Does the interior of a Kuratowski 14-set in a finite space always have cardinality 1? 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.
Let $c$ denote complement and $i$ interior: If $X$ is finite and $A$ is a Kuratowski 14 set, is it always true that $|A^{i}|=|A^{ci}|=1$?
More generally: if $X$ is any topological space and $A$ is a finite Kuratowski 14-set, is always true that $|A^{i}|$ and $|A^{ci}| are equal to 1?
EDIT: The asnwer is affirmative only when $|X|=7$. What if we relax the condition and request that only one of $|A^{i}|$ or $|A^{ci}|$ is equal to 1?
(There is a related question)
 A: The answer is negative.
Let $X=\{1,2,3,4,5,6,7,8\}$ with base of topology $\mathcal B=\{\{1\},\{1,2\},\{4,5\},\{6,7\},\{7\},\{8\},X\}$ and $A=\{1,4,6,8\}$. Then we get the following sets ($k$ - closure, $c$ - complement):
$$\begin{align}A &= \{1,4,6,8\}\\
cA &= \{2,3,5,7\}\\
kcA &= \{2,3,4,5,6,7\}\\
ckcA &= \{1,8\}\\
kckcA &= \{1,2,3,8\}\\
ckckcA &= \{4,5,6,7\}\\
kckckcA &= \{3,4,5,6,7\}\\
ckckckcA &= \{1,2,8\}\\
kA &= \{1,2,3,4,5,6,8\}\\
ckA &= \{7\}\\
kckA &= \{3,6,7\}\\
ckckA &= \{1,2,4,5,8\}\\
kckckA &= \{1,2,3,4,5,8\}\\
ckckckA &= \{6,7\}\end{align}$$
As you see, the interior $ckcA=\{1,8\}$ of $A$ has cardinality $2$. This example was constructed from a standard example by adding the element $8$ to the space, to the base of topology and to the $14$-set. We may construct examples where $ckcA$ has arbitrary  non-zero cardinality by instead adding some other set of new elements and proclaiming it to be open (i.e. adding it to the base).
A: Suppose that $A$ is a (finite) Kuratowski 14-set in some topological space $X$.  Let $Z$ be any collection of objects not contained in $X$.  Set $Y = X \cup Z$, and topologise $Y$ by declaring the points of $Z$ to be isolated, and the open subsets of $X$ to be open in $Y$.  (Y is the topological sum of $X$ and the discrete topology on $Z$.)
Note that for $B \subseteq Y$ we have that \begin{gather}
\mathrm{Int}_Y ( B ) = ( B \cap Z ) \cup \mathrm{Int}_X ( B \setminus Z ); \\
\mathrm{cl}_Y ( B ) = ( B \cap Z ) \cup \mathrm{cl}_X ( B \setminus Z ).
\end{gather}
From this observation, it is easy to show that $A \cup Z$ is a Kuratowski 14-set in $Y$, and $Z \subseteq \mathrm{Int}_Y (A \cup Z)$.
In fact, given any $B \subseteq Z$ we have that $A \cup B$ is a Kuratowski 14-set in $Y$,  $B \subseteq \mathrm{Int}_Y (A \cup B)$ and $Z \setminus B \subseteq \mathrm{Int}_Y ( Y \setminus ( A \cup B ) )$.
