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:

Hasse diagrams Kuratowski closure complement 14 sets

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

Hasse diagrams Kuratowski closure complement 14 sets

(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.


2 Answers 2


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 $B$.

  • $\begingroup$ The given inequality for cardinality at the end works only when set $A$ is finite? $\endgroup$
    – Viki 183
    Apr 29, 2019 at 21:21
  • $\begingroup$ @Viki183 Thanks for pointing that out. I adjusted the cardinalities to make the argument apply to all spaces. $\endgroup$ Apr 29, 2019 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 $B=\{b,c\}$: $$ i(B)=iki(B)=ik(B)=\{b\}, k(B)=\{b,c,d\}, ki(A)=kik(A)=\{b,d\}.$$

  • $\begingroup$ 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. $\endgroup$ Apr 27, 2019 at 20:46
  • $\begingroup$ 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? $\endgroup$
    – Viki 183
    Apr 27, 2019 at 21:03
  • $\begingroup$ 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. $\endgroup$ Apr 27, 2019 at 21:38
  • $\begingroup$ 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? $\endgroup$ Apr 27, 2019 at 22:01
  • $\begingroup$ 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. $\endgroup$ Apr 29, 2019 at 20:58

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