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PII. S0161171203207262 http://ijmms.hindawi.com

© Hindawi Publishing Corp.

SOME PROPERTIES OF MAXIMAL OPEN SETS

FUMIE NAKAOKA and NOBUYUKI ODA Received 11 July 2002

Some fundamental properties of maximal open sets are obtained, such as decom- position theorem for a maximal open set. Basic properties of intersections of max- imal open sets are established, such as the law of radical closure.

2000 Mathematics Subject Classification: 54A05, 54D99.

1. Introduction. A proper nonempty open subsetUof a topological space Xis said to be amaximal open set if any open set which containsU isX or U. In [2], we study minimal open sets. Although the definition of the maximal open set is obtained by “dualizing” the definition of the minimal open set, the properties of them are quite different, as we see in this paper, especially the results in the last two sections. The purpose of this paper is to prove some fundamental properties of maximal open sets and establish a part of the foundation of the theory of maximal open sets in topological spaces.

InSection 2, we prove some basic results which are necessary for the subse- quent arguments. We obtain a relation among maximal open sets inTheorem 2.5. At the end of this section, we show that for any proper nonempty cofinite open subsetV, there exists, at least, one maximal open setUwhich contains V(Theorem 2.7).

InSection 3, we study some relations among closure, interior, and maximal open sets. As an application, we prove a result about a preopen set (Theorem 3.11).

Letᐁ= {Uλ|λ∈Λ}be a set of some maximal open setsUλ. Then, we refer to the intersection= ∩λ∈ΛUλas theradicalofᐁ. In the last two sections, we study various properties of radicals.

InSection 4, we prove fundamental properties of radicals of maximal open sets. We establish a very useful decomposition theorem for a maximal open set inTheorem 4.7.Theorem 4.7will be applied to proveTheorem 4.8.Theorem 4.9gives a sufficient condition for the set of all maximal open sets. In the rest of this section, we study the case when radicals are closed sets.

InSection 5, we consider the closure of the radicals of maximal open sets.

We establish “The law of radical closure” inTheorem 5.4.

2. Maximal open sets. Let(X,τ)be a topological space.

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Definition2.1. A proper nonempty open subsetU ofX is said to be a maximal open setif any open set which containsUisXorU.

Lemma2.2. (1)LetUbe a maximal open set andW an open set. Then,U∪ W=XorW⊂U.

(2)LetUandVbe maximal open sets. Then,U∪V=XorU=V.

Proof. (1) LetWbe an open set such thatU∪WX. SinceUis a maximal open set andU⊂U∪W, we haveU∪W=U. Therefore,W⊂U.

(2) IfU∪VX, thenU⊂VandV⊂Uby (1). ThereforeU=V.

Proposition2.3. LetUbe a maximal open set. Ifxis an element ofU, then for any open neighborhoodWofx,W∪U=XorW⊂U.

Proof. ByLemma 2.2(1), we have the result.

Theorem2.4. LetUα,Uβ, andUγbe maximal open sets such thatUαUβ. IfUα∩Uβ⊂Uγ, thenUα=Uγ orUβ=Uγ.

Proof. We see that Uα∩Uγ=Uα

Uγ∩X

=Uα Uγ

Uα∪Uβ

(byLemma 2.2(2))

=Uα Uγ∩Uα

Uγ∩Uβ

= Uα∩Uγ

Uγ∩Uα∩Uβ

=

Uα∩Uγ

Uα∩Uβ

(byUα∩Uβ⊂Uγ

=Uα Uγ∪Uβ

.

(2.1)

Hence we haveUα∩Uγ=Uα∩(Uγ∪Uβ). IfUγUβ, thenUγ∪Uβ=X, and henceUα∩Uγ=Uα; namely,Uα⊂Uγ. SinceUαandUγare maximal open sets, we haveUα=Uγ.

Theorem2.5. LetUα,Uβ, andUγbe maximal open sets, which are different from each other. Then,

Uα∩Uβ⊂Uα∩Uγ. (2.2)

Proof. IfUα∩Uβ⊂Uα∩Uγ, then we see that Uα∩Uβ

Uβ∩Uγ

Uα∩Uγ

Uβ∩Uγ

(2.3)

hence,

Uβ Uα∪Uγ

Uα∪Uβ

∩Uγ. (2.4)

SinceUα∪Uγ=X=Uα∪Uβ, we haveUβ⊂Uγ. It follows thatUβ=Uγ, which contradicts our assumption.

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Proposition2.6. LetUbe a maximal open set andxan element ofU. Then,

U= ∪{W|W is an open neighborhood ofxsuch thatW∪UX}. (2.5) Proof. ByProposition 2.3and the fact thatUis an open neighborhood of x, we have

U⊂ ∪{W|Wis an open neighborhood ofxsuch thatW∪UX} ⊂U. (2.6) Therefore, we have the result.

Finally, we prove an existence theorem of maximal open sets for special cases. We refer to the complement of any finite subset as acofinitesubset.

Theorem2.7. LetVbe a proper nonempty cofinite open subset. Then, there exists, at least, one (cofinite) maximal open setUsuch thatV⊂U.

Proof. IfVis a maximal open set, we may setU=V. IfVis not a maximal open set, then there exists an (cofinite) open setV1such thatVV1X. IfV1

is a maximal open set, we may setU=V1. IfV1is not a maximal open set, then there exists an (cofinite) open setV2such thatVV1V2X. Continuing this process, we have a sequence of open sets

VV1V2···Vk···. (2.7) SinceVis a cofinite set, this process repeats only finitely. Then, finally, we get a maximal open setU=Vnfor some positive integern.

3. Closure, interior, and maximal open sets. We begin with the following theorem.

Theorem3.1. LetUbe a maximal open set andxan element ofX−U. Then, X−U⊂Wfor any open neighborhoodWofx.

Proof. Sincex∈X−U, we haveW⊂Ufor any open neighborhoodW of x. Then,W∪U=XbyLemma 2.2(1). Therefore,X−U⊂W.

Corollary3.2. LetUbe a maximal open set. Then, either of the following (1) and (2) holds:

(1) for eachx∈X−Uand each open neighborhoodWofx,W=X; (2) there exists an open setW such thatX−U⊂WandWX.

Proof. If (1) does not hold, then there exists an elementx ofX−U and an open neighborhood W of x such that WX. By Theorem 3.1, we have X−U⊂W.

Corollary3.3. LetUbe a maximal open set. Then, either of the following (1) and (2) holds:

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(1) for eachx∈X−Uand each open neighborhoodWofx, we haveX−UW;

(2) there exists an open setW such thatX−U=WX.

Proof. Assume that (2) does not hold. Then, byTheorem 3.1, we haveX− U⊂W for each x∈X−U and each open neighborhood W of x. Hence, we haveX−UW.

Theorem3.4. LetUbe a maximal open set. Then,Cl(U)=XorCl(U)=U. Proof. SinceUis a maximal open set, only the following cases (1) and (2) occur byCorollary 3.3:

(1) for eachx∈X−Uand each open neighborhoodWofx, we haveX−UW: letxbe any element ofX−U andW any open neighborhood ofx. SinceX−UW, we haveW∩Ufor any open neighborhoodW of x. Hence,X−U⊂Cl(U). SinceX=U∪(X−U)⊂U∪Cl(U)=Cl(U)⊂X, we have Cl(U)=X;

(2) there exists an open setWsuch thatX−U=WX: sinceX−U=W is an open set,Uis a closed set. Therefore,U=Cl(U).

Theorem3.5. LetU be a maximal open set. Then,Int(X−U)=X−U or Int(X−U)= ∅.

Proof. ByCorollary 3.3, we have either (1) Int(X−U)= ∅or (2) Int(X− U)=X−U.

Theorem 3.6. LetU be a maximal open set andS a nonempty subset of X−U. Then,Cl(S)=X−U.

Proof. SinceS⊂X−U, we haveW∩Sfor any elementxofX−U and any open neighborhoodWofxbyTheorem 3.1. Then,X−U⊂Cl(S). Since X−U is a closed set and S⊂X−U, we see that Cl(S)⊂Cl(X−U)=X−U. Therefore,X−U=Cl(S).

Corollary 3.7. Let U be a maximal open set andM a subset ofX with UM. Then,Cl(M)=X.

Proof. SinceUM⊂X, there exists a nonempty subsetS ofX−U such thatM=U∪S. Hence, we have Cl(M)=Cl(S∪U)=Cl(S)∪Cl(U)⊃(X−U)∪ U=XbyTheorem 3.6. Therefore, Cl(M)=X.

Theorem3.8. LetUbe a maximal open set and assume that the subsetX−U has two elements at least. Then,Cl(X−{a})=Xfor any elementaofX−U.

Proof. Since UX− {a} by our assumption, we have the result by Corollary 3.7.

Theorem3.9. LetUbe a maximal open set andNa proper subset ofXwith U⊂N. Then,Int(N)=U.

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Proof. IfN=U, then Int(N)=Int(U)=U. OtherwiseNU, and hence UN. It follows thatU⊂Int(N). SinceUis a maximal open set, we have also Int(N)⊂U. Therefore, Int(N)=U.

Theorem3.10. LetU be a maximal open set andS a nonempty subset of X−U. Then,

X−Cl(S)=Int(X−S)=U. (3.1) Proof. SinceU⊂X−SXby our assumption, we have the result by The- orems3.6and3.9.

A subsetMof a space(X,τ)is called apreopenset ifM⊂Int Cl(M). Then, Corollary 3.7implies the following result.

Theorem3.11. LetU be a maximal open set andMany subset of Xwith U⊂M. Then,Mis a preopen set.

Proof. IfM =U, thenM is an open set. Therefore, M is a preopen set.

Otherwise,UM, then Int Cl(M)=IntX=X⊃MbyCorollary 3.7. Therefore, Mis a preopen set.

Corollary3.12. LetUbe a maximal open set. Then,X−{a}is a preopen set for any elementaofX−U.

Proof. Since U X− {a} by our assumption, we have the result by Theorem 3.11.

4. Fundamental properties of radicals

Definition4.1. LetUλbe a maximal open set for any elementλofΛ. Let ᐁ= {Uλ|λ∈Λ};= ∩λ∈ΛUλis called theradicalofᐁ.

The intersection of all maximal ideals of a ring᏾is called the (Jacobson) radical of᏾ [1, 3]. Following this terminology in the theory of rings, we use the terminology “radical” for the intersection of maximal open sets.

The symbolΛ means difference of index sets; namely,Λ\Γ=Λ−Γ, and the cardinality of a setΛis denoted by|Λ|in the following arguments.

Theorem4.2. Assume that|Λ| ≥2. LetUλbe a maximal open set for any elementλofΛandUλUµfor any elementsλandµofΛwithλµ.

(1)Letµbe any element ofΛ. Then,X−∩λ∈Λ\{µ}Uλ⊂Uµ. (2)Letµbe any element ofΛ. Then,∩λ∈Λ\{µ}Uλ∅.

Proof. Letµbe any element ofΛ. (1) ByLemma 2.2(2), we haveX−Uµ⊂Uλ

for any elementλofΛwithλµ. Then,X−Uµ⊂ ∩λ∈Λ\{µ}Uλ. Therefore, we haveX−∩λ∈Λ\{µ}Uλ⊂Uµ.

(2) Ifλ∈Λ\{µ}Uλ= ∅, we haveX=Uµby (1). This contradicts our assumption thatUµis a maximal open set. Therefore, we haveλ∈Λ\{µ}Uλ.

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Corollary4.3. LetUλbe a maximal open set for any elementλofΛand UλUµfor any elementsλandµofΛwithλµ. If|Λ| ≥3, thenUλ∩Uµ for any elementsλandµofΛwithλµ.

Proof. ByTheorem 4.2(2), we have the result.

Theorem 4.4. LetUλ be a maximal open set for any elementλof Λand UλUµfor any elementsλandµofΛwithλµ. Assume that|Λ| ≥2. Letµ be any element ofΛ. Then,∩λ∈Λ\{µ}Uλ⊂Uµ⊂ ∩λ∈Λ\{µ}Uλ.

Proof. Letµ be any element of Λ. Ifλ∈Λ\{µ}Uλ⊂Uµ, then we see that X=(X− ∩λ∈Λ\{µ}Uλ)∪ ∩λ∈Λ\{µ}Uλ⊂Uµ byTheorem 4.2(1). This contradicts our assumption. IfUµ⊂ ∩λ∈Λ\{µ}Uλ, then we haveUµ⊂Uλ, and henceUµ=Uλ

for any elementλ ofΛ\ {µ}. This contradicts our assumption thatUµUλ

whenλµ.

Corollary4.5. LetUλbe a maximal open set for any elementλofΛand UλUµ for any elementsλandµ ofΛwithλµ. If Γ is a proper nonempty subset ofΛ, then∩λ∈Λ\ΓUλ⊂ ∩γ∈ΓUγ⊂ ∩λ∈Λ\ΓUλ.

Proof. Letγbe any element ofΓ. We seeλ∈Λ\ΓUλ= ∩λ∈((Λ\Γ)∪{γ})\{γ}Uλ UγbyTheorem 4.4. Therefore we seeλ∈Λ\ΓUλ⊂ ∩γ∈ΓUγ. On the other hand, sinceγ∈ΓUγ= ∩γ∈Λ\(Λ\Γ)Uγ⊂ ∩λ∈Λ\ΓUλ, we haveγ∈ΓUγ⊂ ∩λ∈Λ\ΓUλ.

Theorem 4.6. LetUλ be a maximal open set for any elementλof Λand UλUµ for any elementsλandµ ofΛwithλµ. If Γ is a proper nonempty subset ofΛ, thenλ∈ΛUλγ∈ΓUγ.

Proof. By Corollary 4.5, we have λ∈ΛUλ = (∩λ∈Λ\ΓUλ)∩(∩γ∈ΓUγ)

γ∈ΓUγ.

Theorem 4.7(a decomposition theorem for maximal open set). Assume that|Λ| ≥2. LetUλbe a maximal open set for any elementλofΛandUλUµ

for any elementsλandµofΛwithλµ. Then, for any elementµofΛ, Uµ=

λ∈ΛUλ

X−∩λ∈Λ\{µ}Uλ

. (4.1)

Proof. Letµbe an element ofΛ. ByTheorem 4.2(1), we have λ∈ΛUλ

X−∩λ∈Λ\{µ}Uλ

=

λ∈Λ\{µ}Uλ

∩Uµ

X−∩λ∈Λ\{µ}Uλ

=

λ∈Λ\{µ}Uλ

X−∩λ∈Λ\{µ}Uλ

Uµ

X−∩λ∈Λ\{µ}Uλ

=Uµ

X−∩λ∈Λ\{µ}Uλ

=Uµ.

(4.2)

Therefore, we haveUµ=(∩λ∈ΛUλ)∪(X−∩λ∈Λ\{µ}Uλ).

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Theorem4.8. LetUλbe a maximal open set for any elementλof a finite set ΛandUλUµfor any elementsλandµofΛwithλµ. If∩λ∈ΛUλis a closed set, thenUλis a closed set for any elementλofΛ.

Proof. By Theorem 4.7, we have Uµ = (∩λ∈ΛUλ)∪(X− ∩λ∈Λ\{µ}Uλ) = (∩λ∈ΛUλ)∪(∪λ∈Λ\{µ}(X−Uλ)). SinceΛis a finite set, we see thatλ∈Λ\{µ}(X− Uλ)is a closed set. Hence,Uµis a closed set by our assumption.

As an application ofTheorem 4.7, we give another proof ofTheorem 4.6.

Another proof ofTheorem4.6. SinceΛ⊋Γ ≠, there exists an ele- mentν ofΛsuch thatν∈Γ and an elementµ ofΓ. If|Γ| =1, then we have

λ∈ΛUλ⊂Uµ. Ifλ∈ΛUλ=Uµ, then we haveUµ⊂Uλfor any elementλofΛ. SinceUλis a maximal open set for any elementλofΛ, we haveUµ=Uλ, which contradicts our assumption. Hence, we haveλ∈ΛUλUµ. If|Γ| ≥2, then by Theorem 4.7, we have

Uν=

λ∈ΛUλ

X−∩λ∈Λ\{ν}Uλ

, Uµ=

γ∈ΓUγ

X−∩γ∈Γ\{µ}Uγ

. (4.3)

If λ∈ΛUλ= ∩γ∈ΓUγ, thenγ∈ΓUγ = ∩λ∈ΛUλ⊂ ∩λ∈Λ\{ν}Uλ⊂ ∩γ∈ΓUγ. Hence, we haveλ∈Λ\{ν}Uλ= ∩γ∈ΓUγ. Therefore,λ∈Λ\{ν}Uλ= ∩γ∈ΓUγ⊂ ∩γ∈Γ\{µ}Uγ. Hence, we see thatUν⊃Uµ. It follows thatUν=Uµwithνµ. This contradicts our assumption.

Theorem4.9. Assume that|Λ| ≥2. LetUλbe a maximal open set for any element λ of Λ andUλUµ for any elements λ and µ of Λ withλµ. If

λ∈ΛUλ= ∅, then{Uλ|λ∈Λ}is the set of all maximal open sets ofX.

Proof. If there exists another maximal open set Uν of X, which is not equal to Uλ for any element λ of Λ, then ∅ = ∩λ∈ΛUλ = ∩λ∈(Λ∪{ν})\{ν}Uλ. By Theorem 4.2(2), we see that λ∈(Λ∪{ν})\{ν}Uλ. This contradicts our assumption.

Example4.10. If each point{x}is closed (e.g.,Xis a Hausdorff space or a cofinite space or a cocountable space), thenX−{a}is a maximal open set for any elementaofX. Moreover, we see that{X− {a} |a∈X}is the set of all maximal open sets ofXbyTheorem 4.9, sincea∈X(X−{a})= ∅.

Proposition4.11. LetAandBbe subsets ofX. IfA∪B=X,A∩Bis a closed set, andAis an open set, thenBis a closed set.

Proof. SinceX−A⊂B, then we see that (A∩B)∪(X−A)=

A∪(X−A)

B∪(X−A)

=B∪(X−A)=B. (4.4) SinceA∩BandX−Aare closed sets, we see thatBis a closed set.

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Proposition4.12. LetUλbe an open set for any elementλofΛandUλ Uµ=Xfor any elementsλandµofΛwithλµ. If∩λ∈ΛUλis a closed set, then

λ∈Λ\{µ}Uλis a closed set for any elementµofΛ.

Proof. Letµbe any element ofΛ. SinceUλ∪Uµ=Xfor any elementλof Λwithλµ, we have

Uµ

λ∈Λ\{µ}Uλ

= ∩λ∈Λ\{µ}

Uµ∪Uλ

=X. (4.5)

SinceUµ∩(∩λ∈Λ\{µ}Uλ)= ∩λ∈ΛUλis a closed set by our assumption,λ∈Λ\{µ}Uλ

is a closed set byProposition 4.11.

Theorem4.13. LetUλbe a maximal open set for any elementλofΛand UλUµ for any elementsλ andµ ofΛwithλµ. If λ∈ΛUλis a closed set, then∩λ∈Λ\{µ}Uλis a closed set for any elementµofΛ.

Proof. ByLemma 2.2(2), we haveUλ∪Uµ=Xfor any elementsλandµof Λwithλµ. ByProposition 4.12, we have thatλ∈Λ\{µ}Uλis a closed set.

If the assumption ofProposition 4.12does not hold, then the condition that

λ∈ΛUλ is a closed set does not always imply thatλ∈Λ\{µ}Uλ is closed. The following is an example.

Example4.14. LetX= {a,b,c,d,e}with topologyθ= {∅,{a},{d},{a,d}, {b,d},{c,d},{a,b,d},{a,c,d},{b,c,d},{a,b,c,d},{b,c,d,e},X}, U1 = {a}, U2= {a,b,d}, andU3= {a,c,d}. Then,U1∩U2∩U3=U1 is a closed set. It follows thatU1∪U2=U2X,U1∪U3=U3X,U2∪U3= {a,b,c,d}X. We see thatU2∩U3= {a,d}is not a closed set.

5. More about radicals of maximal open sets. In this section, we study the closure of radicals. We begin with a proposition.

Proposition5.1. LetUλbe a set for any elementλofΛ. IfCl(∩λ∈ΛUλ)=X, thenCl(Uλ)=Xfor any elementλofΛ.

Proof. We see thatX=Cl(∩λ∈ΛUλ)⊂Cl(Uλ). It follows that Cl(Uλ)=X for any elementλofΛ.

Theorem5.2. LetUλbe a maximal open set for any elementλof a finite set Λ. IfCl(∩λ∈ΛUλ)X, then there exists an elementλofΛsuch thatCl(Uλ)=Uλ. Proof. Assume that Cl(Uλ)=Xfor any elementλofΛ. Letµbe an element ofΛ. Sinceλ∈Λ\{µ}Uλis an open set, we have

Cl

λ∈ΛUλ

=Cl

λ∈Λ\{µ}Uλ

∩Uµ

λ∈Λ\{µ}Uλ

Cl Uµ

=

λ∈Λ\{µ}Uλ

∩X

= ∩λ∈Λ\{µ}Uλ.

(5.1)

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Hence, Cl(∩λ∈Λ\{µ}Uλ)⊂Cl(∩λ∈ΛUλ). On the other hand, we see thatλ∈ΛUλ

λ∈Λ\{µ}Uλ, and hence Cl(∩λ∈ΛUλ)⊂Cl(∩λ∈Λ\{µ}Uλ). It follows that Cl(∩λ∈ΛUλ)

= Cl(∩λ∈Λ\{µ}Uλ). Then, by induction on the element of Λ, we see that Cl(∩λ∈ΛUλ)=Cl(Uλ)=Xfor an elementλofΛ. This contradicts our assump- tion that Cl(∩λ∈ΛUλ)X. Therefore, we see that there exists an elementλof Λsuch that Cl(Uλ)=Uλ.

Theorem 5.2is not true whenΛis not a finite set, as we see by the following example. This example also shows that ifΛis not a finite set, thenTheorem 4.8 is not always true.

Example 5.3. Let X =Rn, the n-dimensional Euclidean space. Let Ux = X−{x}for any elementx∈X. Then,Uxis a maximal open set and we have

Cl

x∈XUx

=Cl(∅)= ∅X. (5.2)

However, Cl(Ux)=Xfor any elementxofX.

The radicals of maximal open sets have the following outstanding property.

Theorem 5.4(the law of radical closure). LetΛ be a finite set andUλ a maximal open set for each elementλofΛ. LetΓ be a subset ofΛsuch that

Cl Uλ

=Uλ for anyλ∈Γ, Cl

Uλ

=X for anyλ∈Λ\Γ. (5.3)

Then,Cl(∩λ∈ΛUλ)= ∩λ∈ΓUλ(=Xif Γ= ∅).

Proof. IfΓ= ∅, then we have the result byTheorem 5.2. OtherwiseΓ≠, and hence we see that

Cl

λ∈ΛUλ

=Cl

λ∈ΓUλ

λ∈Λ\ΓUλ

λ∈ΓUλ

Cl

λ∈Λ\ΓUλ

=

λ∈ΓUλ

∩X= ∩λ∈ΓUλ

(5.4)

by Theorem 5.2 and the fact that λ∈ΓUλ is an open set. It follows that Cl(∩λ∈ΛUλ) = Cl(Cl(∩λ∈ΛUλ)) Cl(∩λ∈ΓUλ). On the other hand, we see that λ∈ΛUλ⊂ ∩λ∈ΓUλ, and hence Cl(∩λ∈ΛUλ)⊂Cl(∩λ∈ΓUλ). It follows that Cl(∩λ∈ΛUλ)=Cl(∩λ∈ΓUλ). The radicalλ∈ΓUλis a closed set sinceUλis a closed set for anyλ∈Γ by our assumption. Therefore, we see that Cl(∩λ∈ΛUλ)=

λ∈ΓUλ.

As an application ofTheorem 5.4, we give another proof ofTheorem 4.8.

Another proof ofTheorem4.8. LetΓ be a subset ofΛsuch that Cl

Uλ

=Uλ for anyλ∈Γ, Cl

Uλ

=X for anyλ∈Λ\Γ. (5.5)

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We suppose that the radical λ∈ΛUλ is a closed set. We see that Γ ≠by Theorem 5.4. Then,λ∈ΛUλ=Cl(∩λ∈ΛUλ)= ∩λ∈ΓUλfor the subsetΓ ofΛby Theorem 5.4. Then, we see thatΛ=Γ byTheorem 4.6.

References

[1] N. Jacobson,The radical and semi-simplicity for arbitrary rings, Amer. J. Math.67 (1945), 300–320.

[2] F. Nakaoka and N. Oda,Some applications of minimal open sets, Int. J. Math. Math.

Sci.27(2001), no. 8, 471–476.

[3] J. J. Rotman,An Introduction to Homological Algebra, Pure and Applied Mathe- matics, vol. 85, Academic Press, New York, 1979.

Fumie Nakaoka: Department of Applied Mathematics, Faculty of Sciences, Fukuoka University, Nanakuma, Jonan-ku, Fukuoka 814-0180, Japan

E-mail address:[email protected]

Nobuyuki Oda: Department of Applied Mathematics, Faculty of Sciences, Fukuoka University, Nanakuma, Jonan-ku, Fukuoka 814-0180, Japan

E-mail address:[email protected]

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according to the following timetable:

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Guest Editors

José Roberto Castilho Piqueira,

Telecommunication and Control Engineering Department, Polytechnic School, The University of São Paulo, 05508-970 São Paulo, Brazil;

[email protected]

Elbert E. Neher Macau,

Laboratório Associado de Matemática Aplicada e Computação (LAC), Instituto Nacional de Pesquisas Espaciais (INPE), São Josè dos Campos, 12227-010 São Paulo, Brazil ; [email protected]

Celso Grebogi,

Center for Applied Dynamics Research, King’s College, University of Aberdeen, Aberdeen AB24 3UE, UK; [email protected]

Hindawi Publishing Corporation http://www.hindawi.com

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