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Let $\mathbb{P}=\{p_1,p_2,\cdots ,p_i,\cdots \}$ be the set of prime numbers (or more generally a set of pairwise co-prime elements). Let us denote $A_p^{a,b}=\{p^{an+b}m\mid n\in \mathbb{N}\cup \left\{0\right\};m\in \mathbb{N},p\mathrm{does}\mathrm{not}\mathrm{divide}m\}$, where $a\in \mathbb{N},b\in \mathbb{N}\cup \left\{0\right\}$. Then for arbitrary finite set $B$, $B\subset \mathbb{P}$ holds $$d\left(\bigcap _{p_i\in B}{A}_{p_i}^{a_i,b_i}\right)=\prod _{p_i\in B}d\left(A_{p_i}^{a_i,b_i}\right),$$ and $$d\left(A_{p_i}^{a_i,b_i}\right)=\frac{\frac{1}{p_i^{b_i}}\left(1-\frac{1}{p_i}\right)}{1-\frac{1}{p_i^{a_i}}}.$$ If we denote $$A=\left\{\frac{\frac{1}{p^b}\left(1-\frac{1}{p}\right)}{1-\frac{1}{p^a}}\mid p\in \mathbb{P},a\in \mathbb{N},b\in \mathbb{N}\cup \left\{0\right\}\right\},$$ where $\mathbb{P}$ is the set of all prime numbers, then for closure of set $A$ holds $$\mathrm{cl}A=A\cup B\cup \{0,1\},$$ where $B=\left\{\frac{1}{p^b}\left(1-\frac{1}{p}\right)\mid p\in \mathbb{P},b\in \mathbb{N}\cup \left\{0\right\}\right\}$.