbox{ for all }n\in\mathbb{N}\}$ for $m\in\mathbb{N}$. Fix $m\in\mathbb{N}$.

We will show that $A_{m}$ is closed. Indeed, take a sequence $(s_{n})_{n}\in A_{m}$ convergent to some $s\in S$. Fix $n\in\mathbb{N}.$ Take $k_{0}\in\mathbb{N}$ such that for all $k\geq k_{0}$ and $i\leq n$ $s_{k}(i)=s(i)$. Then $\left\|\sum_{i=1}^{n}x_{s(i)}\right\|=\left\|\sum_{i=1}^{n}x_{s_{k}(i)}\right\|\leq m$ since $s_{k}\in A_{m}.$ Consequently, $s\in A_{m}.$

Now, we will show that $A_{m}$ has an empty interior, which suffices to infer that $A_{m}$ is nowhere dense. Consider basic open set

[TABLE]

for some increasing sequence $(s_{1},\ldots,s_{k})$ of positive integers. Find the smallest $l\in\mathbb{N}$ with $s^{\prime}(l)>s_{k}$ and define $u:=(s_{1},\ldots,s_{k},s^{\prime}(l),s^{\prime}(l+1),s^{\prime}(l+2),\ldots).$ Then $u\in U\setminus A_{m},$ which proves that no open set is contained in $A_{m}.$

For the second part, suppose there is $p^{\prime}\in P\setminus F(Fin,(x_{n}))$ and write $F(Fin,(x_{n}))=\bigcup_{m\in\mathbb{N}}D_{m},$ where $D_{m}=\{p\in P\colon\left\|\sum_{i=1}^{n}x_{p(i)}\right\|\leq m\mbox{ for all }n\in\mathbb{N}\}$ for $m\in\mathbb{N}$. Similar calculations as above show that all $D_{m}$’s are closed. Fix $m\in\mathbb{N}$.

We will check that $D_{m}$ is nowhere dense. Consider basic open set

[TABLE]

for some injective sequence $(p_{1},\ldots,p_{k})$ of positive integers. Pick the smallest $l_{1}\in\mathbb{N}$ such that $p^{\prime}[\{1,\ldots,l_{1}-1\}]\supset\{p_{1},\ldots,p_{k}\}$ and $j_{1}>l_{1}$ with $\|x_{p_{1}}+\ldots+x_{p_{k}}+x_{p^{\prime}(l_{1})}+\ldots+x_{p^{\prime}(j_{1})}\|>m$. Next, extend a sequence $(p_{1},\ldots,p_{k},p^{\prime}(l_{1}),\ldots,p^{\prime}(j_{1}))$ to bijection $p_{1}\colon\{1,\ldots,k_{1}\}\to\{1,\ldots,k_{1}\}$ for some $k_{1}\in\mathbb{N}$. Now, pick the smallest $l_{2}\in\mathbb{N}$ such that $p^{\prime}[\{1,\ldots,l_{2}-1\}]\supset\{1,\ldots,k_{1}\}$ and $j_{2}>l_{2}$ with $\|x_{{p_{1}}_{1}}+\ldots+x_{{p_{1}}_{k_{1}}}+x_{p^{\prime}(l_{2})}+\ldots+x_{p^{\prime}(j_{2})}\|>m$. Again, extend a sequence $({p_{1}}_{1},\ldots,{p_{1}}_{k_{1}},p^{\prime}(l_{2}),\ldots,p^{\prime}(j_{2}))$ to a bijection $p_{2}\colon\{1,\ldots,k_{2}\}\to\{1,\ldots,k_{2}\}$ for some $k_{2}\in\mathbb{N}.$ This inductive procedure produces $p\in V\setminus D_{m}.$ ∎

Now, we will move to the case of ideal boundedness.

Theorem 2.3**.**

Suppose $\sum x_{n}$ is a series in a Banach space with $\liminf\|x_{n}\|=0$ and $\mathcal{I}$ is an ideal with the Baire property. If $E(Fin,(x_{n}))\neq S$ (equivalently if $F(Fin,(x_{n}))\neq P$), then $E(\mathcal{I},(x_{n})),F(\mathcal{I},(x_{n}))$ are meager in $S$ and $P$, respectively.

Proof.

Let $(n_{k})_{k\in\mathbb{N}}$ be a sequence from Talagrand’s characterisation and $I_{k}:=[n_{k},\ldots,n_{k+1}-1)\cap\mathbb{N}$. Take $u\in S$ such that $\sum x_{u(n)}$ is unbounded.

For each $m\in\mathbb{N}$ define

[TABLE]

Note that $\bigcap_{m\in\mathbb{N}}B_{m}\subset S\setminus E(\mathcal{I},(x_{n})).$ Indeed, take any $M>0$ and take any $m_{1}\in\mathbb{N}$ with $m_{1}>M$. Since $s\in B_{m_{1}}$ then there is $k(m_{1})>m_{1}$ with $\left\|\sum_{i=1}^{l}x_{s(i)}\right\|>m_{1}$ for all $l\in I_{k(m_{1})}.$ Take any integer $m_{2}>n_{k(m_{1})+1}-1$ and note that since $s\in B_{m_{2}}$ then we have $\left\|\sum_{i=1}^{l}x_{s(i)}\right\|>m_{2}>m_{1}$ for all $l\in I_{k(m_{2})}.$ Such procedure generates an increasing sequence $(m_{j})_{j\in\mathbb{N}}$ of positive integers with $\left\|\sum_{i=1}^{l}x_{s(i)}\right\|>m_{j}>m_{1}$ for all $l\in I_{k(m_{j})}$ and $j\in\mathbb{N}$ which shows that $\left\{n\in\mathbb{N}\colon\left\|\sum_{i=1}^{l}x_{s(i)}\right\|>m_{1}>M\right\}$ contains infinitely many intervals of the form $I_{k}.$ Therefore since $M>0$ is arbitrary we infer that (thanks to Lemma 1.1) $s\notin E(\mathcal{I},(x_{n})).$

Below we will show that each $B_{m}$ contains an open dense subset of $S$. Thereby, since $S$ is compleletely metrisable we get that $\bigcup_{m\in\mathbb{N}}B_{m}$ is comeager, and finally $E(\mathcal{I},(x_{n}))$ is meager.

Fix $m\in\mathbb{N}$ and basic open set $U$ in $S$:

[TABLE]

for some increasing sequence $(s_{1},\ldots,s_{r})$ of positive integers, where $r>m.$ Since $\sum_{n=1}^{\infty}x_{u(n)}$ is unbounded then $\sum_{i=r+1}^{\infty}x_{u(i)}$ is unbounded, too. Hence, we can pick $l_{r}>r$ with

[TABLE]

Find the smallest $k\in\mathbb{N}$ with $n_{k}>l_{r}$ and $\sum_{i=l_{r}+1}^{n_{k+1}-1}\|x_{v_{i}}\|<1$ for some increasing $(v_{l_{r}+1},\ldots,v_{n_{k+1}-1})$ such that $v_{l_{r}+1}>u(l_{r})$ (it is possible since $\liminf\|x_{n}\|=0$). Then

[TABLE]

for each $j\in I_{k}.$ Put

[TABLE]

and note that $\bigcup_{U\in\mathcal{B}}V_{U}$ is a dense (in $S$) open subset of $B_{m}$, where $\mathcal{B}$ is a base of topology in $S$ consisting of sets of the form $\left\{w\in S\colon w\textrm{ extends }(z_{1},\ldots,z_{r})\right\}$ for some finite increasing sequence $(z_{1},\ldots,z_{r}).$

Now we will discuss the case of rearrangements. One can show that $\bigcap_{m\in\mathbb{N}}C_{m}\subset P\setminus F(\mathcal{I},(x_{n}))$, where

[TABLE]

for each $m\in\mathbb{N}.$ Now, we fix $m\in\mathbb{N}$ and we will check that $C_{m}$ contains an open basic set. Consider

[TABLE]

for some $r>m.$ Since $E(Fin,(x_{n}))\neq S$, then there is an unbounded rearrangement (see Lemma 2.1) $\sum_{i=1}^{\infty}x_{t(i)}$ of $\sum_{i=1}^{\infty}x_{i}$ ($t\in P$). Put $z:=\max\{r,\max\{p_{1},\ldots,p_{r}\}\}$ and note that $\sum_{i=z+1}^{\infty}x_{t(i)}$ is unbounded, too. Find $m_{r}>z$ with $\|\sum_{i=1}^{r}x_{p_{i}}+\sum_{i=z+1}^{m_{r}}x_{t(i)}\|>m+1.$ Pick the smallest $k\in\mathbb{N}$ with $n_{k}>r+m_{r}-z$ and find an increasing sequence $(v_{m_{r}+1},\ldots,v_{n_{k+1}-1-r+z})$ with $v_{m_{r}+1}>\max\{z,m_{r},\max\{t(z+1),\ldots,t(m_{r})\}\}$ and $\sum_{i=m_{r}+1}^{n_{k+1}-1-r+z}\|x_{v_{i}}\|<1.$ Then

[TABLE]

for each $j\in I_{k}.$ Then

[TABLE]

The conclusion is similar as the final reasoning for case of subseries. ∎

Remark 1**.**

In Theorem 2.3 we can change the assumption $\liminf\|x_{n}\|=0$ to $\limsup\|x_{n}\|=\infty$, with the same thesis. It follows from the fact that in this case one can construct subseries $\sum_{n=1}^{\infty}x_{u(n)}$ of a series $\sum x_{n}$ that both sequences $(\|x_{u(n)}\|)_{n},(\|\sum_{i=1}^{n}x_{u(i)}\|)_{n}$ tends increasingly to $\infty.$

Remark 2**.**

In a case of subseries determined by $0-1$ sequences we can omit condition $\liminf\|x_{n}\|=0.$

Recall that Bessaga-Pełczyński $c_{0}-$Theorem ([5],Theorem 6.4.1, p.85) states that a Banach space $X$ contains no isomorphic copy of $c_{0}$ iff every weakly absolutely convergent series is unconditionally convergent [iff every weakly absolutely convergent series is convergent].

Lemma 2.4**.**

([3],Theorem 6, p.44 ($1.\Leftrightarrow 4.$)) A series $\sum x_{n}$ in a Banach space $X$ is weakly absolutely convergenct iff there is $C>0$ such that for each $n\in\mathbb{N}$ and finite sequence $t\colon\{1,\ldots,n\}\to\{-1,0,1\}$ we have $\|\sum_{i=1}^{n}t(i)x_{i}\|\leq C$.

Theorem 2.5**.**

Suppose that $\sum x_{n}$ is not unconditionally convergent series in a Banach space containing no copy of $c_{0}$. Then $E(Fin,(x_{n})),F(Fin,(x_{n}))$ are meager.

Proof.

Suppose $E(Fin,(x_{n}))$ or $F(Fin,(x_{n}))$ is nonmeager. Then by Theorem 2.2 $E(Fin,(x_{n}))=S,F(Fin,(x_{n}))=P$ and by Lemma 2.1 all subseries and rearrangements are bounded by a common constant, say $M>0$. Note that for each sequence $t\colon\{1,\ldots,n\}\to\{-1,0,1\}$ we can write $\|\sum_{i=1}^{n}t(i)x_{i}\|=\|\sum_{i=1}^{n}\chi_{A}(i)x_{i}+\sum_{i=1}^{n}\chi_{B}(i)t(i)x_{i}\|\leq\|\sum_{i=1}^{n}\chi_{A}(i)x_{i}\|+\|\sum_{i=1}^{n}-\chi_{B}(i)t(i)x_{i}\|\leq 2M$, if we put $A:=t^{-1}[\{1\}],B:=\{1,\ldots,n\}\setminus A$, because $\chi_{A},-\chi_{B}t\colon\{1,\ldots,n\}\to\{0,1\}.$ It follows from Lemma 2.4 that $\sum x_{n}$ is weakly absolutely convergent and by Bessaga-Pełczyński $c_{0}-$Theorem $\sum x_{n}$ is unconditionally convergent. ∎

Using an above fact and Example 3. in [1] we get following result.

Corollary 2.6**.**

A Banach space contains no copy of $c_{0}$ if and only if for each series which is not unconditionally convergent both sets $E(Fin,(x_{n})),F(Fin,(x_{n}))$ are meager in $S,P$, respectively.

Note that the fact that $E(Fin,(x_{n}))$ is meager or a series $\sum x_{n}$ is weakly unconditionally convergent was proven ([8],Corollary 2.2) using theory of summability methods of a series.

Theorem 2.7**.**

Suppose $\mathcal{I}$ is an ideal with the Baire property. A Banach space $X$ contains no copy of $c_{0}$ if and only if for each series $\sum x_{n}$ in $X$ with $\liminf\|x_{n}\|=0$ which is not unconditionally convergent both sets $E(\mathcal{I},(x_{n})),F(\mathcal{I},(x_{n}))$ are meager in $S,P$, respectively.

Proof.

If Banach space $X$ contains a copy of $c_{0}$ then we use a construction from Example 3. in [1]. Now, suppose that $X$ contains no copy of $c_{0}$ and $E(\mathcal{I},(x_{n}))$ or $F(\mathcal{I},(x_{n}))$ is nonmeager. Suppose that $E(Fin,(x_{n}))\neq S$. Then by Lemma 2.1 and Theorem 2.3 both sets $E(\mathcal{I},(x_{n})),F(\mathcal{I},(x_{n}))$ are meager, contradiction. Then $E(Fin,(x_{n}))=S$. By Lemma 2.1 $F(Fin,(x_{n}))=P$ (hence obviously $E(\mathcal{I},(x_{n}))=S,F(\mathcal{I},(x_{n}))=P.$). Corollary 2.6 implies that $\sum x_{n}$ is conditionally convergent. ∎

Define $A(\mathcal{I},(x_{n}))=\left\{t\in\{0,1\}^{\mathbb{N}}\colon\left(\sum_{i=1}^{n}t(i)x_{i}\right)_{n}\textrm{ is }\mathcal{I}-\textrm{bounded}\right\}.$ In a similar way as above, thanks to Remark 2, we get following characterisation.

Corollary 2.8**.**

Suppose $\mathcal{I}$ is an ideal with the Baire property. A Banach space $X$ contains no copy of $c_{0}$ if and only if for each series $\sum x_{n}$ in $X$ which is not unconditionally convergent the set $A(\mathcal{I},(x_{n}))$ is meager in $\{0,1\}^{\mathbb{N}}$.

Acknowledgement. I would like to thank Gilles Godefroy for fruitful conversation.