\section{Introduction}


Given an infinite real matrix $A=(a_{n,k})$, we say that $A$ sums a real sequence $x$ if $Ax=(\sum_k a_{n,k}x_k)$ is well defined and convergent. 
The matrix $A$ is regular if it maps convergent sequences into convergent sequences, preserving the corresponding limits. 

The classical Steinhaus' theorem states that a regular matrix $A$ cannot sum all $\{0,1\}$-valued sequences, see e.g. \cite{MR795251}. This can be rewritten equivalently as follows: a regular matrix $A$ cannot sum all subsequences of a divergent $\{0,1\}$-valued sequence $x$. We recall also that Hahn's theorem \cite[Theorem 2.4.5]{MR1817226} states that a matrix $A$ sums all $\{0,1\}$-valued sequences if and only if it sums all bounded sequences. On this direction, Buck \cite{MR9208, MR81983} showed that a regular matrix $A$ cannot sum all subsequences of a given divergent sequence $x$. 

Let $\Sigma$ be the set of strictly increasing functions $\sigma: \mathbf{N}\to \mathbf{N}$, so that a subsequence of a sequence $x$ is uniquely identified by some $\sigma \in \Sigma$, and write $\sigma(x):=(x_{\sigma(n)})$. Accordingly, Buck's result tells us that, if $A$ is regular and $x$ is divergent, then the set
\begin{equation}\label{eq:definitionsigmaxA}
\Sigma_{x,A}:=\{\sigma \in \Sigma: A\sigma(x) \in c\}
\end{equation}
cannot be equal to $\Sigma$. Finally, Keogh and Petersen \cite{MR92024} proved the following:
\begin{thm}\label{thm:keoghpetersen}
Let $x$ be a divergent sequence and $A$ be a regular matrix. Then $\Sigma_{x,A}$ is meager. 
\end{thm}
This is meaningful since $\Sigma$ is a $G_\delta$-subset of the Polish space $\mathbf{N}^{\mathbf{N}}$, hence it is a Polish space as well by Alexandrov's theorem; in particular, $\Sigma$ is not meager in itself.
Special instances of Theorem \ref{thm:keoghpetersen} have been recently rediscovered in \cite{MR4049166, MR3809152, MR3216022}. Related results can be found in \cite[Theorem 2.3]{MR13435} and \cite{MR404918, MR425410, MR261220, MR1123336}.

Note that, if $x$ is convergent and $A$ is regular, then $\Sigma_{x,A}$ is equal to the whole $\Sigma$. Therefore, by Theorem \ref{thm:keoghpetersen}, $x$ is convergent if and only if there exists a regular matrix $A$ such that $\Sigma_{x,A}$ is not meager. 
The aim of this work is to show that Theorem \ref{thm:keoghpetersen} holds even if the space of convergent sequences $c$ in the definition of $\Sigma_{x,A}$ in \eqref{eq:definitionsigmaxA} is replaced by a much bigger set, e.g., the space of statistically convergent sequences (see Section \ref{sec:main}). 
Our methods are different from those employed by Keogh and Petersen, and rely on more recent result on filter games and ideal Baire classes. 


\section{Main result}\label{sec:main}

Let $\mathcal{I}$ be an ideal on the positive integers $\mathbf{N}$, that is, a hereditary family of subsets of $\mathbf{N}$ closed under finite unions; 
moreover, it is assumed that $\mathcal{I}$ contains the family of finite sets $\mathrm{Fin}$ and it is different from $\mathcal{P}(\mathbf{N})$. 
Let 
$\mathcal{I}^\star:=\{S\subseteq \mathbf{N}: S^c\in \mathcal{I}\}$ be its dual filter. 
Every subset of $\mathcal{P}(\mathbf{N})$ is endowed with the relative Cantor-space topology. 
In particular, we may speak about $F_\sigma$-subsets of $\mathcal{P}(\mathbf{N})$, meager ideals, etc. 
In addition, let $\mathcal{Z}$ be the ideal of asymptotic density zero sets, that is, 
$$
\textstyle 
\mathcal{Z}:=\left\{S\subseteq \mathbf{N}: \lim_n |S\cap [1,n]|/n=0\right\}.
$$
For our purposes, it will be important also to recall the definition of the ideal $\mathrm{Fin}\times \mathrm{Fin}$, which can be represented as an ideal on $\mathbf{N}$ by
$$
\textstyle 
\mathrm{Fin}\times \mathrm{Fin}:=\left\{S\subseteq \mathbf{N}: \forall^\infty k\in \mathbf{N}, \{n \in S: \nu_2(n)=k\}\in \mathrm{Fin}\right\},
$$
where $\nu_2(n)$ stands for the $2$-adic valuation of $n\in \mathbf{N}$ (that is, $\nu_2(n)=\max\{k\ge 0: 2^k \text{ divides }n\}$). 
We recall that $\mathcal{Z}$ is a $F_{\sigma\delta}$-ideal, while $\mathrm{Fin}\times \mathrm{Fin}$ is $F_{\sigma\delta\sigma}$. 
Given ideals $\mathcal{I}, \mathcal{J}$ on $\mathbf{N}$, we say that $\mathcal{I}$ contains an isomorphic copy of $\mathcal{J}$ if there exists an injection $\iota: \mathbf{N} \to \mathbf{N}$ such that $S \in \mathcal{J}$ if and only if $\iota^{-1}[S] \in \mathcal{I}$ for all $S\subseteq \mathbf{N}$; see e.g. \cite{MR1711328}. 

Let $\omega$ be the vector space of all real sequences, together with its classical subspaces $\ell_\infty$, $c$, $c_0$, $\ell_1$, $c_{00}$ of bounded, convergent, convergent to $0$, absolutely summable, and eventually $0$ sequences, respectively. 
In addition, given an ideal $\mathcal{I}$ on $\mathbf{N}$, we define also the following subspaces of $\omega$: 
\begin{displaymath}
\begin{split}
\ell_\infty(\mathcal{I}) &:=\{\,\mathcal{I}\text{-bounded sequences }\}, \\
c(\mathcal{I})& :=\{\,\mathcal{I}\text{-convergent sequences }\}, \\
c_b(\mathcal{I})&:= \{\,\mathcal{I}\text{-convergent bounded sequences }\}. \\
\end{split}
\end{displaymath}
Note that $c=c(\mathrm{Fin})=c_b(\mathrm{Fin})$ 
and $\ell_\infty=\ell_\infty(\mathrm{Fin})$. Here, a sequence $x \in \omega$ is $\mathcal{I}$-bounded if there exists $k$ such that $\{n\in \mathbf{N}: |x_n|\ge k\} \in \mathcal{I}$. 
Moreover, $x$ is $\mathcal{I}$-convergent to $\eta \in \mathbf{R}$, shortened as $\mathcal{I}\text{-}\lim x=\eta$, provided that $\{n \in \mathbf{N}: |x_n-\eta|>\varepsilon\} \in \mathcal{I}$ for all $\varepsilon>0$. 
Every subspace of $\ell_\infty$ will be endowed with the topology induced by the supremum norm. 
In the literature, $\mathcal{Z}$-convergence is usually called \emph{statistical convergence}.

At this point, for each infinite real matrix $A=(a_{n,k})$, let $\omega_A$ be the set of sequences $x$ such that $Ax$ is well defined, so that
$$
\textstyle 
\omega_A:=\left\{x \in \omega: \sum_ka_{n,k}x_k \text{ converges for all }n\right\}.
$$
For all nonempty $X,Y\subseteq \omega$, let $(X,Y)$ be the matrix class of matrices $A$ which map sequences in $X$ into sequences in $Y$, i.e.,
$
\textstyle 
(X,Y):=\{A: x \in \omega_A \text{ and } Ax \in Y \text{ for all }x \in X\}.
$ 
Given ideals $\mathcal{I}, \mathcal{J}$ on $\mathbf{N}$, a matrix $A$ is $\bm{(\mathcal{I}, \mathcal{J})}$\textbf{-regular} if it is maps $\mathcal{I}$-convergent bounded sequences into $\mathcal{J}$-convergent bounded sequences, preserving the corresponding ideal limits, that is,  
$$
\textstyle 
A \in (c_b(\mathcal{I}),c_b(\mathcal{J}))
\quad \text{ and } \quad 
\mathcal{I}\text{-}\lim x=\mathcal{J}\text{-}\lim Ax
\,\,\text{ for all }
 x \in c_b(\mathcal{I}).
$$
Note that $(\mathrm{Fin}, \mathrm{Fin})$-regular matrices are simply the classical regular matrices. 
Probably the most important regular matrix is the Cesàro matrix $C_1=(a_{n,k})$ defined by $a_{n,k}=\frac{1}{n}$ if $k\le n$ and $a_{n,k}=0$ otherwise. 
Classes of $(\mathcal{I}, \mathcal{J})$-regular matrices have been recently used and characterized in \cite{ConnorLeo}.
In particular, the following result extends the classical Silverman--Toeplitz characterization, see \cite[Theorem 1.2 and Theorem 1.3]{ConnorLeo}:
\begin{thm}\label{thm:connotleofinJ}
A matrix $A$ is $(\mathrm{Fin},\mathcal{I})$-regular if and only if:
\begin{enumerate}[label={\rm (\textsc{R}\arabic{*})}]
\item \label{item:R1} $\sup_n \sum_k|a_{n,k}|<\infty$\textup{;}
\item \label{item:R2} $\mathcal{I}\text{-}\lim_n a_{n,k}=0$ for all $k \in \mathbf{N}$\textup{;}
\item \label{item:R3} $\mathcal{I}\text{-}\lim_n \sum_k a_{n,k}=1$\textup{.}
\end{enumerate}
\end{thm}

Finally, for each real sequence $x$, matrix $A$, and ideal $\mathcal{I}$, define 
$$
\textstyle 
\Sigma_{x,A}(\mathcal{I}):=\left\{\sigma \in \Sigma: A\sigma(x) \in c(\mathcal{I})\right\}.
$$
Note that $\Sigma_{x,A}=\Sigma_{x,A}(\mathrm{Fin})$. 
Our main result follows (for the proof, see Section \ref{sec:mainproof}):
\begin{thm}\label{thm:mainTauberianFinI}
Let $x$ be a divergent sequence, $\mathcal{I}$ be a Borel ideal on $\mathbf{N}$ which does not contain an isomorphic copy of $\mathrm{Fin}\times \mathrm{Fin}$, and $A$ be a $(\mathrm{Fin}, \mathcal{I})$-regular matrix. Then $\Sigma_{x,A}(\mathcal{I})$ is meager. 
\end{thm}

Some remarks are in order. First of all, Theorem \ref{thm:mainTauberianFinI} holds for $(\mathrm{Fin}, \mathcal{I})$-regular matrices, hence in particular it holds for regular matrices. Secondly, all $F_{\sigma\delta}$-ideals satisfy the above hypothesis, see Remark \ref{rmk:caseinclusions} below. 
In particular, Theorem \ref{thm:mainTauberianFinI} holds for 
the ideal $\mathcal{Z}$ of asymptotic density zero sets, 
for the ideals generated by nonnegative regular matrices (see \cite[Proposition 13]{MR3405547}),
for the ideal $\mathrm{Fin}$ (hence, providing another proof of Theorem \ref{thm:keoghpetersen}), and for the ideal of Banach density zero sets. 
Lastly, if $x$ is convergent then $\Sigma_{x,A}(\mathcal{I})=\Sigma$. 

Putting everything together, we obtain:
\begin{cor}\label{cor:characterization}
A real sequence $x$ is convergent if and only if there exists a regular matrix $A$ such that $\{\sigma \in \Sigma: A\sigma(x) \text{ is statistically convergent}\,\}$ is not meager. 
\end{cor}

In particular, letting $x$ be the sequence $(0,1,0,1,\ldots)$, 
we obtain that $\{\sigma(x): \sigma \in \Sigma\}=\{0,1\}^\mathbf{N}$ and $\Sigma_{x,A}(\mathcal{Z})\neq \Sigma$, which give us a stronger version of Steinhaus' theorem:
\begin{cor}\label{cor:steinhaus}
For each regular matrix $A$, there exists a $\{0,1\}$-valued sequence $x$ such that $Ax$ is not statistically convergent.
\end{cor}


On a similar direction, letting $A$ be the infinite identity matrix, it follows that, a real sequence $x$ is convergent if and only if $\{\sigma \in \Sigma: \sigma(x) \in c(\mathcal{I})\}$ is not meager, provided that $\mathcal{I}$ is an ideal on $\mathbf{N}$ as in Theorem \ref{thm:mainTauberianFinI}. For the latter class of ideals, this gives us a generalization of \cite[Corollary 2.7]{MR4165727}, which states that $\lim x=\eta$ if and only if $\{\sigma \in \Sigma: \mathcal{I}\text{-}\lim \sigma(x)=\eta\}$ is not meager, provided that $\mathcal{I}$ is a meager ideal and $\mathcal{I}\text{-}\lim x=\eta$ (note that $\eta$ is explicit). 
Other Tauberian theorems related to statistical convergence can be found in \cite{
MR4052262,
MR2324333, MR1659877, MR1653457, MR1002541, MR1260176}. 

Finally, note that Theorem \ref{thm:mainTauberianFinI} does not hold without any restriction on the ideal $\mathcal{I}$. 
Indeed, if $A$ is a $(\mathrm{Fin},\mathcal{I})$-regular matrix, then it satisfies condition \ref{item:R1} of Theorem \ref{thm:connotleofinJ}, so that $A \in (\ell_\infty,\ell_\infty)$. 
Thus, if $x\in \ell_\infty\setminus c$ and $\mathcal{I}$ is a maximal ideal, then every subsequence $\sigma(x)$ is bounded, hence $A\sigma(x) \in c_b(\mathcal{I})$. 
In particular, for each bounded divergent sequence $x$, there exist an ideal $\mathcal{I}$ and a $(\mathrm{Fin}, \mathcal{I})$-regular matrix $A$ such that $\Sigma_{x,A}(\mathcal{I})$ is not meager.







\section{Preliminaries}\label{sec:preliminaries}

Given an ideal $\mathcal{I}$, let $G(\mathcal{I})$ be the following game introduced by Laflamme in \cite{MR1367134}: at stage $n \in \mathbf{N}$ player I chooses a set $A_n \in \mathcal{I}^\star$ and, then, player II chooses a nonempty finite set $F_n\subseteq A_n$. At the end of the game, player II is declared the winner if $\bigcup_n F_n \notin \mathcal{I}$.
Moreover, a sequence $(F_k)$ of nonempty finite sets is said to be a $\mathcal{I}^\star$\emph{-universal set} if each $A \in \mathcal{I}^\star$ contains some $F_k$. We say that $\mathcal{I}^\star$ is $\omega$\emph{-diagonalizable by} $\mathcal{I}^\star$\emph{-universal sets} if there exists an infinite matrix $(F_{n,k})$ of nonempty finite sets such that each row is a $\mathcal{I}^\star$-universal set and, moreover, for each $A \in \mathcal{I}^\star$ there exist $n,m \in \mathbf{N}$ such that $F_{n,k} \cap A\neq \emptyset$ for all $k>m$. 
\begin{thm}\label{thm:laflamme}
Let $\mathcal{I}$ be an ideal on $\mathbf{N}$. Then the following are equivalent:
\begin{enumerate}[label={\rm (\textsc{c}\arabic{*})}]
\item \label{item:c1} $\mathcal{I}^\star$ is $\omega$-diagonalizable by $\mathcal{I}^\star$-universal sets\textup{;}
\item \label{item:c2} player II has a winning strategy in the game $G(\mathcal{I})$. 
\end{enumerate}
If, in addition, $\mathcal{I}$ is a Borel ideal, then they are also equivalent to\textup{:}
\begin{enumerate}[label={\rm (\textsc{c}\arabic{*})}]
\setcounter{enumi}{2}
\item \label{item:c3} $\mathcal{I}$ does not contain an isomorphic copy of $\mathrm{Fin}\times \mathrm{Fin}$\textup{;}
\item \label{item:c4} $\mathcal{I}$ is $F_\sigma$-separated from its dual filter $\mathcal{I}^\star$ \textup{(}that is, there exists an $F_\sigma$-set $\mathcal{K}$ such that $\mathcal{I}\subseteq \mathcal{K}$ and $\mathcal{K}\cap \mathcal{I}^\star=\emptyset$\textup{)}.
\end{enumerate}
\end{thm}
\begin{proof}
See \cite[Theorem 2.16(ii)]{MR1367134} for \ref{item:c1} $\Longleftrightarrow$ \ref{item:c2}. For the other equivalences, see \cite{MR2491780}.
\end{proof}

\begin{rmk}\label{rmk:caseinclusions}
As it has been shown in \cite[Corollary 1.5]{MR1758325}, all $F_{\sigma\delta}$-ideals satisfy condition \ref{item:c4}. Related results on condition \ref{item:c4} can be found in \cite[Proposition 3.6]{MR3090420} and \cite[Theorem 2.1]{MR3968131}.
\end{rmk}


At this point, for each ideal $\mathcal{I}$ and topological space $X$, define the $\mathcal{I}$-Baire one class 
$$
\mathcal{B}_1^{\mathcal{I}}(X):=\left\{f \in \mathbf{R}^X: \exists (f_n) \in C(X)^{\mathbf{N}}, \forall x \in X,\,\, f(x)=\mathcal{I}\text{-}\lim f_n(x)\right\},
$$
where $C(X)$ stands for the space of real-valued continuous functions on $X$. Note that $\mathcal{B}_1^{\mathcal{I}}(X)$ coincides with the classical Baire one class $\mathcal{B}_1(X)$ if $\mathcal{I}=\mathrm{Fin}$. However, it has been shown in \cite[Proposition 8]{MR2491780}, the same holds if $X$ is a complete metric space and $\mathcal{I}$ is an ideal for which player II has a winning strategy in the game $G(\mathcal{I})$, cf. also \cite{MR2520152} for related results. 
Some years later, 
Filip\'{o}w and Szuca proved in \cite{MR2899832}, in particular, that the hypothesis of completeness on $X$ is unnecessary to obtain the latter conclusion:
\begin{thm}\label{thm:filipowszuca}
Let $\mathcal{I}$ be an ideal on $\mathbf{N}$ such that $\mathcal{I}^\star$ is $\omega$-diagonalizable by $\mathcal{I}^\star$-universal sets. Moreover, let $X$ be a perfectly normal topological space. Then $\mathcal{B}_1^{\mathcal{I}}(X)=\mathcal{B}_1(X)$.
\end{thm}
\begin{proof}
See \cite[Theorem 3.2]{MR2899832}.
\end{proof}
This 
has been also obtained in \cite[Corollary 12]{MR2491780} for nonpathological analytic P-ideals $\mathcal{I}$ and arbitrary topological spaces $X$. 
We recall also the classical Baire classification theorem.
\begin{thm}\label{thm:baireclassification}
Let $X$ be a metrizable space and fix a Baire one function $f \in \mathcal{B}_1(X)$. Then the set of points of continuity of $f$ is a comeager $G_\delta$-set.
\end{thm}
\begin{proof}
See \cite[Theorem 24.14]{MR1321597}.
\end{proof}

Finally, we will use the well-known characterization of meager ideals due to Talagrand. 
\begin{thm}\label{prop:talagrand}
Let $\mathcal{I}$ be an ideal on $\mathbf{N}$. Then the following are equivalent: 
\begin{enumerate}[label={\rm (\textsc{m}\arabic{*})}]
\item \label{item:m1} $\mathcal{I}$ is meager\textup{;}
\item \label{item:m2} there exists $\sigma \in \Sigma$ such that $A \notin \mathcal{I}$ whenever $\mathbf{N} \cap [\sigma(n),\sigma(n+1)) \subseteq A$ for infinitely many $n \in \mathbf{N}$\textup{;}
\item \label{item:m3} $\mathcal{I}$ is $F_\sigma$-separated from the Fr\'{e}chet filter $\mathrm{Fin}^\star$ \textup{(}that is, there exists a sequence $(F_n)$ of closed sets such that $\mathcal{I}\subseteq \bigcup_n F_n$ and $F_n \cap \mathrm{Fin}^\star=\emptyset$ for all $n\in \mathbf{N}$\textup{)}\textup{.}
\end{enumerate}
\end{thm}
\begin{proof}
See \cite[Proposition 3.1]{MR4165727} and \cite[Theorem 2.1]{MR579439}.
\end{proof}

At this point, we split the intermediate results into two cases: the first one assumes that $x$ is a bounded divergent sequence, the second one that $x$ is unbounded.

\subsection{Bounded case} 


\begin{lem}\label{lem:continuity}
Fix two sequences $x \in \ell_\infty$ and $a \in \ell_1$. 
Then the  map 
$$
\textstyle 
\Sigma\to \mathbf{R}: \sigma\mapsto a\cdot \sigma(x)
$$
is uniformly continuous. 
\end{lem}
\begin{proof}
The claim holds trivially for $x=0$, hence suppose hereafter that $x\neq 0$, so that $\|x\|>0$. 
Note that the topology on $\Sigma$ is metrizable by $d: \Sigma\times \Sigma \to \mathbf{R}$ defined as
$$
\forall \sigma_1, \sigma_2 \in \Sigma, \quad d(\sigma_1,\sigma_2)=\sum_{i \in\, \mathrm{Im}(\sigma_1) \bigtriangleup \,\mathrm{Im}(\sigma_2)}\frac{1}{2^i}.
$$
Fix $\varepsilon>0$. Since $a \in \ell_1$, there exists $k_0 \in \mathbf{N}$ such that $\sum_{k> k_0}|a_k|<\frac{\varepsilon}{2\|x\|}$. 
At this point, define $\delta:=1/2^{k_0}$ and note that, if there exists $k\le k_0$ such that $k \in \mathrm{Im}(\sigma_1) \bigtriangleup \mathrm{Im}(\sigma_2)$ then $d(\sigma_1,\sigma_2) \ge 1/2^k\ge \delta$. Therefore, for each $\sigma_1,\sigma_2 \in \Sigma$ with $d(\sigma_1,\sigma_2)<\delta$, we obtain that the least element of 
$\mathrm{Im}(\sigma_1) \bigtriangleup \mathrm{Im}(\sigma_2)$ is greater than $k_0$, which implies that 
$$
\textstyle 
|a\cdot \sigma_1(x)-a\cdot \sigma_2(x)|
=\left|\sum_{k>k_0}a_k(x_{\sigma_1(k)}-x_{\sigma_2(k)}))\right|
\le 2\|x\|\sum_{k>k_0}|a_k|<\varepsilon.
$$
This concludes the proof.
\end{proof}


\begin{cor}\label{cor:continuitypartialmaps}
Fix a matrix $A \in (c_0, \ell_\infty)$ and a sequence $x \in \ell_\infty$. Then the map 
$$
\textstyle 
\Sigma \to \mathbf{R}:\sigma \mapsto \sum_k a_{n,k}x_{\sigma(k)}
$$
is uniformly continuous for each $n \in \mathbf{N}$.
\end{cor}
\begin{proof}
Thanks to \cite[Theorem 2.3.5]{MR1817226}, $A$ belongs to $(c_0, \ell_\infty)$ if and only if $\sup_n \sum_k |a_{n,k}|<\infty$. The claim follows by Lemma \ref{lem:continuity}.
\end{proof}


\begin{lem}\label{lem:discontinuousverythwre}
Fix a bounded divergent sequence $x \in \ell_\infty\setminus c$, an ideal $\mathcal{I}$ on $\mathbf{N}$, and $(\mathrm{Fin}, \mathcal{I})$-matrix $A$. 
Then $\Sigma_{x,A}(\mathcal{I})$ is dense and the map 
\begin{equation}\label{eq:mapTj}
\textstyle 
T: \Sigma_{x,A}(\mathcal{I}) \to \mathbf{R}: \sigma \mapsto \mathcal{I}\text{-}\lim_n \sum_k a_{n,k}x_{\sigma(k)}
\end{equation}
is everywhere discontinuous.
\end{lem}
\begin{proof}
Let $B\subseteq \Sigma$ an arbitrary open ball. Observe that there exist $\sigma_1,\sigma_2\in B$ such that $\lim \sigma_1(x)=\alpha$ and $\lim \sigma_2(x)=\beta$, where $\alpha:=\limsup_n x_n$ and $\beta:=\liminf_n x_n$ are finite and distinct since $x \in \ell_\infty\setminus c$. 
Considering that $A$ is $(\mathrm{Fin}, \mathcal{I})$-regular, we obtain that $\alpha=\mathcal{I}\text{-}\lim A\sigma_1(x)$ and $\beta=\mathcal{I}\text{-}\lim A\sigma_2(x)$. 
In particular, $B\cap \Sigma_{x,A}(\mathcal{I})\neq \emptyset$, hence $\Sigma_{x,A}(\mathcal{I})$ is dense. 

In addition, $\sup_{\sigma, \sigma^\prime \in B \cap \Sigma_{x,A}(\mathcal{I})}|T(\sigma)-T(\sigma^\prime)|\ge \alpha-\beta$ for every open ball $B\subseteq \Sigma$, therefore $T$ cannot be continuous at any point of $\Sigma_{x,A}(\mathcal{I})$. 
\end{proof}


\begin{rmk}\label{rmk:borelmeasurability}
With the same hypotheses of Lemma \ref{lem:discontinuousverythwre}, assume that $\mathcal{I}$ is a Borel ideal. 
Then $\Sigma_{x,A}(\mathcal{I})$ is Borel and the map $T$ defined in \eqref{eq:mapTj} is Borel measurable. 
For, note that necessarily $A \in (c_0, \ell_\infty)$. Hence the map $T$ is the $\mathcal{I}$-pointwise limit of the sequence of functions $T_n: \sigma \mapsto \sum_k a_{n,k}x_k$ restricted to $\Sigma_{x,A}(\mathcal{I})$, which are continuous for each $n \in \mathbf{N}$ thanks to Corollary \ref{cor:continuitypartialmaps}. In particular, each of them is Borel measurable. The claim follows by \cite[Lemma 1 and Lemma 2]{MR2390885}. 
However, as  remarked in \cite[Example 1]{MR2390885}, some assumptions on $\mathcal{I}$ are needed to ensure that the $\mathcal{I}$-pointwise limit of measurable functions is still measurable.
\end{rmk}














\begin{thm}\label{thm:conclusionmaincategorybounded}
Fix a bounded divergent sequence $x \in \ell_\infty\setminus c$, an ideal $\mathcal{I}$ on $\mathbf{N}$ such that $\mathcal{I}^\star$ is $\omega$-diagonalizable by $\mathcal{I}^\star$-universal sets, and a $(\mathrm{Fin}, \mathcal{I})$-regular matrix $A$. 

Then $\Sigma_{x,A}(\mathcal{I})$ is meager. 
\end{thm}
\begin{proof}
Set $X:=\Sigma_{x,A}(\mathcal{I})$ and let $T$ be the map defined in \eqref{eq:mapTj}. 
Thanks to Corollary \ref{cor:continuitypartialmaps},  $T \in \mathcal{B}_1^{\mathcal{I}}(X)$. 
Since $X$ is metrizable and $\mathcal{I}^\star$ is $\omega$-diagonalizable by $\mathcal{I}^\star$-universal sets, then $T \in \mathcal{B}_1(X)$ by Theorem \ref{thm:filipowszuca}. 
At this point, it follows by Theorem \ref{thm:baireclassification} that the set of continuity points of $T$ is a comeager subset of $X$. 
Hence, by Lemma \ref{lem:discontinuousverythwre}, $X$ is a dense subset of $\Sigma$ which is meager in itself. 
We conclude that $X$ is meager in $\Sigma$. 
\end{proof}



\subsection{Unbounded case.}


\begin{thm}\label{lem:domaindefinition}
Fix sequences $x \in \omega\setminus \ell_\infty$ and $a \in \omega\setminus c_{00}$. Then 
$$
\textstyle 
\left\{\sigma \in \Sigma: \left(\sum_{k\le n} a_k x_{\sigma(k)}\right) \in \ell_\infty\right\}
$$ 
is meager. 

\end{thm}
\begin{proof}
Let $E$ be the claimed set. Observe that $E=\bigcup_{m}E_m$, where 
$$
\textstyle 
\forall m \in \mathbf{N}, \quad E_m:=\left\{\sigma \in \Sigma: \left|\sum_{k\le n} a_k x_{\sigma(k)}\right|\le m \text{ for all }n\right\}.
$$
We are going each $E_{m}$ is nowhere dense. Note that each $E_m$ is closed. Thus, let us assume for the sake of contradiction that there exists $m_0$ such that $E_{m_0}$ has nonempty interior, hence there exist positive integers $t_1<\cdots<t_j$ such that $\sigma \in E_{m_0}$ whenever $\sigma(s)=t_s$ for all $s=1,\ldots,j$. 
Since $a \notin c_{00}$, there exists a minimal integer $i_0\ge j+1$ such that $a_{i_0}\neq 0$.  
In addition, since $x \notin \ell_\infty$, there exists an integer $t_0 \ge t_j+i_0$ such that
\begin{equation}\label{eq:contradictionabsolutevalues}
\textstyle 
|a_{i_0}x_{t_0}|\ge m_0+1+\left|\sum_{k\le j} a_k x_{t_k}\right|.
\end{equation}

Finally, define  $\sigma_0: \mathbf{N}\to \mathbf{N}$ as follows:
\begin{enumerate}[label=(\roman*)]
\item \label{item:sigma1} $\sigma_0(s)=t_s$ for all $s=1,\ldots,j$;
\item \label{item:sigma2} $\sigma_0(j+s)=t_j+s$ for all $s=1,\ldots, i_0-j-1$;
\item \label{item:sigma3} $\sigma_0(i_0+s)=t_0+s$ for all integers $s\ge 0$.
\end{enumerate}
Note that $\sigma_0$ is strictly increasing, hence $\sigma_0 \in \Sigma$. Moreover, thanks to \ref{item:sigma1}, $\sigma_0 \in E_{m_0}$. On the other hand, by \ref{item:sigma3} we have $\sigma_0(i_0)=t_0$. And by the minimality of $i_0$ we get $a_{j+s}=0$ for all $s=1,\ldots,i_0-j-1$. Putting everything together with \eqref{eq:contradictionabsolutevalues}, we obtain that 
$$
\textstyle 
\left|\sum_{k\le i_0}a_k x_{\sigma_0(k)}\right|=\left|a_{i_0}x_{t_0}+\sum_{k\le j}a_kx_{t_k}\right|\ge m_0+1,
$$
which would imply that $\sigma_0 \notin E_{m_0}$. This contradiction concludes the proof. 
\end{proof}

A matrix $A$ is said \emph{row finite} if every row is in $c_{00}$, that is, for each $n$ there exists $k_0 \in \mathbf{N}$ such that $a_{n,k}=0$ for all $k\ge k_0$.
\begin{cor}\label{cor:unboundedcase1}
Let $A$ be a matrix which is not row finite, and fix $x \in \omega \setminus \ell_\infty$. 

Then 
$
\textstyle 
\{\sigma \in \Sigma: \sigma(x) \in \omega_A\}
$ 
is meager.
\end{cor}
\begin{proof}
By hypothesis, there exists $r_0 \in \mathbf{N}$ such that $a_{r_0,k}\neq 0$ for infinitely many $k$. 
Hence
\begin{displaymath}
\begin{split}
\textstyle
\{\sigma \in \Sigma: \sigma(x) \in \omega_A\}&\textstyle = \bigcap_r \left\{\sigma \in \Sigma: \left(\sum_{k\le n}a_{r,k}x_{\sigma(k)}\right) \in c\right\}\\
&\textstyle 
\subseteq 
\left\{\sigma \in \Sigma: \left(\sum_{k\le n}a_{r_0,k}x_{\sigma(k)}\right) \in \ell_\infty\right\}.
\end{split}
\end{displaymath}
The conclusion follows by Theorem \ref{lem:domaindefinition}.
\end{proof}

\begin{thm}\label{thm:unboundedrowfinite}
Let $\mathcal{I}$ be a meager ideal on $\mathbf{N}$. Fix an unbounded sequence $x \in \omega \setminus \ell_\infty$ and a row finite matrix $A$ such that 
\begin{equation}\label{eqitem:U1}
\textstyle 
\forall w \in \mathbf{N}, \quad 
Z_w:=\{n \in \mathbf{N}: a_{n,k}=0 \text{ for all }k\ge w\} \in \mathcal{I}.
\end{equation}

Then 
$
\textstyle 
\{\sigma \in \Sigma: A\sigma(x) \in \ell_\infty(\mathcal{I})\}
$  
is meager. 
\end{thm}
\begin{proof} 
Thanks to Theorem \ref{prop:talagrand}, there exists a sequence $(F_n)$ of closed sets in $\mathcal{P}(\mathbf{N})$ such that $\mathcal{I}\subseteq \bigcup_n F_n$ and $F_n \cap \mathrm{Fin}^\star=\emptyset$ for all $n\in \mathbf{N}$. 
Fix $w \in \mathbf{N}$. 
Since $Z_w\in\mathcal{I}$ by condition \eqref{eqitem:U1}, it follows that the family $\mathcal{J}_w:=\{U\setminus Z: U\in \mathcal{I}\}$ is an ideal on $T_w:=\mathbf{N}\setminus Z_w$ such that $\mathcal{J}_w\subseteq \bigcup_n G_n$ and $G_n \cap \mathrm{Fin}^\star=\emptyset$ for all $n \in \mathbf{N}$, where $G_n:=F_n \cap \mathcal{P}(T_w)$ is closed in $\mathcal{P}(T_w)$. 
It follows, again by Theorem \ref{prop:talagrand}, that $\mathcal{J}_w$ is a meager ideal on $T_w$. 
Hence there exists a partition $\{I_{w,1}, I_{w_2},\ldots\}$ of $T_w$ into nonempty finite subsets such that a set $U\subseteq T_w$ does not belong to $\mathcal{J}_w$ whenever $I_{w,n} \subseteq U$ for infinitely many $n$.

At this point, let $S$ be the claimed set. Since $T_w \in \mathcal{I}^\star$ for each $w \in \mathbf{N}$, we obtain that
\begin{displaymath}
\begin{split}
\textstyle 
\forall w \in \mathbf{N}, \quad 
S&\textstyle =\bigcup_m\{\sigma \in \Sigma: \{n \in \mathbf{N}: |\sum_ka_{n,k}x_{\sigma(k)}|\ge m\}\in \mathcal{I}\}\\
&\textstyle 
=\bigcup_m\{\sigma \in \Sigma: \{n \in T_w: |\sum_{k} a_{n,k}x_{\sigma(k)}|\ge m\}\in \mathcal{J}_w\}\\
&\textstyle 
\subseteq \bigcup_m\{\sigma \in \Sigma: \{n \in I_{w,q}: |\sum_k a_{n,k}x_{\sigma(k)}|\ge m\}\neq I_{w,q} \text{ for all large }q\}\\
&\textstyle 
= \bigcup_m\bigcup_p \bigcap_{q\ge p}\{\sigma \in \Sigma: \{n \in I_{w,q}: |\sum_ka_{n,k}x_{\sigma(k)}|\ge m\}\neq I_{w,q}\}\\
&\textstyle 
= \bigcup_m\bigcup_p \bigcap_{q\ge p} \bigcup_{J\subsetneq I_{w,q}} \{\sigma \in \Sigma: \{n \in I_{w,q}: |\sum_ka_{n,k}x_{\sigma(k)}|\ge m\}=J\}.
\end{split}
\end{displaymath}
Note that all the inner sets are closed, so that also each
$$
\textstyle 
S_w(m,p):=\bigcap_{q\ge p} \{\sigma \in \Sigma: \forall q\ge p, \exists n \in I_{w,q}, |\sum_ka_{n,k}x_{\sigma(k)}|< m\}
$$
is closed. Since $S\subseteq \bigcup_{m,p}S_w(m,p)$, it is sufficient to show that there exists $w \in \mathbf{N}$ such that each $S_w(m,p)$ has empty interior: this would imply that $S$ is contained in a countable union of nowhere dense sets. 

To this aim, suppose for the sake of contradiction that there exist $m_0,p_0 \in \mathbf{N}$ and positive integers $t_{1}<\cdots<t_{j_0}$ such that $\sigma \in S(m_0,p_0)$ whenever $\sigma(s)=t_{s}$ for all $s=1,\ldots,j_0$. 
Set $w_0:=j_0+1$. 
By condition \eqref{eqitem:U1}, the set $T_{w_0}$ belongs to $\mathcal{I}^\star$, hence it is nonempty. 
Note that, by construction, $r_n \ge w_0$ for all $n\in T_{w_0}$. 
Define $n_0:=\min T_{w_0}$, let $p_1$ be the positive integer for which $n_0 \in I_{w_0,p_1}$, and finally set
$$
\textstyle
\alpha:=\min\{|a_{n,k}|: a_{n,k} \neq 0 \text{ and }n \in I_{w_0,q_0}\}, \,\,
\text{ where }q_0:=\max\{p_0, p_1+1\}.
$$
Note that $\alpha$ is well defined since $A$ is row finite, and set $k_0:=\max\{r_n: n \in I_{w_0,q_0}\}$. 


Finally, define $\sigma_0: \mathbf{N}\to \mathbf{N}$ recursively as follows:
\begin{enumerate}[label=\textup{(}\roman*\textup{)}]
\item \label{item:cond1} $\sigma_0(s)=t_s$ for all $s=1,\ldots,j_0$;
\item \label{item:cond2} for each $s=j_0+1,\ldots,k_0$, if $\sigma_0(1)<\cdots<\sigma_0(s-1)$ are already defined, then $\sigma_0(s)$ is an integer $h>\sigma_0(s-1)$ such that
$$
\textstyle 
|x_h|\ge \frac{1}{\alpha}\left(m_0+\max\left\{\left|\sum_{k< s}a_{n,k}x_{\sigma_0(k)}\right|: n \in I_{w_0,q_0}\right\}\right).
$$
\item \label{item:cond3} $\sigma_0(k_0+s)=\sigma_0(k_0)+s$ for all $s \in \mathbf{N}$.
\end{enumerate}
Considering that $\sigma_0$ is strictly increasing by construction, it follows by \ref{item:cond1} that $\sigma_0 \in S_{w_0}(m_0,p_0)$. 
At the same time, since $j_0<r_n\le k_0$ for all $n \in I_{w_0,q_0}$, we obtain by \ref{item:cond2} that
\begin{displaymath}
\begin{split}
\textstyle 
\forall n\in I_{w_0,q_0}, \quad |\sum_ka_{n,k}x_{\sigma_0(k)}|
&\textstyle =|\sum_{k\le r_n}a_{n,k}x_{\sigma_0(k)}|\\
&\textstyle \ge |a_{n,r_n} x_{\sigma_0(r_n)}|-\left|\sum_{k< r_n}a_{n,k}x_{\sigma_0(k)}\right|\\
&\textstyle \ge \alpha |x_{\sigma_0(r_n)}|- \max\left\{\left|\sum_{k< r_n}a_{i,k}x_{\sigma_0(k)}\right|: i \in I_{w_0,q_0}\right\}\ge m_0.
\end{split}
\end{displaymath}
This implies that $\sigma_0 \notin S_{w_0}(m_0,p_0)$, hence we obtained the desired contradiction. 
\end{proof}

\section{Proof of Theorem \ref{thm:mainTauberianFinI}}\label{sec:mainproof}

\begin{proof}[Proof of Theorem \ref{thm:mainTauberianFinI}] 
First, suppose that $x$ is a bounded divergent sequence. Then $\Sigma_{x,A}(\mathcal{I})$ is meager by Theorem \ref{thm:laflamme} and Theorem \ref{thm:conclusionmaincategorybounded}. 

Secondly, suppose that $x$ is unbounded. If $A$ is not row finite, then $\Sigma_{x,A}(\mathcal{I})\subseteq \{\sigma \in \Sigma: \sigma(x) \in \omega_A\}$, which is meager by Corollary \ref{cor:unboundedcase1}. 
Otherwise, suppose hereafter that $A$ is a row finite $(\mathrm{Fin}, \mathcal{I})$-regular matrix. 
Note that $\mathcal{I}$ is a Borel ideal, hence it is meager. 
Thanks to Theorem \ref{thm:connotleofinJ}, $A$ satisfies conditions \ref{item:R1}--\ref{item:R3}. 
For each $w \in \mathbf{N}$, it follows by \ref{item:R2} and \ref{item:R3} that $\mathcal{I}\text{-}\lim_n\sum_{k\ge w}a_{n,k}=1$. 
In particular, $Z_w$ is contained in $\{n \in \mathbf{N}: \sum_{k\ge w}a_{n,k}=0\}$, which belongs to $\mathcal{I}$. 
This implies that condition \eqref{eqitem:U1} in Theorem \ref{thm:unboundedrowfinite} holds. 
It follows that $\Sigma_{x,A}(\mathcal{I})$ is contained in $\{\sigma \in \Sigma: A\sigma(x) \in \ell_\infty(\mathcal{I})\}$, which is meager by Theorem \ref{thm:unboundedrowfinite}. 
\end{proof}


\section{Concluding remarks}

It doesn't come as a surprise that, under suitable hypotheses on the matrix $A$ and the ideal $\mathcal{I}$, the set $\Sigma_{x,A}(\mathcal{I})$ is either meager or the whole $\Sigma$. Indeed, by a known topological $0$-$1$ law, see e.g. \cite[Theorem 8.47]{MR1321597}, a tail subset of $\Sigma$ with the Baire property is either meager or comeager. This applies also in our case. For, let $\mathcal{I}$ be a Borel ideal and $A$ be a $(\mathrm{Fin}, \mathcal{I})$-regular matrix. Then, it follows by Remark \ref{rmk:borelmeasurability} that $\Sigma_{x,A}(\mathcal{I})$ is Borel, hence it has the Baire property. Moreover, if $\sigma_1(n)=\sigma_2(n)$ for all but finitely many $n$, then $y\in c_0$ where $y_n:=x_{\sigma_2(n)}-x_{\sigma_1(n)}$ for all $n$. Hence, if $\sigma_1 \in \Sigma_{x,A}(\mathcal{I})$ then 
$$
\textstyle 
\mathcal{I}\text{-}\lim A\sigma_2(x)=\mathcal{I}\text{-}\lim A\sigma_1(x)+\mathcal{I}\text{-}\lim Ay=\mathcal{I}\text{-}\lim A\sigma_1(x).
$$
This implies that also $\sigma_2 \in \Sigma_{x,A}(\mathcal{I})$, proving that $\Sigma_{x,A}(\mathcal{I})$ is also a tail set. 

In the same direction of \cite{MR404918, MR382890, MR425410}, we leave as open question for the interest reader to check whether the analogues of Theorem \ref{thm:mainTauberianFinI} hold for permutations and strechings of the sequence $x$.

Finally, our main result seems to be related also to a conjecture of DeVos \cite{MR501124} which can be reformulated as follows: if $E$ is an FK-space (that is, a locally convex vector space of $\omega$ which is also Fr\'{e}chet and with continuous coordinates) containing $c_{00}$, then $\{0,1\}^{\mathbf{N}}\subseteq E$ if and only if $E \cap \{0,1\}^{\mathbf{N}}$ is not meager. However, it seems quite unlikely that spaces of the type $\{x \in \omega: Ax \in c(\mathcal{I})\}$ may provide a counterexample to the latter conjecture. Indeed,  it has been shown by Connor in \cite[Theorem 3.3]{MR954458} that, even for the well-behaved ideal $\mathcal{I}=\mathcal{Z}$, the unique FK-space containing $c(\mathcal{Z})$ is $\omega$.



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