Combining Interior, Complement, and Closure Operators

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We have three natural operations on a subset of a space with a topology: interior, closure, and complement. It is a surprising and interesting fact that the number of distinct sets one may produce using these operations on a subset of a topological space is limited. In fact:

Theorem:

1. (Kuratowski, 1922) For any given subset A of a topological set, there are, at most, 14 distinct sets that may be constructed by applying any permutation of the interior, closure, and complement operations on A.
2. There exists a subset of the real line, with the standard topology, which yields 14 distinct sets when operated on by every permutation of the interior, closure, and complement operations.

Proof of 1:

We will represent the interior, closure, complement operation as $\circ, c, -,$ respectively. A composition of operations will be written with the left-most operation to be performed first: i.e. '$\circ\; c - c$' corresponds to the operation $A \rightarrow (Cl((Int(A))^c))^c$.

We require the following relationships between the operators:

1. $\circ \; \circ = \circ$
2. $- \; - = -$
3. $c \; c = id$
4. $c \; - = \circ \; c$
5. $c \; \circ = - \; c$
6. $- \; \circ \; - \; \circ = - \; \circ$

The proof of the first three relations are trivial. The last three are left as an exercise (hint: it may help to begin by proving that $\circ = c - c$). With these relationships, we may easily reach our result. Consider the below state transition diagram. Each grey box represents a set. If the box is labelled, say, XYZ, the box represents the set A with these operators applied to it: Z(Y(X(A))). Following an arrow labelled X from a state box A leads to another state box which represents the image of the set represented by A under the operator X. A set A upon which no operations have been applied (or alternatively, the complement has been applied twice) starts in state 'nil'. If for any of the three operators under consideration, a state box does not have an arrow leading out labelled as such, it is implied that the operator does not change that set.

A verification of the proof involves checking each arrow (and missing, implied arrow) to ensure that the diagrammatically indicated relationship between the sets at each end of the arrow holds. Only the above listed relationships are needed for this.

There are only 14 states in the diagram and an operator on any one of them again yields one of these 14. Therefore, there are at most 14 distinct sets that may be constructed through the application of the interior, closure, and complement operators to a subset of a topological set.

Proof of 2:

Consider the following set:

$A = [0, 1] \; \cup (2, 3) \; \cup [(4, 5) \cap \mathbb{Q}] \; \cup [(6, 8) \setminus \{7\}] \; \cup \{9\}$

One may check explicitly that the images of A under the 14 operators in the above diagram are 14 distinct sets. The sets are graphically depicted below: