We call a subset $A = \{a_1, a_2, a_3, \dots\}$ of $\mathbb N$ with $a_1 < a_2 < \dots $ transparent if $a_{k+1}  a_k$ goes to $\infty$ as $k \rightarrow \infty$. Is the following true? For every finite set $P$ of prime numbers, the set $A_P := \{p_1 \cdots p_s : p_i \in P\}$ is transparent.
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Proving that $a_{k+1}a_k\to \infty$ reduces to proving that $$a_{k+1}a_k=n$$ has finitely many solutions for every $n$. If you let $S$ be the set of prime divisors of $n$ union $P$, then this follows from the Sunit theorem. 

