Covering Rough Sets and Formal Topology – A Uniform Approach Through Intensional and Extensional Constructors

Abstract

Approximation operations induced by coverings are reinterpreted through a set of four “constructors” defined by simple logical formulas. The very logical definitions of the constructors make it possible to readily understand the properties of such operators and their meanings.

A Appendix

The following result is the basis of the calculus of constructors by means of Boolean matrices:

Duality between the constructors. In any SRS P, $\langle e\rangle$ and [e] are dual; $\langle i\rangle$ and [i] are dual.

Proof

$-R^\smile (-A)=\{x:R(x)\subseteq A\}$ (aka: $-\langle e\rangle (-A)=[e](A)$).

Same for the intensional pair of constructors.    $\Box$

Direct proof of Lemma 2.(2). The proof comes straightforwardly from (14). Indeed, $\bigcap \{-X\in {\mathcal Z}:X\cap A=\emptyset \}=\bigcap \{-R^\smile (m):R^\smile (m)\subseteq -A\}= \{g:\forall m((R^\smile (m)\subseteq -A)\Longrightarrow g\notin R^\smile (m))\}= \{g:\forall m(g\in R^\smile (m)\Longrightarrow R^\smile (m)\nsubseteq -A)\}= \{g:\forall m(m\in R(g)\Longrightarrow R^\smile (m)\cap A\ne \emptyset )\}$. From Lemma 1, the latter set equals $\{g:\forall m(m\in R(g)\Longrightarrow m\in R(A))\}=\{g:R(g)\subseteq R(A)\}$.

Direct proof of Theorem 12.(7). Since R is a preorder, for any $X\subseteq U$, $[e](X)\subseteq X$ so that $[e]\langle e\rangle [e](X)\subseteq [e]\langle e\rangle (X)$. Suppose now $x\in [e]\langle e\rangle (X)$. Then $x\in [e](R^\smile (X))$ so that $R(x)\subseteq R^\smile (X)$. If $x\notin [e]\langle e\rangle [e](X)$ then $R(x)\nsubseteq R^\smile (\{y:R(y)\subseteq X\})$. Therefore, $\exists z$ such that $z\in R(x)$ and $z\notin R^\smile (\{y:R(y)\subseteq X\})$. Therefore, $\forall y(R(y)\subseteq X\Longrightarrow y\notin R(z))$, so that $\forall y(y\in R(z)\Longrightarrow R(y)\nsubseteq X)$. But R is transitive. So $R(y)\subseteq R(z)$. It follows $R(z)\nsubseteq X$, which leads to a contradiction because $z\in R(x)$ so that $R(z)\subseteq R(x)$ which implies $R(x)\nsubseteq X$.

Direct proof of Lemma 7.(2):

Direct proof of Lemma 7.(5):

Moreover, $(uC)_6(A)=\bigcup \{K:K\cap A\ne \emptyset \}$. Thus $x\in (uC)_6(A)$ iff $\exists K(x\in K\wedge K\cap A\ne \emptyset )$ iff $\exists K(x\in R^\smile (K)\wedge R^\smile (\{K\})\cap A\ne \emptyset )$ iff $\exists K(x\in \langle e\rangle (K)\wedge K\in \langle i\rangle (A))$ iff $x\in \langle e\rangle \langle i\rangle (A)$.

Direct proof of Lemma 12: $\langle i\rangle (X)=X$ iff $R(X)=X$ iff $\bigcup \{R(x):x\in X\}=X$ which implies $\forall x\in X(R(x)\subseteq X)$. Let $y\notin X$. Since R is reflexive, $x\in R(x)$, so that $R(x)\nsubseteq X$. In sum, $x\in X\Longrightarrow R(x)\subseteq X$ and $x\notin X\Longrightarrow R(x) \nsubseteq X$. We conclude that $x\in X$ iff $R(x)\subseteq X$ which amounts to saying $R(X)=\{x:R(x)\subseteq X\}=[e](X)$. Notice that seriality, trivially, is not enough. Indeed seriality does not prevent from the existence of a $g\in X$ such that $R(g)\cap X=\emptyset$ - think of the relation $\{\langle a,b\rangle ,\langle b,b\rangle \}$ on the set $\{a,b\}$ and put $X=\{a\}$.

Alternative proof of Theorem 13 (point (3) of Corollary 11 below):

Lemma 18

Let C be unary and $\mathbf{C}^\blacktriangledown =\{K\in \mathbf{C}:\exists x\wedge md(x)=\{K\}\}$. Then $R_{\mathbf{C}^\blacktriangledown }=R_\mathbf{C}$.

Proof

Since C is unary, $\mathbf{P}(\mathbf{C^\blacktriangledown })=\langle U, C^\blacktriangledown , R^\blacktriangledown \rangle$, where $R^\blacktriangledown$ is R restricted to $U\times \mathbf{C}^\blacktriangledown$, is a specialised covering of U. So, let us assume $y\in R_{\mathbf{C}^\blacktriangledown }(x)$. Therefore, $\forall K\in \mathbf{C}^\blacktriangledown (x\in K\Longrightarrow y\in K)$. In particular this occurs for $\{K\}=md(x)$. Then for all $K\in \mathbf{C}(x\in K\Longrightarrow y\in K)$ so that $y\in R_\mathbf{C}(x)$. Thus $R_{\mathbf{C}^\blacktriangledown }\subseteq R_\mathbf{C}$. Vice-versa, since $\mathbf{C}^\blacktriangledown \subseteq \mathbf{C}$, $R_\mathbf{C}\subseteq R_{\mathbf{C}^\blacktriangledown }$.    $\Box$

Corollary 11

For any unary covering C:

• (1) $\mathbf{S}_{int}(\mathbf{P}(\mathbf{C}))=\mathbf{S}_{int}(\mathbf{P}(\mathbf{C}^\blacktriangledown ))$;

• (2) $\mathbf{S}_{int}(\mathbf{P}(\mathbf{C}))=\mathbf{S}_{\mathcal C}(\mathbf{P}(R_{\mathbf{C}^\blacktriangledown }))$;

• (3) $\mathbf{S}_{int}(\mathbf{P}(\mathbf{C}))$ is a topological space.

Proof

(1) From Theorem 9: $\preceq _{{int}^{\mathbf{P}(\mathbf{C})}}=R_{\mathbf{C}}$ and $\preceq _{{int}^{\mathbf{P}(\mathbf{C^\blacktriangledown })}}=R_{\mathbf{C}^\blacktriangledown }$. But from Lemma 18, $R_\mathbf{C}=R_{\mathbf{C}^\blacktriangledown }$. Therefore, $\preceq _{{int}^{\mathbf{P}(\mathbf{C})}}=\preceq _{{int}^{\mathbf{P}(\mathbf{C^\blacktriangledown })}}$ so that $\mathbf{S}_{int}(\mathbf{P}(\mathbf{C}))=\mathbf{S}_{int}(\mathbf{P}(\mathbf{C}^\blacktriangledown ))$. (2) From Corollary 7, $int^{\mathbf{P}(\mathbf{C}^\blacktriangledown )}=int^{\mathbf{P}(R_{\mathbf{C}^\blacktriangledown })}$, because $\mathbf{P}(\mathbf{C}^\blacktriangledown )$ is a specialised covering whenever C is unary. In view of (1) one concludes $\mathbf{S}_{int}(\mathbf{P}(\mathbf{C}))=\mathbf{S}_{int}(\mathbf{P}(\mathbf{C}^\blacktriangledown ))= \mathbf{S}_{\mathcal C}(\mathbf{P}(R_{\mathbf{C}^\blacktriangledown }))$. (3) is obtained from (1) and Theorem 13.    $\Box$

Corollary 11.(3) is a proof, by means of the constructor approach, of Theorem 12 of [17].

The following Lemma helps clarifying the isomorphism stated above:

Lemma 19

$\forall K\in \mathbf{C}\cap - \mathbf{C}^\blacktriangledown$, $\exists {\mathcal K}\subseteq \mathbf{C}^\blacktriangledown$ such that $K=\bigcup {\mathcal K}$.

Proof

Let $K\in \mathbf{C}\cap - \mathbf{C}^\blacktriangledown$ and $x\in K$. Clearly, $K_x\varsubsetneq K$, for minimality of $K_x$. Therefore, $\bigcup \{K_x:x\in K\}\subseteq K$. But by definition, $x\in K_x$. Therefore, $\bigcup \{K_x:x\in K\}=K$.    $\Box$

Example 11

Let $\mathbf{C}=\{\{a\},\{b\},\{a,b\},\{a,c\}\}$. $\mathbf{C}^\blacktriangledown =\{\{a\},\{b\},\{a,c\}\}$.

Theorem 16

If C is a representative covering of a set U, then C is the set of coprime elements of $\mathbf{S}_{int}(\mathbf{P}(\mathbf{C}))$.

Proof

Since C is representative, for all $K\in \mathbf{C}$, $K=K_x$ for some $x\in U$. Therefore, for all $K'\varsubsetneq K$, $x\notin K'$, by definition of $K_x$. It follows that if ${\mathcal K}\subseteq \mathbf{C}$ and $K\notin {\mathcal K}$, then $K\ne \bigcup {\mathcal K}$.    $\Box$

Justification of the definition of an “irredundant covering”. We recall that in [13] a covering C of a set U is called “irredundant” if for any $K\in \mathbf{C}$, $\mathbf{C}\cap -\{K\}$ is no longer a covering of U.

Theorem 17

C is an irredundant covering if and only if it is representative and reduced.

Proof

(A) If C is redundant, then either it is not reduced or it is not representative. Assume $K\in \mathbf{C}$ is redundant. Then for all $x\in K$, there is a $K'\in \mathbf{C}$ such that $x\in K'$. If $K\subseteq K'$ or $K'\subseteq K$, then C is not reduced. If it is reduced, then $K\cap K'\notin \mathbf{C}$, because $K\cap K'\subseteq K$ and $K\cap K'\subseteq K'$. It follows that for all $x\in K$, $K\ne K_x$. Hence K is not represented. (B) If C is either non-reduced or non-representative, then it is redundant. Trivially, if C is not reduced, and $K\varsubsetneq K'$, then K is redundant. Suppose C is reduced but there exists $K\in \mathbf{C}$ which is not represented. Then for all $x\in K$, $K\ne K_x$. It follows that for all $x\in K$, there is a $K'\ne K$ such that $x\in K'$. We conclude that K is redundant.    $\Box$

The composition of two relations $R\subseteq X\times Y$ and $Z\subseteq Y\times W$, is defined as $R\otimes Z=\{\langle x,w\rangle : \exists y\in Y(\langle x,y\rangle \in R\wedge \langle y,w\rangle \in Z\}$.

In [13] it is stated that if R is a preorder, then $R^\smile \otimes R$, is a tolerance relation. We know that $R_\mathbf{C}$ is a preorder. Now we prove that if in addition C is reduced, then $R^\smile _\mathbf{C}\otimes R_\mathbf{C}(x)=R_\mathbf{C}(x)$ for all $x\in U^\blacktriangleleft$. Since $R^\smile _\mathbf{C}\otimes R_\mathbf{C}(x)$ is a tolerance relation, as a corollary we have that if C in addition is specialised then $R_\mathbf{C}$ is an equivalence relation. From that we can deduce that $\mathbf{S}_{int}(\mathbf{P}(\mathbf{C}))$ is a Boolean algebra (i.e. Theorem 15).

Lemma 20

For any covering C of U, for all $x,y\in U$, $\langle x,y\rangle \in R^\smile _\mathbf{C}\otimes R_\mathbf{C}$ iff $\exists z(\langle z,x\rangle \in R_\mathbf{C}\wedge \langle z,y\rangle \in R_\mathbf{C})$.

The proof is trivial.

Lemma 21

For any covering C of U, $R^\smile _\mathbf{C}\otimes R_\mathbf{C}\supseteq R_\mathbf{C}$.

Proof

Let $\langle x,y\rangle \in R_\mathbf{C}$. From reflexivity, $\langle x,x\rangle \in R_\mathbf{C}$, so that from Lemma 20, $\langle x,y\rangle \in R^\smile _\mathbf{C}\otimes R_\mathbf{C}$.    $\Box$

Theorem 18

For any reduced covering C of U, $R^\smile _\mathbf{C}\otimes R_\mathbf{C}(x)=R_\mathbf{C}(x)$ for all $x\in U^\blacktriangleleft$.

Proof

Suppose $x\in U^\blacktriangleleft$ and $\langle x,y\rangle \in R^\smile _\mathbf{C}\otimes R_\mathbf{C}$. Then $\exists z$ such that $x,y\in R_\mathbf{C}(z)=n(z)=\bigcap md(z)$. Since x is representative there is a $K_x\in \mathbf{C}$ and since $x\in n(x)$, $K_x\subseteq K$ for all $K\in md(z)$. But C is reduced, so that it must be $K=K_x$ for all $K\in md(z)$. That is, $md(z)=\{K_x\}$, so that $R_\mathbf{C}(z)=R_\mathbf{C}(x)$. It follows that $y\in R_\mathbf{C}(x)$ and one concludes $R^\smile _\mathbf{C}\otimes R_\mathbf{C}\subseteq R_\mathbf{C}$. In view of Lemma 21 the proof is complete.    $\Box$