Thinnable Ideals and Invariance of Cluster Points

TL;DR

This paper introduces thinnable ideals on positive integers, demonstrating that the set of their cluster points remains invariant across almost all subsequences, and provides a new characterization of ideal convergence.

Contribution

It defines a broad class of thinnable ideals and proves the invariance of cluster points for sequences in metric spaces, improving existing convergence characterizations.

Findings

Cluster points are invariant under subsequences for thinnable ideals.

The set of cluster points is the same for almost all subsequences in measure.

A new characterization of ideal convergence is established.

Abstract

We define a class of so-called thinnable ideals $\mathcal{I}$ on the positive integers which includes several well-known examples, e.g., the collection of sets with zero asymptotic density, sets with zero logarithmic density, and several summable ideals. Given a sequence $(x_n)$ taking values in a separable metric space and a thinnable ideal $\mathcal{I}$, it is shown that the set of $\mathcal{I}$-cluster points of $(x_n)$ is equal to the set of $\mathcal{I}$-cluster points of almost all its subsequences, in the sense of Lebesgue measure. Lastly, we obtain a characterization of ideal convergence, which improves the main result in [Trans. Amer. Math. Soc. 347 (1995), 1811--1819].

Equations

$${\mathcal{I}}_0:=\left\{S\subseteq \mathbf{N}:\underset{n\to \infty }{\lim }\frac{|S\cap [1,n]|}{n}=0\right\},$$

$$\{n:x_n\in U\}\notin \mathcal{I}$$

$$A_B:=\{a_b:b\in B\}\text{ and }kA:=\{ka:a\in A\}.$$

$${\mathcal{I}}_f:=\left\{S\subseteq \mathbf{N}:\sum _{n\in S}f(n)<\infty \right\}.$$

$$\underset{n\to \infty }{\mathrm{liminf}}\frac{{\sum }_{i\in [1,n]}f(i)}{{\sum }_{i\in [1,kn]}f(i)}\ne 0\text{ for all }k\in \mathbf{N}$$

$${\mathcal{E}}_f:=\left\{S\subseteq \mathbf{N}:\underset{n\to \infty }{\lim }\frac{{\sum }_{i\in S\cap [1,n]}f(i)}{{\sum }_{i\in [1,n]}f(i)}=0\right\}.$$

$${\sum }_{n\ge 1}f(a_{b_n})\ge O(1)+{\sum }_{n\ge 1}f(k{b}_n)=\infty ,$$

$$\sum _{n\equiv 1\mathrm{mod}k}f(b_n)\ge \sum _{n\equiv 2\mathrm{mod}k}f(b_n)\ge \cdots \ge \sum _{n\equiv 0\mathrm{mod}k}f(b_n)\ge \sum _{\begin{array}{c}n\equiv 1\mathrm{mod}k,\\ n\ne 1\end{array}}f(b_n)$$

$$\sum _{n\ge 1}f(b_{a_n})\ge O(1)+\sum _{n\ge 1}f(b_{kn})=\infty ,$$

$${\sum }_{a_{b_n}\le x}f(a_{b_n})\ge O(1)+{\sum }_{b_n\le x/k}f(k{b}_n).$$

$${\sum }_{b_n\le x/k}f(k{b}_n)\ne o\left({\sum }_{i\le x/k}f(i)\right);$$

$${\sum }_{b_n\le x/k}f(k{b}_n)\ne o\left({\sum }_{i\le x}f(i)\right),$$

$$\frac{f(b_{a_1})+\cdots +f(b_{a_n})}{f(1)+\cdots +f(b_{a_n})}\ge \frac{O(1)+f(b_k)+\cdots +f(b_{kn})}{f(1)+\cdots +f(b_{kn})}\nrightarrow 0,$$

$$\frac{f(x_1)+\cdots +f(x_n)}{f(1)+\cdots +f(x_n)}\ge \frac{f(y_1)+\cdots +f(y_n)}{f(1)+\cdots +f(y_n)}$$

$${\mathcal{I}}_{\alpha }:=\left\{S\subseteq \mathbf{N}:{\mathrm{d}}_{\alpha }^{\star }(S)=0\right\},\text{ where }{\mathrm{d}}_{\alpha }^{\star }(S)=\underset{n\to \infty }{\mathrm{limsup}}\frac{{\sum }_{i\in S\cap [1,n]}{i}^{\alpha }}{{\sum }_{i\in [1,n]}{i}^{\alpha }}.$$

$$\frac{\alpha +1}{y_n^{\alpha +1}}\sum _{i\le n}{y}_i^{\alpha }\le \frac{\alpha +1}{y_n^{\alpha +1}}\sum _{i\in ((1-\lambda )y_n,y_n]}{i}^{\alpha }\le \left(1-(1-\lambda {)}^{\alpha +1}\right))(1+o(1))<\frac{2}{3}{\mathrm{d}}_{\alpha }^{\star }(Y)$$

$$\frac{1}{x_n^{\alpha +1}}\sum _{i\le n}{x}_i^{\alpha }\ge \frac{1}{x_n^{\alpha +1}}\sum _{i\le \lambda y_n}{i}^{\alpha }\ge \frac{1}{x_n^{\alpha +1}}\sum _{i\le \lambda x_n}{i}^{\alpha }\ge \frac{{\lambda }^{\alpha +1}}{2}$$

$$\frac{1}{a_{b_n}^{\alpha +1}}\sum _{k\le n}{\left(a_{b_k}\right)}^{\alpha }\ge \frac{1}{{\left(\frac{1}{c}+\epsilon \right)}^{\alpha +1}{b_n}^{\alpha +1}}\left(O(1)+\sum _{n_0\le k\le n}{\left(\frac{1}{c}-\epsilon \right)}^{\alpha }{b_k}^{\alpha }\right).$$

$$\begin{array}{rl}\frac{{\mathrm{d}}_{\alpha }^{\star }(A_B)}{\alpha +1}& =\underset{n\to \infty }{\mathrm{limsup}}\frac{1}{a_{b_n}^{\alpha +1}}\sum _{k\le n}{\left(a_{b_k}\right)}^{\alpha }\ge \underset{n\to \infty }{\mathrm{limsup}}\frac{\kappa }{{b_n}^{\alpha +1}}\left(O(1)+\sum _{n_0\le k\le n}\kappa {b_k}^{\alpha }\right)\\ & ={\kappa }^2\underset{n\to \infty }{\mathrm{limsup}}\frac{1}{{b_n}^{\alpha +1}}\sum _{n_0\le k\le n}{b_k}^{\alpha }={\kappa }^2\frac{{\mathrm{d}}_{\alpha }^{\star }(B)}{\alpha +1}>0.\end{array}$$

$${\mathcal{I}}_{\mathfrak{p}}:=\left\{S\subseteq \mathbf{N}:{\mathfrak{p}}^{\star }(S)=0\right\},\text{ where }{\mathfrak{p}}^{\star }(S)=\underset{s\to 1^{-}}{\lim }\underset{n\to \infty }{\mathrm{limsup}}\frac{|S\cap [ns,n]|}{(1-s)n}.$$

$${\mathcal{I}}_f=\{I\subseteq \mathbf{N}:I\cap 2\mathbf{N}\in \mathrm{Fin}\}.$$

$$\sum _{i\ge 1}{d}_i(\omega )2^{-i},$$

$$\lambda \left(\left\{\omega \in (0,1]:{\Gamma }_x(\mathcal{I})={\Gamma }_{x\upharpoonright \omega }(\mathcal{I})\right\}\right)=1.$$

$$\Omega :=\left\{\omega \in (0,1]:\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{i=1}^n{d}_i(\omega )=\frac{1}{2}\right\}.$$

$${\Gamma }_{x\upharpoonright \omega }(\mathcal{I})\subseteq {\Gamma }_x(\mathcal{I})\text{ for all }\omega \in \Omega .$$

$$\{n:x_n\in U\}\supseteq \{n_k:x_{n_k}\in U\}\notin \mathcal{I}$$

$$\lambda \left(\left\{\omega \in (0,1]:{\Gamma }_x(\mathcal{I})\subseteq {\Gamma }_{x\upharpoonright \omega }(\mathcal{I})\right\}\right)=1.$$

$$\Theta (\ell ,U_m):=\left\{\omega \in (0,1]:\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{j=1}^n{d}_{i_j}(\omega )=\frac{1}{2}\right\}$$

$$\{k:x_{n_k}\in U_m\}=\{k:n_k\in I\}\le \{n_k:n_k\in I\}=I_J.$$

$$\lambda \left(\left\{\omega \in (0,1]:\ell \in {\Gamma }_{x\upharpoonright \omega }(\mathcal{I})\right\}\right)=1.$$

Full text

Lastly, we obtain a characterization of ideal convergence, which improves the main result in [Trans. Amer. Math. Soc. 347 (1995), 1811–1819].

1. Introduction

It is well known that the set of ordinary limit points of “almost every” subsequence of a real sequence $(x_{n})$ coincides with the set of ordinary limit points of the original sequence, in the sense of Lebesgue measure, see Buck [5]. In the same direction, we prove its analogue for ideal cluster points.

To this aim, let $\mathcal{I}$ be an ideal on the positive integers $\mathbf{N}$, that is, a family of subsets of $\mathbf{N}$ closed under taking finite unions and subsets of its elements. It is assumed that $\mathcal{I}$ contains the collection $\mathrm{Fin}$ of finite subsets of $\mathbf{N}$ and it is different from the whole power set of $\mathbf{N}$. Note that the collection of subsets with zero asymptotic density, i.e.,

[TABLE]

is an ideal. Let also $x=(x_{n})$ be a sequence taking values in a topological space $X$. We denote by $\Gamma_{x}(\mathcal{I})$ the set of $\mathcal{I}$-cluster points of $x$, that is, the set of all $\ell\in X$ such that

[TABLE]

for all neighborhoods $U$ of $\ell$. Statistical cluster points (that is, $\mathcal{I}_{0}$-cluster points) of real sequences were introduced by Fridy [8], cf. also [7, 9, 11]. However, it is worth noting that ideal cluster points have been studied much before under a different name. Indeed, as it follows by [11, Theorem 4.2], they correspond to classical “cluster points” of a filter $\mathscr{F}$ on $\mathbf{R}$ (depending on $x$), cf. [4, Definition 2, p.69].

As anticipated, the main question addressed here is to find suitable conditions on $X$ and $\mathcal{I}$ such that the set of $\mathcal{I}$-cluster points of a sequence $(x_{n})$ is equal to the set of $\mathcal{I}$-cluster points of “almost all” its subsequences. Finally, we obtain a characterization of ideal convergence. Related results were obtained in [1, 6, 15, 16, 17, 18].

2. Thinnability

Given $k\in\mathbf{N}$ and infinite sets $A,B\subseteq\mathbf{N}$ with canonical enumeration $\{a_{n}:n\in\mathbf{N}\}$ and $\{b_{n}:n\in\mathbf{N}\}$, respectively, we write $A\leq B$ if $a_{n}\leq b_{n}$ for all $n\in\mathbf{N}$ and define

[TABLE]

Definition 2.1**.**

An ideal $\mathcal{I}$ is said to be weakly thinnable if $A_{B}\notin\mathcal{I}$ whenever $A\subseteq\mathbf{N}$ admits non-zero asymptotic density and $B\notin\mathcal{I}$.

If, in addition, also $B_{A}\notin\mathcal{I}$ and $X\notin\mathcal{I}$ whenever $X\leq Y$ and $Y\notin\mathcal{I}$, then $\mathcal{I}$ is said to be thinnable.

Definition 2.2**.**

An ideal $\mathcal{I}$ is said to be strechable if $kA\notin\mathcal{I}$ for all $k\in\mathbf{N}$ and $A\notin\mathcal{I}$.

The terminology has been suggested from the related properties of finitely additive measures on $\mathbf{N}$ studied in [21]. In this regard, $\mathrm{Fin}$ is thinnable and strechable.

This is the case of several other ideals:

Proposition 2.3**.**

Let $f:\mathbf{N}\to(0,\infty)$ be a definitively non-increasing function such that $\sum_{n\geq 1}f(n)=\infty$. Define the summable ideal

[TABLE]

Then $\mathcal{I}_{f}$ is thinnable provided $\mathcal{I}_{f}$ is strechable.

In addition, suppose that

[TABLE]

and define the Erdős–Ulam ideal

[TABLE]

Then $\mathscr{E}_{f}$ is thinnable provided $\mathscr{E}_{f}$ is strechable.

Proof.

Let us suppose that $A=\{a_{n}:n\in\mathbf{N}\}$ admits asymptotic density $c>0$ and $B=\{b_{n}:n\in\mathbf{N}\}\notin\mathcal{I}_{f}$, that is, $\sum_{n\geq 1}f(b_{n})=\infty$. Define the integer $k:=\lfloor 1/c\rfloor+1\geq 2$ and note that $\sum_{n\geq 1}f(kb_{n})=\infty$ by the fact that $\mathcal{I}_{f}$ is strechable. Then $a_{n}=\frac{1}{c}n(1+o(1))$ as $n\to\infty$, which implies

[TABLE]

i.e., $A_{B}\notin\mathcal{I}_{f}$, hence $\mathcal{I}_{f}$ is weakly thinnable. Moreover, observe that

[TABLE]

and note that the first sum is finite if and only if the last sum is finite. Since $I\notin\mathcal{I}_{f}$, then all the above sums are infinite, which implies that

[TABLE]

i.e., $B_{A}\notin\mathcal{I}_{f}$. Lastly, given infinite sets $X,Y\subseteq\mathbf{N}$ with $X\leq Y$ and $X\in\mathcal{I}_{f}$, we have $\sum_{y\in Y}f(y)\leq\sum_{x\in X}f(x)<\infty$. Therefore $\mathcal{I}_{f}$ is thinnable.

The proof of the second part is similar, where (2) is replaced by

[TABLE]

Moreover, $B\notin\mathscr{E}_{f}$ implies $kB\notin\mathscr{E}_{f}$ by the hypothesis of strechability, i.e.,

[TABLE]

thanks to (1), we conclude that

[TABLE]

hence $A_{B}\notin\mathscr{E}_{f}$, which shows that $\mathscr{E}_{f}$ is weakly thinnable. In addition, we get

[TABLE]

so that $B_{A}\notin\mathscr{E}_{f}$, where the last $\not\to$ comes from a reasoning similar to (3). Finally, given infinite subsets $X,Y\subseteq\mathbf{N}$ with canonical enumeration $\{x_{n}:n\in\mathbf{N}\}$ and $\{y_{n}:n\in\mathbf{N}\}$, respectively, such that $X\leq Y$ and $X\in\mathscr{E}_{f}$, it holds

[TABLE]

for all $n\in\mathbf{N}$ therefore $Y\in\mathscr{E}_{f}$. ∎

Given a real $\alpha\geq-1$, let $\mathcal{I}_{\alpha}$ be the collection of subsets with zero $\alpha$-density, that is,

[TABLE]

Proposition 2.4**.**

All ideals $\mathcal{I}_{\alpha}$ are thinnable.

Proof.

If $\alpha\in[-1,0]$, the claim follows by Proposition 2.3 (we omit details). Hence, let us suppose hereafter than $\alpha>0$. Fix infinite sets $X,Y\subseteq\mathbf{N}$ with canonical enumerations $\{x_{n}:n\in\mathbf{N}\}$ and $\{y_{n}:n\in\mathbf{N}\}$, respectively, such that $Y\notin\mathcal{I}_{\alpha}$. Then, there exist an infinite set $S$ such that $|Y\cap[1,y_{n}]|\geq\lambda y_{n}$ for all $n\in S$, where $\lambda:=1-\left(1-\frac{1}{2}\mathrm{d}_{\alpha}^{\star}(Y)\right)^{\frac{1}{\alpha+1}}>0$. Indeed, in the opposite case, we would have that

[TABLE]

for all sufficiently large $n$. Since $|Y\cap[1,n]|\leq|X\cap[1,n]|$ for all $n$, we conclude that

[TABLE]

for all large $n\in S$, so that $X\notin\mathcal{I}_{\alpha}$.

At this point, fix sets $A,B\subseteq\mathbf{N}$ with canonical enumerations $\{a_{n}:n\in\mathbf{N}\}$ and $\{b_{n}:n\in\mathbf{N}\}$, respectively, such that $A$ admits asymptotic density $c>0$ and $B\notin\mathcal{I}_{\alpha}$. Fix also $\varepsilon>0$ sufficiently small and note that there exists $n_{0}=n_{0}(\varepsilon)\in\mathbf{N}$ such that $(\nicefrac{{1}}{{c}}-\varepsilon)n\leq a_{n}\leq(\nicefrac{{1}}{{c}}+\varepsilon)n$ for all $n\geq n_{0}$. In particular, it follows that

[TABLE]

Therefore, setting $\kappa:=\min\left\{\left(\frac{1}{c}+\varepsilon\right)^{-\alpha-1},\left(\frac{1}{c}-\varepsilon\right)^{\alpha}\right\}>0$, we obtain

[TABLE]

This proves that $A_{B}\notin\mathcal{I}_{\alpha}$. Finally, let $k$ be an integer greater than $\nicefrac{{1}}{{c}}$ and note that $B_{A}\leq B_{k\mathbf{N}}\setminus S$, for some finite set $S$. By the previous observation, it is sufficient to show that $B_{k\mathbf{N}}\notin\mathcal{I}_{\alpha}$ and this is straightforward by an analogous argument of (3). ∎

To mention another example, let $\mathcal{I}_{\mathfrak{p}}$ be the Pólya ideal, i.e.,

[TABLE]

Among other things, the upper Pólya density $\mathfrak{p}^{\ast}$ has found a number of remarkable applications in analysis and economic theory, see e.g. [19], [13] and [14].

Corollary 2.5**.**

The Pólya ideal $\mathcal{I}_{\mathfrak{p}}$ is thinnable.

Proof.

The upper Pólya density $\mathfrak{p}^{\ast}$ is the pointwise limit of the real net of the upper $\alpha$-densities $\mathrm{d}_{\alpha}^{\star}$ defined in (4), see [12, Theorem 4.3].

Fix infinite sets $X,Y\subseteq\mathbf{N}$ with canonical enumerations $\{x_{n}:n\in\mathbf{N}\}$ and $\{y_{n}:n\in\mathbf{N}\}$, respectively, such that $Y\notin\mathcal{I}_{\mathfrak{p}}$. Then, there exists $\alpha>0$ such that $\mathrm{d}_{\alpha}^{\star}(Y)>0$ and, thanks to Proposition 2.4, we get $\mathrm{d}_{\alpha}^{\star}(X)>0$ as well. This implies that $X\notin\mathcal{I}_{\mathfrak{p}}$. Other properties can be shown similarly. ∎

Lastly, it is worth noting that there exist summable ideals which are not weakly thinnable: for instance, let $\mathcal{I}_{f}$ be the ideal defined by $f(2n)=1$ and $f(2n-1)=0$ for all $n\in\mathbf{N}$, so that

[TABLE]

Set $A:=\mathbf{N}\setminus\{1\}$ and $B:=2\mathbf{N}$. Then, $A$ has asymptotic density $1$, $B\notin\mathcal{I}_{f}$, and $A_{B}=2\mathbf{N}+1\in\mathcal{I}_{f}$. Therefore $\mathcal{I}_{f}$ is not weakly thinnable.

3. Main Results

Consider the natural bijection between the collection of all subsequences $(x_{n_{k}})$ of $(x_{n})$ and real numbers $\omega\in(0,1]$ with non-terminating dyadic expansion

[TABLE]

where $d_{i}(\omega)=1$ if $i=n_{k}$, for some integer $k$, and $d_{i}(\omega)=0$ otherwise, cf. [3, Appendix A31] and [15]. Accordingly, for each $\omega\in(0,1]$, denote by $x\upharpoonright\omega$ the subsequence of $(x_{n})$ obtained by omitting $x_{i}$ if and only if $d_{i}(\omega)=0$.

Moreover, let $\lambda:\mathscr{M}\to\mathbf{R}$ denote the Lebesgue measure, where $\mathscr{M}$ stands for the completion of the Borel $\sigma$-algebra on $(0,1]$. Our main result follows:

Theorem 3.1**.**

Let $\mathcal{I}$ be a thinnable ideal and $(x_{n})$ be a sequence taking values in a first countable space $X$ where all closed sets are separable. Then

[TABLE]

Proof.

Let $\Omega$ be the set of normal numbers, that is,

[TABLE]

It follows by Borel’s normal number theorem [3, Theorem 1.2] that $\Omega\in\mathscr{M}$ and $\lambda(\Omega)=1$. Then, it is claimed that

[TABLE]

To this aim, fix $\omega\in\Omega$ and denote by $(x_{n_{k}})$ the subsequence $x\upharpoonright\omega$. Let us suppose for the sake of contradiction that $\Gamma_{x\upharpoonright\omega}(\mathcal{I})\setminus\Gamma_{x}(\mathcal{I})\neq\emptyset$ and fix a point $\ell$ therein. Then, the set of indexes $\{n_{k}:k\in\mathbf{N}\}$ has asymptotic density $\nicefrac{{1}}{{2}}$ and, for each neighborhood $U$ of $\ell$, it holds $\{k:x_{n_{k}}\in U\}\notin\mathcal{I}$. This implies

[TABLE]

by the hypothesis that $\mathcal{I}$ is, in particular, weakly thinnable. Therefore $\{n:x_{n}\in U\}\notin\mathcal{I}$, which is a contradiction since $\ell$ would be also a $\mathcal{I}$-cluster point of $x$. This proves (6).

To complete the proof, it is sufficient to show that

[TABLE]

This is clear if $\Gamma_{x}(\mathcal{I})$ is empty. Otherwise, note that $\Gamma_{x}(\mathcal{I})$ is closed by [11, Lemma 3.1(iv)], hence there exists a non-empty countable dense subset $L$. Fix $\ell\in L$ and let $(U_{m})$ be a decreasing local base of neighborhoods at $\ell$. Fix also $m\in\mathbf{N}$ and define $I:=\{n:x_{n}\in U_{m}\}$ which does not belong to $\mathcal{I}$; in particular, $I$ is infinite and we let $\{i_{n}:n\in\mathbf{N}\}$ be its enumeration. Again by Borel’s normal number theorem,

[TABLE]

belongs to $\mathscr{M}$ and has Lebesgue measure $1$. Fix $\omega$ in the above set and denote by $(x_{n_{k}})$ the subsequence $x\upharpoonright\omega$. Hence, the set $J:=\{n:i_{n}\in\{n_{k}:k\in\mathbf{N}\}\}$ admits asymptotic density $\nicefrac{{1}}{{2}}$ and, by the thinnability of $\mathcal{I}$, we get $I_{J}\notin\mathcal{I}$. Lastly, note that

[TABLE]

Therefore $\{k:x_{n_{k}}\in U_{m}\}\notin\mathcal{I}$. In addition, $\Theta(\ell):=\bigcap_{m\geq 1}\Theta(\ell,U_{m})$ belongs to $\mathscr{M}$ and has Lebesgue measure $1$. This implies that

[TABLE]

(See also [20, Theorem 1] for the case $\mathcal{I}=\mathrm{Fin}$.) At this point, since $L$ is countable, we get $\lambda\left(\left\{\omega\in(0,1]:L\subseteq\Gamma_{x\upharpoonright\omega}(\mathcal{I})\right\}\right)=1.$ Claim (7) follows by the fact that also $\Gamma_{x\upharpoonright\omega}(\mathcal{I})$ is closed by [11, Lemma 3.1(iv)], so that each of these $\Gamma_{x\upharpoonright\omega}(\mathcal{I})$ contains the closure of $L$, i.e., $\Gamma_{x}(\mathcal{I})$. ∎

Remark 3.2**.**

Separable metric spaces $X$ satisfy the hypotheses of Theorem 3.1. Indeed, $X$ is first countable and every closed subset $F$ of $X$ is separable. To prove the latter, let $A$ be a countable dense subset of $X$ and note that

[TABLE]

is a base for $F$, where $B(a,r)$ is the open ball with center $a$ and radious $r$. Then, a set which picks one point for every set in $\mathscr{F}$ is a countable dense subset of $F$.

As a consequence of Proposition 2.4, Theorem 3.1, and Remark 3.2, we obtain:

Corollary 3.3**.**

Let $x$ be a sequence taking values in a separable metric space. Then the set of statistical cluster points of $x$ is equal to the set of statistical cluster points of almost all its subsequences (in the sense of Lebesgue measure).

Similarly, setting $\mathcal{I}=\mathrm{Fin}$, we recover Buck’s result [5]:

Corollary 3.4**.**

Let $x$ be a sequence taking values in a separable metric space. Then the set of ordinary limit points of $x$ is equal to the set of ordinary limit points of almost all its subsequences (in the sense of Lebesgue measure).

Lastly, we recall that a sequence $x=(x_{n})$ taking values in topological space $X$ converges (with respect to an ideal $\mathcal{I}$) to $\ell\in X$, shortened as $x\to_{\mathcal{I}}\ell$, if

[TABLE]

for all neighborhoods $U$ of $\ell$. In this regard, Miller [15, Theorem 3] proved that a real sequence $x$ converges statistically to $\ell$, i.e., $x\to_{\mathcal{I}_{0}}\ell$, if and only if almost all its sequences converge statistically to $\ell$.

This is extended in the following result. Here, we say that an ideal $\mathcal{I}$ is invariant if, for each $A\subseteq\mathbf{N}$ with positive asymptotic density, it holds $A_{B}\notin\mathcal{I}$ if and only if $B\notin\mathcal{I}$ (in particular, $\mathcal{I}$ is weakly thinnable). This condition is strictly related with the so-called “property (G)” defined in [2].

Theorem 3.5**.**

Let $\mathcal{I}$ be an invariant ideal and $x$ be a sequence taking values in a topological space. Then $x\to_{\mathcal{I}}\ell$ if and only if

[TABLE]

Proof.

First, let us suppose that $x\to_{\mathcal{I}}\ell$ and let $U$ be a neighborhood of $\ell$. Let $\Omega$ be set of normal numbers defined in (5), fix $\omega\in\Omega$, and denote by $(x_{n_{k}})$ the subsequence $x\upharpoonright\omega$. Then $I:=\{n:x_{n}\notin U\}\in\mathcal{I}$ and $A:=\{n_{k}:k\in\mathbf{N}\}$ has asymptotic density $\nicefrac{{1}}{{2}}$. Define $B:=\{k:x_{n_{k}}\notin U\}=\{k:n_{k}\in I\}$. Since $\mathcal{I}$ is, in particular, weakly thinnable and $A_{B}=\{n_{k}:x_{n_{k}}\notin U\}\in\mathcal{I}$, it follows that $B\in\mathcal{I}$, i.e., $x\upharpoonright\omega\to_{\mathcal{I}}\ell$.

Conversely, note that $\lambda(\Omega\cap(1-\Omega))=1$. Hence, there exists $\omega\in\Omega$ such that $x\upharpoonright\omega\to_{\mathcal{I}}\ell$ and $x\upharpoonright(1-\omega)\to_{\mathcal{I}}\ell$. It easily follows that $x\to_{\mathcal{I}}\ell$. Indeed, denoting by $(x_{n_{k}})$ and $(x_{m_{r}})$ the subsequences $x\upharpoonright\omega$ and $x\upharpoonright(1-\omega)$, respectively, we have that, for each neighborhood $U$ of $\ell$, it holds $\{k:x_{n_{k}}\notin U\}\in\mathcal{I}$ and $\{r:x_{m_{r}}\notin U\}\in\mathcal{I}$. Since $\{n_{k}:k\in\mathbf{N}\}$ and $\{m_{r}:r\in\mathbf{N}\}$ form a partition of $\mathbf{N}$, then

[TABLE]

The claim follows by the hypothesis that $\mathcal{I}$ is invariant. ∎

It is not possible to extend Theorem 3.5 on the class of all ideals: indeed, it has been shown in [2, Example 2] that there exists an ideal $\mathcal{I}$ and a real sequence $x$ such that $x\to_{\mathcal{I}}\ell$ and, on the other hand, $\lambda\left(\left\{\,\omega\in(0,1]:x\upharpoonright\omega\to_{\mathcal{I}}\ell\,\right\}\right)=0$.

Acknowledgments

The author is grateful to Piotr Miska (Jagiellonian University, PL) and Marek Balcerzak (Łódź University of Technology, PL) for several useful comments.

Note added in proof

It turns out that the topological analogue of Theorem 3.1 is quite different, providing a non-analogue between measure and category. Indeed, it has been shown in [10] that, if $x$ is a sequence in a separable metric space, then $\{\omega\in(0,1]:\Gamma_{x}(\mathcal{I}_{0})=\Gamma_{x\upharpoonright\omega}(\mathcal{I}_{0})\}$ is not a first Baire category set if and only if every ordinary limit point of $x$ is also a statistical cluster point of $x$, that is, $\Gamma_{x}(\mathrm{Fin})=\Gamma_{x}(\mathcal{I}_{0})$.