Kuratowski closure-interior families in a given topological space Suppose $X$ is a topological space containing a subset $A$ such that
$$\tag{$*$}iA=ikA=ikiA\subsetneq kiA=kikA\subsetneq kA=A$$
where $k$ is closure and $i$ is interior. That is, $A$ satisfies the following Hasse diagram, where sets in a given diagram are equal iff they have the same color:

Must $X$ then contain a subset $B$ satisfying this diagram?

(The set $B$ is required to satisfy every relation in $(*)$ except the last one and not equal any of the other sets in the Hasse diagram.)
Labeling the entries in the table on page 21 of Gardner and Jackson from 1 to 30 in order from top to bottom, let the label associated with a given subset be called its Kuratowski character.
Using this nomenclature, the question above reduces to:
If a topological space $X$ contains a subset with Kuratowski character 23, must it then also contain a subset with Kuratowski character 14?
I recently posted a similar question here. Both questions are motivated by this overarching question.
 A: The answer is yes. Let $x\in kiA\setminus iA,$ $B=A\setminus\{x\},$ and $c$ denote complement. Since $x\in kiA,$ every neighborhood of $x$ intersects $iA$. Hence $x$ has no neighborhood contained in $cA\cup\{x\}.$ Thus $cA\cup\{x\}$ is not open. Therefore $A\setminus\{x\}$ is not closed. Thus we have $A\setminus\{x\}\subsetneq k(A\setminus\{x\})\subset kA=A,$ which implies $k(A\setminus\{x\})=A.$ Hence $$\tag1B\subsetneq kB=A.$$ The inclusion $\{x\}\subset ciA$ implies $iA\subset c(\{x\}),$ thus $iA\subset ic(\{x\}).$ Hence $$\tag2iB=i(A\setminus\{x\})=iA\cap ic(\{x\})=iA.$$ By $(1)$ we have $$\tag3ikB=iA.$$ Thus $kikB=kiA.$ By $(2),$ $kiB=kiA.$ Thus $$\tag4kiB=kikB.$$ By $(2)$ we also have $$\tag5ikiB=ikiA=iA.$$ Thus, $(2),$ $(3),$ and $(5)$ imply $$\tag6iB=ikB=ikiB.$$ Since $oB=oA$ for $o\in\{iki,ki,kik,k\},$ the inequalities $ikiA\subsetneq kiA$ and $kikA\subsetneq kA$ are also satisfied by $B.$ Hence $$\tag7iB=ikB=ikiB\subsetneq kiB=kikB\subsetneq kB.$$ It remains to show that $B\neq iB$ and $B\neq kiB.$ Since $x\in kiA=kiB$ and $x\not\in B,$ we get $B\neq kiB.$ Since $iA\subsetneq kiA\subsetneq kA=A$ it follows that $|A\setminus iB|=|A\setminus iA|\geq2.$ But $|A\setminus B|=1,$ hence $B\neq iB.$ $\blacksquare$
============ added May 9 2019 =============
It's worth noting that the converse also holds. Given $B$, we apply the general identities $kkE=kE$ and $ikikE=ikE$ to verify that $kB$ satisfies $(*){:}$ $$i(kB)=ik(kB)=iki(kB)\subsetneq ki(kB)=kik(kB)\subsetneq k(kB)=kB.$$ Thus $X$ contains $A$ iff $X$ contains $B$.
A: The answer is yes. The family $\mathcal O=\{\emptyset,\{a\},\{b\},\{a,b\},\{a,c\},\{a,b,c\},X\}$ is a topology on the set $X=\{a,b,c,d\}$. For set $A$ take $A=\{b,c,d\}$: $$ i(A)=iki(A)=ik(A)=\{b\}, A=k(A)=\{b,c,d\}, ki(A)=kik(A)=\{b,d\}.$$ For set $B$ take $B=\{b,c\}$: $$ i(B)=iki(B)=ik(B)=\{b\}, k(B)=\{b,c,d\}, ki(A)=kik(A)=\{b,d\}.$$
