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(4.3a) Two overlapped sets with

tolerance relation

(4.3b) Two disjoint set with tol-

erance relation

(4.3c) Inclusion and tolerance

relation

FIGURE 4.3: Tolerance relation between sets; Space X is partitioned into squares, Sets

A(solid line), set Y dashed lines have tolerance relation. A∩Y are colored in gray

The relation
∼
=
X,ϕ,ε
is a special case of the tolerance near set relation
∼
=
B,ε
(Peters, 2009)

(see, also, the weak nearness relation in (Peters and Wasilewski, 2009)). For conciseness,
∼
=

is used to denote
∼
=
X,ϕ,ε
. In a manner similar to what L. Polkowski has done in defining the

equivalence class of a rough set A⊂X, we introduce the following equation as a tolerance

class for a rough subset of the universe, A⊂X, with A 6= A:

A
/
∼

={Y⊆U|A∩Y 6= φ}.

(4.22)

=

In other words we are suggesting that two rough sets of universe A, Y⊂X have tolerance

relation(
∼
=
) with each other iff A∩Y 6= φ. Notice that equation 4.21 is defined on elements

in a set X. By contrast, equation 4.22 is defined for two sets.

Proposition 4.4.1
∼
= is a tolerance relation.

Proof. To show
∼
= is a tolerance relation, we have to show that it is reflexive and symmetric:

•if A = Y⇒A = Y therefore A∩Y 6= φ; So A
∼
=
Y . This means that
∼
=
is a

reflexive relation, A
∼
=
A.

•if A
∼
= Y⇒A∩Y 6= φ⇒Y∩A 6= φ⇒Y
∼
= Y ; therefore
∼
= is a symmetric

relation.

€

Proposition 4.4.2 Equation4.20 is the specialization of equation 4.22.

Proof. We have to show that the equivalence class of a set A is a special case of its toler-

ance class. In other words, an equivalence class A
/∼
is included in A
/
∼

. Let Y⊂X and

=

. Since Y∈A
/∼
, we have A = Y and A = Y .

We only need the first part; by A = Y we get A∩Y = A 6= φ. Therefore A
/∼
is included

in A
/
∼

Y∈A
/∼
, we have to show that Y∈A
/
∼

=

.
€

=

Figure 4.3 are showing three different sets that have tolerance relation with set A⊂X. In

figure 4.3b the set Y⊂X is disjoint from the set A with A∩Y = φ, but has a tolerance

relation with A because A∩Y 6= φ.

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