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Let $\mathbb {P} = \lbrace p_1, p_2, \dots , p_i, \dots \rbrace $ be the set of prime numbers (or more generally a set of pairwise co-prime elements). Let us denote $A_p^{a,b} = \lbrace p^{an+b}m \mid n \in \mathbb {N} \cup \lbrace 0\rbrace ;m \in \mathbb {N},
p \mathrm {\, does \, not \, divide \,} m \rbrace $, where $a \in \mathbb {N}, b \in \mathbb {N} \cup \lbrace 0\rbrace $. 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 \[A = \left\lbrace \frac{\frac{1}{p^b}\left(1 - \frac{1}{p}\right)}{1 - \frac{1}{p^a}} \mid p \in \mathbb {P}, a \in \mathbb {N}, b \in \mathbb {N} \cup \lbrace 0\rbrace \right\rbrace ,\] where $\mathbb {P}$ is the set of all prime numbers, then for closure of set $A$ holds \[\mathop {\rm cl}A = A \cup B \cup \lbrace 0, 1\rbrace ,\] where $B = \left\lbrace \frac{1}{p^b}\left(1 - \frac{1}{p}\right) \mid p \in \mathbb {P}, b \in \mathbb {N} \cup \lbrace 0\rbrace \right\rbrace $. (English) |
