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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)

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2 Answers

up vote 2 down vote accepted

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 $Z \setminus B \subseteq \mathrm{Int}_Y ( Y \setminus ( A \cup B ) )$.

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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).

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