TITLE: What is the smallest cardinality of a Kuratowski 14-set?
QUESTION [18 upvotes]: A subset $A$ of a topological space $X$ is called a Kuratowski 14-set if exactly 14 different sets (including $A$) can be obtained from $A$ by alternately taking closures and complements.
Are there any examples of 14-sets having finite cardinality? In a finite space, so far the best I can produce is 10 sets from a 3-element subset.
What is the smallest possible cardinality of a 14-set?

REPLY [3 votes]: I guess there is/was some demand for this answer to be undeleted.  Oh well, here it is un-spoilered.

There are no Kuratowski 14-sets of size 2.  
Suppose that $A = \{ a , b \}$ is a Kuratowski 14-set in some topological space $X$.  Note that, in particular, the following seven sets are distinct:  

$A$;  
$\overline{ A }$;  
$\mathop{Int} ( A )$;  
$\mathop{Int} ( \overline{ A } )$;  
$\overline{ \mathop{Int} ( A ) }$;  
$\overline{ \mathop{Int} ( \overline{ A } ) }$; and  
$\mathop{Int} ( \overline{ \mathop{Int} ( A ) } )$.  

(This comes from the fact that $\mathop{Int} ( B ) = X \setminus \overline{ X \setminus B }$ for all $B \subseteq X$.  In fact, these seven sets being distinct is equivalent to $A$ being a Kuratowski 14-set -- as the other seven are just the complements of these -- but that is of little importance for the moment.)  
Note that as $\mathop{Int} ( A ) \subsetneqq A$, it follows that $\mathop{Int} ( A )$ is one of $\emptyset$, $\{ a \}$, or $\{ b \}$. Since $\overline{ \mathop{Int} ( A ) } \neq \mathop{Int} ( A )$, it follows that $\mathop{Int} ( A )$ must be nonempty, and therefore without loss of generality we may assume that $\mathop{Int} ( A ) = \{ a \}$.  
Claim: $A \subseteq \mathop{Int} ( \overline{ A } )$.  
Proof of claim:  First note that, clearly, $\{ a \} = \mathop{Int} ( A ) \subseteq \mathop{Int} ( \overline{ A } )$.  
Next note that $\mathop{Int} ( \overline{ \mathop{Int} ( A ) } ) \subsetneqq \mathop{Int} ( \overline{ A } )$.  From this it follows that $U = \mathop{Int} ( \overline{ A } ) \setminus \overline{ \mathop{Int} ( A ) }$ is a nonempty open set.  (Clearly $U$ is open.  If $U = \emptyset$, then $\mathop{Int} ( \overline{ A } ) \subseteq \overline{ \mathop{Int} ( A ) }$, and so $\mathop{Int} ( \overline{ A } ) \subseteq \mathop{Int} ( \overline{ \mathop{Int} ( A ) } )$, contradicting the fact above.)  Trivially, $U \subseteq \overline{ A } = \overline{ \{ a , b \} } = \overline{ \{ a \} } \cup \overline{ \{ b \} }$, but is disjoint from $\overline{ \mathop{Int} ( A ) } = \overline{ \{ a \} }$, and so it must be that $U \subseteq \overline{ \{ b \} }$.  So then $b \in U \subseteq \mathop{Int} ( \overline{ A } )$ $\dashv$  
Note that we now have that $$\overline{ A } \subseteq \overline{ \mathop{Int} ( \overline{ A } ) } \subseteq \overline{ ( \overline{A} ) } = \overline{A},$$ contradicting our assumption that $A$ is a Kuratowski 14-set!