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Kuratowski 14-Set, Kuratowski’s Theorem

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1. Introduction

The problem of Kuratowski 14-sets was first proposed in 1922 by the famous Polish mathematician Kazimierz Kuratowski who discovered that at most 14 distinct sets can be generated by applying the closure operator and the com- plementation operator on a subset A of a topological space X in any order repeatedly (see [1] ). We now call a subset A in a topological space a K-set if A can generate 14 distinct sets by taking closure and complementation on A in any order. To find an example of a K-set in the real line is posed to many mathematics students all over the world, usually in the class of elementary general topology. The students can easily check that the set $A=\left(\mathrm{0,1}\right)\cup \left(\mathrm{1,2}\right)\cup \left\{3\right\}\cup \left[\mathrm{4,5}\right]\cap Q$ is a K- set (see [2] ). Another interesting example of a K-set in the real line is the set $A=\left\{1/n|n=1,2,3,\cdots \right\}\cup \left(3,4\right)\cup \left\{4,5\right\}\cup \left[6,7\right]\cup \left(7,8\right)\cap Q$ [3] [5] .

Now, we denote the complementation of the set A by $c\left(A\right)$ . As $c\left(c\left(A\right)\right)=A=e\left(A\right)$ , where e is the identity operator, and so $c^2=e$ . We des- cribe this fact by saying that the operator c has the involution property, more- over if $A\subseteq B$ , then $c\left(A\right)\supseteq c\left(B\right)$ . Also,we use $f\left(A\right)$ to denote the closure of the set A. It means the smallest closed set containing A. According to Kura- towski, the closure operator $f$ has the following properties:

i) $f(\varnothing )=\varnothing$ , (the closure of a void set).

ii) $f\left(A\right)\supseteq A$ , (the expansion property).

iii) $f\left(f\left(A\right)\right)=f\left(A\right)$ , (the idempotent property).

iv) $f\left(A\cup B\right)=f\left(A\right)\cup f\left(B\right)$ , (the additivity property).

In point set topology, we usually use $A^{-}$ to denote the closure of a set A. However, for the sake of convenience, in dealing with the algebraic closure of a set A, we simply use $f\left(A\right)$ to denote the closure of the set A. Thus, both $A^{-}$ and $f\left(A\right)$ are used to denote the closure of a set A. Now, we also use the notation $A^{\circ }$ and $i\left(A\right)$ to denote the interior of a set A. For the sake of sim- plicity, We write B as the complement of the set A.It is clear that $i\left(A\right)$ is the complete of $f\left(B\right)$ . With the applications of the algebraic closure operator, It is well known that the interior of the set A is the set $i\left(A\right)=cf\left(B\right)=A^{\circ }$ . Since B is used to denote $c\left(A\right)$ , we now list the family of 14 sets generated by A in Table 1.

In order to illustrate the inclusion relationship of the above 14 sets, we give the following relationship diagram. As the relationship diagram would look quite bulky if we put all the set inclusion symbols in the diagram. Thus, for the sake of simplicity, we use “®” to mean the inclusion relationship in the diagram. This nice relationship diagram was first given in Kuratowski in [1] (Diagram 1).

Diagram 1. The inclusion relationship of the 14 sets.

The papers [3] - [8] are closely related to the Kuratowski’s theorem. The paper of Gardner and Jackson in [9] gave detailed description and information of the Kuratowski 14 set problem. The papers [3] [10] - [23] also studied and investigated the problems which are closely related to the closure and interior operators of a given set.

2. Algebraic Operator and the Algebraic Interior Operators

Throughout this section, we use $f$ and $i$ to denote the algebraic closure operator, respectively. According to Kuratowski [3] , the algebraic closure operator has the following properties:

i) Monotonic property, i.e., if $B\subseteq A$ , then $f\left(B\right)\subseteq f\left(A\right)$ ,

ii) Expansive property, i.e., $A\subseteq f\left(A\right)$ ,

iii) Idempotent property, i.e., $f\left(f\left(A\right)\right)=f\left(A\right),f^2=f$ ,

iv) Additive property, i.e., $f\left(A\cup B\right)=f\left(A\right)\cup f\left(B\right)$ .

On the other hand, the algebraic interior operator $i$ has the following pro- perties:

a) Monotonic property, i.e., if $B\subseteq A$ , then $i\left(B\right)\subseteq i\left(A\right)$ ,

b) Shrinking property, i.e., $i\left(A\right)\subseteq A$ ,

c) Idempotent property, i.e., $i\left(i\left(A\right)\right)=i\left(A\right)$ , i.e., $i^2=i$ ,

d) Additive property, i.e., $i\left(A\cup B\right)=i\left(A\right)\cup i\left(B\right)$ .

Consider $A\subseteq B$ . By using the above properties of algebraic closure operator and the algebraic interior operator, we obtain immediately the following crucial theorem.

Theorem 2.1. Let $f\mathrm{,}i$ be the closure and interior operators. Then we derive the following identities, ${\left(fi\right)}^2=fi$ and ${\left(if\right)}^2=if$ .

Proof. By using the shrinking property of $i$ , we have $ifi\le fi$ . By applying the monotonic property and the idempotent property of $f$ , we can easily deduce the equations ${\left(fi\right)}^2=fifi\le f\left(fi\right)=f^{2}i=fi$ . On the other hand, because $i\le fi$ we have $i=i^2\le ifi$ , consequently, we derive that $fi={\left(fi\right)}^2$ . Recall that $i=cfc$ and $f=cic$ , by acting these two identities in the above equations, we obtain $if={\left(if\right)}^2$ . Thus, our theorem is proved. ,

Corollary 2.2. Because $f\left(A\right)=cif\left(A\right)$ and $i\left(A\right)=cfc\left(A\right)$ . We immediately see in the above Table 1, we have $A_4=A_8$ , $B_8=B_4$ . Since the sets $A_1$ , $A_2$ , $A_3$ , $A_4$ , $A_5$ , $A_6$ , $A_7$ , and $B_1$ , $B_2$ , $B_3$ , $B_4$ , $B_5$ , $B_6$ , $B_7$ are possibly distinct. We therefore conclude that at most 14 distinct sets can be generated by taking closure and complementation operator on the set $A$ , repeatedly in any order.

Let $A\mathrm{,}B$ be subsets of a given set $M$ . Then, by applying the complementation operator $c$ on these two sets. we have the following properties:

a) The inversion property, If $A\subseteq B$ , then $c\left(A\right)\supseteq c\left(B\right)$ , we simply denote this situation by $c\downarrow$ .

b) The involution property, $c\left(c\left(A\right)\right)=A$ , i.e., $c^2=e$ , the identity.

c) The separation property. Since $c\left(A\right)=M-A$ , $c\left(A\right)\cap A=\varnothing$ , we simply denote this case by $c\wedge e=0$ .

Now, we consider the complementation operator c on a bounded lattice $\left(L,\wedge ,\vee ,0,1\right)$ , where 0 is the minimum element of the lattice L and 1 is the greatest element of the lattice L. In general, the bounded lattice can be regarded as a power set of a give set L, if $\alpha \left(x\right)\le \beta \left(x\right)$ , then we write $\alpha \le \beta$ . The $0\in L$ can always be regarded as the zero function. We notice that if $c\uparrow$ , then $c\left(g\wedge h\right)\le cg\wedge ch$ , $c\left(g\vee h\right)\ge cg\vee ch$ . Moreover, if $g\le h$ and $\alpha \uparrow$ , then we always have $\alpha g\le \alpha h$ .

For the complementation operator c, we have the following theorem.

Theorem 2.3. Let $c$ be a complementation operator and $e$ the identity operator on a bounded lattice $\left(L,\wedge ,\vee ,0,1\right)$ . Then $c\left(1\right)=0$ and $c\left(0\right)=1$ .

Proof. The proof is quite easy. For the sake of the interest of the readers, we provide here two proofs.

Proof i). Because $c^2=e$ , c is clearly surjective and injective. Thus, for $0\in L$ , by the inversion property of c, i.e., $c\downarrow$ , we have $0=c\left(x_L\right)\ge c\left(1\right)$ and hence $c\left(1\right)=0$ . Now, by the involution property of c, we deduce that $1=c\left(c\left(1\right)\right)=c\left(0\right)$ .

Proof ii). Suppose that $c\left(x\right)\ne 0$ for any $x\in L$ . Then we have $c\left(c\left(x\right)\right)\ne 0$ . This contradicts to $x=0$ , $c\left(c\left(0\right)\right)=e\left(0\right)=0$ . Hence, there exists $x_L\in L$ such that $c\left(x_L\right)=0$ . Consequently, $0\le x_l\le 1$ , $0=c\left(x_L\right)\ge c\left(1\right)$ . Therefore, $c\left(1\right)=0$ . The proof is now completed.

3. The Boundary of a Subset in a Topological Space

For any subset A in a topological space, the boundary of A is defined by $b\left(A\right)=A^{-}\cap c{\left(A\right)}^{-}$ . Hence, the boundary operator of A is denoted by $b=f\wedge fc$ . Similarly, we can define the rim operator, $r=e\wedge fc$ and the side operator by $s=f\wedge c$ . Obviously, the boundary operator, the rim operator and the side operator are closely related on a distributive lattice. It is noted that we always have $b=r\vee s$ , $e=i\vee r$ . It is obvious that the boundary, the rim and the side of a subset A in a topological space are different concepts. The following example is an example of the above three notations.

Example 3.1. Let $A=\left[2,3\right)$ be a closed-open interval in the real line $R$ . Then according to the definitions of the boundary, the rim and the side of the given set $A$ in a topological space $X$ , we immediately see that $b\left(A\right)=\left\{2,3\right\}$ , $r\left(A\right)=\left\{2\right\}$ and $s\left(A\right)=\left\{3\right\}$ . It is clear that the above three concepts are dis- tinct concepts, in fact, the rim and the side of a subset $A$ in $X$ are parts of the boundary of $b\left(A\right)$ .

It is noteworthy that the sequence $s\mathrm{,}{s}^2\mathrm{,}{s}^3\mathrm{,}\cdots \mathrm{,}{s}^n\mathrm{,}\cdots$ of powers of s is sometimes infinite(for $i\ne j$ , $s^i\ne s^j$ ).

In the following example, we construct an infinite sequence of side operator.

The construction of such an infinite sequence of side operator in a subset of positive integers is quite technical.

Example 3.2. We first consider the set $N=\left\{\mathrm{1,2,3},\cdots \right\}$ , $F\left(N\right)=\left\{A\subseteq N\right\}$ . Now, we make $F\left(N\right)$ a bounded lattice $\left(L\mathrm{,}\wedge \mathrm{,}\vee \mathrm{,0,1}\right)$ . In this bounded lattice $L$ , “0” is the empty set and “1” is the set $N$ itself. The operation “ $\vee$ ” can be regarded as the set union and the operation “ $\wedge$ ” can be regarded as the set intersection. Consider the subsets in this bounded lattice $\left(L\mathrm{,}\wedge \mathrm{,}\vee \mathrm{,0,1}\right)$ .

Let $N$ be the set of all positive integers. For all $n\in N$ , let $F_n=\left\{n,n+1,\cdots \right\}$ , denoted by $\left[n\mathrm{,}\infty \right)$ with $F_{\infty }=\varnothing$ . Consider the set

$F_{\Sigma }=\left\{F_{\infty },F_1,F_2,\cdots ,F_n,F_{n+1},\cdots \right\}.$

Now for $A\subseteq N$ . Define $f\left(A\right)=\cap \left\{F_i:F_i\in F_{\Sigma }\text{and}{F}_i\supseteq A\right\}$ . Then, we can

easily verify that $f$ is an algebraic closure operator of $\mathcal{P}\left(N\right)\to \mathcal{P}\left(N\right)$ , where

$\mathcal{P}\left(N\right)$ is the power set of the set $N$ . We first take $N=\left\{1,3,5,\cdots \right\}$ . Then we

have $s\left(A\right)=\left\{2,4,6,\cdots \right\}$ , and $s^2\left(A\right)=\left\{\mathrm{3,5,7,}\cdots \right\}$ ; $s^3\left(A\right)=\left\{4,6,8,\cdots \right\}$ , and so

$s^{2k-1}\left(A\right)=\left\{2^k,2^k+2,2^k+4,\cdots \right\}$ , $s^{2k}\left(A\right)=\left\{2^k+1,2^k+3,\cdots \right\}$ .

Thus, for $i\ne j$ , we have $s^i\left(A\right)\ne s^j\left(A\right)$ . This shows that $\left\{s\mathrm{,}{s}^2\mathrm{,}{s}^3\mathrm{,}\cdots \right\}$ is an infinite sequence.

We now call an expansive operator g a quasi algebraic closure operator if $g^2\ne g$ but $g^2=g^3$ . Obviously, the quasi algebraic closure operator is a generalized algebraic closure operator. Naturally, one would ask whether the Kuratowski’s theorem still holds for the quasi algebraic closure operator?

In the following example, we show that Kuratowski’s theorem does not hold for the quasi algebraic closure operator.

Example 3.3. For any arbitrary integer $n$ , consider the subset $S=\left\{\left(n\mathrm{,}n+1\right)\right\}\cup \left\{\mathcal{R}\backslash \left(n\mathrm{,}n+1\right)\right\}\mathrm{.}$

We first let $\mathcal{P}\left(\mathcal{R}\right)$ be the power set of the real line $\mathcal{R}$ .

Define the function $g_0\mathrm{<mo>\; :\; </mo>}S\to \mathcal{P}\left(\mathcal{R}\right)$ by

$g_0\left(n,n+1\right)=\left(-\infty ,n+1\right]\cup \left[n+1,\infty \right)=\mathcal{R}\backslash \left(n+1,n+2\right),$

and

$g_0\left(\mathcal{R}\backslash \left(n,n+1\right)\right)=\mathcal{R}.$

Now, we define a function $g\mathrm{<mo>\; :\; </mo>}\mathcal{P}\left(\mathcal{R}\right)\to \mathcal{P}\left(\mathcal{R}\right)$ as follows:

$g\left(x\right)=x\cup \left\{\cup g_0\left(y\right)\mathrm{<mi>\; </mi>\; <mo>\; |\; </mo>\; <mi>\; </mi>}y\subseteq x\mathrm{,}y\in S\right\}\mathrm{,}$

where, $x\subseteq \mathcal{R}$ . Then, after verification, we can easily see that if $x_1\subset x_2$ , then $g\left(x_1\right)\subseteq g\left(x_2\right)$ , also $x\subseteq g\left(x\right)$ . Thus, g is an expansive operator.

By the definition of the operator g again, we can verify that $g^2\ne g$ but $g^2=g^3$ . Hence, g is indeed a quasi algebraic closure operator.

Let $A=\left(0,1\right)$ . Then by applying the operator g on the set $A=\left(0,1\right)$ , we have $g\left(\left(\mathrm{0,1}\right)\right)=\left(-\infty \mathrm{,1}\right]\cup \left[\mathrm{2,}\infty \right)$ .

Applying c and g by repetition on the above equality, we obtain the following sequence of sets:

$\begin{array}{l}cg\left(\left(0,1\right)\right)=\left(1,2\right),\\ ⋮\\ {\left(cg\right)}^2\left(\left(0,1\right)\right)=(\left(2,3\right)),\\ {\left(cg\right)}^k\left(\left(0,1\right)\right)=(\left(k,k+1\right)),\\ ⋮\end{array}$

Thus, it is clear that ${\left({\left(cg\right)}^k\right)}_{k=1}^{\infty }$ is an infinite sequence of sets.

This example illustrates that the theorem of Kuratowski does not hold anymore for quasi algebraic closure operators.

For the boundary operator and the rim operator, we have the following theorem. (see [24] )

Theorem 3.4. (i) $b^3=b^2$ (ii) $r^2=r$ .

Proof of i):

$\begin{array}{c}{b}^2=b\left(b\right)=f\left(b\right)\wedge fcb=f\left(f\wedge fc\right)\wedge fcb\le f^2\wedge fc\wedge fcb\\ \le f\wedge fc\wedge \ast \le f\wedge fc=b\end{array}$

Thus we have $b^3=b^2\left(b\right)\le b\left(b\right)=b^2$ . $b^2\wedge fcb=\left(fb\wedge fcb\right)\wedge fcb$ . On the other hand, by $b^2\le b$ , we deduce that $c{b}^2\ge cb$ , $fc{b}^2\ge fcb$ and $b^3=b\left(b^2\right)=f{b}^2\wedge fc{b}^2\ge f{b}^2\wedge fcb\ge b^2\wedge fcb=\left(fb\wedge fcb\right)\wedge fcb=fb\wedge fcb=b\left(b\right)$ Math_209#. Hence, we have proved that $b^2=b^3$ .

Proof of ii): Clearly, $r^2=r\left(r\right)=r\wedge fcr\le r$ . Also by $r=e\wedge fc$ , we have $r\le fc$ and thereby we obtain $r^2=r\left(r\right)=r\wedge fcr\ge r\wedge fc=e\wedge fc\wedge fc=e\wedge fc=r$ . Thus, we deduce that $r^2\le r\le r^2$ , and consequently $r^2=r$ .

Remark. The amalgamation of the boundary operator with the other algebraic operators also produces a number of useful equalities. (see [25] )

Proposition 3.5. (i) $b^{2}f=bf$ .

Proof. By $bf=f^2\wedge fcf=f\wedge fcf$ , $f\wedge fcf\le fc$ , $f\wedge fcf\le fcf$ . Hence we have $c\left(fcfcf\right)\ge cf$ . Consequently,

$c\left(f\wedge fccf\right)\ge f\left(cfcfcf\right)\ge fcffcfcf\ge fcf\mathrm{.}$

This implies that $fcbf\ge fcf$ . Thus, $b^{2}f=b\left(bf\right)=fbf\wedge fcbf=bf\wedge fcbf=f\wedge fcfcbf=f\wedge fcf=b$ .

Remark 3.6. Using the equalities $i=cfc$ , $f=cic$ , $r=f\wedge fc$ , $c^2=e$ we can easily deduce that $bc=b$ , $bf=bic$ , $b^{2}i=bi$ , $ibi=0$ (that is, the zero operator), $i{b}^2=0$ , $ir=ri=0$ , $bf=fr$ , $b^2=rb$ , $b^{2}r=rbr=rfr$ , $frf=rf$ , $re=s$ , $sc=r$ , $\cdots$ , etc.

Finally, we consider some questions which are closely related to the theorem of Kuratowski.

Let $A\mathrm{,}B$ be subsets of a space M. Then we call an operator $c^{\prime }$ an algebraic quasi-complmentation operator if $c^{\prime }$ satisfies the separation property $c^{\prime }\left(A\right)=M-A$ , $c^{\prime }\left(A\right)\cap A=\varnothing$ , simply denoted it by $c^{\prime }\wedge e=0$ ; also $c^{\prime }$ sa- tisfies the inversion property, i.e., if $A\subseteq B$ , then $c\left(A\right)\supseteq c\left(B\right)$ , we simply denote this situation by $c^{\prime }\downarrow$ , but $c^{\prime }$ does not satisfies the involution property, $c\left(c\left(A\right)\right)\ne A$ , i.e., $c^2\ne e$ , the identity

We now pose the following question.

Question 3.7. a) Let $A$ be a given subset in a topological space $X$ . How many distinct sets can be generated by applying the algebraic closure operator and the algebraic quasi-complementation operator successfully on $A$ in any order?

b) Likewisely, we call a shrinking operator $j$ an algebraic quasi-interior operator if $j^2=j^3$ but not $j^2=j$ .Can we construct an algebraic quasi- interior open set for the set $A$ , that $j\left(A\right)=A$ ?

Now, we ask when will an algebraic quasi-interior operator $j$ and the complementation operator $c$ generate a finite semigroup under the usual operator composition?

In order to help answer the above questions, we construct an algebraic quasi-complementation operator $c^{\prime }$ .

Let $R$ be a real line. For any element $x\in R$ , define $c^{\prime }\left(x\right)=\left(-\infty ,-x\right]$ , $c^{\prime }\left(-x\right)=\left[x,\infty \right)$ and $c^{\prime }\left(0\right)=\left\{0\right\}$ . Obviously, $c^{\prime }\wedge e=0$ ; and $c^{\prime }\downarrow$ because if $x\le y$ ,then $c^{\prime }\left(x\right)\ge c^{\prime }\left(y\right)$ , but ${c^{\prime }}^2\ne e\mathrm{.}$ Thus, $c^{\prime }$ is indeed an algebraic quasi- complementation operator.

By using the idea of algebraic quasi closure operator described in Example 2.6, one can therefore construct an algebraic quasi interior operator. Hence, by the idea of the construction of an algebraic quasi interior operator, the reader should be able to answer the above questions.

4. Further Relationship of Algebraic Closure and the Algebraic Interior Operations

We notice that an equivalent formulation of the Kuratowski’s theorem is that at most 7 distinct sets can be obtained by the composition of the algebraic closure operator $f$ and the algebraic interior operator $i$ . In order to show clearly the mutually inclusion relationship of these 7 distinct sets composed by $f$ and $i$ on a given set $A$ . we give again the following Hasse diagram (Diagram 2).

Diagram 2. The inclusion relationship of 7 distinct sets.

For a given subset A in a topological space X, we call the sets generated from A by successive applications of the algebraic closure and complementation operations in any order on A the relatives of A. Thus, by Kuratowski’s theorem, a K-set A always have 14 relative sets.

It is natural to ask whether there exists a subset in a topological space with less than 14 relative sets? The following theorem gives an answer to the above question.

Theorem 4.1. If the rim of a set $A$ is nowhere dense, then by successive applications of closure operator and complementation operator, in any order, on $A$ , $A$ has at most 10 relative sets.

Proof. We first denote the interior of the set $A$ by $i\left(A\right)$ . Then $A=i\left(A\right)\cup r\left(A\right)$ . Consequently, by using the algebraic closure operator, we have $f\left(A\right)=fi\left(A\right)\cup fr\left(A\right)$ . Now, $if\left(A\right)=i\left(fi\left(A\right)\cup fr\left(A\right)\right)\subseteq fi\left(A\right)\cup ifr\left(A\right)$ . Be- cause the rim of the set $A$ , say $r\left(A\right)$ is nowhere dense, we have $fr\left(A\right)=\varnothing$ and so, $f\left(A\right)\subseteq f\left(A\right)$ . Thus, we see immediately that $i\left(if\left(A\right)\right)\subseteq ifi\left(A\right)$ , that is, $f\left(A\right)\subseteq ifi\left(A\right)$ . Conversely, $A\supseteq i\left(A\right)$ implies that $f\left(A\right)\supseteq ifi\left(A\right)$ . Hence, we deduce that

$f\left(A\right)=ifi\left(A\right).$ (1)

Now we put $A_1=A,A_2=f\left(A\right)$ and so on. For $n\ge 2$ , we put $A_{2n}={\left(fc\right)}^{n-1}f\left(A\right)$ , for $n>1$ , put $A_{2n+1}={\left(cf\right)}^n\left(A\right)$ . Also, similarly, we put $B_1=c\left(A\right)$ . For $n\ge 1$ , $B_{2n}={\left(fc\right)}^n\left(A\right)$ , $B_{2n+1}=c{\left(fc\right)}^n\left(A\right)$ . By using Kurato- wski’s theorem again, we always have $B_8=B_4$ , $A_8=A_4$ . By the equation ob- tained above, we see that $A_5=B_8=B_4$ . Thus A has at most 10 distinct relative sets.

In the literature, P. Halmos [12] first called a set A a regular open set if $f\left(A\right)=A$ . Dually, he called a set B regular closed if $fi\left(B\right)=B$ . If X is a topological space containing a K-set, then we always have ${\left(if\right)}^2=if$ , ${\left(fi\right)}^2=fi$ . This means that the space X simultaneously contains a regular closed set and a regular open set.

The following theorem is also an interesting theorem.

Theorem 4.2. If the rim of a set $A$ is dense in a regular closed subset of a topological space $X$ , then $A$ has at most 12 distinct relative sets by taking the algebraic closure operator $f$ and the the complementation operators $c$ repeatedly in any order on $A$ .

Proof. Let $T$ be a regular closed set. Then by definition, we have $fi\left(T\right)=T$ . Since $A=i\left(A\right)\cup r\left(A\right)$ , by the additive property of the algebraic closure, we have $f\left(A\right)=fi\left(A\right)\cup fr\left(A\right)=fi\left(A\right)\cup T$ . This is because we assume that $r\left(A\right)$ is dense in $T$ . Now, we have $f\left(A\right)\supseteq i\left(fi\left(A\right)\right)\cup i\left(T\right)$ . This leads to $fif\left(A\right)\supseteq fif\left(A\right)\cup T=fi\left(A\right)\cup T$ , and thereby $fif\left(A\right)\supseteq f\left(A\right)$ . Trivially, $fif\left(A\right)\subseteq f\left(A\right)$ . Hence, we have proved that $fif\left(A\right)=f\left(A\right)$ , that is, $A_6=A_2$ . By the result of Kuratowski’s theorem, we always have $B_8=B_4$ , $A_8=A_4$ . Thus, we can easily see that $A$ has at most 12 distinct relative.

It was stated by D. Sherman [22] that 13 distinct operators can be produced by the combination of the algebraic closure operation $f$ , the algebraic interior operation $i$ and the intersection operation “ $\wedge$ ” on a given set $A$ . This result is the answer to the well known problem $p\left(f\mathrm{,}i\mathrm{,}\wedge \right)$ . If the union operation “ $\vee$ ” is also allowed to use, then D.Sherman in [22] has shown that at most 35 different operators can be derived.This problem is called the $p\left(f\mathrm{,}i\mathrm{,}\wedge \mathrm{,}\vee \right)$ pro- blem. The solution of the $p\left(f\mathrm{,}i\mathrm{,}\wedge \right)$ problem and its dual problem $p\left(f\mathrm{,}i\mathrm{,}\vee \right)$ are shown in the following Hasse diagrams [23] (Diagram 3).

Diagram 3. The solution of the $p\left(f\mathrm{,}i\mathrm{,}\wedge \right)$ problem and its dual problem $p\left(f\mathrm{,}i\mathrm{,}\vee \right).$

Proof. Let $T$ be a regular closed set. Then $fi\left(T\right)=T$ . Since $A=i\left(A\right)\cup r\left(A\right)$ , by the additive property of the relation “ $\wedge$ ” on a given set $A$ . This result is the answer to the well known problem $p\left(f\mathrm{,}i\mathrm{,}\wedge \right)$ . If the union operation “ $\vee$ ” is also allowed to use, then at most 35 different operators can be formed. This problem is now called the $p\left(f\mathrm{,}i\mathrm{,}\wedge \mathrm{,}\vee \right)$ problem. The solution of $p\left(f\mathrm{,}i\mathrm{,}\wedge \right)$ and $p\left(f\mathrm{,}i\mathrm{,}\vee \right)$ are shown in the following Hasse diagrams [23] (Diagram 4 and Diagram 5).

Diagram 4. $p\left(f\mathrm{,}i\mathrm{,}\wedge \right).$

Diagram 5. $p\left(f\mathrm{,}i\mathrm{,}\vee \right).$

It is clear that the $p\left(f\mathrm{,}i\mathrm{,}\vee \right)$ problem is the dual of the $p\left(f\mathrm{,}i\mathrm{,}\wedge \right)$ problem.

One can verify the following subset $T$ of the real line $R$

$T=\left[\left\{1/n,n\in N\right\}\right]\cup \left[2,4\right]-\left\{3+1/n\right\}\cup \left[\left[5,7\right]\cap \left(Q\cup \left(6+1/2n\text{π},6+1/\left(2n-1\right)\text{π}\right)\right)\right]$

is a Shermann set, that is, it can generate 13 distinct sets by successively applications of the algebraic closure operator,the algebraic interior operator and the set intersection operations on T in any order (see also [22] ).

5. An Example of Kuratowski 14 Sets in the Set of Positive Integers

Let $M$ be the set of all positive integers and $A\subseteq M$ . Then Hammer in 1960 defined $h\left(A\right)={\displaystyle {\cup }_{n=1}^{\infty }{A}^n}$ , where $A^n$ is the set product of $n$ copies of $A$ . Clearly, $h\left(A\right)\supseteq A$ and $h\left(h\left(A\right)\right)=h\left(A\right)$ . Thus, $h$ is an expansive and ide- mpotent operator, we now call $h$ the Hammer closure. Now, we define $i=chc$ and call it the Hammer interior operator. It is clear that $i$ is also a shrinking and idempotent operator.

The following interesting question was asked by Hammer in 1960 [12] . Do $h$ and $c$ generated exactly 14 distinct subsets by composition of $h$ and $c$ in any order, in the set of positive integers? By using the expansive property of $h$ and the shrinking property of $i=chc$ , we can easily prove that ${\left(ih\right)}^2=ih$ , ${\left(hi\right)}^2=hi$ . Write $h=cic$ and $i=chc$ . Then we obtain the following 14 distinct sets listed in Table 1.

We now give the following example to answer the open problem of Hammer in 1960. This example was given by Shum in 1996. (see [26] ).

Example 5.1. The following subset of the set of positive integers

$A=\left\{5,2^2\times 3^2,7,2\times 5^2,2^2\times 5^3,3\times 7^2,3^2\times 7^3\right\}$

is a K-set.

We list bellow the family of all the 14 relatives of the set $A$ . This would help the readers to check the validity of the example.

$A_1=A=e\left(A\right)=\left\{5,2^2\times 3^2,7,2\times 5^2,2^2\times 5^3,3\times 7^2,3^2\times 7^3\right\};$

$A_2=h\left(A\right)=A^{-}=A_1\cup \left\{5\times 7,2^2\times 5^4,2^2\times 3^2\times 5\times 7,2^2\times 3^2\times 5^3\times 7^3,\cdots \right\}\cup \cdots$ ;

$A_3=ch\left(A\right)=B^{\circ }=\left\{1,2,3,2\times 3,2^2\times 7^3,3^2\times 5^3,2^2\times 7,3^2\times 5,\cdots \right\}\cup \cdots$ ;

$A_4=h\left(h\left(A\right)\right)=B^{\circ -}=A_3\cup \left\{2^2\times 3^2,2^2\times 3^2\times 5^3\times 7^3,2^2\times 3^2\times 5\times 7,\cdots \right\}\cup \cdots$ ;

$A_5=chch\left(A\right)=A^{-\circ }=\left\{5,7,2^2\times 5^3,3^2\times 3,2\times 5^2,3\times 7^2,\cdots \right\}\cup \cdots$ ;

$A_6=hchch\left(A\right)=A^{-\circ -}=A_5\cup \left\{5\times 7,2^2\times 3^2,2^2\times 3^2\times 5^3\times 7^3,2\times 3\times 5^2\times 7^2,\cdots \right\}\cup \cdots$ ;

$A_7=chch\left(A\right)=A^{\circ -\circ }=\left\{1,2,3,2^2\times 3^2\times 5\times 7,\cdots \right\}$ ;

$B_1=c\left(A\right)=\left\{1,2,3,7^2,2\times 3,2\times 5,2^2\times 5^4,2^2\times 5^3,3\times 7^3,\cdots \right\}$ ;

$B_2=hc\left(A\right)=B^{-}=B_1\cup \left\{2^2\times 3^2,3\times 7^2,3\times 7^3,3^2\times 7^3,\cdots \right\}\cdots$ ;

$B_3=chc\left(A\right)=A^{\circ }=\left\{5,7,2\times 5,3\times 7,2\times 5^3,3^2\times 5^3,2\times 7^3\right\}\cdots$ ;

$B_4=hchc\left(A\right)=A^{\circ -}=B_3\cup \left\{5\times 7,2^2\times 3^2\times 5^3\times 7^3,2^2\times 5^4,3\times 7^3,\cdots \right\}\cdots$ ;

$B_5=chchc\left(A\right)=B^{-\circ }=\left\{1,2,3,2\times 3,3^2\times 7,2^2\times 5^3,3^2\times 7^3\times 2\times 3\times 5\times 7,\cdots \right\}\cdots$ ;

$B_6=hchchc\left(A\right)=B^{-\circ -}=B_5\cup \left\{2^2\times 3^2\times 5^3\times 7^3,2^2\times 3^2\times 5\times 7,\cdots \right\}\cdots$ ;

$B_7=chchchc\left(A\right)=B^{\circ -\circ }=\left\{5,7,5\times 7,3\times 7^3,\cdots \right\}$ .

The relationship of the above 14 sets can be seen in Diagram 6.

Diagram 6. The relationship of the 14 sets.

Recall that for a subset $A$ in the set of positive integers $Z_{+}$ , the Hammer closure of $A$ is defined by $h\left(A\right)={{\displaystyle \cup }}_{n=1}^{\infty }$ , where $A^n$ is the set product of n copies of $A$ . It is natural to ask how to construct a K-set in the set $Z_{+}$ ? The following lemma gives some ideas and hints.

Lemma 5.2. Let $p\mathrm{,}q\mathrm{,}r\mathrm{,}s$ be prime numbers and $i\left(A\right)=chc\left(A\right)$ . If $p^n\in i\left(A\right)$ , then $p\in i\left(A\right)$ , where $n>1$ .

Proof. If $p\cancel{\in }i\left(A\right)=chc\left(A\right)$ , then $p\in hc\left(A\right)$ . This implies that $p^n\in hc\left(A\right)$ and so we have $p^n\notin chc\left(A\right)=i\left(A\right)$ , a contradiction.

Lemma 5.3. Assume that $i\left(A\right)=\left\{p,p^n\right\}$ . Then $n\le 3$ .

Proof. If $n\ge 4$ , then by our assumption, we have $p^{n-2}\notin i\left(A\right)=chc\left(A\right)$ , and so $p^2\notin chc\left(A\right)$ . This leads to $p^{n-2}\in hc\left(A\right)$ , and hence $p^n=p^{n-2}{p}^2\in hc\left(A\right)$ . Consequently, we deduce that $p^n\notin chc\left(A\right)=i\left(A\right)$ , a contradiction.

By using the arguments $a\notin i\left(A\right)$ , $b\in i\left(A\right)$ , $ab\notin i\left(A\right)$ repeatedly, we de- duce the following lemma.

Lemma 5.4. 1) Let $m$ be a composite number and $p$ is a prime number. If $i\left(A\right)=\left\{p,m\right\}$ , then $p\mathrm{<mo>\; |\; </mo>}m$ ;

2) If $i\left(A\right)=\left\{p,p^{n}q\right\}$ or $i\left(A\right)=\left\{p\mathrm{,}p{q}^n\right\}$ , then $n=1$ ;

3) If $i\left(A\right)=\left\{p,p^2,n\right\}$ , $n$ is a composite number, then $n=p^3,p^4,p^5,ps$ or $p^{2}s$ , where $s$ is a prime number but is not equal to $p$ ;

4) If $p^{2}q\in i\left(A\right)$ , $p^{n-1}\notin i\left(A\right)$ , then $p\in i\left(A\right)$ ;

5) $i\left(A\right)=\left\{p,pqr\right\}$ is impossible;

6) Let $d\left(m\right)$ be the prime decomposition length of the composite number $m$ . If $i\left(A\right)=\left\{p,m\right\}$ , then $d\left(m\right)\le 2$ ;

7) $i\left(A\right)=\left\{m,n\right\}$ is impossible if $m\mathrm{,}n$ are both composite numbers.

Proof. The proofs of this lemma are routine but quite tedious, we hence omit the details.

By summing up the above lemmas, we state the following theorem.

Theorem 5.5. If $\left|i\left(A\right)\right|\le 3$ , then $A$ is not a K-set.

Proof. The proof is to check $\left|i\left(A\right)\right|=1,2,3$ case by case. For example, we consider $i\left(A\right)=\left\{p\right\}$ , $i\left(A\right)=\left\{p,q\right\}$ , $i\left(A\right)=\left\{p,q,r\right\}$ , $i\left(A\right)=\left\{p,m,n\right\}$ or $i\left(A\right)=\left\{m,n,u\right\}$ until all possible cases are exhausted and are impossible. There- fore we conclude that $A$ is not a K-set if $\left|i\left(A\right)\right|\le 3$ .

Theorem 5.6. Let $A$ be a subset of the set of all positive integers and $p\mathrm{,}q\mathrm{,}r\mathrm{,}s$ are prime numbers. Then we have the following statements:

1) If $A$ is a K-set, then $A$ contains at least five elements and the elements of $A$ can be composed of not more than three distinct prime numbers.

2) If $A$ is a K-set with five elements, then $i\left(A\right)$ must be one of the following forms:

a) $i\left(A\right)=\left\{p,r,pr,pqr\right\}$ ;

b) $i\left(A\right)=\left\{p,pq,r,rs\right\}$ ;

c) $i\left(A\right)=\left\{p,pq,pr,pqr\right\}$ ;

d) $i\left(A\right)=\left\{p,pq,p^{\alpha +1},p^{\alpha +1}q|\alpha \ge 1\right\}$ .

For more information concerning the construction of a K-set, the reader is referred to K. P. Shum [26] .

In view of Diagram 5, there are four distinct sets, namely the sets $B_4=A^{\circ -}$ , $A_4=A^{-\circ -}$ , $B_7=A^{\circ -\circ }$ between the sets $A_2=A^{-}$ and $B_3=A^{\circ }$ . Likewisely, there are also four distinct sets, namely, the sets $B_6=B^{-\circ -}$ , $A_4=B^{\circ -}$ , $B_5=B^{-\circ }$ and $A_7=B^{\circ -\circ }$ between the sets $B_2=B^{-}$ and $A_3=B^{\circ }$ .

Now, we perform the usual set operations of union intersection and subtraction on the above sets, we obtain the following equalities:

1) $A^{\circ }=A^{\circ }$ ;

2) $A^{\circ -\circ }=A^{\circ }\cup \left(A^{\circ -\circ }\backslash A^{\circ }\right)$ ;

3) $A^{\circ -}=A^{\circ }\cup \left(A^{\circ -\circ }\backslash A\right)\cup \left(A^{\circ -}\cap A^{-\circ }\backslash A^{\circ -\circ }\right)\cup \left(A^{\circ -}\backslash A^{-\circ }\right)$ ;

4) $A^{-\circ }=A^{\circ }\cup \left(A^{\circ -\circ }\backslash A^{\circ }\right)\cup \left(A^{\circ -}\cap A^{-\circ }\backslash A^{\circ -\circ }\right)\cup \left(A^{-\circ }\backslash A^{\circ -}\right)$ ;

5) $\begin{array}{l}{A}^{-\circ }=A^{\circ }\cup \left(A^{\circ -\circ }\backslash A^{\circ }\right)\cup \left(A^{\circ -}\cap A^{-\circ }\backslash A^{\circ -\circ }\right)\cup \left(A^{-\circ }\backslash A^{\circ -}\right)\cup \left(A^{\circ -}\backslash A^{-\circ }\right)\\ \cup \left(A^{-\circ -}\backslash A^{\circ -}\cap A^{-\circ }\right)；\end{array}$

6) $A=A^{\circ }\cup \left(A\backslash A^{\circ }\right)$ ;

7) $A^{-}=A^{-\circ -}\cup \left(A^{-}\backslash A^{-\circ -}\right)$ .

Because the boundary of a set $A$ in the topological space $X$ is defined by $b\left(A\right)=A^{-}\cap c{\left(A\right)}^{-}=f\left(A\right)\cap fc\left(A\right)$ .

Since the rim of a set $A$ is defined by $rb\left(A\right)=A\cup fc\left(A\right)$ and so $rb\left(A\right)\subseteq A^{-}\cap c{\left(A\right)}^{-}=b\left(A\right)$ . This means that the rim of $A$ , $rb\left(A\right)$ is a part of the boundary $b\left(A\right)$ .

By using the equation $i=cfc$ and $f=cic$ , we immediately see that the gain boundary of $A$ $Gb\left(A\right)=A^{\circ -\circ }\backslash A^{\circ }=ifi\left(A\right)\cap ci\left(A\right)=ifi\left(A\right)\cap c\left(cfc\left(A\right)\right)\subseteq f\left(A\right)\cap fc\left(A\right)=b\left(A\right)$ . Likewisely, we can also see that the inner boundary of $A$ , $I\left(A^{\circ -}\cap A^{-\circ }\backslash A^{\circ -\circ }\right)\subseteq f\left(A\right)\cap fc\left(A\right)=b\left(A\right)$ ; also, the public boundary of $A$ , $Pb\left(A\right)=A^{\circ -}\backslash A^{-\circ }=fi\left(A\right)\cap ifif\left(A\right)\subseteq fi\left(A\right)\cap cf\left(A\right)\subseteq fi\left(A\right)\cap cf\left(A\right)=b\left(A\right)$ ; the neutral boundary $Nb\left(A\right)=A^{-\circ }\backslash A^{\circ -}\subseteq f\left(A\right)\cap fc\left(A\right)=b\left(A\right)$ ; the outer boun- dary of $A$ , $Ob\left(A\right)=A^{-\circ -}\backslash A^{\circ -}\cap A^{-\circ }\subseteq f\left(A\right)\cap fc\left(A\right)=b\left(A\right)$ ; the loss boun- dary of $A$ $Lb\left(A\right)=A^{-}\backslash A^{-\circ -}=f\left(A\right)\cap c\left(fif\left(A\right)\right)\subseteq f\left(A\right)\cap ifc\left(A\right)\subseteq f\left(A\right)\cap fc\left(A\right)=b\left(A\right).$

Now, by using the above concepts of the components of the boundary of a set $A$ , we obtain the following equalities related to the above components of the boundaries and the 7 Kuratowski sets. (see [26] )

1) $A=A^{\circ }\cap Rb\left(A\right)$ ;

2) $A^{\circ }$ ;

3) $A^{\circ -\circ }=A^{\circ }\cup Gb\left(A\right)$ ;

4) $A^{\circ -}=A^{\circ }\cup Gb\left(A\right)\cup Ib\left(A\right)\cup Pb\left(A\right)$ ;

5) $A^{-\circ }=A^{\circ }\cup Gb\left(A\right)\cup Ib\left(A\right)\cup Nb\left(A\right)$ ;

6) $A^{-\circ -}=A^{\circ }\cup Gb\left(A\right)\cup Ib\left(A\right)\cup Nb\left(A\right)\cup Pb\left(A\right)\cup Ob\left(A\right)$ ;

7) $A^{-}=A^{\circ }\cup Gb\left(A\right)\cup Ib\left(A\right)\cup Nb\left(A\right)\cup Lb\left(A\right)$ .

We now consider the Structure of the boundary of a set $A$ in a topological space.

The boundary of a set $A$ and all its components in a topological space can be interpreted in the following “envelope diagram” (see [27] ) (Diagram 7).

Diagram 7. The boundary of a set A and all its components in a topological space.

In Diagram 6, $A^{\circ }$ can be regarded as a troop of soldiers and $B^{\circ }$ is the troop of the enemies of $A^{\circ }$ .

In the diagram, $Pb\left(A\right)$ is the public boundary at which the troops $A^{\circ }$ and $B^{\circ }$ are in combatomg.

$Nb\left(A\right)$ can be regarded as the ceased fire zone or the negotiation area between the troops $A^{\circ }$ and $B^{\circ }$ , and therefore, we call this part of boundary the neutral boundary of $A$ .

$Ib\left(A\right)$ , the inner boundary is the final defence frontier of $A^{\circ }$ , that is, it can be regarded as the Mignot defence line of the troop $A^{\circ }$ .

$Gb\left(A\right)$ , the gain boundary of $A^{\circ }$ , it can be interpreted as the picket posts of the troop $B^{\circ }$ and these posts will be seized and occupied by the troop $A^{\circ }$ when the war begins.

$Ob\left(A\right)$ , the outer boundary of $A^{\circ }$ which is the inner boundary of the troop $B^{\circ }$ . Roughly speaking, the inner boundary of $A^{\circ }$ can be regarded as the skin of $A^{\circ }$ and the outer boundary can be regarded as the clothes of $A^{\circ }$ . Then one can easily distinct the inner boundary and the outer boundary of a set $A$ in a given space $X$ .

$Lb\left(A\right)$ , the loss boundary of $A^{\circ }$ which can be described as the gain boundary of the troop $B^{\circ }$ .

$Rb\left(A\right)$ , the rim boundary of $A$ . Of course,the rim of a set $A$ is a part of the boundary of the set $A$ but it is not necessarily belongs to the given set $A$ . In considering the boundary of a set $A$ , many people always overlook that there are rims of the set $A$ .

Thus,in terms of the closure operator $f$ , the interior operator $i$ and the complementation operator $c$ , we can express the above components of the boundary of the set $A$ in the topological space $X$ as follows:

We always use the equalities $i=cfc$ and $f=cic$ to describe the different components of the boundary of the set $A$ . These different components of the boundary of the set $A$ are the followings:

$Pb\left(A\right)=fi\left(A\right)\cap fcf\left(A\right)=fi\left(A\right)\cap fic\left(A\right)=B_4\cap A_4;$

$Nb\left(A\right)=cfi\left(A\right)\cap fic\left(A\right)=cfcfc\left(A\right)\cap cfcfc\left(A\right)=B_5\cap B_5;$

$Ib\left(A\right)=fi\left(A\right)\cap cfcf\left(A\right)\cap fcfcfc\left(A\right)=B_4\cap A_5\cap B_6;$

$Ob\left(A\right)=fic\left(A\right)\cap cfcfc\left(A\right)\cap fcfcf\left(A\right)=A_4\cap B_5\cap A_6;$

$Gb\left(A\right)=ifi\left(A\right)\cap fc\left(A\right)=B_7\cap B_2;$

$Lb\left(A\right)=f\left(A\right)\cap cfif\left(A\right)=A_7\cap A_2;$

$Rb\left(A\right)=A\cap fc\left(A\right)=A_1\cap B_2.$

After we have displayed all the possible components of the boundary of a set $A$ , we can easily find out the conditions for a set $A$ which leads to the set $A$ to be a K-set. In fact, we have the following three different types of sets.

Type I. A set $A$ is called a standard set if $Pb\left(A\right)\mathrm{,}Nb\left(A\right)\mathrm{,}Ib\left(A\right)\mathrm{,}Ob\left(A\right)\mathrm{,}$ $Gb\left(A\right)\mathrm{,}Lb\left(A\right)$ and $Rb\left(A\right)$ are all non-empty sets.

One can easily observe that a set $A$ is a K-set if and only if $A$ is a standard set.

It may happen that $Ib\left(A\right)=Ob\left(A\right)=Nb\left(A\right)\ne \varnothing$ .

Type II. A set $A$ is called an abnormal set if $Pb\left(A\right)=\varnothing$ . It can be observed that an abnormal set $A$ is a K-set if and only if $Nb\left(A\right)\mathrm{,}Ib\left(A\right)\mathrm{,}Ob\left(A\right)\mathrm{,}Gb\left(A\right)\mathrm{,}Lb\left(A\right)$ and $Rb\left(A\right)$ are all non-empty sets.

Type III. A set $A$ is called a normal set if $Pb\left(A\right)\ne \varnothing$ . It is clear that a set $A$ is a K-set if and only if $Nb\left(A\right)\mathrm{,}Ib\left(A\right)\mathrm{,}Ob\left(A\right)\mathrm{,}Gb\left(A\right)\mathrm{,}Lb\left(A\right)$ and $Rb\left(A\right)$ are all non-empty, moreover, $Ib\left(A\right)$ and $Ob\left(A\right)$ may be possibly empty.

Remark 5.7. Because $Ib\left(A\right)\subseteq Nb\left(A\right)$ and $Ob\left(A\right)\subseteq Nb\left(A\right)=A_3\cap B_5$ . In this case, the set $A$ is also a standard K-set.

Below are some examples of normal set and abnormal set.

Example 5.8. $A=\left\{2,5,2\times 5,2\times 3\times 5,2^2\times 3^2\right\}$ is a normal set. In this set set, we can check that $Pb\left(A\right)=\left\{2^2\times 3^2\times 5^2,\cdots \right\}$ , $Nb\left(A\right)=\left\{2^4\times \mathrm{3,}\cdots \right\}$ , $Ob\left(A\right)=\varnothing$ , $Ib\left(A\right)=\varnothing$ , $Gb\left(A\right)=\left\{2^2\times 3\times 5,\cdots \right\}$ , $Lb\left(A\right)=\left\{2^3\times 3^2,\cdots \right\}$ , $Rb\left(A\right)=\left\{2^2\times 3^2\right\}$ .

Therefore $A$ is a normal K-set.

Example 5.9. The $A=\left\{2,2\times 3,5,2\times 7,3\times 7\right\}$ is a abnormal set.

It is clear to see that $i\left(A\right)=\left\{2,2\times 3,5,2\times 7\right\}$ . Since $3\notin i\left(A\right)$ , $7\notin i\left(A\right)$ , we know that $3\times 7\notin A\cap fc\left(A\right)=Rb\left(A\right)$ . Because we can easily see that $\mathrm{3,7,5}\times \mathrm{7,3}\times 5$ and $2\times 3\times \mathrm{7,}\cdots$ are all in the interior of $c\left(A\right)$ . Thus $Pb\left(A\right)=fi\left(A\right)\cap fi\left(c\left(A\right)\right)=\varnothing$ . This shows that $A$ is an abnormal set.