### Background


Let $A$ be a subset of a topological space $X$. An old problem asks, by applying various combinations of closure and complement operations, how many distinct subsets of $X$ can you describe?


The answer is 14, which follows from the observations that $Cl(Cl(A))=Cl(A)$, $\neg(\neg (A)=A$ and the slightly harder fact that
$$ Cl(\neg (Cl(\neg (Cl(\neg (Cl(A))))=Cl(\neg (Cl(A))$$
where $Cl(A)$ is the closure of $A$ and $\neg (A)$ is the complement. This makes every expression in $Cl$ and $\neg$ equivalent to one of 14 possible expressions, and all that remains is to produce a specific choice of $A$ which makes all 14 possiblities distinct.
This problem goes by the name [Kuratowski's closure-complement problem](http://en.wikipedia.org/wiki/Kuratowski%2527s_closure-complement_problem), since it was first stated and solved by Kuratowski in 1922.


### The Problem


A very similar problem recently came up in a discussion that was based on a topological model for modal logic (though the logical connection is unrelated to the basic question). The idea was to take a subset $A$ of $\mathbb{R}$, and to consider all possible expressions on $A$ consisting of closure, complement, and intersection. To be clear, we are allowed to take the complement or closure of any subset we have already constructed, and we are allowed to intersect any two subsets we have already constructed.


The question: **Is this collection of subsets always finite?**


A potentially harder question: **If there are multiple starting subsets $A\_1$, $A\_2$, $A\_3$... (finite in number), is this collection of subsets always finite?**


The first question is essentially the Kuratowski question, with the added operation of intersection. There is also the closely related (but slightly stronger) question of whether there are a finite number of *formally* distinct expressions on an indeterminant subset $A$ (or a collection of subsets $A\_1$, $A\_2$...).


### Some Thoughts


My guess to both questions is yes, but the trick is showing it. I can take an example of a set $A$ which realizes all 14 possibilities from Kuratowski's problem, and show that the collection of distinct subsets I can construct from it is finite. However, just because such a set captures all the interesting phenomena which can happen when closing and complementing, doesn't mean this example is missing a property that is only important when intersecting.


It also seems difficult to approach this problem formally. The problem is that there are many non-trivial intersections of the 14 expressions coming from Kuratowski's theorem. Then, each of these intersections could potentially have its own new set of 14 possible expressions using closures and complements. In examples, the intersections don't contribute the full number of 14 new sets, but its hard to show this aside from case-by-case analysis.



The Closure-Complement-Intersection Problem
From one set, you can generate infinitely many sets. 


Let A be a closed set of infinite [Cantor-Bendixon rank](http://en.wikipedia.org/wiki/Derived_set_(mathematics)). That is, the successive finite Cantor-Bendixon derivatives A', A'', and so on are all distinct. The rank of an element x in A is the least n such that x becomes isolated in An, where A0 = A and An+1 = A', the set of limit points of An. Thus, the rank 0 elements are the isolated points of A, and the rank 1 elements are the limits of these points which are not limits of limit points of A, and so on.


Now, let B be the subset of A consisting of the elements of A having even rank. This includes all isolated points of A, but not their limits that are not limits of limits, but does include the limits of limits (as long as they are not limits-of-limits-of-limits) and so on. Define the operation B+ = cl(B) - B, which is obtainable from your operations. Note that cl(B) = A, and that B+ consists of all elements of A having odd rank in A. Similarly, B++ = cl(B+) - B+ consists of all elements having odd rank in cl(B+) = A', which is exactly those elements of A having even rank at least 2 in A. And so on. The set B+n will consist of all elements of A having rank at least n+1, which have even/odd rank depending on the parity of n. Since A has infinite rank, these sets will all be distinct.


Thus, the set B generates infinitely many distinct sets B, B+, B++, B+++, etc. 


