$\bullet$ 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 $\{n\in\mathbb N:|x_n-\alpha|\geq\epsilon\}\in \mathcal I$.


$\mathcal I_f=\{A\subset\mathbb N: A \text{ is finite}\}$ 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?


2 Answers 2


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 $I$.


A slightly different argument using the sequence $x=(1,0,1,0,1,0,\dots)$.$\newcommand{\I}{\mathcal I}\newcommand{\Ilim}{\operatorname{\I-lim}}\newcommand{\Flim}{\operatorname{\mathcal F-lim}}\newcommand{\Glim}{\operatorname{\mathcal G-lim}}$

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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.