# MINIMAL CLOSED SETS AND MAXIMAL CLOSED SETS

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1 MINIMAL CLOSED SETS AND MAXIMAL CLOSED SETS FUMIE NAKAOKA AND NOBUYUKI ODA Received 20 December 2005; Revised 28 May 2006; Accepted 6 August 2006 Some properties of minimal closed sets and maximal closed sets are obtained, which are dual concepts of maximal open sets and minimal open sets, respectively. Common properties of minimal closed sets and minimal open sets are clarified; similarly, common properties of maximal closed sets and maximal open sets are obtained. Moreover, interrelations of these four concepts are studied. Copyright 2006 Hindawi Publishing Corporation. All rights reserved. 1. Introduction Some properties of minimal open sets and maximal open sets are studied in [1, 2]. In this paper, we define dual concepts of them, namely, maximal closed set and minimal closed set. These four types of subsets appear in finite spaces, for example. More generally, minimal open sets and maximal closed sets appear in locally finite spaces such as the digital line. Minimal closed sets and maximal open sets appear in cofinite topology, for example. We have to study these four concepts to understand the relations among their properties. Some of the results in [1, 2] can be dualized using standard techniques of general topology. But we have to study the dual results carefully to understand the duality which we propose in this paper and in [2]. For example, considering the interrelation of these four concepts,we see that some results in [1, 2] can be generalized further. In [1] we called a nonempty open set U of X a minimal open set if any open set which is contained in U is or U. But to consider duality, we have to consider only proper nonempty open set U of X, as in the following definitions: a proper nonempty open subset U of X is said to be a minimal open set if any open set which is contained in U is or U. A proper nonempty open subset U of X is said to be a maximal open set if any open set which contains U is X or U. In this paper, we will use the following definitions. A proper nonempty closed subset F of X is said to be a minimal closed set if any closed set which is contained in F is or F. A proper nonempty closed subset F of X is said to be a maximal closed set if any closed set which contains F is X or F. Hindawi Publishing Corporation International Journal of Mathematics and Mathematical Sciences Volume 2006, Article ID 18647, Pages 1 8 DOI /IJMMS/2006/18647

2 2 Minimal closed sets and maximal closed sets Let F be a subset of a topological space X. Then the following duality principle holds: 1 F is a minimal closed set if and only if F is a maximal open set; 2 F is a maximal closed set if and only if F is a minimal open set. In Sections 2 and 3, we will show some results on minimal closed sets and maximal closed sets. In Section 4, we will obtain some further results on minimal open sets and maximal open sets. The symbol Λ \ Γ means difference of index sets, namely, Λ \ Γ = Λ Γ, and the cardinalityofasetλ is denoted by Λ in the following arguments. A subset M ofatopological space X is called a pre-open set if M IntClM and a subset M is called a pre-closed set if M is a pre-open set. 2. Minimal closed sets In this section, we prove some results on minimal closed sets. The following result shows that a fundamental result of [1, Lemma 2.2] also holds for closed sets; it is the dual result of [2, Lemma 2.2]. Lemma Let F be a minimal closed set and N a closed set. Then F N = or F N. 2 Let F and S be minimal closed sets. Then F S = or F = S. The proof of Lemma 2.1 is omitted, since it is obtained by an argument similar to the proofof[1, Lemma 2.2]. Now, we generalize the dual result of [2, Theorem 2.4]. Theorem 2.2. Let F and be minimal closed sets for any element λ of Λ. 1 If F, then there exists an element λ of Λ such that F =. 2 If F for any element λ of Λ, then F =. Proof. 1 Since F,wegetF = F = F. If F for any element λ of Λ,thenF = for any element λ of Λ by Lemma 2.12, hence we have = F = F. This contradicts our assumption that F is a minimal closed set. Thusthereexistsanelementλof Λ such that F =. 2 If F, then there exists an element λ of Λ such that F.By Lemma 2.12, we have = F, which is a contradiction. Corollary 2.3. Let be a minimal closed set for any element λ of Λ and F μ for any elements λ and μ of Λ with λ μ.ifγ is a proper nonempty subset of Λ, then \Γ γ Γ F γ =. The following result is a generalization of the dual result of [2, Theorem 2.5]. Theorem 2.4. Let and F γ be minimal closed sets for any element λ of Λ and γ of Γ.Ifthere exists an element γ of Γ such that F γ for any element λ of Λ, then γ Γ F γ. Proof. Suppose that an element γ of Γ satisfies F γ for any element λ of Λ.If γ Γ F γ,thenwegetf γ.bytheorem 2.21, there exists an element λ of Λ such that F γ =, which is a contradiction. The dual result of [2, Theorem 4.6] is stated as follows.

3 F. Nakaoka and N. Oda 3 Theorem 2.5. Let be a minimal closed set for any element λ of Λ and F μ for any elements λ and μ of Λ with λ μ. IfΓ is a proper nonempty subset of Λ, then γ Γ F γ. Proof. Let κ be any element of Λ \ Γ. ThenF κ γ Γ F γ = γ ΓF κ F γ = and F κ = F κ = F κ.if γ Γ F γ =,thenwehave =F κ. This contradicts our assumption that F κ is a minimal closed set. The following theorem is the dual result of [2, Theorem 4.2] and is the key to the proof of Theorem2.7, which is closelyconnectedwith [2, Theorem 4.7]. Theorem 2.6. Assume that Λ 2 and let be a minimal closed set for any element λ of Λ and F μ for any elements λ and μ of Λ with λ μ. Then F μ and hence X for any element μ of Λ. Proof. By Corollary 2.3,we have the result. Theorem 2.7 recognition principle for minimal closed sets. Assume that Λ 2 and let be a minimal closed set for any element λ of Λ and F μ for any elements λ and μ of Λ with λ μ.thenforanyelementμ of Λ, F μ = Proof. Let μ be an element of Λ.ByTheorem 2.6,wehave = F μ = = F μ = F μ. F μ As an application of Theorem 2.7, we prove the following result which is the dual result of [2, Theorem 4.8]. Theorem 2.8. Let be a minimal closed set for any element λ of a finite set Λ and F μ for any elements λ and μ of Λ with λ μ.if is an open set, then is an open set for any element λ of Λ. Proof. Let μ be an element of Λ. ByTheorem 2.7, wehavef μ = =. By our assumption, it is seen that isanopenset.hencef μ is an open set.

4 4 Minimal closed sets and maximal closed sets As an immediate consequence of Theorem 2.6, we have the following result which is the dual of [2, Theorem 4.9]. Theorem 2.9. Assume that Λ 2 and let be a minimal closed set for any element λ of Λ and F μ for any elements λ and μ of Λ with λ μ.if = X, then { λ Λ} is the set of all minimal closed sets of X. Let ={ λ Λ} be a set of minimal closed sets. We call = the coradical of. The following result about coradical is obtained by [2, Theorem 4.13]. Theorem Let be a minimal closed set for any element λ of Λ and F μ for any elements λ and μ of Λ with λ μ. If is an open set, then is an open set for any element μ of Λ. Theorem Let F be a minimal closed set. Then, IntF = F or IntF =. Proof. If we put U = F in [2, Theorem 3.5], then we have the result. If S is a proper subset of a minimal closed set, then IntS = and hence S is a preclosed set cf. [2, Corollary 3.7]. Theorem Let be a minimal closed set for any element λ of a finite set Λ. If Int, then there exists an element λ of Λ such that Int =. Proof. Since Int, there exists an element x of Int. Then there exists an open set U such that x U and hence there exists an element μ of Λ such that x F μ.bytheorems2.6 and 2.7, x U = F μ.sinceu X isanopenset,weseethatx is an element of IntF μ and hence IntF μ. Therefore there exists an element μ of Λ such that IntF μ = F μ by Theorem Remark If Λ is an infinite set, then Theorem 2.12 does not hold. For example, see [2, Example 5.3]. Theorem 2.14 the law of interior of coradical. Let be a minimal closed set for any element λ of a finite set Λ and F μ for any elements λ and μ of Λ with λ μ. LetΓ be a subset of Λ such that Int = Fλ for any λ Γ, Int = for any λ Λ \ Γ. 2.3 Then Int = λ Γ = if Γ =. Proof. By [2, Theorem 5.4], we have the result. 3. Maximal closed sets The following lemma is the dual of Lemma 2.1. Lemma Let F be a maximal closed set and N a closed set. Then F N = X or N F. 2 Let F and S be maximal closed sets. Then F S = X or F = S.

5 F. Nakaoka and N. Oda 5 Let be a maximal closed set for any element λ of Λ. Let ={ λ Λ}. Wecall = the radical of. Corollary 3.2. Let F and be maximal closed sets for any element λ of Λ. IfF for any element λ of Λ, then F = X. Proof. By Lemma 3.12, we have the result. Theorem 3.3. Let F and be maximal closed sets for any element λ of Λ.If F, then there exists an element λ of Λ such that F =. Proof. Since F, wegetf = F = F. If F = X for any element λ of Λ,thenwehaveX = F = F. This contradicts our assumption that F is a maximal closed set. Then there exists an element λ of Λ such that F X. By Lemma 3.12, we have the result. Corollary 3.4. Let and F γ be maximal closed sets for any elements λ Λ and γ Γ. If γ Γ F γ, then for any element λ of Λ, there exists an element γ Γ such that = F γ. Corollary 3.5. Let F α, F β,andf γ be maximal closed sets which are different from each other. Then F α F β F α F γ. 3.1 Theorem 3.6. Let be a maximal closed set for any element λ of Λ and F μ for any elements λ and μ of Λ with λ μ. IfΓ is a proper nonempty subset of Λ, then \Γ γ Γ F γ \Γ. Proof. Let γ be any element of Γ. If \Γ F γ,thenweget = F γ for some λ Λ \ Γ by Theorem 3.3. This contradicts our assumption. Therefore we have \Γ γ Γ F γ. On the other hand, since γ Γ F γ = γ Λ\Λ\Γ F γ \Γ,wehave γ Γ F γ \Γ. Theorem 3.7. Let be a maximal closed set for any element λ of Λ and F μ for any elements λ and μ of Λ with λ μ. IfΓ is a proper nonempty subset of Λ, then γ Γ F γ. Proof. Let κ be any element of Λ \ Γ.ThenF κ γ Γ F γ = γ ΓF κ F γ = X and F κ = F κ = F κ.if γ Γ F γ =,thenwehavex = F κ. This contradicts our assumption that F κ is a maximal closed set. Theorem 3.8. Assume that Λ 2 and let be a maximal closed set for any element λ of Λ and F μ for any elements λ and μ of Λ with λ μ. Then F μ and hence for any element μ of Λ. Proof. Let μ be any element of Λ.ByLemma 3.12, we have F μ for any element λ of Λ with λ μ.then F μ. Therefore we have F μ. If =,wehavex = F μ. This contradicts our assumption that F μ is a maximal closed set. Therefore we have.

6 6 Minimal closed sets and maximal closed sets Theorem 3.9 decomposition theorem for maximal closed sets. Assume that Λ 2 and let be a maximal closed set for any element λ of Λ and F μ for any elements λ and μ of Λ with λ μ. Then for any element μ of Λ, F μ =. 3.2 Proof. Let μ be an element of Λ.ByTheorem 3.8,wehave = F μ = = F μ = F μ. F μ 3.3 Theorem Let be a maximal closed set for any element λ of Λ and F μ for any elements λ and μ of Λ with λ μ. If is an open set, then is an open set for any element λ of Λ. Proof. By Theorem 3.9, we have F μ = =. Then isanopenset.hencef μ is an open set by our assumption. Theorem Assume that Λ 2 and let be a maximal closed set for any element λ of Λ and F μ for any elements λ and μ of Λ with λ μ.if =, then { λ Λ} is the set of all maximal closed sets of X. Proof. If there exists another maximal closed set F ν of X which is not equal to for any element λ of Λ, then = = {ν}\{ν}.bytheorem 3.8, weget {ν}\{ν}. This contradicts our assumption. Theorem Assume that Λ 2 and let be a maximal closed set for any element λ of Λ and F μ for any elements λ and μ of Λ with λ μ. IfInt =, then { λ Λ} is the set of all maximal closed sets of X. Proof. If there exists another maximal closed set F of X which is not equal to for any element λ of Λ, then F by Theorem 3.8. It follows that Int Int F = F. This contradicts our assumption. The proof of the following lemma is immediate and is omitted. Lemma Let A and B be subsets of X.IfA B = X, anda B is an open set and A is a closed set, then B is an open set.

7 F. Nakaoka and N. Oda 7 Proposition Let be a closed set for any element λ of Λ and F μ = X for any elements λ and μ of Λ with λ μ. If is an open set, then is an open set for any element μ of Λ. Proof. Let μ be any element of Λ. Since F μ = X for any element λ of Λ with λ μ, we have F μ = F μ = X. SinceF μ = is an open set by our assumption, is an open set by Lemma Theorem Let be a maximal closed set for any element λ of Λ and F μ for any elements λ and μ of Λ with λ μ. If is an open set, then is an open set for any element μ of Λ. Proof. By Lemma 3.12, we have F μ = X for any elements λ and μ of Λ with λ μ. By Proposition 3.14,wehave is an open set. 4. Minimal open sets and maximal open sets In this section, we record some results on minimal open sets and maximal open sets, which are not proved in [1, 2]. The proofs are omitted since they are obtained by dual arguments. Let ={U λ λ Λ} be a set of minimal open sets. We call = U λ the coradical of. By an argument similar to Theorem 2.2, we have the following result. Theorem 4.1. Let U and U λ be minimal open sets for any element λ of Λ. 1 If U U λ, then there exists an element λ of Λ such that U = U λ. 2 If U U λ for any element λ of Λ, then U λ U =. Corollary 4.2. Let U λ be a minimal open set for any element λ of Λ and U λ U μ for any elements λ and μ of Λ with λ μ.ifγ is a proper nonempty subset of Λ, then \Γ U λ γ Γ U γ =. The following results are shown by the arguments similar to the proofs of Theorems 2.4, 2.5, and2.6, respectively. Theorem 4.3. Let U λ and U γ be minimal open sets for any elements λ Λ and γ Γ. If there exists an element γ of Γ such that U λ U γ for any element λ of Λ, then γ Γ U γ U λ. Theorem 4.4. Let U λ be a minimal open set for any element λ of Λ and U λ U μ for any elements λ and μ of Λ with λ μ. IfΓ is a proper nonempty subset of Λ, then γ Γ U γ U λ. Theorem 4.5. Assume that Λ 2 and let U λ be a minimal open set for any element λ of Λ and U λ U μ for any elements λ and μ of Λ with λ μ. Then U μ U λ and hence U λ X for any element μ of Λ. Weobtainthefollowingresultforminimalopensets cf. Theorems2.7 and3.9. Theorem 4.6 recognition principle for minimal open sets. Assume that Λ 2 and let U λ be a minimal open set for any element λ of Λ and U λ U μ for any elements λ and μ of Λ

8 8 Minimal closed sets and maximal closed sets with λ μ. Then for any element μ of Λ, U μ = U λ U λ. 4.1 Theorem 4.7. Let U λ be a minimal open set for any element λ of any set Λ and U λ U μ for any elements λ and μ of Λ with λ μ. If U λ is a closed set, then U λ is a closed set for any element λ of Λ cf. Theorems 2.8 and Theorem 4.8. Assume that Λ 2 and let U λ be a minimal open set for any element λ of Λ and U λ U μ for any elements λ and μ of Λ with λ μ. If U λ = X, then {U λ λ Λ} is the set of all minimal open sets of X cf. Theorems 2.9 and Theorem 4.9. Assume that Λ 2 and let U λ be a minimal open set for any element λ of Λ and U λ U μ for any elements λ and μ of Λ with λ μ. IfCl U λ = X, then {U λ λ Λ} is the set of all minimal open sets of X cf. Theorem Example 4.10 the digital line. The digital line is the set Z of the intergers equipped with the topology τ having a family of subsets S ={{2k 1,2k,2k +1} k Z} as a subbase. We consider a set of minimal open sets {U k ={2k +1} k Z}. ThenCl k Z U k = Cl{2k +1 k Z} = Z and hence {U k ={2k +1} k Z} is the set of all minimal open sets in Z,τ. Finally we consider maximal open sets. The following results are the duals of Theorems 2.21 and 2.4. Theorem Let U and U λ be maximal open sets for any element λ of Λ.IfU U λ, then there exists an element λ of Λ such that U = U λ. Theorem Let U λ and U γ be maximal open sets for any elements λ of Λ and γ of Γ. If there exists an element γ of Γ such that U λ U γ for any element λ of Λ, then γ Γ U γ U λ. References [1] F. Nakaoka and N. Oda, Some applications of minimal open sets, International Journal of Mathematics and Mathematical Sciences , no. 8, [2], Some properties of maximal open sets, International Journal of Mathematics and Mathematical Sciences , no. 21, Fumie Nakaoka: Department of Applied Mathematics, Fukuoka University, Nanakuma Jonan-ku, Fukuoka , Japan address: Nobuyuki Oda: Department of Applied Mathematics, Fukuoka University, Nanakuma Jonan-ku, Fukuoka , Japan address:

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