Geometric Lattice Structure of Covering-Based Rough Sets through Matroids

Theorem 3.1. Let E be a set and M(ℱ) a transversal matroid induced by a family of ℱ = {F₁, F₂, …, F_m} subsets of E, where F_i ≠ ∅  (1 ≤ i ≤ m). cl_ℱ(∅) = ∅ if and only if ℱ is a covering of E.

Proof. “⇐”: According to the definition of transversal matroid, any partial transversal is an independent set of transversal matroid. Since ℱ is a covering, any single-point set is an independent set. Based on the definition of closure operator of a matroid, we have cl_M(ℱ)(∅) = ∅.

“⇒”: Since cl_M(ℱ)(∅) = ∅, any single-point set is an independent set, that is, for all x ∈ E, there exists i_x ∈ {1,2, …, m} such that $$. Hence, $$. Thus $$. For all i ∈ {1,2, …, m}, F_i ≠ ∅ and $$, hence ℱ is a covering.

Lemma 3.2. Let 𝒞 be a covering of E. For all x ∈ E, cl_M(𝒞)({x}) is an atom of ℒ(M(𝒞)).

Proof. Since 𝒞 is a covering, we know any single-point set is an independent set. Thus for all x ∈ E, r_M(𝒞)(cl_M(𝒞)({x})) = r_M(𝒞)({x}) = 1. As we know, the atoms of a lattice are precisely the elements of height one. Combining with Lemma 2.10 and the fact that cl_M(𝒞)({x}) is a closed set, we know cl_M(𝒞)({x}) is an atom of ℒ(M(𝒞)).

Definition 3.3. Let 𝒞 = {K₁, K₂, …, K_m} be a covering of a finite set E = {x₁, x₂, …, x_n}. We define the following.

• (i)

  $$.

• (ii)

  $$.

Remark 3.4. For all i ∈ {1,2, …, s} and x ∈ A_i, there exists only one block such that x belongs to it, and there exist at least two blocks such that y belongs to them for all y ∈ B.

Proposition 3.5. Let 𝒞 be a covering of E. {A₁, A₂, …, A_s}∪{{x}  : x ∈ B} forms a partition of E.

Proof. Let P = {A₁, A₂, …, A_s}∪{{x}  : x ∈ B}. According to Definition 3.3, we know $$. Now we need to prove for all P₁, P₂ ∈ P, P₁∩P₂ = ∅. According to the definition of A, if P₁, P₂ ∈ A, then P₁∩P₂ = ∅. If P₁, P₂ ∈ {{x}  : x ∈ B}, then P₁∩P₂ = ∅ because P_i and P_j are different single-points. If P₁ ∈ A and P₂ ∈ {{x} : x ∈ B}, then P₁∩P₂ = ∅ because $$, $$ and P₂⊆B.

Proposition 3.6. 𝒞 is a partition of E if and only if B = ∅.

Proof. According to the definition of A and B, the necessity is obvious. Now we prove the sufficiency. If 𝒞 is not a partition, then there exist K_i, K_j ∈ 𝒞 such that K_i∩K_j ≠ ∅. Thus there exists x ∈ E such that x ∈ K_i∩K_j, that is, there exist at least K_i, K_j ∈ 𝒞 such that x belongs to them, hence x ∈ B. That contradicts the assumption that B = ∅.

Theorem 3.7. Let 𝒞 be a covering of E. {A₁, A₂, …, A_s}∪{{x}  : x ∈ B} is the set of atoms of lattice ℒ(M(𝒞)).

Proof. According to the definition of A_i, we may as well suppose $$. Based on 𝒞 being a covering and the definition of transversal matroid, we know any single-point set is an independent set, thus for all x ∈ A_i, {x} is an independent set. For all y ∈ A_i and y ≠ x, we know x, y ∈ K_h and x, y ∉ K_j for all 1 ≤ j ≤ m and j ≠ h, thus x and y cannot be chosen from different blocks in the covering 𝒞. That shows that {x, y} is not an independent set according to the definition of transversal matroid. Hence, {x} is a maximal independent set included in A_i, that is, r_M(𝒞)(A_i) = 1. Next, we need to prove A_i is a closed set. Since A_i⊆cl_M(𝒞)(A_i), we need to prove cl_M(𝒞)(A_i)⊆A_i, that is, y ∉ A_i implies y ∉ cl_M(𝒞)(A_i). If y ∉ A_i, based on the fact that 𝒞 is a covering and the definition of A_i, then there exists j ≠ h such that y ∈ K_j. Thus {x, y}(x ∈ A_i) is an independent set. That implies y ∉ cl_M(𝒞)(A_i), thus cl_M(𝒞)(A_i) = A_i. Hence, A_i ∈ ℒ(M(𝒞)). Combining with r_M(𝒞)(A_i) = 1, we know for all 1 ≤ i ≤ m, A_i is an atom of lattice ℒ(M(𝒞)).

According to the definition of transversal matroid and the fact that 𝒞 is a covering, any single-point set is an independent set. Thus for all x ∈ B, r_M(𝒞)({x}) = 1. For all y ∈ E and y ≠ x, if y ∈ B, then there exist at least two blocks containing y according to the definition of B. We may as well suppose y ∈ K_k, K_t and x ∈ K_l, K_p, where {K_k, K_t} may be the same as {K_l, K_p}. Based on this, {x, y} is an independent set. This implies y ∉ cl_M(𝒞)({x}). If y ∉ B, then we may as well suppose y ∈ A_i, thus y ∈ K_h for the definition of A_i, where K_h may be the same with K_l or K_p. Based on this, x and y can be chosen from different blocks in covering 𝒞, thus {x, y} is an independent set. That implies y ∉ cl_M(𝒞)({x}). From above discussion, we have cl_M(𝒞)({x}) = {x}. Hence, {x} ∈ ℒ(M(𝒞)) for all x ∈ B. Combining with r_M(𝒞)({x}) = 1, we know {x} is an atom of lattice ℒ(M(𝒞)) for all x ∈ B.

Next, we will prove the set of atoms of lattice ℒ(M(𝒞)) cannot be anything but {A₁, A₂, …, A_s}∪{{x}  : x ∈ B}. According to Lemma 3.2, we know {cl_M(𝒞)({x})  : x ∈ E} is the set of atoms of lattice ℒ(M(𝒞)). Similar to the proof of the second part, we know that if x ∈ B then cl_M(𝒞)({x}) = {x}. If x ∉ B, then x belongs to one of elements in A. We may as well suppose x ∈ A_i. Combining A_i is an atom with ∅⊆cl_M(𝒞)({x})⊆cl_M(𝒞)(A_i) = A_i, we have cl_M(𝒞)({x}) = A_i. Hence, {A₁, A₂, …, A_s}∪{{x}  : x ∈ B} is the set of atoms of lattice ℒ(M(𝒞)).

Lemma 3.8. Let 𝒞 be a covering of E. For all A_i ∈ A, if |A_i | ≥ 2, then x and y are parallel in M(𝒞) for all x, y ∈ A_i.

Proof. According to the definition of A_i, we may as well suppose $$, where 1 ≤ h ≤ m. For all x, y ∈ A_i, then x, y ∈ K_h, and for all j ∈ {1,2, …, m} and j ≠ h, we have x ∉ K_j and y ∉ K_j. Thus {x, y} is not an independent set. Based on the definition of transversal matroid and the fact that 𝒞 is a covering, any single-point set is an independent set. Thus {x} or {y} is an independent set. Hence, x, y are parallel in M(𝒞).

Proposition 3.9. Let 𝒞 be a covering of E. M(𝒞) is a simple matroid if and only if |A_i | = 1 for all A_i ∈ A.

Proof. “⇒”: Since M(𝒞) is a simple matroid, it does not contain parallel elements. If there exists A_i ∈ A such that |A_i | ≠ 1, then |A_i | ≥ 2 because A_i ≠ ∅. According to Lemma 3.8, we know for all x, y ∈ A_i, x, y are parallel which contradicts the assumption that M(𝒞) is a simple matroid. Hence, for all A_i ∈ A, |A_i | = 1.

“⇐”: According to the definition of parallel element, if |A_i | = 1 for all A_i ∈ A, then M(𝒞) does not contain parallel elements; otherwise, we may as well suppose x, y are parallel, then there exists only one block which contains x, y. Hence, there exists A_i ∈ A such that x, y ∈ A_i, that is, |A_i | ≥ 2. This contradicts the fact that |A_i | = 1 for all A_i ∈ A. Based on the definition of transversal matroid and the fact that 𝒞 is a covering, any single-point set is an independent set, thus M(𝒞) does not contain loops. Hence, M(𝒞) does not contain parallel elements and loops which implies that M(𝒞) is a simple matroid.

Corollary 3.10. Let 𝒫 = {P₁, P₂, …, P_m} be a partition of E. 𝒫 is the set of atoms of lattice ℒ(M_𝒫).

Corollary 3.11. Let 𝒫 = {P₁, P₂, …, P_m} be a partition of E. M(𝒫) is a simple matroid if and only if |P_i | = 1 for all P_i ∈ P.

Proposition 3.12. Let 𝒞 be a covering of E. For all x, y ∈ E, cl_M(𝒞)({x, y}) covers cl_M(𝒞)({x}) if and only if there does not exist A_i ∈ A such that x, y ∈ A_i.

Proof. “⇐”: For all x, y ∈ E, {x}⊆{x, y}, then cl_M(𝒞)({x})⊆cl_M(𝒞)({x, y}) and 1 = r_M(𝒞)(cl_M(𝒞)({x})) ≤ r_M(𝒞)(cl_M(𝒞)({x, y})) = r_M(𝒞)({x, y})≤|{x, y}| = 2. Now we need to prove r_M(𝒞)(cl_M(𝒞)({x, y})) = 2. If r_M(𝒞)(cl_M(𝒞)({x, y})) = r_M(𝒞)({x, y}) = 1 = r_M(𝒞)({x}), then {x, y} ∉ ℐ(𝒞), that is, there is only one block contains x, y. It means that there exists A_i ∈ A such that x, y ∈ A_i. That contradicts the hypothesis. Hence, r_M(𝒞)(cl_M(𝒞)({x, y})) = 2, that is, cl_M(𝒞)({x, y}) covers cl_M(𝒞)({x}).

“⇒”: For all x, y ∈ E, if there exists A_i ∈ A such that x, y ∈ A_i, then there is only one block contains x, y, thus x, y ∉ ℐ(𝒞), hence {x, y}⊆cl_M(𝒞)({x}). That implies cl_M(𝒞)({x, y})⊆cl_M(𝒞)({x}) which contradicts the assumption that cl_M(𝒞)({x, y}) covers cl_M(𝒞)({x}).

Theorem 3.13. Let M be a matroid and ℒ(M) the set of all closed sets of M.

• (1)

  For all X, Y ∈ ℒ(M), (X, Y) is a modular pair of ℒ(M) if and only if r_M(X ∪ Y)+r_M(X∩Y) = r_M(X)+r_M(Y).

• (2)

  For all X ∈ ℒ(M), X is a modular element of ℒ(M) if and only if r_M(X ∪ Y)+r_M(X∩Y) = r_M(X)+r_M(Y), for all Y ∈ ℒ(M).

Proof. (1) According to Lemmas 2.8 and 2.10, we know (X, Y) is a modular pair of ℒ(M) if and only if r_M(X) + r_M(Y) = r_M(X∨Y) + r_M(X∧Y) = r_M(cl_M(X ∪ Y)) + r_M(X∩Y) = r_M(X ∪ Y) + r_M(X∩Y).

(2) It comes from the definition of modular element and (1).

Theorem 3.14. Let 𝒞 be a covering of E. For all i, j ∈ Γ.

• (1)

  (A_i, A_j) is a modular pair of ℒ(M(𝒞)).

• (2)

  A_i is a modular element of ℒ(M(𝒞)).

Proof. (1) Since 𝒞 is a covering, cl_M(𝒞)(∅) = ∅. A_i and A_j are atoms, so A_i∩A_j = ∅. According to Theorem 3.13, we need to prove r_M(𝒞)(A_i ∪ A_j) + r_M(𝒞)(A_i∩A_j) = r_M(𝒞)(A_i) + r_M(𝒞)(A_j), that is, r_M(𝒞)(A_i ∪ A_j) = 2. According to the submodular inequality of r_M(𝒞), we have r_M(𝒞)(A_i ∪ A_j) + r_M(𝒞)(A_i∩A_j) ≤ r_M(𝒞)(A_i) + r_M(𝒞)(A_j), that is, 1 ≤ r_M(𝒞)(A_i ∪ A_j) ≤ 2. If r_M(𝒞)(A_i ∪ A_j) = 1 = r_M(𝒞)(A_i), then A_j⊆cl_M(𝒞)(A_i) = A_i which contradicts that A_i∩A_j = ∅.

(2) A_i is a modular element of ℒ(M(𝒞)) if and only if r_M(𝒞)(A_i ∪ A)+r_M(𝒞)(A_i∩A) = r_M(𝒞)(A_i)+r_M(𝒞)(A) for all A ∈ ℒ(M(𝒞)).

Case  1. If A_i and A are comparable, that is, A_i⊆A, then r_M(𝒞)(A_i ∪ A)+r_M(𝒞)(A_i∩A) = r_M(𝒞)(A_i)+r_M(𝒞)(A).

Case  2. If A_i and A are not comparable, there are two cases. One is that A is an atom of ℒ(M(𝒞)), the other is that A is not an atom of ℒ(M(𝒞)). If A is an atom of ℒ(M(𝒞)), then we obtain the result from (1). If A is not an atom of ℒ(M(𝒞)), then A∩A_i = ∅. Hence, r_M(𝒞)(A)≤r_M(𝒞)(A ∪ A_i)≤r_M(𝒞)(A)  +  1. If r_M(𝒞)(A ∪ A_i) = r_M(𝒞)(A), then A_i⊆cl_M(𝒞)(A) = A which contradicts that A_i∩A = ∅. Hence, r_M(𝒞)(A ∪ A_i) = r_M(𝒞)(A) + 1.

In a word, for all A ∈ ℒ(M(𝒞)), r_M(𝒞)(A_i ∪ A) + r_M(𝒞)(A_i∩A) = r_M(𝒞)(A_i) + r_M(𝒞)(A), that is, A_i is a modular element of ℒ(M(𝒞)) for all i ∈ Γ.

Corollary 3.15. Let 𝒫 be a partition of E. For all P_i, P_j ∈ 𝒫:

• (1)

  (P_i, P_j) is a modular pair of ℒ(M(𝒫)).

• (2)

  P_i is a modular element of ℒ(M(𝒫)).

Lemma 3.16 (see [19].)Let ℒ_E be a lattice of subsets of a set E such that ℒ_E is closed under intersection, and contains ∅ and E. Suppose that f is a nonnegative, integer-valued, submodular function on ℒ_E for which f(∅) = 0. Let ℐ(ℒ_E, f) = {X⊆E : f(X)≥|X∩T | ,   for  all  T ∈ ℒ_E}. ℐ(ℒ_E, f) is the collection of independent sets of a matroid on E.

Theorem 3.17. Let 𝒞 be a covering of E. We define ℐ(ℒ(M(𝒞)), r_M(𝒞)) = {X⊆E : r_M(𝒞)(Y)≥|X∩Y|,  for  all  Y ∈ ℒ(M(𝒞))}, then M(E, ℐ(ℒ(M(𝒞), r_M(𝒞))) is a matroid.

Lemma 3.18. Let M be a matroid. X is an independent set of M if and only if for all closed set Y of M, r_M(Y)≥|X∩Y|.

Proof. “⇒”: Since for all closed set Y, X∩Y⊆X and X is an independent set, X∩Y is an independent set of M according to the independent set axiom of a matroid. Hence, we have r_M(Y) ≥ r_M(X∩Y)≥|X∩Y|.

“⇐”: For all closed set Y, r_M(Y)≥|X∩Y|. Especially, for Y = cl_M(X), we have r_M(X) = r_M(cl_M(X)) = r_M(Y)≥|X∩Y | = |X|. Hence, X is an independent set of matroid M.

Theorem 3.19. Let 𝒞 be a covering of E. For all X⊆E, ℐ(ℒ(M(𝒞), r_M(𝒞)) = ℐ(M(𝒞)) and r_ℒ(M(𝒞))(X) = r_M(𝒞)(X).

Proof. According to Lemma 3.18, we know that ℐ(ℒ(M(𝒞), r_M(𝒞)) = {X⊆E : r_M(𝒞)(Y)≥|X∩Y | ,   for  all  Y ∈ ℒ(M(𝒞))} and ℐ(𝒞) = {X⊆E : r_M(𝒞)(X) = |X|} are equivalent, that is, M(E, ℐ(ℒ(M(𝒞), r_M(𝒞))) and M(E, ℐ(𝒞)) are equivalent. So does r_ℒ(M(𝒞))(X) = r_M(𝒞)(X) for all X⊆E.

Lemma 3.20 (see [25].)If 𝒫 is a partition of E and M is the matroid, then cl_M = R^* for all X⊆E.

Lemma 3.21. Let 𝒫 be a partition of E. For all X, Y ∈ ℒ(M(𝒫)), X∨Y = X ∪ Y.

Proof. X∨Y = cl_M(𝒫)(X ∪ Y) = R^*(X ∪ Y) = R^*(X) ∪ R^*(Y) = cl_M(𝒫)(X) ∪ cl_M(𝒫)(Y) = X ∪ Y.

Proposition 3.22. Let 𝒫 be a partition of E. For all X⊆E, ℐ(ℒ(M(𝒫)), r_M(𝒫)) = ℐ(M(𝒫)) and r_ℒ(M(𝒫))(X) = r_M(𝒫)(X).

Proof. We need to prove only ℐ(ℒ(M(𝒫)), r_M(𝒫)) = ℐ(M(𝒫)). For all X ∈ ℐ(ℒ(M(𝒫)), r_M(𝒫)), then r_M(𝒫)(Y)≥|X∩Y| for all Y ∈ ℒ(M(𝒫)). Since P_i ∈ ℒ(M(𝒫)), |X∩P_i | ≤ r_M(𝒫)(P_i) = 1. Thus X ∈ ℐ(M(𝒫)). Hence, ℐ(ℒ(M(𝒫), r_M(𝒫))⊆ℐ(M(𝒫)). According to Lemmas 2.9 and 3.21, for all Y ∈ ℒ(M(𝒫)), there exists K⊆{1,2, …, m} such that Y = ⋁_i∈KP_i = ⋃_i∈K P_i. Thus r_M(𝒫)(Y) = |K| and |X∩Y | = |X∩(⋃_i∈K P_i)| = |⋃_i∈K (X∩P_i)| = ∑_i∈K|X∩P_i|. For all X ∈ ℐ(M(𝒫)), |X∩P_i | ≤ 1. Hence, |X∩Y | = ∑_i∈K|X∩P_i|≤|K | = r_M(𝒫)(Y), that is, ℐ(M(𝒫))⊆ℐ(ℒ(M(𝒫)), r_M(𝒫)).

Proposition 4.1. Let 𝒞 be a covering of E. SH has the following properties:

• (1)

  SH(∅) = ∅,

• (2)

  X⊆SH(X) for all X⊆E,

• (3)

  SH(X ∪ Y) = SH(X) ∪ SH(Y) for all X, Y⊆E,

• (4)

  x ∈ SH({y})⇔y ∈ SH({x}) for all x, y ∈ E,

• (5)

  X⊆Y⊆E⇒SH(X)⊆SH(Y),

• (6)

  for all x, y ∈ E, y ∈ SH(X ∪ {x}) − SH(X), then x ∈ SH(X ∪ {y}).

Proof. (1)–(5) were proven in [30, 32, 33]. Here we prove only (6). According to (3), we know SH(X ∪ {x}) − SH(X) = SH(X) ∪ SH({x}) − SH(X) = SH({x}) − SH(X). If y ∈ SH(X ∪ {x}) − SH(X), then y ∈ SH({x}). According to (4) and (5), we have x ∈ SH({y})⊆SH(X ∪ {y}).

Proposition 4.2. Let 𝒞 be a covering of E. For all X⊆E, SH(SH(X)) = SH(X) if and only if {I(x) : x ∈ E} induced by 𝒞 forms a partition of E.

Proof. “⇐”: According to (2), (5) of Proposition 4.1, we have SH(X)⊆SH(SH(X)). Now we prove SH(SH(X))⊆SH(X). For all x ∈ SH(SH(X)), there exists y ∈ SH(X) such that x ∈ I(y). Since y ∈ SH(X), there exists z ∈ X such that y ∈ I(z). According to the definition of I(y), we know y ∈ I(y), thus I(z)∩I(y) ≠ ∅. For {I(x) : x ∈ E} forms a partition, I(z) = I(y). Since x ∈ I(y), x ∈ I(z), that is, x ∈ SH(X), thus SH(SH(X))⊆SH(X).

“⇒”: In order to prove {I(x)  : x ∈ E} forms a partition, we need to prove that for all x, y ∈ E, if I(x)∩I(y) ≠ ∅, then I(x) = I(y). If I(x)∩I(y) ≠ ∅, then there exists z ∈ I(x)∩I(y). For SH(SH({x})) = ∪{I(u)  : u ∈ I(x)} and z ∈ I(x), then I(z)⊆SH(SH({x})) = SH({x}) = I(x). Based on the definition of I(z) and z ∈ I(x), we have x ∈ I(z), thus I(x)⊆SH(SH({z})) = SH({z}) = I(z). Hence, I(x) = I(z). Similarly, we can obtain I(y) = I(z), thus I(x) = I(z) = I(y).

Theorem 4.3. Let 𝒞 be a covering of E. SH is a closure operator of a matroid if and only if {I(x)  : x ∈ E} induced by 𝒞 forms a partition of E.

Proof. It comes from Propositions 4.1 and 4.2 and (2), (5), and (6) of Proposition 2.3.

Definition 4.4. Let 𝒞 be a covering of E. We define ℐ^′ = {I⊆E:|I∩I(x_i)| ≤ 1,   for  all  i ∈ {1,2, …, s}}.

Proposition 4.5. Let 𝒞 be a covering of E. If {I(x)  : x ∈ E} induced by 𝒞 forms a partition of E, then M(E, ℐ_SH(𝒞)) is a matroid and ℐ_SH(𝒞) = ℐ^′.

Proof. Let ℐ_cl = {I⊆E : for  all  x ∈ I, x ∉ cl (I − {x})}. we know that if an operator cl  satisfies (1)–(4) of Proposition 2.3, M(E, ℐ_cl) is a matroid. {I(x)  : x ∈ U} induced by 𝒞 forms a partition, hence, M(E, ℐ_SH(𝒞)) is a matroid. Since SH(I) = ⋃_y∈I I(y), SH(I − {x}) = ⋃_y∈I−{x} I(y). According to the definition of I(x), we know x ∈ I(x). On one hand, for all I ∈ ℐ_SH(𝒞), we know that for all x ∈ I, x ∉ SH(I − {x}), that is, for all y ∈ I and y ≠ x, x ∉ I(y). If I ∉ ℐ^′, that is, there exists 1 ≤ i ≤ s such that |I∩I(x_i)| ≥ 2, then we may as well suppose there exist u, v such that u, v ∈ I(x_i) and u, v ∈ I. Since u ∈ I(u), v ∈ I(v) and {I(x) : x ∈ E} forms a partition, I(u) = I(v) = I(x_i). Based on that, we know there exists u ∈ I and u ≠ v such that u ∈ I(v), that implies contradiction. Hence, I ∈ ℐ^′, that is, ℐ_SH(𝒞)⊆ℐ^′. On the other hand, if I ∉ ℐ_SH(𝒞), then there exists x ∈ I such that x ∈ SH(I − {x}) = ⋃_y∈I−{x} I(y). That implies that there exists y ∈ I and y ≠ x such that x ∈ I(y). Since x ∈ I(x) and {I(x)  : x ∈ U} forms a partition, I(x) = I(y). Thus x, y ∈ I∩I(x), that implies |I∩I(x)| ≥ 2, that is, I ∉ ℐ^′. Hence, ℐ^′⊆ℐ_SH(𝒞).

Proposition 4.6. Let 𝒞 be a covering of E. If {I(x)  : x ∈ E} = {I(x₁), I(x₂), …, I(x_s)} induced by 𝒞 forms a partition of E, then

• (1)

  X is a base of M(E, ℐ_SH(𝒞)) if and only if |X∩I(x_i)| = 1 for all i ∈ {1,2, …, s}. Moreover, M(E, ℐ_SH(𝒞)) has |I(x₁)||I(x₂)| ⋯ |I(x_s)| bases.

• (2)

  For all X⊆E, r_SH(X) = |{I(x_i) : I(x_i)∩X ≠ ∅,   i = 1,2, …, s}|.

• (3)

  X is a dependent set of M(E, ℐ_SH(𝒞)) if and only if there exists I(x_i)∈{I(x) : x ∈ E} such that |I(x_i)∩X | > 1.

• (4)

  X is a circuit of M(E, ℐ_SH(𝒞)) if and only if there exists I(x_i)∈{I(x) : x ∈ E} such that X⊆I(x_i) and |X | = 2.

Proof. (1) According to the definition of base of a matroid, we know that X is a base of M(E, ℐ_SH(𝒞))⇔X is a maximal independent set of M(E, ℐ_SH(𝒞))⇔|X∩I(x_i)| = 1 for all i ∈ {1,2, …, s} because ℐ_SH(𝒞) = ℐ^′. Since X is a base of M(E, ℐ_SH(𝒞)) and I(x₁), I(x₂), …, I(x_s) are different, M(E, ℐ_SH(𝒞)) has |I(x₁)||I(x₂)| ⋯ |I(x_s)| bases.

(2) According to the definition of rank function, we know r_SH(X) = |B_X | = |{I(x_i) :|B_X∩I(x_i)| = 1}|≤|{I(x_i) : X∩I(x_i) ≠ ∅}|, where B_X is a maximal independent set included in X. Now we just need to prove the inequality |{I(x_i) :|B_X∩I(x_i)| = 1}|<|{I(x_i)  : X∩I(x_i) ≠ ∅}| does not hold; otherwise, there exists 1 ≤ i ≤ s such that I(x_i)∩X ≠ ∅ and I(x_i)∩B_X = ∅ because B_X ∈ ℐ_SH(𝒞). Thus there exists e_i ∈ I(x_i)∩X such that B_X ∪ {e_i}⊆X and B_X ∪ {e_i} ∈ ℐ_SH(𝒞). That contradicts the assumption that B_X is a maximal independent set included in X. Hence, r_SH(X) = |{I(x_i) : I(x_i)∩X ≠ ∅,   i = 1,2, …, s}|.

(3) According to the definition of dependent set, we know that X is a dependent set ⇔X ∉ ℐ_SH(𝒞)⇔ there exists i ∈ {1,2, …, s} such that |X∩I(x_i)| > 1.

(4) “⇒”: As we know, a circuit is a minimal dependent set. X is a circuit of M(E, ℐ_SH(𝒞)), then there exists i ∈ {1,2, …, s} such that |I(x_i)∩X | = 2. Now we just need to prove |X | = 2; otherwise, we may as well suppose X = {x, y, z} where x, y ∈ I(x_i)∩X. Thus we can obtain |(X − {z})∩I(x_i)| = 2, that is, X − {z} ∉ ℐ_SH(𝒞). That contradicts the minimality of circuit. Combining |X | = 2 with |I(x_i)∩X | = 2, we have X⊆I(x_i).

“⇐”: Since |X | = 2, we may as well suppose X = {x, y}, and |X∩I(x_i)| = 2 because there exists I(x_i)∈{I(x)  : x ∈ E} such that X⊆I(x_i), thus X is a dependent set. For all j ∈ {1,2, …, s}, |{x}∩I(x_j)| ≤ 1 and |{y}∩I(x_j)| ≤ 1 which implies {x} and {y} are independent sets, hence X is a circuit of M(E, ℐ_SH(𝒞)).

Proposition 4.7. Let 𝒞 be a covering of E. If {I(x) : x ∈ E} = {I(x₁), I(x₂), …, I(x_s)} forms a partition of E, then

• (1)

  I(x₁), I(x₂), …, I(x_s) are all atoms of ℒ(M_SH).

• (2)

  For all x, y ∈ E and x ≠ y, there does not exist I(z)∈{I(x₁), I(x₂), …, I(x_s)} such that x, y ∈ I(z) if and only if SH({x, y}) covers SH({x}) or SH({y}).

• (3)

  For all i, j ∈ {1,2, …, s}, (I(x_i), I(x_j)) is a modular pair of ℒ_SH(M(𝒞)).

• (4)

  For all i ∈ {1,2, …, s}, I(x_i) is a modular element of ℒ_SH(M(𝒞)).

Proof. (1) comes from Corollary 3.10, Theorem 4.3 and Proposition 4.5. Based on Proposition 3.12 and Theorem 4.3, we can obtain (2). According to Corollary 3.15, Theorem 4.3 and Proposition 4.5, it is easy to obtain (3) and (4).

Lemma 4.8. Let 𝒞 be a covering of E and K ∈ 𝒞. If K is an immured element, then I(x) is the same in 𝒞 as in 𝒞 − {K}.

Proof. If x ∉ K, then I_𝒞(x) = I_𝒞−{K}(x). If x ∈ K, then $$. Since K is an immured element, there exists x ∈ K^′ such that K⊆K^′. Thus $$. Hence, I(x) is the same in 𝒞 as in 𝒞 − {K}.

Proposition 4.9. Let 𝒞 be a covering of E. If exclusion(𝒞) is a partition of E, then {I(x)  : x ∈ E} induced by 𝒞 also forms a partition of E.

Proof. Since exclusion(𝒞) is a partition of E, {I(x)  : x ∈ E} induced by exclusion(𝒞) forms a partition. Suppose {K₁, K₂, …, K_s} is the set of all immured elements of 𝒞. According to Lemma 4.8, we know for all x ∈ E, I(x) is the same in exclusion(𝒞) as in exclusion(𝒞)∪{K₁}. Thus {I(x)  : x ∈ E} induced by exclusion(𝒞)∪{K₁} forms a partition of E. And the rest may be deduced by analogy, we know that for all x ∈ E, I(x) is the same in exclusion(𝒞) as in 𝒞, thus {I(x)  : x ∈ E} induced by 𝒞 forms a partition of E.

Proposition 4.10. Let 𝒞 be a covering of E. {I(x)  : x ∈ E} induced by 𝒞 forms a partition of E if and only if 𝒞 satisfies (TRA) condition: For all x, y, z ∈ E,   x, z ∈ K₁ ∈ 𝒞,   y, z ∈ K₂ ∈ 𝒞, there exists K₃ ∈ 𝒞 such that x, y ∈ K₃.

Proof. “⇐”: For all x, y ∈ E, I(x)∩I(y) = ∅ or I(x)∩I(y) ≠ ∅. If I(x)∩I(y) ≠ ∅, then there exists z ∈ I(x) and z ∈ I(y). According to the definition of I(x) and I(y), there exist K₁, K₂ such that x, z ∈ K₁ and y, z ∈ K₂. According to the hypothesis, we know there exists K₃ ∈ 𝒞 such that x, y ∈ K₃. Now we need to prove only I(x) = I(y). For all u ∈ I(x), there exists K ∈ 𝒞 such that u, x ∈ K. Since x, y ∈ K₃, there exists K^′ ∈ 𝒞 such that u, y ∈ K^′, that is, u ∈ I(y), thus I(x)⊆I(y). Similarly, we can prove I(y)⊆I(x). Hence, I(x) = I(y), that is, {I(x)  : x ∈ E} forms a partition of E.

“⇒”: For all x, y, z ∈ E,   x, z ∈ K₁ ∈ 𝒞 and y, z ∈ K₂ ∈ 𝒞, we can obtain z ∈ I(x) and z ∈ I(y). That implies I(x)∩I(y) ≠ ∅. Since {I(x)  : x ∈ E} forms a partition of E, I(x) = I(y). Thus there exists K₃ ∈ 𝒞 such that x, y ∈ K₃.

Theorem 4.11. Let 𝒞 be a covering of E. 𝒞 satisfies (TRA) condition if and only if SH induced by 𝒞 is a closure operator of a matroid.

Proof. It comes from Theorem 4.3 and Proposition 4.10.

Proposition 4.12 (see [26].)Let 𝒞 be a covering of E. XH has the following properties:

• (1)

  XH(E) = E,

• (2)

  XH(∅) = ∅,

• (3)

  X⊆XH(X) for all X⊆E,

• (4)

  XH(X ∪ Y) = XH(X) ∪ XH(Y) for all X, Y⊆E,

• (5)

  XH(XH(X)) = XH(X) for all X⊆E,

• (6)

  for all X⊆Y⊆E⇒XH(X)⊆XH(Y).

Proposition 4.13. Let 𝒞 be a covering of E. For all x, y ∈ E and X⊆E, XH satisfies

Proof. ⇒: For all x, y ∈ E, if N(x)∩N(y) ≠ ∅, then there exists z ∈ N(x)∩N(y). Let X = ∅. According to (2) of Proposition 4.12, we know that if y ∈ XH({x}) then x ∈ XH(y), that is, if x ∈ N(y) then y ∈ N(x). Since z ∈ N(x), N(z)⊆N(x). According to the assumption, we also have x ∈ N(z), that is, N(x)⊆N(z). Thus N(x) = N(z). Similarly, z ∈ N(y), we have N(z) = N(y). Thus N(x) = N(z) = N(y). Hence, {N(x)  : x ∈ E} forms a partition of E.

⇐: Since XH(X ∪ Y) = XH(X) ∪ XH(Y) for all X, Y⊆U, y ∈ XH(X ∪ {x}) − XH(X) = XH(X) ∪ XH({x}) − XH(X) = XH({x}) − XH(X). Now we prove x ∈ XH({y}). Since y ∈ XH({x}), x ∈ N(y). Because the fact that {N(x)  : x ∈ E} forms a partition of E, x ∈ N(x) and x ∈ N(y), we have N(x) = N(y), thus y ∈ N(x), that is, x ∈ XH({y}). Hence x ∈ XH({y})⊆XH(X ∪ {y}).

Theorem 4.14. Let 𝒞 be a covering of E. {N(x)  : x ∈ E} induced by 𝒞 forms a partition of E if and only if XH is a closure operator of a matroid.

Proof. It comes from (3), (5), and (6) of Propositions 4.12 and 4.13.

Definition 4.15. Let 𝒞 be a covering of E. We define ℐ^′′ = {I⊆E :|I∩N(x_i)| ≤ 1,   for  all  i ∈ {1,2, …, t}}.

Proposition 4.16. Let 𝒞 be a covering of E. If {N(x)  : x ∈ E} induced by 𝒞 forms a partition of E, then M(E, ℐ_XH(𝒞)) is a matroid and ℐ_XH(𝒞) = ℐ^′′.

Proposition 4.17. Let 𝒞 be a covering of E. If {N(x)  : x ∈ E} = {N(x₁), N(x₂), …, N(x_t)} induced by 𝒞 forms a partition of E, then

• (1)

  X is a base of M(E, ℐ_XH(𝒞)) if and only if |X∩N(x_i)| = 1  (1 ≤ i ≤ t), and M(E, ℐ_XH(𝒞)) has |N(x₁)||N(x₂)| ⋯ |N(x_s)| bases.

• (2)

  For all X⊆E, r_XH(X) = |{N(x_i)  : N(x_i)∩X ≠ ∅,   1 ≤ i ≤ t}|.

• (3)

  X is a circuit of M(E, ℐ_XH(𝒞)) if and only if there exists N(x_i)∈{N(x)  : x ∈ E} such that X⊆N(x_i) and |X | = 2.

• (4)

  X is a dependent set of M(E, ℐ_XH(𝒞)) if and only if there exists N(x_i)∈{N(x) : x ∈ E} such that |N(x_i)∩X | > 1.

• (5)

  {N(x₁), N(x₂), …, N(x_t)} is the set of all atoms of lattice ℒ_XH(M(𝒞)).

• (6)

  For all x, y ∈ E and x ≠ y, there does not exist N(z)∈{N(x₁), N(x₂), …, N(x_t)} such that x, y ∈ N(z) if and only if XH({x, y}) covers XH({x}) or XH({y}).

• (7)

  For all i, j ∈ {1,2, …, t}, (N(x_i), N(x_j)) is a modular pair of lattice ℒ_XH(M(𝒞)).

• (8)

  For all i ∈ {1,2, …, t}, N(x_i) is a modular element of lattice ℒ_XH(M(𝒞)).

Lemma 4.18. Let 𝒞 be a covering on E and K be reducible in 𝒞. For all x ∈ U, N(x) is the same in 𝒞 as in 𝒞 − {K}.

Proof. For all x ∈ E, Md(x) is the same for covering 𝒞 and covering 𝒞 − {K}, so N(x) = ∩Md(x) is the same for the covering 𝒞 and covering 𝒞 − {K}.

Proposition 4.19. Let 𝒞 be a covering of E. If reduct(𝒞) is a partition of E, then {N(x)  : x ∈ E} induced by 𝒞 is also a partition of E.

Proof. Since reduct(𝒞) is a partition of E, {N(x)  : x ∈ E} induced by reduct(𝒞) forms a partition of E. Suppose {K₁, K₂, …, K_s} is the set all reducible elements of 𝒞. According to Lemma 4.18, we know that for all x ∈ E, N(x) is the same in reduct(𝒞) as in reduct(𝒞)∪{K₁}, thus {N(x)  : x ∈ E} induced by reduct(𝒞)∪{K₁} forms a partition of E. And the rest may be deduced by analogy, then we can obtain for all x ∈ E, N(x) is the same in reduct(𝒞) as in 𝒞, thus {N(x)  : x ∈ E} induced by 𝒞 forms a partition of E.

Proposition 4.20. Let 𝒞 be a covering of E. If 𝒞 satisfies (EQU) condition: For all K ∈ 𝒞, for all x, y ∈ K, the number of blocks which contain x is equal to that of blocks which contain y, then {N(x)  : x ∈ E} induced by 𝒞 forms a partition of E.

Proof. For all x, y ∈ E, if N(x)∩N(y) ≠ ∅, then there exists z ∈ E such that z ∈ N(x) and z ∈ N(y), that is, the blocks which contain x also contain z and the blocks which contain y also contain z. Hence, there exist $$ such that x, z ∈ K_i and $$. Without loss of generality, we suppose {K₁, K₂, …, K_s} is the set of all blocks which contain x and $$ is the set of all blocks which contain y. Since the number of blocks which contain z is equal to that of blocks which contain x, {K₁, K₂, …, K_s} is the set of all blocks which contain z, thus $$. Hence, N(x)⊆N(y). Similarly, we can prove N(y)⊆N(x). Hence, N(x) = N(y), that is, {N(x)  : x ∈ E} forms a partition of E.

Corollary 4.21. Let 𝒞 be a covering of E. If reduct(𝒞) is a partition of E, then XH is a closure operator of a matroid.

Corollary 4.22. Let 𝒞 be a covering on E. If 𝒞 satisfies (EQU) condition, then XH is a closure operator of a matroid.

Proposition 5.1. Let 𝒞 be a covering of E. If SH induced by 𝒞 is a closure operator, then ℐ_SH(𝒞)⊆ℐ(𝒞) and ℒ_SH(M(𝒞))⊆ℒ(M(𝒞)).

Proof. Since SH induced by 𝒞 is a closure operator, {I(x)  : x ∈ E} = {I(x₁), I(x₂), …, I(x_s)} forms a partition of E. For all I ∈ ℐ_SH(𝒞), suppose I = {i₁, i₂, …, i_α}  (α ≤ s) such that $$ and $$. According to the definition of I(x), there exists $$ such that $$. Since {I(x₁), I(x₂), …, I(x_s)} forms a partition of E, thus $$ are different blocks. According to the definition of transversal matroid, we have I ∈ ℐ. Hence, ℐ_SH(𝒞)⊆ℐ(𝒞).

For all X ∈ ℒ_SH(M(𝒞)), X = SH(X) = ⋃_x∈X I(x). Now we need to prove X ∈ ℒ(M(𝒞)), that is, cl_M(𝒞)(X) = X = {x∣r_M(𝒞)(X) = r_M(𝒞)(X ∪ {x})}. Since X⊆cl_M(𝒞)(X), if cl_M(𝒞)(X) ≠ X, then cl_M(𝒞)(X)  ⊈  X, that is, there exists y ∉ X such that r_M(𝒞)(X) = r_M(𝒞)(X ∪ {y}). Suppose T = {t₁, t₂, …, t_t}  (t ≤ s) is a maximal independent set included in X, then {t₁, t₂, …, t_t}⊆X = ⋃_x∈X I(x) and there exist different K₁, K₂, …, K_t such that for all i ∈ {1,2, …, t},   t_i ∈ K_i. Since y ∉ X, I(x)∩I(y) = ∅ for all x ∈ X. Based on {I(x₁), I(x₂), …, I(x_s)} forms a partition of E, there exists K⊆I(y) such that K₁, K₂, …, K_t,   K are different blocks and y ∈ K, thus T ∪ {y} is a maximal independent set included in X ∪ {y}. Hence, we have r_M(𝒞)(X ∪ {y}) = r_M(𝒞)(X) + 1 which contradicts r_M(𝒞)(X) = r_M(𝒞)(X ∪ {y}). Thus we can obtain cl_M(𝒞)(X) = X.

Proposition 5.2. Let 𝒞 be a covering of E. If SH induced by 𝒞 is a closure operator, then for all x ∈ E,   I(x) ∈ ℒ(M(𝒞)).

Proof. Since SH induced by 𝒞 is a closure operator, {I(x)  : x ∈ E} forms a partition of E. Thus, for all x ∈ E, SH(I(x)) = ⋃_y∈I(x) I(y) = I(x), that is, I(x) ∈ ℒ_SH(M(𝒞)). According to Proposition 5.1, I(x) ∈ ℒ(M(𝒞)).

Example 5.3. Let 𝒞 = {{1,2}, {1,3}, {2,3}, {4,5}}. I(1) = I(2) = I(3) = {1,2, 3}, I(4) = I(5) = {4,5}. Let T = {1,2}. T ∈ ℐ(𝒞) but T ∉ ℐ_SH(𝒞) because |T∩I(1)| = 2. Hence, ℐ(𝒞)  ⊈  ℐ_SH(𝒞). ℒ_SH(M(𝒞)) = {∅, {1,2, 3}, {4,5}, {1,2, 3,4, 5}}. ℒ(M(𝒞)) = {∅, {1}, {2}, {3}, {4,5}, {1,2}, {1,3}, {2,3}, {1,4, 5}, {2,4, 5}, {3,4, 5}, {1,2, 3}, {1,2, 4,5}, {1,3, 4,5}, {2,3, 4,5}, {1,2, 3,4, 5}}. It is obvious that ℒ(M(𝒞))  ⊈  ℒ_SH(M(𝒞)). The structures of ℒ_SH(M(𝒞)) and ℒ(M(𝒞)) are showed in Figure 1.

Remark 5.4. Let 𝒞 be a covering of E. Although XH induced by 𝒞 is a closure operator, it has no relationship between ℐ_XH(𝒞) and ℐ(𝒞), and has no relationship between ℒ_XH(M(𝒞)) and ℒ(M(𝒞)).

Example 5.5. Let 𝒞 = {K₁, K₂, K₃, K₄} be a covering of E = {a, b, c, d, e, f, g, h, i}, where K₁ = {a, b, i}, K₂ = {a, b, c, d, e, f}, K₃ = {f, g, h}, K₄ = {c, d, e, g, h, i}. Then N(a) = N(b) = {a, b}, N(c) = N(d) = N(e) = {c, d, e}, N(f) = {f}, N(g) = N(h) = {g, h} and N(i) = {i}. Let T = {a, c, f, g, i}. It is clear that T ∈ ℐ_XH(𝒞), but T ∉ ℐ(𝒞) because |T∩K₂ | = 2, thus ℐ_XH(𝒞)  ⊈  ℐ(𝒞). Let T^′ = {a, c, d}. It is clear that T^′ ∈ ℐ(𝒞), but T^′ ∉ ℐ_XH because |T^′∩N(c)| = 2, thus ℐ(𝒞)  ⊈  ℐ_XH(𝒞). Let X = {a, b, i}. X ∈ ℒ_XH(M(𝒞)) for XH(X) = X. However, X ∉ ℒ(M(𝒞)) for cl_M(𝒞)(X) = {a, b, c, d, e, i}. Let X = {a}. X ∈ ℒ(M(𝒞)) for cl_M(𝒞)(X) = X. However, X ∉ ℒ_XH(M(𝒞)) for XH(X) = {a, b} ≠ X.

Proposition 5.6. Let 𝒞 be a covering of E. If XH and SH induced by 𝒞 are closure operators, then ℐ_SH(𝒞)⊆ℐ_XH(𝒞) and ℒ_SH(M(𝒞))⊆ℒ_XH(M(𝒞)).

Proof. If XH and SH induced by 𝒞 are closure operators, then {N(x)  : x ∈ E} and {I(x)  : x ∈ E} form a partition of E, respectively. For all x ∈ X, N(x) = ⋂_x∈K K⊆⋃_i∈K K = I(x), thus {N(x)  : x ∈ E} is finer than {I(x)  : x ∈ E}. Based on this, we can obtain ℐ_SH(𝒞)⊆ℐ_XH(𝒞).

For all X ∈ ℒ_SH(M(𝒞)), X = SH(X) = ⋃_x∈X I(x). Since x ∈ N(x)⊆I(x), X = ⋃_x∈X {x}⊆⋃_x∈X N(x)⊆⋃_x∈X I(x) = X, thus, X = ⋃_x∈X N(x), that is, X ∈ ℒ_XH(M(𝒞)). Hence, ℒ_SH(M(𝒞))⊆ℒ_XH(M(𝒞)).

Example 5.7. From Example 5.3, we know N(1) = {1}, N(2) = {2}, N(3) = {3}, N(4) = N(5) = {4,5} and I(1) = I(2) = I(3) = {1,2, 3}, I(4) = I(5) = {4,5}. Let I = {1,2}. It is clear that I ∈ ℐ_XH(M(𝒞)) but I ∉ ℐ_SH(M(𝒞)) because |I∩I(1)| = 2. Hence ℐ_XH(M(𝒞))  ⊈  ℐ_SH(M(𝒞)). ℒ_SH(M(𝒞)) = {∅, {1,2, 3}, {4,5}, {1,2, 3,4, 5}} and ℒ_XH(M(𝒞)) = {∅, {1}, {2}, {3}, {4,5}, {1,2}, {1,3}, {2,3}, {1,4, 5}, {2,4, 5}, {3,4, 5}, {1,2, 3}, {1,2, 4,5}, {1,3, 4,5}, {2,3, 4,5}, {1,2, 3,4, 5}}. It is obvious that ℒ_XH(M(𝒞))  ⊈  ℒ_SH(M(𝒞)). The structures of them are showed in Figure 1.

Theorem 5.8. If 𝒞 is a partition of E, then ℐ_SH(𝒞) = ℐ_XH(𝒞) = ℐ(𝒞) and ℒ_SH(M(𝒞)) = ℒ_XH(M(𝒞)) = ℒ(M(𝒞)).

Proof. Since 𝒞 is a partition of E, I(x) = N(x) = K(x ∈ K) for all x ∈ E, and for all X⊆E, SH(X) = XH(X) = R^*(X). Thus ℐ_SH(𝒞) = ℐ_XH(𝒞) = ℐ(𝒞) and ℒ_SH(M(𝒞)) = ℒ_XH(M(𝒞)) = ℒ(M(𝒞)).

Theorem 5.9. Let ℱ be a family of subset of E and K ∈ ℱ. ℐ(ℱ − {K})⊆ℐ(ℱ).

Proof. For all I ∈ ℐ(ℱ − {K}), we may as well suppose I = {i₁, i₂, …, i_t} where i₁, i₂, …, i_t ∈ E. According to the definition of transversal matroid, there exist different blocks K₁, K₂, …, K_t ∈ ℱ satisfy K_i ≠ K and i_j ∈ K_j for all 1 ≤ i, j ≤ t. Thus I ∈ ℐ(ℱ).

Example 5.10. Let ℱ = {K₁, K₂, K₃} be a family of subset of E = {1,2, 3,4}, where K₁ = {1,2}, K₂ = {1,3}, K₃ = {3}. ℐ(ℱ) = 2^E, ℐ(ℱ − {K₃}) = {∅, {1}, {2}, {3}, {1,3}, {1,2}{2,3}}. Hence, ℐ(ℱ)  ⊈  ℐ(ℱ − {K}).

Corollary 5.11. Let 𝒞 be a covering of E and K ∈ 𝒞. If K is reducible, then ℐ(𝒞 − {K})⊆ℐ(𝒞).

Corollary 5.12. Let 𝒞 be a covering of E. ℐ(reduct(𝒞))⊆ℐ(𝒞).

Corollary 5.13. Let 𝒞 be a covering of E and K ∈ 𝒞. If K is an immured element, then ℐ(𝒞 − {K})⊆ℐ(𝒞).

Corollary 5.14. Let 𝒞 be a covering of E. ℐ(exclusion(𝒞))⊆ℐ(𝒞).

Theorem 5.15. Let ℱ be a family of subset of E and for all K ∈ ℱ. ℒ(M(ℱ − {K}))⊆ℒ(M(ℱ)).

Proof. First, we prove cl_ℱ({x})⊆cl_ℱ−{K}({x}) for all x ∈ E. For all y ∉ cl_ℱ−{K}({x}), {x, y} ∈ ℐ_ℱ−{K}⊆ℐ_ℱ. Thus y ∉ cl_ℱ({x}) which implies cl_ℱ({x})⊆cl_ℱ−{K}({x}).

Second, we need to prove that any atom of ℒ(M(ℱ − {K})) is a closed set of ℒ(M(ℱ)), that is, cl_ℱ(cl_ℱ−{K}({x})) = cl_ℱ−{K}({x}). Since cl_ℱ−{K}({x})⊆cl_ℱ(cl_ℱ−{K}({x}))⊆cl_ℱ−{K}(cl_ℱ−{K}({x})) = cl_ℱ−{K}({x}), cl_ℱ(cl_ℱ−{K}({x})) = cl_ℱ−{K}({x}).

Third, we need to prove ℒ(M(ℱ − {K}))⊆ℒ(M(ℱ)). For all X⊆ℒ(M(ℱ − {K})), $$ because ℒ(M(ℱ − {K})) is a atomic lattice, thus X ∈ ℒ(M(ℱ)).

Example 5.16. Based on Example 5.10, we have ℒ(M(ℱ)) = 2^E and ℒ(M(ℱ − {K₃})) = {∅, {1}, {2}, {3}, {1,2, 3}}. It is clear that ℒ(M(ℱ))  ⊈  ℒ(M(ℱ − {K₃})).

Corollary 5.17. Let 𝒞 be a covering of E and K ∈ 𝒞. If K is reducible, then ℒ(M(𝒞 − {K}))⊆ℒ(M(𝒞)).

Corollary 5.18. Let 𝒞 be a covering of E. ℒ(M(reduct(𝒞)))⊆ℒ(M(𝒞)).

Corollary 5.19. Let 𝒞 be a covering of E and K ∈ 𝒞. If K is an immured element, then ℒ(M(𝒞 − {K}))⊆ℒ(M(𝒞)).

Corollary 5.20. Let 𝒞 be a covering of E. ℒ(M(exclusion(𝒞)))⊆ℒ(M(𝒞)).

Example 5.21. Let E = {1,2, 3} and 𝒞 = {K₁, K₂, K₃} where K₁ = {1,2}, K₂ = {1,3}, K₃ = {1,2, 3}. I(1) = I(2) = I(3) = {1,2, 3}, thus {I(x)  : x ∈ E} forms a partition of E. Hence, SH is a closure operator induced by 𝒞. It is clear that K₃ is a reducible element, 𝒞 − {K₃} = {K₁, K₂}. Then the indiscernible neighborhoods induced by 𝒞 − {K₃} are I(1) = {1,2, 3}, I(2) = {1,2}, I(3) = {1,3}. we find that {I(x)  : x ∈ E} cannot form a partition of E. Hence, SH is not a closure operator induced by 𝒞 − {K₃}.

Theorem 5.22. Let 𝒞 be a covering of E and K an immured element of 𝒞. If SH induced by 𝒞 is a closure operator, then SH induced by covering 𝒞 − {K} is also a closure operator. Moreover, ℐ_SH(𝒞) = ℐ_SH(𝒞 − {K}) and ℒ_SH(M(𝒞)) = ℒ_SH(M(𝒞 − {K})).

Proof. It comes from Lemma 4.8 and Theorem 4.3.

Corollary 5.23. Let 𝒞 be a covering of E. If SH induced by 𝒞 is a closure operator, then SH induced by exclusion(𝒞) is also a closure operator. Moreover, ℐ_SH(𝒞) = ℐ_SH(exclusion(𝒞)) and ℒ_SH(M(𝒞)) = ℒ_SH(M(exclusion(𝒞))).

Theorem 5.24. Let 𝒞 be a covering of E and K be a reducible element. If XH induced by 𝒞 is a closure operator, then XH induced by covering 𝒞 − {K} is also a closure operator. Moreover, ℐ_XH(𝒞) = ℐ_XH(𝒞 − {K}) and ℒ_XH(M(𝒞)) = ℒ_XH(M(𝒞 − {K})).

Proof. Since XH induced by 𝒞 is a closure operator, {N(x) : x ∈ E} induced by 𝒞 forms a partition. Based on the definition of ℐ_XH(𝒞) and Lemma 4.18, XH induced by 𝒞 − {K} is also a closure operator and ℐ_XH(𝒞) = ℐ_XH(𝒞 − {K}).

Corollary 5.25. Let 𝒞 be a covering of E. If XH induced by 𝒞 is a closure operator, then XH induced by reduct(𝒞) is a closure operator. Moreover, ℐ_XH(𝒞) = ℐ_XH(reduct(𝒞)) and ℒ_XH(M(𝒞)) = ℒ_XH(M(reduct(𝒞))).

Example 5.26. Suppose K₁ = {1}, K₂ = {1,2}, K₃ = {2,3}, K₄ = {3}, K₅ = {1,2, 3} and 𝒞₁ = {K₁, K₂, K₃, K₄}. Then N(1) = {1}, N(2) = {2} and N(3) = {3}, thus {N(x) : x ∈ E} forms a partition of E. Hence, XH is a closure operator. It is clear that K₁ is an immured element, and the neighborhoods induced by 𝒞 − {K₁} are N(1) = {1,2}, N(2) = {2} and N(3) = {3}, thus {N(x) : x ∈ E} cannot form a partition of E. Hence, 𝒞 − {K₁} is not a covering which makes XH be a closure operator.