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From one set, you can generate infinitely many sets.

Let A be a closed set of infinite Cantor-Bendixon rank. 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. Similarly, B++ = cl(B+) - B+ consists of all elements having odd rank in B+, which is exactly those elements of A having even rank at least 2. And so on. The set B+ⁿ will consist of all elements of A having rank at least n, 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.

From one set, you can generate infinitely many sets.

Let A be a closed set of infinite Cantor-Bendixon rank. 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.

I think that I can prove that from one set, you can generate infinitely many sets.

Let A be a closed set of infinite Cantor-Bendixon rank. 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. Similarly, B++ = cl(B+) - B+ consists of all elements having odd rank in B+, which is exactly those elements of A having even rank at least 2. And so on. The set B+ⁿ will consist of all elements of A having rank at least n, which have even/odd rank depending on the parity of n. Since A has infinite rank, these sets will all be distinct.

From one set, you can generate infinitely many sets.

Let A be a closed set of infinite Cantor-Bendixon rank. 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. Similarly, B++ = cl(B+) - B+ consists of all elements having odd rank in B+, which is exactly those elements of A having even rank at least 2. And so on. The set B+ⁿ will consist of all elements of A having rank at least n, 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.

You seem just to be generating the Boolean subalgebra of the power set generated by your family. If you start with one nonempty non-everything set, you will get 4 sets: the set, its complement, the empty set and the whole space. In general, if you start with a finite collection of sets, then form all possible intersections, to generate the smallest possible nonempty sets, which are pairwise-disjoint. Throw in the complement of the union of them, and you have in effect a collection of n atoms. You can now generate 2ⁿ sets by taking the union of any of these atoms, and this collection is closed under union, intersection and complement, and includes the original sets. So a finite set will generate a finite set this way. This problem has nothing to do with the reals and works with subsets of any set. You have a collection of subsets of a given set X. The power set P(X) is a Boolean algebra with the operations of union, intersection and complement. Your question is asking about the finitely generated subalgebras of P(X). But every finitely genereated Boolean algebra is isomorphic to a finite power set (of the atoms), and has size 2ⁿ, where n is the number of atoms.

I think that I can prove that from one set, you can generate infinitely many sets.

Let A be a closed set of infinite Cantor-Bendixon rank. 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. Similarly, B++ = cl(B+) - B+ consists of all elements having odd rank in B+, which is exactly those elements of A having even rank at least 2. And so on. The set B+ⁿ will consist of all elements of A having rank at least n, which have even/odd rank depending on the parity of n. Since A has infinite rank, these sets will all be distinct.