# Ideal characterization of almost convergence

$$\bullet$$ A real sequence $$x=(x_n)_n$$ is called convergent to $$\alpha$$ in usual sense if for any $$\epsilon>0$$ the set $$\{n\in\mathbb N:|x_n-\alpha|\geq\epsilon\}$$ is finite.

$$\bullet$$ A real sequence $$x=(x_n)_n$$ is called statistically convergent to $$\alpha$$ if for any $$\epsilon>0$$ the set $$\{n\in\mathbb N:|x_n-\alpha|\geq\epsilon\}$$ has natural density $$0$$. The natural density $$d$$ of $$A\subset\mathbb N$$ is defined by $$d(A)=\lim\limits_{n\to\infty}\frac{|A\cap\{1,2,\dots,n\}|}{n}$$, (provided the limit exists) where $$|A|$$ denotes the cardinality of $$A$$.

$$\bullet$$ A bounded real sequences $$x=(x_n)_n$$ is said to be almost convergent to $$\alpha$$ if all the Banach limit functionals give an unique value for the sequence $$x$$.

A family $$\mathcal I$$ of subsets of $$\mathbb N$$ is said to be an ideal in $$\mathbb N$$ if

(i) $$A,B\in \mathcal I$$ $$\implies$$ $$A\cup B\in \mathcal I$$

(ii) $$A\in \mathcal I$$ and $$B\subset A$$ $$\implies$$ $$B\in \mathcal I$$

$$\bullet$$ A real sequence $$x=(x_n)_n$$ is called $$\mathcal I$$-convergent to $$\alpha$$ in usual sense if for any $$\epsilon>0$$ the set $$\{n\in\mathbb N:|x_n-\alpha|\geq\epsilon\}\in \mathcal I$$.

$$\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots$$

$$\mathcal I_f=\{A\subset\mathbb N: A \text{ is finite}\}$$ and $$\mathcal I_d=\{A\subset\mathbb N: d(A)=0\}$$ become ideals in $$\mathbb N$$. Moreover, $$\mathcal I_f$$-convergence and $$\mathcal I_d$$-convergence coincide with usual convergence and statistical convergence respectively. But, what is the ideal in the case of almost convergence?

My Question : Find the ideal $$\mathcal I$$ for which $$\mathcal I$$-convergence coincides with the almost convergence. Is it available in literature?

Such an ideal does not exist.

Indeed, suppose the contrary, and let $$I$$ be such an ideal. The sequence $$(x_n)=(1,0,1,0,\dots)$$ is almost convergent, and therefore $$I$$-convergent, to $$1/2$$. So, $$\mathbb N=\{n\in\mathbb N\colon|x_n-1/2|\ge1/2\}\in I,$$ and hence $$I$$ is the powerset of $$\mathbb N$$. So, every sequence is $$I$$-convergent, and therefore almost convergent, to every real limit, which is of course absurd.

This consideration also shows that the Cesàro convergence -- which is implied by the almost-convergence -- is also not the $$I$$-convergence, for any ideal $$I$$.


This sequence is almost convergent to $$1/2$$. At the same time, it is not difficult to show that if this sequence has $$\I$$-limit of some ideal $$\I$$, then the $$\I$$-limit can only be $$0$$ or $$1$$.

• We can use the fact that $$\I$$-limit of a sequence is a cluster point of that sequence. (This holds for any admissible ideal, i.e., for any ideal which contains all finite sets. If we allow also non-admissible ideals, then we can get cluster points or terms of the sequence as limits.)
• For $$\I$$-convergence we have multiplicativity, i.e., $$\Ilim (x_ny_n)=\Ilim x_n\cdot\Ilim y_n$$. In particular, for our sequence $$x$$ we have $$x^2=x$$. Consequently, if $$L$$ is an $$\I$$-limit, then we get $$L^2=L$$.

This is basically just a reformulation of Lorenz's criterion for almost convergent sequence, but since you're looking at connection between almost convergent ideals, I'll mention that a sequence is almost convergent to $$L$$ if and only if $$\Flim_n \Glim_k \frac{x_k+\dots+x_{k+n-1}}n=L$$ for any free ultrafilters $$\mathcal F$$, $$\mathcal G$$.

You can check the section on almost convergent sequences of the paper Jerison, Meyer, The set of all generalized limits of bounded sequences, Can. J. Math. 9, 79-89 (1957). ZBL0077.31004, MR83697. I have notes on some part of the paper where the results are expressed using ultrafilters (rather than using nets in Stone-Čech compactification) available here: notes (Wayback Machine), slides (Wayback Machine).