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.

Are there any examples of 14-sets having finite cardinality? In a finite space, so far the best I can produce is 10 sets from a 3-element subset.

What is the smallest possible cardinality of a 14-set?

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    $\begingroup$ Perhaps you are thinking of exercise 21 page 102 in "Topology" 2nd ed. by Munkres. In part a) we are asked to prove Kuratowski's original result. If I am not mistaken an answer for b) is given by the set $\{0\}\cup(1,2)\cup(2,3)\cup(Q\cap(3, 4))$. The question I posted here was motivated precisely by this part b). I would like to find the smallest possible set that generates 14 differents sets. $\endgroup$
    – John
    Aug 23, 2012 at 22:11
  • $\begingroup$ You may find of interest the following 36 page survey The Kuratowski Closure-Complement Theorem by B.J. Gardner and M. Jackson, 2007. $\endgroup$ Aug 27, 2012 at 17:14
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    $\begingroup$ Well, for two days now I've been hemming and hawing over posting my solution to the 2-element Kuratowski 14-set question. I actually did post it, and deleted almost immediately after (but I guess the 10K+ folks here know that). If there is enough demand for it, I can undelete it, but otherwise I have convinced myself (finally) to wait until November. $\endgroup$
    – user642796
    Aug 28, 2012 at 7:29
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    $\begingroup$ @Arthur I made the same decision when I read the proposer's answer below. $\endgroup$
    – John
    Aug 28, 2012 at 7:59
  • $\begingroup$ @John: I'm very glad to hear you have a solution. My mind is completely at ease now. $\endgroup$
    – user642796
    Aug 28, 2012 at 8:05

4 Answers 4


This is Problem 1898 in the latest (June) issue of Mathematics Magazine:

"A subset E of a topological space X is called a 'Kuratowski 14-set' if 14 distinct sets can be obtained by repeatedly applying closure and complement to E in some order. It is known that Kuratowski 14-sets E with |E| = 3 exist. Do any exist with |E| < 3?"

Solutions can be emailed to [email protected]. The acceptance deadline is November 1st for possible publication in the journal.

Given this problem's status as an open journal problem, I feel constrained in saying much at all about its solution. I am the one who proposed it, so I do have a proof.

As a minor, indirect hint, I will at least say that Herda and Metzler's paper (referred to in Gerry Myerson's answer) is one of the very best places to look for ideas.

------------- post-deadline, complete answer added 6 Nov 2012 -------------

Here is my proof:

The answer is no. Let $cE$ denote the complement of $E$, $iE$ the interior of $E$, and $kE$ the closure of $E.$ Let $K(E)$ denote the family of sets generated by $E$ under $k$ and $c.$ Since $i = ckc$ and $k = cic$, $K(E)$ equals the family generated by $E$ under $i$ and $c.$ It is well known that $$K(E)=\{E,iE,ciE,iciE,ciciE,iciciE,ciciciE,cE,icE,cicE,icicE,cicicE,icicicE,cicicicE\},$$ where some of the sets in the list may be equal to one another.

Assume $E=\{a,b\}\subset X$ with $a\neq b.$ Claim |$K(E)$| $\leq12.$ This is obvious when $iE=E$ or $iE=\emptyset$, so we may assume without loss of generality that $iE=\{a\}.$ Note that $a\in icicE$, since $iE\subset ikE=icicE.$

Suppose $b\in icicE.$ Then $E\subset icicE.$ Applying $ic$ to both sides produces $icicicE\subset icE.$ The reverse containment always holds, so $icicicE=icE.$ Since |$K(E)$| is always even, the claim follows.

Suppose $b\not\in icicE.$ Assume there exists a point $x$ such that $x\in icicE\cap iciE.$ Since $x\in icicE$, there exists an open set $U$ such that $x\in U\subset cicE.$ Since $x\in iciE$, there exists an open set $V$ such that $x\in V\subset ciE\ (=X\setminus\{a\}).$ Since $b\not\in icicE$ it follows that $b\not\in U.$ Thus $x\in U\cap V\subset cE$, which contradicts $x\in cicE.$ This contradiction implies $icicE\subset ciciE.$ Applying $i$ to both sides produces $icicE\subset iciciE.$ The reverse containment always holds, so $iciciE=icicE.$ This proves the claim.

The claim obviously implies no singleton can generate 14 sets under closure and complement (since "cloning" its element would produce a 14-set of cardinality 2). The empty seed set generates only two sets. Therefore no Kuratowski 14-sets exist with cardinality less than 3.

It turns out that 12 sets are possible when starting with a 2-point seed set. This can be done in a 6-point space, with the number 6 being minimal by results in Anusiak and Shum, Remarks on finite topological spaces, Colloq. Math. 23 (1971), 217–223, MR0326635 (48 #4978).

Lately I have been working on a free reference website aimed at indexing anything-and-everything Kuratowski 14. Though the site is still in a very unfinished state, I feel I have enough to do the launch less than two weeks from now. (I will mention the URL here immediately upon launch.)

The page's main attractions at launch are going to be (a) a useful C program that generates set families under various operations in various spaces, (b) several English translations of major papers (including Kuratowski's 1922 groundbreaker, originally in French), and (c) references to some "new" old papers I came across during some deep Google- and Bing-searching.

As before, I shall conclude with a hint to a question. This time the question is "What is the title of my page going to be?" Here's the hint: a big motivator for its imminent launch is its alliterative, [North American] holiday-themed title. (If you come up with any guess at all, it's almost certainly correct...)

Talk again soon!

------------- post-launch, new website link added 14 Nov 2012 -------------

The title of my page is Kuratowski's Closure-Complement Cornucopia. Here is the link:


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    $\begingroup$ This sort of situation interested me, so I opened up a meta question on how we should react to ongoing journal questions. $\endgroup$
    – davidlowryduda
    Aug 26, 2012 at 1:34

Herda and Metzler, Closure and interior in finite topological spaces, Colloq. Math. 15 (1966) 211–216, MR0202101 (34 #1975) prove that you can do it with 7 points, and not with any fewer. The result can also be found in Berman and Jordan, The Kuratowski closure-complement problem, Amer. Math. Monthly 82 (1975), no. 8, 841–842, MR0388305 (52 #9142).

EDIT: The 7 refers to the number of elements of $X$; examples are given where the set $A$, the actual 14-set, has only 3 elements.

  • $\begingroup$ The second article seems to show that $|A|=3$ is possible. Do you know by any chance if $|A|=2$ can be achieved? $\endgroup$
    – Dejan Govc
    Aug 24, 2012 at 1:08
  • $\begingroup$ @Dejan, are you saying there's a 3-point 14-set? $\endgroup$ Aug 24, 2012 at 1:55
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    $\begingroup$ @Dejan, I see what you mean: they give a 3-point 14-set in a 7-point topological space. No, I don't know whether a 2-point 14-set is possible, but I haven't gone through the papers in detail. Maybe the methods in the papers are powerful enough to answer your question. $\endgroup$ Aug 24, 2012 at 2:01
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    $\begingroup$ @Gerry, I finally had a chance to get my hands on Herda and Metzler´s paper that you mention. I may be wrong, but it seems to me that Theorem 2 is not quite complete. I think that there are 6 topologies, not 5. They seem to have forgotten to insert a 4 into the set $\{1,2,6,7\}$ formed by taking the union of $\{1,2,7\}$ on the left 2nd row from the bottom, and $\{1,6,7\}$ on the right, in Figure 3. $\endgroup$
    – John
    Aug 29, 2012 at 18:41
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    $\begingroup$ @John, good catch: Herda and Metzler did overlook this case. Their omission was pointed out in a footnote on p. 220 of Anusiak and Shum, Remarks on Finite Topological Spaces, Colloq. Math. 23 (1971) 217-223. $\endgroup$ Sep 1, 2012 at 23:41

As I point out in a comment, these questions came to mind while I was doing a related exercise in Munkres' "Topology". I am pleased to see that it is a problem in the Magazine!

I will give here a partial answer that does not affect the problem in the magazine.

Statement: For any given $n\ge 3$ there is a finite topological space containing a 14-set of cardinality $n$.

Proof: For $n\ge 4$, let $m=n+3$, the point set $X=\{1,2,3,\ldots, m\}$, and $\mathcal{B}=\{\emptyset, X, \{1\}, \{6\}, \{1,2\}, \{3,4\}, \{5,6\}\}$ as the basis for the topology on $X$. Then the subset $A=\{1,3,5\}$ is a 14-set of cardinality 3. Therefore, $A'$ (the complement of $A$ in $X$) is a 14-set of cardinality $m-3=n$, as desired.

For $n=3$, let $m=7$ in the previous setting and $A$ is the desired 14-set.

Moreover, we can obtain 14-sets of any desired infinite cardinality as follows: Let $Y$ be a set of cardinality $k\ge \aleph_{0}$, and let $X=\{1,2,3,4,5,6,7\}\cup Y$ with basis for the topology $\mathcal{B}$ as before. Then the set $A$ is a 3-element 14-set in $X$, and $A'$ is a 14-set of cardinality $k$. For example, if $Y=\mathbb{N}$ then $A'$ is a countably infinite 14-set.

Answer added Nov. 18:

The smallest possible cardinality of a Kuratowski 14-set $E$ is $3$. Let $k$, and $i$ denote closure and interior, respectively. Then $E$ is a Kuratowski 14-set iff $E$, $E^{i}$, $E^{iki}$, $E^{ki}$, $E^{ik}$, $E^{kik}$, and $E^{k}$ are all non-empty, distinct sets strictly contained in $X$. If $E$ is empty the answer is trivial. If $|E|=1$ then $E^{i}$ equals either $E$ or is empty. If $E=\{a,b\}$ then we may assume that $E^{i}=\{a\}$. But then $a\notin B=E^{ki}-E^{ik}$, which is an open set contained in $E^{k}$ and so it must intersect $E$ non-trivially. Hence, $b\in B$ from which we obtain $E^{kik}=E^{k}$.

Remark: Since a Kuratowski 14-set $E$ of cardinality 2 or less does not exist, we may ask what is the maximum number of different sets obtainable from $E$ by taking closures and complements.

If $|E|=1$ we have two possibilities:

First, if $E^{i}=E$ then $E^{ik}=E^{kik}=E^{k}$ and $E^{iki}=E^{ki}$. Therefore, at most 6 distinct sets can be obtained from $E$ by taking closures and complements. This bound is possible if $|X|\ge 3$. For example, $X=\{1,2,3\}$ with basis $\{\{3\},X\}$ and $E=\{1\}$.

Second, if $E^{i}=\emptyset$ then $E^{ik}=E^{iki}=\emptyset$ and $E^{kik}=E^{k}$. Therefore, at most $8$ distinct sets can be obtained from $E$ by taking closures and complements. This bound is possible if $|X|\ge 4$. For example, $X=\{1,2,3,4\}$ with basis $\{\{4\},\{1,2\},X\}$ and $E=\{1\}$.

If $E=\{a,b\}$ we may assume that $E^{i}=\{a\}$. But then $a\notin B=E^{ki}-E^{ik}$, which is an open set contained in $E^{k}$ and so it must intersect $E$ non-trivially. Hence, $b\in B$ from which we obtain $E^{kik}=E^{k}$. Therefore, at most 12 distinct sets can be obtained from $E$ by taking closures and complements. This bound is possible if $|X|\ge 6$. For example $X=\{1,2,3,4,5,6\}$ with basis $\{\{1\},\{6\},\{1,2\},\{4,5\},X\}$ and $E=\{1,4\}$.

  • $\begingroup$ All nice proofs. The sentence "Hence, $b\in B$ from which we obtain $E^{kik}=E^k$." follows since $E^i$ is contained in each of the other six Kuratowski sets including $E^{ki}$. Hence $E\subset E^{ki}$, which gives $E^k\subset E^{kik}$. $\endgroup$ Nov 21, 2012 at 16:09

I guess there is/was some demand for this answer to be undeleted. Oh well, here it is un-spoilered.

There are no Kuratowski 14-sets of size 2.

Suppose that $A = \{ a , b \}$ is a Kuratowski 14-set in some topological space $X$. Note that, in particular, the following seven sets are distinct:

  1. $A$;
  2. $\overline{ A }$;
  3. $\mathop{Int} ( A )$;
  4. $\mathop{Int} ( \overline{ A } )$;
  5. $\overline{ \mathop{Int} ( A ) }$;
  6. $\overline{ \mathop{Int} ( \overline{ A } ) }$; and
  7. $\mathop{Int} ( \overline{ \mathop{Int} ( A ) } )$.

(This comes from the fact that $\mathop{Int} ( B ) = X \setminus \overline{ X \setminus B }$ for all $B \subseteq X$. In fact, these seven sets being distinct is equivalent to $A$ being a Kuratowski 14-set -- as the other seven are just the complements of these -- but that is of little importance for the moment.)

Note that as $\mathop{Int} ( A ) \subsetneqq A$, it follows that $\mathop{Int} ( A )$ is one of $\emptyset$, $\{ a \}$, or $\{ b \}$. Since $\overline{ \mathop{Int} ( A ) } \neq \mathop{Int} ( A )$, it follows that $\mathop{Int} ( A )$ must be nonempty, and therefore without loss of generality we may assume that $\mathop{Int} ( A ) = \{ a \}$.

Claim: $A \subseteq \mathop{Int} ( \overline{ A } )$.

Proof of claim: First note that, clearly, $\{ a \} = \mathop{Int} ( A ) \subseteq \mathop{Int} ( \overline{ A } )$.

Next note that $\mathop{Int} ( \overline{ \mathop{Int} ( A ) } ) \subsetneqq \mathop{Int} ( \overline{ A } )$. From this it follows that $U = \mathop{Int} ( \overline{ A } ) \setminus \overline{ \mathop{Int} ( A ) }$ is a nonempty open set. (Clearly $U$ is open. If $U = \emptyset$, then $\mathop{Int} ( \overline{ A } ) \subseteq \overline{ \mathop{Int} ( A ) }$, and so $\mathop{Int} ( \overline{ A } ) \subseteq \mathop{Int} ( \overline{ \mathop{Int} ( A ) } )$, contradicting the fact above.) Trivially, $U \subseteq \overline{ A } = \overline{ \{ a , b \} } = \overline{ \{ a \} } \cup \overline{ \{ b \} }$, but is disjoint from $\overline{ \mathop{Int} ( A ) } = \overline{ \{ a \} }$, and so it must be that $U \subseteq \overline{ \{ b \} }$. So then $b \in U \subseteq \mathop{Int} ( \overline{ A } )$ $\dashv$

Note that we now have that $$\overline{ A } \subseteq \overline{ \mathop{Int} ( \overline{ A } ) } \subseteq \overline{ ( \overline{A} ) } = \overline{A},$$ contradicting our assumption that $A$ is a Kuratowski 14-set!

  • $\begingroup$ I can't understand how it is concluded in the 'Proof of claim:' that $U \subseteq \overline{ \{ b \} }$ implies $b \in U$. Will you please explain? $\endgroup$
    – Silent
    Oct 28, 2017 at 21:06
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    $\begingroup$ @Silent since $U$ is open we have $\big(b\not\in U\big)\Longleftrightarrow\big(\{b\}\cap U=\varnothing\big)\Longrightarrow\big(\overline{\{b\}}\subset X\setminus U\big)\Longleftrightarrow\big(\overline{\{b\}}\cap U=\varnothing\big)\Longrightarrow\big(U\not\subseteq\overline{\{b\}}\big)$. Hence $U\subseteq\overline{\{b\}}$ implies $b\in U$. $\endgroup$ Oct 29, 2017 at 13:09
  • $\begingroup$ You're welcome. Addendum: even though it was clear from the context, I should have explicitly said "since $U$ is nonempty and open" (since the last implication requires that $U$ be nonempty). $\endgroup$ Oct 29, 2017 at 14:12

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