Arithmetic closed subsets

Question

Let $S, S_1$ be subsets of the positive numbers $\mathbb{N}$. We say (as usual) that $S$ is multiplicative closed if $x \in S$ and $y \in S$ implies $xy \in S.$ We say also that $S_1$ is arithmetic closed if $x \in S_1$ and $y \in S_1,$ and $\gcd(x,y)=1,$ implies $xy \in S_1.$

Let $T$ be a subset of the positive integers containing $1$ and at least another element. The smallest multiplicative closed set that contains $T$ (say $C(T)$ ) i.e., the intersection of all multiplicative closed sets that contain $T,$ is also equal to the set containing $1$ and containing all finite products of powers of primes $p^{a_p}$ (say), where $p$ is prime and $a_p >0$ is minimal with the property $p^{a_p} \in T.$

Question: How to describe (analogously ?) the smallest arithmetic closed set that contains $T$ (say $A(T)$ ) i.e., the intersection of all arithmetic closed sets that contain $T.$

Example: If $T$ contains exactly $1$ and all $3 \cdot 2^n$ with $n>0$ then $C(T)$ is the set of $1$ and all products $2^a \cdot 3^b$ with $a >0$ and $b > 0,$ while

$$ A(T) = T. $$

Answer 1

It does not matter much if one includes 1 or not. In the case that $T$ is finite and the members are square free you have the question: Given a family $T$ of subsets of a finite set $U$, what can we say about the smallest subset of the power-set $2^U$ which is closed under disjoint unions? I'm not sure how much there is to say. When $T$ is finite, also $A(T)$ is finite and a subset of the set of divisors of the least common multiple of the members of $T$. For $T$ infinite not so much changes.

For fixed $k \gt 1,$ we have $ A(\lbrace t^k \mid t \in T \rbrace)=\lbrace a^k \mid a \in A(T) \rbrace$ but $A(kT)=kT.$