Kuratowski's closure-complement problem

Remarks

The problem is solved: at most $14$ distinct sets arise, counting $A$ itself, and the bound $14$ is attained in $\R$. The interior of $A$ is the complement of the closure of the complement of $A$. With $k$ for closure and $i$ for interior, every set that arises is one of $A, kA, iA, ikA, kiA, kikA, ikiA$ or the complement of one of them. The key identity is $kikik A = kik A$, which stops the list.

In $\R$, the set $$A = (0, 1) \cup (1, 2) \cup \set{3} \cup ([4, 5] \cap \Q)$$ gives $14$ distinct sets. For it, $kA = [0, 2] \cup \set{3} \cup [4, 5]$, $iA = (0, 1) \cup (1, 2)$, $ikA = (0, 2) \cup (4, 5)$, $kiA = [0, 2]$, $kikA = [0, 2] \cup [4, 5]$ and $ikiA = (0, 2)$. These seven sets are distinct and bounded, and their seven complements are unbounded, so all $14$ are distinct.