# On relation between statistical ideal and ideal generated by a modulus function

### Abstract

Ideal on an arbitrary non-empty set $\Omega$ it's a non-empty family of subset $\mathfrak{I}$ of the set $\Omega$ which satisfies the following axioms: $\Omega \notin \mathfrak{I}$, if $A, B \in \mathfrak{I}$, then $A \cup B \in \mathfrak{I}$, if $A \in \mathfrak{I}$ and $D \subset A$, then $D \in \mathfrak{I}$. The ideal theory is a very popular branch of modern mathematical research. In our paper we study some classes of ideals on the set of all positive integers $\mathbb{N}$, namely the ideal of statistical convergence $\mathfrak{I}_s$ and the ideal $\mathfrak{I}_f$ generated by a modular function $f$. Statistical ideal it's a family of subsets of $\mathbb{N}$ whose natural density is equal to 0, i.e. $A \in \mathfrak{I}_s$ if and only if $\displaystyle\lim\limits_{n \rightarrow \infty}\frac{\#\{k \leq n: k \in A\}}{n} = 0$. A function $f:\mathbb{R}^+ \rightarrow \mathbb{R}^+$ is called a modular function, if $f(x) = 0$ only if $x = 0$, $f(x + y) \leq f(x) + f(y)$ for all $x, y \in\mathbb{R}^+$, $f(x) \le f(y)$ whenever $x \le y$, $f$ is continuous from the right 0, and finally $\lim\limits_{n \rightarrow \infty} f(n) = \infty$. Ideal, generated by the modular function $f$ it's a family of subsets of $\mathbb{N}$ with zero $f$-density, in other words, $A \in \mathfrak{I}_f$ if and only if $\displaystyle\lim\limits_{n \rightarrow \infty}\frac{f(\#\{k \leq n: k \in A\})}{f(n)} = 0$. It is known that for an arbitrary modular function $f$ the following is true: $\mathfrak{I}_f \subset \mathfrak{I}_s$. In our research we give the complete description of those modular functions $f$ for which $\mathfrak{I}_f = \mathfrak{I}_s$. Then we analyse obtained result, give some partial cases of it and prove one simple sufficient condition for the equality $\mathfrak{I}_f = \mathfrak{I}_s$. The last section of this article is devoted to examples of some modulus functions $f, g$ for which $\mathfrak{I}_f = \mathfrak{I}_s$ and $\mathfrak{I}_g \neq \mathfrak{I}_s$. Namely, if $f(x) = x^p$ where $p \in (0, 1]$ we have $\mathfrak{I}_f = \mathfrak{I}_s$; for $g(x) = \log(1 + x)$, we obtain $\mathfrak{I}_g \neq \mathfrak{I}_s$. Then we consider more complicated function $f$ which is given recursively to demonstrate that the conditions of the main theorem of our paper can't be reduced to the sufficient condition mentioned above.### Downloads

### References

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*Visnyk of V. N. Karazin Kharkiv National University. Ser. Mathematics, Applied Mathematics and Mechanics*,

*95*, 23-30. https://doi.org/10.26565/2221-5646-2022-95-02

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