Kuratowski's closure-complement problem

Letting ${\displaystyle S}$ denote an arbitrary subset of a topological space, write ${\displaystyle kS}$ for the closure of ${\displaystyle S}$, and ${\displaystyle cS}$ for the complement of ${\displaystyle S}$. The following three identities imply that no more than 14 distinct sets are obtainable:

1. ${\displaystyle kkS=kS}$. (The closure operation is idempotent.)
2. ${\displaystyle ccS=S}$. (The complement operation is an involution.)
3. ${\displaystyle kckckckcS=kckcS}$. (Or equivalently ${\displaystyle kckckckS=kckckckccS=kckS}$, using identity (2)).

The first two are trivial. The third follows from the identity ${\displaystyle kikiS=kiS}$ where ${\displaystyle iS}$ is the interior of ${\displaystyle S}$ which is equal to the complement of the closure of the complement of ${\displaystyle S}$, ${\displaystyle iS=ckcS}$. (The operation ${\displaystyle ki=kckc}$ is idempotent.)

A subset realizing the maximum of 14 is called a 14-set. The space of real numbers under the usual topology contains 14-sets. Here is one example:

${\displaystyle (0,1)\cup (1,2)\cup \{3\}\cup {\bigl (}[4,5]\cap \mathbb {Q} {\bigr )},}$

where ${\displaystyle (1,2)}$ denotes an open interval and ${\displaystyle [4,5]}$ denotes a closed interval. Let ${\displaystyle X}$ denote this set. Then the following 14 sets are accessible:

1. ${\displaystyle X}$, the set shown above.
2. ${\displaystyle cX=(-\infty ,0]\cup \{1\}\cup [2,3)\cup (3,4)\cup {\bigl (}(4,5)\setminus \mathbb {Q} {\bigr )}\cup (5,\infty )}$
3. ${\displaystyle kcX=(-\infty ,0]\cup \{1\}\cup [2,\infty )}$
4. ${\displaystyle ckcX=(0,1)\cup (1,2)}$
5. ${\displaystyle kckcX=[0,2]}$
6. ${\displaystyle ckckcX=(-\infty ,0)\cup (2,\infty )}$
7. ${\displaystyle kckckcX=(-\infty ,0]\cup [2,\infty )}$
8. ${\displaystyle ckckckcX=(0,2)}$
9. ${\displaystyle kX=[0,2]\cup \{3\}\cup [4,5]}$
10. ${\displaystyle ckX=(-\infty ,0)\cup (2,3)\cup (3,4)\cup (5,\infty )}$
11. ${\displaystyle kckX=(-\infty ,0]\cup [2,4]\cup [5,\infty )}$
12. ${\displaystyle ckckX=(0,2)\cup (4,5)}$
13. ${\displaystyle kckckX=[0,2]\cup [4,5]}$
14. ${\displaystyle ckckckX=(\infty ,0)\cup (2,4)\cup (5,\infty )}$

Despite its origin within the context of a topological space, Kuratowski's closure-complement problem is actually more algebraic than topological. A surprising abundance of closely related problems and results have appeared since 1960, many of which have little or nothing to do with point-set topology.[3]

The closure-complement operations yield a monoid that can be used to classify topological spaces.[4]

• The Kuratowski Closure-Complement Theorem Archived 2022-02-12 at the Wayback Machine by B. J. Gardner and Marcel Jackson
• The Kuratowski Closure-Complement Problem by Mark Bowron