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Let the multiplicative closure operator $h$ be defined on $\mathcal{P}(\mathbb{N})$ by setting $hA$ equal to the smallest set containing $A$ that is closed under multiplication. In L. F. Meyers' solution to Problem E $1941$ on pp. $408$-$9$ of the April $1968$ American Mathematical Monthly (https://www.jstor.org/stable/2313449) it is shown that the set $A=\{2,3,14,15,25\}$ generates $14$ distinct subsets under $h$ and the complement operator $c$ on $\mathbb{N}.$ An editorial comment at the end then asks:

What is the size of the smallest set having the desired property?

I've never seen this problem mentioned anywhere else, so would guess it is still unsolved.

The "Answer added Nov. 18" section of this answer proves that no Kuratowski 14-set has cardinality $1$ or $2$ in a topological space $X{:}$

https://math.stackexchange.com/a/186428/32209

The proof also works for a general closure operator (any operator that is extensive, idempotent, and isotonic) not just a topological one, so the answer to the question above is $3$, $4$, or $5.$*

Given the power of modern computers, maybe someone can find a tripleton of natural numbers that generates the maximum possible number of $14$ distinct subsets under $h$ and $c$?

*$h$ is clearly extensive $(A\subseteq hA$ for all $A)$, idempotent $(hh=h)$, and isotonic $(A_1\subseteq A_2\implies hA_1\subseteq hA_2).$ It is not a topological closure operator though because, for example, $6\in h\{2,3\}\setminus(h\{2\}\cup h\{3\}).$

---------------- update added 15 Feb 2024, edited 28 Feb 2024 ----------------

1. It is noteworthy that the set $A$ in the Monthly equals $\{p_1,p_2,p_1p_4,p_2p_3,p_3^2\}$ where $p_1,p_2,p_3,p_4$ are the first four primes. We could obviously substitute any four distinct primes and get the same result.

2. In light of the proof above, it may have been necessary for Meyers to use all four primes to distinguish $hjhA$ from $jhA.$

3. Before taking a close look at this question, I thought the answer would turn out to be $3.$ Now I suspect it is $5.$

4. The problem editors were slightly off target when they wrote "Another proof [that $14$ is the maximum number of sets obtainable under $h$ and $c$] is given in J. L. Kelley, General Topology, p. $57.$" because it ignores the distinction between the (topological) closure operator in Kelley and the more general one here.

5. I just cross-posted this question here with some additional info:

https://stackoverflow.com/questions/78077889/how-many-natural-numbers-are-needed-to-generate-14-distinct-subsets-under-comple

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    $\begingroup$ You need to only find certain $7$ of those distinct sets, right. The rest will be found already. (potentially minimizes computing power needed) $\endgroup$
    – Jakobian
    Feb 13 at 20:30

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