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.

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 $$B$$.

• The given inequality for cardinality at the end works only when set $A$ is finite? Apr 29 '19 at 21:21
• @Viki183 Thanks for pointing that out. I adjusted the cardinalities to make the argument apply to all spaces. Apr 29 '19 at 21:41

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 $$B=\{b,c\}$$: $$i(B)=iki(B)=ik(B)=\{b\}, k(B)=\{b,c,d\}, ki(A)=kik(A)=\{b,d\}.$$

• Your example does not suffice to prove that in general the existence of $A$ in a space implies the existence of $B$ in that same space, which is what the question is asking for. Apr 27 '19 at 20:46
• Ok, now I understood what kind of proof you are looking for. Did you try to prove this result for special types of spaces e.g. for open irresolvable spaces? Apr 27 '19 at 21:03
• It is easy to show that any space containing a subset with Kuratowski character 23 has to either be what GJ call "open unresolvable" (OU) or a Kuratowski space. According to computer evidence, in finite spaces you can always remove a certain point from $A$ to get $B$. In fact, it's the same singleton for every $A$. There may be an expression for that singleton that will prove the existence of $B$ in general. I looked briefly but didn't find one. Apr 27 '19 at 21:38
• The singleton that works is always closed, as $\{d\}$ is in your example. So the question is, what is it about $\{d\}$ in your space that guaranteed $A\setminus\{d\}$ would have Kuratowski character 14? Apr 27 '19 at 22:01
• As the proof in my answer shows, the key property is $\{d\}\subset kiA\setminus iA.$ By the way, scratch what I wrote about it being "the same singleton for every $A.$" I had only looked at a few spaces when I wrote that. It's not true. Apr 29 '19 at 20:58