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In this answer I included this lemma:

Suppose $A\subseteq\{1,2,3,\ldots\}$ and $\sum\limits_{n\in A} \dfrac 1 n < \infty$. Then $\sum\limits_{n\in B} \dfrac 1 n <\infty$ where $B$ is the closure of $A$ under multiplication.

It was pointed out in a comment that at one point in the proof I should have had $\text{“}{\le}\text{”}$ rather than $\text{“}{=}\text{”}$ because some of the members of $B$ may occur more than once as products of members of $A.$

So I wonder: Are there any interesting or enlightening results stating necessary or sufficient conditions on a set $A\subseteq\{1,2,3,\ldots\}$ for every member of the multiset of products of members of $A$ to have multiplicity only $1\text{?}$

How to count products: Each member of the closure of $A$ under multiplication looks like this: $$ \prod\left\{ a_1a_2a_3\cdots a_k : a_1 \le a_2 \le a_3\le \cdots \le a_k, \, k\in\{0,1,2,3,\ldots \} \right\} $$ where $\text{“}{\le}\text{”}$ rather than $\text{“}{<}\text{”}$ appears, so a member of $A$ can appear more than once. Thus if $A=\{ 4,6,7,9 \}$ then \begin{align} & 4\times9\times7=252 \\ \text{and } & 6 \times 6 \times 7 = 252 \end{align} count as two separate products, so that the multiplicity of $252$ as a member of the closure is at least $2,$ but $6\times9$ and $9\times6$ are not counted as two separate products. This is consistent with the way things are done when it is asserted that each positive integer has only one prime factorization.

Michael Hardy
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  • An hour later I thinking: Might it be simply a matter of members of $A$ being coprime? More later$,\ldots\qquad$ – Michael Hardy Dec 25 '21 at 22:40
  • It would help to define multiplicity a little more precise. For example: "for every member of the multiset of products of members of $A$ to have multiplicity only 1?" How many elements in $A$ are we allowed to multiply with one another? And if two different products give the same value, do we care if the two products used the same number of terms? – Andreas Lenz Dec 25 '21 at 22:56
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    Coprimitivity of the elements in $A$ is probably sufficient. However not necessary, since $A={4,6}$ seems to have the desired property. – Andreas Lenz Dec 25 '21 at 23:00

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