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On e-spaces and rings of real valued e-continuous functions

S. Afrooz^a, F. Azarpanah^b and N. Hasan Hajee^b

a Faculty of Marine Engineering, Khorramshahr University of Marine Science and Technology, Khorramshahr, Iran([email protected])

bDepartment of Mathematics, Shahid Chamran University of Ahvaz, Ahvaz, Iran([email protected];

[email protected], [email protected])

Communicated by S. Garc´ıa-Ferreira

Abstract

Whenever the closure of an open set is also open, it is callede-open and if a space have a base consisting ofe-open sets, it is callede-space. In this paper we first introduce and studye-spaces ande-continuous func- tions (we call a functionffrom a spaceX to a spaceY ane-continuous atx∈X if for each open setV containingf(x)there is ane-open set containingxwithf(U)⊆V). We observe that the quasicomponent of each point in a spaceX is determined bye-continuous functions onX and it is characterized as the largest set containing the point on which everye-continuous function onXis constant. Next, we study the rings Ce(X)of all real valuede-continuous functions on a spaceX. It turns out thatCe(X)coincides with the ring of real valued clopen continuous functions onX which is aC(Y)for a zero-dimensional space Y whose elements are the quasicomponents ofX. Using this fact we characterize the real maximal ideals ofCe(X)and also give a natural representation of its maximal ideals. Finally we have shown thatCe(X) determines the topology ofX if and only if it is a zero-dimensional space.

2020 MSC:54C40; 54D35.

Keywords: ^e-space;e-continuous function; real maximal ideal; quasicompo- nent; zero-dimensional space.

1. Introduction

Throughout this articleC(X) denotes the ring of all real-valued continuous functions on a completely regular Hausdorff spaceX, andC^∗(X) is the subring of C(X) consisting of all bounded elements. For each f ∈ C(X), the set Z(f) = {x ∈ X : f(x) = 0} is called the zero-set of f. If I is an ideal in C(X), we denote Z[I] ={Z(f) :f ∈I} and TZ[I] =T

f∈IZ(f). Whenever TZ[I] is nonempty, I is called fixed; else, free. Maximal ideals ofC(X) are precisely of the form M^p = {f ∈ C(X) : p ∈ cl_βXZ(f)} for each p ∈ βX, whereβXis the Stone- ˇCech compactification ofX. More generally ifA⊆βX, we denote M^A = {f ∈C(X) :A⊆cl_βXZ(f)}. The fixed maximal ideals of C(X) are the sets M_p ={f ∈C(X) :p∈Z(f)}, forp∈X. The idealsO^p = {f ∈C(X) :p∈int_βXcl_βXZ(f)}, p∈βX in the context of C(X) are lower bounds for prime ideals of C(X) in the sense that for every prime ideal P in C(X), there exists a unique p ∈βX such that O^p ⊆ P ⊆M^p; see Theorem 7.15 in [4]. More generally, for eachA⊆βXthe idealO^Ais defined by the set {f ∈C(X) :A⊆intβXclβXZ(f)}. The reader is referred to [3], [4] and [7] for undefined terms and notations concerning C(X) and the concepts of general topology.

Let us give a brief outline of this article. In the next section we first observe that the intersection of twoe-open sets is a ane-open set and this follows that the set Ce(X) consisting of real valued e-continuous functions on a space X is a ring under pointwise addition and multiplication of functions. Next we show that every open or dense subspace of an e-space is an e-space but not every subspace (even a closed subspace) is necessarily an e-space. Section 3 is devoted to e-continuous functions. We observe in this section that each e- continuous function on a spaceX is constant on every quasicomponent of X and every subset of X with this property is in fact a quasicomponent of X.

This fact help us to define equivalence classes in βX for characterization of maximal and real maximal ideals of Ce(X). The characterization of spaces X for which Ce(X) or C_e^∗(X) (the subring of Ce(X) consisting of bounded ones) coincides with one of the ringsC(X) and C^∗(X) are characterized. For instance it is shown thatCe(X) = C(X) if and only ifX is an e-space (zero- dimensional) and Ce(X) = C^∗(X) if and only if X is a pseudocompact e- space (zero-dimensional). We also observe in this section that the ringsCe(X) coincides with the rings of real valued clopen continuous functions onX which are first introduced in [6] under the name of super continuous functions.

Finally in section 4 we characterize the maximal and real maximal ideals of C_e(X). In [1] it is shown that the ring C_s(X) (=C_e(X)) of real valued super (clopen) continuous functions on a space X is isomorphic with C(X_z) for a zero-dimensional X_z. In that reference, for characterization of maximal ideals of C_s(X) the authors have given a representation for maximal ideals of C(X_z). In this section we present the characterization of maximal ideals of Ce(X) by some equivalent classes in βX so that if we take X as a zero- dimensional space they coincide with usual ones. Real maximal ideals ofCe(X)

are also characterized and we have shown in this section that a spaceXis zero- dimensional if and only if its topology coincides with the weak topology induced byCe(X) (C_e^∗(X)).

2. e-spaces

A set Gin a topological space X is called extremely open (brieflye-open) ifGand clXGare open subsets ofX. We call a subset of a topological space an e-closed if its complement is an e-open. Equivalently, a set is e-closed if and only if it is closed and its interior is also closed. Clearly every closed-open (clopen) set in a topological space is an e-open set, but not conversely. For exampleR\ {0} is ane-open subset ofRwhich is not a clopen set. Moreover, for eachT₁-spaceX, the setX\ {x}is ane-open set for eachx∈X. In fact, if xis an isolated point ofX, thenX\ {x} is clopen, so it ise-open. Otherwise X\ {x}is open, sinceXisT₁and cl_X(X\ {x}) =X is open. Thus every dense open subset of a space is ane-open set. In particular an open subsetGofRis e-open inRif and only if R\Ghas an empty interior.

Using the following lemma, we show that the intersection of every twoe-open sets is ane-open set.

Lemma 2.1. Suppose thatV is an e-open andU is an open subset of a space X, then clX(V ∩U) =clXV ∩clXU.

Proof. Letp∈clXV ∩clXU and W be an arbitrary neighborhood ofp. Since V ise-open, clXV is clopen and therefore it is a neighborhood ofpand hence (W∩clXV)∩U 6=∅. ThusW∩(V∩U)6=∅, which impliesp∈clX(V∩U).

Now the proof of the following result is evident.

Corollary 2.2. In any space, the intersection of every two e-open sets is an e-open set.

Corollary2.2is not true for an arbitrary intersection (union) ofe-open sets.

In fact an arbitrary intersection ofe-open sets, even clopen sets, need not be even an open set. If X is a space and x∈ X is the lone non-isolated point of X, then for each x 6= y ∈ X, the set X \ {y} is e-open (clopen). Now G=T

x6=y∈X(X\ {y}) ={x}which is not even open. Also an arbitrary union ofe-open sets need not be ane-open set. For example if we takeX={¹_n :n∈ N} ∪ {0} as a subspace ofR, then Gn ={_2n¹}, for each n∈N, is clopen and hencee-open butS

n∈NG_nis note-open because its closure is{_2n¹ :n∈N}∪{0}

which is not open inX.

Since the intersection of two e-open sets in a topological space (X, τ) is an e-open set by Corollary2.2, the set of alle-open subsets ofX form a base for a topologyτeonX. Wheneverτecoincide withτ (i.e.,τ =τe), we call the space X ane-space.

Example 2.3. Whenever every open subset of a space has an open closure, i.e., if a space is extremally disconnected, then clearly it is ane-space. In particular, every discrete space is ane-space. For a non-extremally disconnectede-space

we may consider the one-point compactification spaceX={1,¹₂,· · ·,_n¹,· · · } ∪ {0} as a subspace of R. The set B = {{_n¹} : n ∈ N} ∪ {G ⊆ X : X \ Gis finite and 0∈G}consisting ofe-open (clopen) sets is a base forX. Since the closure of the open set{1,¹₃,¹₅,· · · } is not open,X is an e-space which is not extremally disconnected.

Proposition 2.4. The trace of any e-open set on an open subspace and also on a dense subspace is e-open.

Proof. First, suppose that X is an open subspace of Y and let V ⊆ Y be e-open. LetV₀=V ∩X. Then using the previous lemma we have

clXV0= clYV0∩X = clYV ∩clYX∩X= clYV ∩X.

SinceV is ane-open subset ofY, cl_YV is clopen inY and hence cl_XV₀is clopen inX. Now, suppose thatX is a dense subspace ofY and letV ⊆Y bee-open.

If we put againV0=V∩X, it is clear that clYV0= clY(V∩X) = clYV because X is dense inY. Now clXV0=X∩clYV implies that clXV0 is open.

Corollary 2.5. Every open or dense subspace of ane-space is an e-space.

Every subspace (even everye-closed subspace) of ane-space need not be an e-space; see the following example.

Example 2.6. Let

B={U ⊆R:U is open inRwith usual topology and [0,∞)\U is finite}.

ClearlyBmay be a base for a topology onR, sayτ. Since each member ofB is an e-open set with respect to the topologyτ, (R, τ) is an e-space. In fact for each U ∈B, we have clU =R and hence clU is open. Now consider the subspace (−∞,0] of (R, τ) which ise-closed. The collection of all subsets of the form U∩(−∞,0] forms a base for (−∞,0], where U is an open subset of R with usual topology. This implies that the space (−∞,0] as a subspace of (R, τ) has the usual topology which is not ane-space.

From [3], recall that a T1-space X is zero-dimensional if each point of X has a neighborhood base consisting of clopen sets. Equivalently, aT1-spaceX is zero-dimensional if and only if for each x ∈ X and each closed set A not containingx, there exists a clopen set containingxwhich does not meetA. So every zero-dimensional space is a completely regular Hausdorff e-space. The converse is also true by the following proposition.

Proposition 2.7. A space is aT3-e-space if and only if it is zero-dimensional.

Proof. LetX be a T3-e-space. Let Gbe an open set in X and x∈G. Using the regularity of thee-spaceX, there exists an open setH such thatx∈H ⊆ clXH ⊆G. Now, sinceX is ane-space, there is ane-open setKinXsuch that x∈K⊆H and hencex∈K⊆clXK⊆clXH ⊆G, where clXKis clopen, so

X is zero-dimensional.

Example 2.8. Using Proposition2.7, eachT1-(evenT2-)-e-space need not be zero-dimensional. Whenever X is an infinite set with cofinite topology, then clearly X is a T1-e-space which is not even T2, so it is not zero-dimensional.

Moreover everyT₂-e-space is not necessarily a zero-dimensional space. In fact as in Example 14.2 in [7], we letX be the real line with neighborhoods of any nonzero point being as in the usual topology, while neighborhoods of 0 will have the formU\A, whereU is a neighborhood of 0 in the usual topology and A={_n¹ :n= 1,2, . . .}. Now let Y =X∩Qas a subspace ofX. ClearlyY is aT2-space which is notT3 andB={((α, β)\A)∩Q:α, β∈R\Q}is a base forY consisting ofe-open sets. So Y is ane-space and sinceY is notT3, it is not zero-dimensional.

A topological spaceXis said to be an extremelyT1-space (briefly aT₁^e-space) if wheneverxandy are distinct points inX, there is ane-open set containing each not the other. We call a spaceX ane-Hausdorff (or aT₂^e-space), if every two different points ofX can be separated by two disjointe-open sets and from [5] a space X is called ultra Hausdorff if every two different points of X can be separated by two disjoint clopen sets. It is easy to see thatT1-spaces and T₁^e-spaces coincide and the following result states thatT₂^e-spaces also coincide with ultra Hausdorff spaces.

Proposition 2.9. A topological space is e-Hausdorff if and only if it is ultra Hausdorff.

Proof. First we note that wheneverUandV are two disjointe-open subsets of a topological spaceX, then clXU∩clXV =∅by Lemma2.1. Next if a space X is ane-Hausdorff, for each two different pointsxandy inX there exist two disjointe-open subsetsU andV ofX containingxandy respectively. But by Lemma 2.1, clXU and clXV are two disjoint clopen sets containing x and y respectively, henceX is an ultra Hausdorff. Since every clopen set is ane-open

set, the proof of the converse is evident.

Corollary 2.10. Let X be ane-space. Then X is Hausdorff if and only if it is ultra Hausdorff.

Proposition 2.11. Every homeomorphic image of ane-space is an e-space.

Proof. Let X and Y be two homeomorphic spaces, X be an e-space and ϕ: X →Y be an onto homeomorphism. LetV be an open subset ofY andy∈V. Then there is x∈X such thaty=ϕ(x). Sinceϕis continuous ϕ⁻¹(V) is an open subset ofX containingxand hence there exists ane-open subsetU ofX such thatx∈U ⊆ϕ⁻¹(V). But ϕis an open function, so ϕ(U) is open and y∈ϕ(U)⊆V. Now it is enough to show that clYϕ(U) is open, i.e.,ϕ(U) is an e-open set. Using Theorem 7.9 in [7], we have ϕ(clXU) = clYϕ(U) and since ϕis open,ϕ(clXU) is an open subset ofY because clXU is an open subset of X. This shows that clYϕ(U) is open and we are through.

Similar to definitions preceding the Proposition 2.9, we may define the e- compactness of the spaces: A space is called e-compact if every e-open cover

of the space has a finite subcover. Clearly every compact space is e-compact and using the following proposition, everye-compact is countably compact.

Proposition 2.12. If a spaceX is compact, then it ise-compact and whenever X is ane-compactT₁-space, then it is countably compact.

Proof. The first part is evident. For the second part, suppose on the con- trary that X is not countably compact. Then there is an infinite subset A={x1, x2, . . . , xn, . . .} ofX without any cluster point. Now for eachn∈N, the setGn=X\An, whereAn={xn, xn+1, . . .} is ane-open set becauseAn

is closed and clXGn =X\Bn, where Bn ⊆An will be open since Bn is also closed. Clearly∪^∞_n=1Gn=X and no finite number of Gn’s coverX, i.e.,X is

note-compact..

The converse of the second part of the above proposition is not true. To see this consider the space of ordinals [1, ω1), whereω1is the first uncountable ordinal number. It is known that the space [1, ω1) is countably compact. But {[1, α) :α∈[2, ω1)}is a clopen cover for [1, ω1) that can have no finite subcover.

We could not prove or disprove the converse of the first part of the proposition, so we cite it here as a question for interested readers.

Question: Is everye-compact space a compact space?

Let A be a subset of a topological space X. An element x ∈ X is called ane-cluster point ofAif eache-open subset of X containingxmeets A. The set of all e-cluster points ofA is called thee-closure ofA and we denote it by e-clXA. Clearly for each subset A of a spaceX, we have clXA⊆e-clXAand the inclusion may be proper. For instance if we consider the open interval (0,1) inR, then cl_R(0,1) = [0,1], bute-cl_R(0,1) =R.

As a closure of a set, thee-closure of a setAin a spaceX is the intersection of all e-closed subsets ofX containingA. Whenever

S={H ⊆X: H is ane-closed set andA⊆H}, thene-cl_XA=T

H∈SH. In fact if x∈T

H∈SH andx /∈e-cl_XA, then there is ane-open setGcontainingxsuch thatG∩A=∅. nowX\Gis ane-closed set containingAwhich does not containx, a contradiction. The reverse inclusion is also routine. The e-interior is defined similarly and thee-interior of a setA is denoted bye-intXA.

In contrast to the closure of a set which is closed, thee-closure of a set need not be e-closed. To this end, we letX ={0,1,¹₂,¹₃, . . .} be a subspace of R with usual topology and A={¹₂,¹₄, . . .}. Then e-clA=A∪ {0} which is not e-closed.

Proposition 2.13. Let A be an e-compact subset of a space X and X be e-Hausdorff. ThenA=e-clXA.

Proof. Letx∈e-clXA\A. SinceX ise-Hausdorff, for eacha∈Athere exists a clopen setUa not containingxby Proposition2.9. NowC={A∩Ua:a∈A}is ane-open cover of thee-compact subspaceAand hence it has a finite subcover,

say{A∩Ua₁,· · ·, A∩Ua_n}, soA=Sn

i=1A∩Ua_i. ButU =Sn

i=1Ua_i is clopen not containingxand henceX\U ise-open containingxwhich does not meet A, a contradiction. ThereforeA=e-clXA.

3. e-continuous functions

Definition 3.1. Let X and Y be topological spaces and f : X → Y be a function. We say thatf ise-continuous at a pointx∈X if for each open setV inY containingf(x) there exists ane-open setU inX containingxsuch that f(U)⊆V. A functionf :X→Y is callede-continuous if it ise-continuous at each point ofX.

Clearly every e-continuous function is continuous, but the converse is not necessarily true. For instance, the identity function i: R → Ris continuous but note-continuous, because the only nonemptye-open subsets ofRare dense open subsets ofR. WheneverX is ane-space, then every continuous function onX ise-continuous. In fact if f :X →Y is continuous, x∈X and V is an open subset ofY containingf(x), then there exists an open setG containing xsuch thatf(G)⊆V. ButX is ane-space, so there is ane-open subsetU of X withx∈U ⊆G. Thusf(U)⊆V impliesf is indeede-continuous.

As the continuity, thee-continuity of a function may be stated via the inverse images of the open sets under the function and other similar standard condi- tions. The proof of the following proposition is analogous to that of Theorem 7.2 in [7].

Proposition 3.2. Let X and Y be topological spaces and f : X → Y be a function. Then the following statements are equivalent.

(1) f ise-continuous.

(2) f⁻¹(V)is a union ofe-open subsets ofX for each open subsetV ofY. (3) f⁻¹(H)is an intersection ofe-closed subsets ofXfor each closed subset

H of Y.

(4) f(e-clXA)⊆clYf(A).

Proposition 3.3. An e-continuous image of ane-compact space is compact.

Proof. Letf : X → Y be e-continuous from X ontoY andX be e-compact.

LetC ={V_α:α∈S} be an open cover ofY. For eachx∈X, there isα∈S such thatf(x)∈V_α. Sincef ise-continuous, there exists an e-open setG_x in X containingx withf(G_x)⊆V_α. Clearly X =S

x∈XG_x and e-compactness ofX implies thatX=Sn

i=1Gx_i for somex1,· · · , xn∈X. Sincef(Gx_i)⊆Vα_i, we haveY =Sn

i=1Vα_i, i.e., Y is compact.

WheneverX is a topological space, we recall that for eachx∈X, the largest connected subset Cx of X containingx is the component ofx. In fact Cx is the union of all connected subsets ofX containingx. The quasicomponentQx

of x in X is the intersection of all clopen subsets of X which contain x. It is well-known that Cx ⊆Qx for each xand the inclusion may be proper, see Exercise 26B in [7].

If X and Y are topological spaces, then a function f : X → Y is called clopen continuous at x ∈ X if for each neighborhood V of f(x), there is a clopen subset ofX containingxsuch thatf(U)⊆V. The clopen continuous functions and also the rings of real valued clopen continuous functions are studied in [6] and [1] respectively. Clearly every clopen continuous function f :X →Y ise-continuous and wheneverY is aT₃-space, the converse is also true.

Theorem 3.4. Let X and Y be topological spaces and f : X → Y be an e-continuous function. Then the following statements hold.

(1) If Y is a T₂-space, then f is constant on each quasi-component Q_x, x∈X.

(2) If Y is aT₃-space, then f is clopen continuous.

Proof. (1) SinceY is Hausdorff, for each y ∈ Y,where y 6=y0 =f(x0), there exists an open set V in Y such that y₀ ∈ Y \cl_YV and y ∈ V. By the e- continuity off there is ane-open setG_yinX containingx₀such thatf(G_y)⊆ Y \cl_YV. Butf is also continuous, so f(cl_XG_y)⊆cl_Yf(G_y)⊆cl_Y(Y \V) = Y\V. SinceQ_x0 ⊆T

y06=y∈Y cl_XG_yandy /∈f(cl_XG_y) (for otherwise,y∈Y\V which is impossible),y /∈f(Q_x0) for eachy∈Y withy6=y₀. This implies that f(Q₀) =y₀.

(2) Let x∈ X and V be a neighborhood of f(x) in Y. Then there exists an open subsetW ofY containingf(x) withW ⊆cl_YW ⊆V by regularity of Y. Since f is e-continuous, there exists an e-open subsetU of X containing xsuch thatf(U)⊆W. Now f(clXU)⊆clYf(U)⊆clYW ⊆V, because f is continuous. ButU ise-open, so clXU is clopen and we are through.

The converse of part (2) of the above lemma is also true in the sense that Qx for eachxin a spaceX is in fact the largest subset ofX containingxon which everye-continuous function onX is constant.

Proposition 3.5. LetX be a space andY be a Hausdorff space containing at least two points. Then for eachx∈X,

Q_x={y∈X :f(x) =f(y), for each e-continuous functionf :X→Y}.

Proof. Whenever y ∈ Q_x, then f(x) = f(y) for each e-continuous function f :X →Y by Proposition 3.4. Conversely suppose thaty /∈Q_x. Hence there exists a clopen set U containing xbut not y. Now define f : X → Y with f(U) =y₁ andf(X\U) =y₂, wherey₁ andy₂ are two different points of Y. Clearlyf ise-continuous,f(x) =y16=y2=f(y) and we are done.

IfX is a topological space andf :X →Ris ane-continuous function, then using Theorem3.4,f is a clopen continuous, becauseRis aT3-space. Therefore the ring of all real valuede-continuous functionsCe(X) on a topological space X coincides with the ring Cs(X) consisting of real valued clopen continuous functions. It is easy to see thatCe(X)(=Cs(X)) is an ordered ring which is a subring ofC(X) and the following lemma follows thatCe(X) is in fact a lattice ordered ring. Using Theorem3.4, wheneverX is connected, thenCe(X) =R.

We denote byC_e^∗(X) the set of all bounded members ofCe(X) and we call X an e-pseudocompact space if Ce(X) = C_e^∗(X). Clearly every e-compact space ise-pseudocompact by Proposition3.3, but the converse is not true. For exampleRwith usual topology ise-pseudocompact, since it is connected and each member of C_e(X) is constant by Theorem 3.4. But the e-open cover {R\ {n, n+ 1, . . .}}_n∈Nof Rcan have no finite subcover. It is also clear that every pseudocompact space is e-pseudocompact, however the aforementioned example shows that the converse is not necessarily true.

WheneverX is ane-space, then clearlyC(X) coincides withCe(X). By the following proposition, the converse is also true if the spaceX is a completely regular Hausdorff space. First we need the following lemma.

Lemma 3.6. LetX,Y andZ be topological spaces andf :X →Y,g:Y →Z be functions. Then the following statements hold.

(1) If f is e-continuous and g is continuous, then g◦f : X → Z is e- continuous.

(2) Ifg◦f ise-continuous andgis an open one-to-one continuous function, thenf is e-continuous.

Proof. It is evident.

Our proof of the following proposition shows that the parts (1) and (2) are equivalent for any space X. In the other words, for any spaceX the equality C^∗(X) =C_e^∗(X) impliesC(X) =Ce(X).

Proposition 3.7. For a completely regular Hausdorff space X, the following statements are equivalent.

(1) C^∗(X) =C_e^∗(X).

(2) C(X) =Ce(X).

(3) The spaceX is an e-space.

(4) The spaceX is a zero-dimensional space.

Proof. Clearly (2) implies (1) and if (1) holds we let f ∈ C(X) and take a homeomorphismφ:R→(−1,1). Thenφ◦f is boundede-continuous by part (1). Now by Lemma3.6 f will bee-continuous becauseφ is homeomorphism.

Hence C(X) =Ce(X), so (1) also implies (2). Using Proposition2.7, (3) and (4) are equivalent and clearly (3) implies (2). Thus it remains to show that (2) implies (3). We note that the set{cozf :f ∈C(X)}is a base for open subsets ofX by Theorem 3.2 in [4]. Now ifx∈cozf for somef ∈C(X) =C_e(X), then f(x)6= 0, hence for an open setV in Rcontainingf(x) but not 0, there exists ane-open setU containingxsuch thatU ⊆f⁻¹(V)⊆cozf. This implies that the spaceX have a base consisting of e-open sets and we are through.

By Proposition3.7, wheneverX is ane-space, thenC(X) =Ce(X). More- over ifX is alsoe-pseudocompact, thenCe(X) =C_e^∗(X) implies thatC(X) = C_e^∗(X)⊆C^∗(X) and henceX will be pseudocompact. Therefore in the follow- ing proposition, we may replace “pseudocompact” with “e-pseudocompact”.

Proposition 3.8. The following statements for a completely regular Hausdorff spaceX are equivalent.

(1) C(X) =C_e^∗(X).

(2) Ce(X) =C^∗(X).

(3) The spaceX is a pseudocompact e-space.

(4) The spaceX is a pseudocompact zero-dimensional space.

Proof. (1) ⇒ (2). First C_e(X) ⊆ C(X) = C_e^∗(X) ⊆C^∗(X). Next C^∗(X) ⊆ C(X) =C_e^∗(X)⊆Ce(X). HenceCe(X) =C^∗(X).

(2) ⇒ (3). The equalityCe(X) = C^∗(X) implies that C_e^∗(X) = Ce(X)∩ C^∗(X) = C^∗(X). Hence by Proposition 3.7 X is an e-space. On the other hand, by the same proposition we haveCe(X) =C(X). Now using part (2), C(X) =Ce(X) =C^∗(X), soX is pseudocompact.

(3)⇒(4)⇒(1). By Propositions2.7and3.7, the proof is evident.

4. Characterization of maximal and real maximal ideals of rings of real valued e-continuous functions

From now on we need some notations and details of the proof of Theorem 3.1 in [1] for later use. By Theorem 3.1 in [1], C_s(X), the ring of real valued clopen continuous functions on a space X, is a C(Y) for a zero-dimensional spaceY. As we have already observedCe(X) coincides withCs(X), so we use Ce(X) instead ofCs(X) in the statement of our Theorem 4.1 which is the same theorem in [1]. On the other hand, in order to familiar with some notations in the proof of Theorem 3.1 in [1] and their applications, we have to give the sketch of the proof.

Theorem 4.1. For every topological space X, there exists a zero-dimensional spaceXz such thatCe(X)∼=C(Xz).

Proof. Let X_z be the decomposition {Qx : x ∈ X} on X, where Q_x is the quasicomponent of xand take the collection τ consisting of subsetsG of X_z such thatS

Qx∈GQ_xis a union of clopen subset of the spaceX. It is not hard to see that τ is a topology on Xz andXz with this topology is Hausdorff. To see that Xz is zero-dimensional, letH be an open set in Xz and Qy ∈H for somey∈X. Then by definitionT =S

Q_x∈HQxis a union of clopen subsets of X andy∈T. Therefore there is a clopen subsetU ofX such thaty∈U ⊆T. Now takeG={Qz:z∈U}. SinceS

Q_z∈GQz=U andU is clopen inX, the set Gis clopen inXz andQy ∈G⊆H (to see thatG⊆H letQx∈G, then x∈U ⊆T, so there isQz ∈H withx∈Qz ⊆H. ThereforeQx=Qz ∈H).

This shows thatXz is indeed a zero-dimensional space.

Finally we defineϕ:Ce(X)→C(Xz) withϕ(f) =fz for eachf ∈Ce(X), where fz(Qx) = f(x) for each x ∈ X. By a routine proof we observe that fz ∈ C(Xz) for each f ∈ C(X) and it is easy to see that ϕ is a one-to-one homomorphism. To complete the proof it remains to show that ϕis onto. To this end, letg ∈ C(Xz). The function f : X →Rdefined by f(x) =g(Qx), for all x ∈ X is e-continuous. In fact, if x ∈ X, f(x) = g(Qx) = c and

ε > 0 is given, then there is an open set H in Xz containing Qx such that g(H)⊆(c−ε, c+ε). Now it is enough to take thee-open subsetG=S

Q_z∈HQz

ofX. Clearly x∈Gand f(G)⊆(c−ε, c+ε) which implies that f ∈Ce(X).

By definitions of f and ϕ it is clear that ϕ(f) = g and we have thus shown

thatϕis onto.

By the above proof, all members ofC(Xz) will be of the form fz for some f ∈ C(X). Using Corollaries 27.10 and 27.11 in [7], for a locally connected spaceX we haveC_e(X)∼=C(Y), whereY is a discrete space and in particular for a compact locally connected space, we haveC_e(X)∼=Rⁿ for somen∈N.

Now using the definition of the isomorphismϕ, the following result is evident.

Lemma 4.2([1, Lemma 4.1]). An idealI inC_e(X)is fixed if and only ifϕ(I) is a fixed ideal in C(Xz). In particular, ϕ takes fixed maximal ideals to fixed maximal ideals.

By this lemma as in Theorems 4.2 and 4.4 in [1], fixed and free maximal ideals ofCe(X) will be characterized as follows.

Proposition 4.3. ([1], Theorem 4.2) For a topological space X, the fixed max- imal ideals ofCe(X)are precisely of the form

M_Qx={f ∈C_e(X) :Q_x⊆Z(f)}= ( \

y∈Qx

M_y)∩C_e(X) x∈X.

The idealsM_Qx are distinct for distinctQ_x and for each x∈X,C_e(X)/M_Qx is isomorphic with the real fieldR.

Proposition 4.4 ([1, Theorem 4.4]). For every space X, the maximal ideals of Ce(X)are precisely of the form

M^p={f ∈Ce(X) :p∈clβX_zZ(fz)} , p∈βXz

As we observe the free maximal ideals of Ce(X) are characterized via the zero-sets of X_z. These maximal idealsM^p so defined are in fact the maximal ideals ofC(X_z) which are not necessarily distinct for distinctp. For instance whenever p ∈ X, then M^p = M^q, for each q ∈ Q_p. Here we are going to introduce more natural representation of the maximal ideals ofC_e(X) by some equivalence classes in βX which do not depend upon X_z. First we construct equivalence classes in βX similar to Q_x’s in X. Lemma 3.5 will show us the right way to define such classes. For everyp, q∈βX define p≡q if and only if f^β(p) = f^β(q), for each f ∈ C_e^∗(X). Clearly, this defines an equivalence relation on βX. Let Q^p be the equivalence class containing p, for every p∈ βX. In caseX is a completely regular Hausdorff space, we also note that the mapping σ:X →Xz withσ(x) =Qx for eachx∈X has the Stone extension

¯

σ:βX→βXz, by Theorem 6.5 in [4]. This extension map is onto asσ is. to characterize the maximal ideals ofCe(X), we first need the following lemmas.

Lemma 4.5. Let p, q∈βX. Then σ(p) = ¯¯ σ(q)if and only if q∈Q^p.

Proof. For each f ∈Ce(X) we have fz(Qx) = f(x) for each x∈ X and this means thatf ∈C_e^∗(X) if and only iffz∈C^∗(Xz). On the other handf =fz◦σ on X for eachf ∈Ce(X) implies thatf^β =f_z^β◦σ¯ for eachf ∈C_e^∗(X). Now suppose that q /∈ Q^p, then there exists f ∈ C_e^∗(X) such thatf^β(p)6=f^β(q).

This implies that f_z^β◦σ(p)¯ 6=f_z^β◦σ(q) and hence ¯¯ σ(p)6= ¯σ(q). Conversely, let q∈ Q^p but ¯σ(p) 6= ¯σ(q). Then there exists g ∈ C^∗(X_z) withg^β(¯σ(p)) 6=

g^β(¯σ(q). Since every member ofC^∗(X_z) is of the formf_z for somef ∈C_e^∗(X), we may takeg=f_zfor somef ∈C_e^∗(X). Nowf_z^β◦σ(p)¯ 6=f_z^β◦¯σ(q) implies that f^β(p)6=f^β(q), sopis not equivalent toq, i.e.,q /∈Q^p, a contradiction.

Note that,σtakes zero-sets of members ofCe(X) to zero-sets inXzand also the images of any two disjoint zero-sets of members of Ce(X), are disjoint in Xz, sinceσ(Z(f)) =Z(fz) for eachf ∈Ce(X). Also using Lemma3.6,Ce(X) is a lattice ring. In fact iff ∈Ce(X) and we takeg:R→Rwithg(x) =|x|

for eachx∈R, then by Lemma3.6we infer that|f| ∈Ce(X).

Lemma 4.6. Let f ∈ Ce(X). Then σ(p)¯ ∈ clβX_zZ(fz) if and only if Q^p∩ clβXZ(f)6=∅.

Proof. LetQ^p∩clβXZ(f) =∅. Then by the relation defined preceding Lemma 4.5, for everya∈clβXZ(f), there existsga∈C_e^∗(X) such thatg^β_a(p)6=g_a^β(a).

If we take g^β_a(p) = δ and g^β_a(a) = α and define ka = |^g_δ−α^a−α| ∧1, then clearly ka ∈ C_e^∗(X) by the argument preceding the lemma, 0 ≤ ka ≤ 1, ka(a) = 0 and ka(p) = 1. Hence for every a ∈ clβXZ(f) we may assume that 0≤ga ≤1,g_a^β(p) = 1 and g^β_a(a) = 0. Again, lettingha = 2(ga∨¹₂ −¹₂), we have ha ∈ C_e^∗(X), 0 ≤ ha ≤ 1, a ∈ intβXZ(h^β_a) and h^β_a(p) = 1. Since clβXZ(f) is compact, there exists a finite subset {a1, a2, ..., an} of clβXZ(f), such that cl_βXZ(f)⊆Sn

i=1Z(h^β_a

i). ThenZ(f)⊆Z(h), where h=Qn k=1h_ak and evidently h ∈ C_e^∗(X), 0 ≤ h ≤ 1 and h^β(p) = 1. Hence Z(f) and h⁻¹[²₃,1] = Z(g), where g = h∧ ²₃ − ²₃ ∈ Ce(X) are disjoint zero-sets of C_e(X). Therefore by the argument preceding the lemma, σ(Z(f)) = Z(f_z) andσ(Z(g)) =Z(g_z) are disjoint zero-sets inX_z. Moreoverp∈cl_βXZ(g) im- plies that ¯σ(p)∈σ(cl¯ _βXZ(g))⊆cl_βXzσ(Z(g)) = cl¯ _βXzσ(Z(g)) = cl_βXzZ(g_z) and so ¯σ(p)∈/ cl_βXzZ(f_z).

Conversely, suppose that q ∈ Q^p ∩cl_βXZ(f) 6= ∅. Using the previous lemma ¯σ(p) = ¯σ(q), and since q ∈cl_βXZ(f), we have ¯σ(q)∈σ(cl¯ _βXZ(f))⊆ clβX_zσ(Z(f¯ )) = clβX_zσ(Z(f)) = clβX_zZ(fz). Therefore ¯σ(p) ∈ clβX_zZ(fz).

Theorem 4.7. For every completely regular Hausdorff spaceX, the maximal ideals of Ce(X)are precisely of the form

M^Qp={f ∈Ce(X) :Q^p∩clβXZ(f)6=∅}=



 [

q∈Q^p

M^q



∩Ce(X) , p∈βX,

and they are distinct for distinctQ^p.

Proof. Using the previous lemma and Theorem4.4, the proof of the first part is evident. For the second part, letQ^p6=Q^q. Thenq /∈Q^pand hence ¯σ(p)6= ¯σ(q) by Lemma4.5. Now there existsfz∈C^∗(Xz) such that ¯σ(p)∈clβXzZ(fz) and

¯

σ(q)∈/cl_βXzZ(f_z). This means thatQ^p∩clβXZ(f)6=∅butQ^q∩clβXZ(f) =∅ by Lemma4.6. Thereforef ∈M^Qp\M^Qq, i.e.,M^Qp6=M^Qq. We may directly obtain the fixed maximal ideals ofCe(X) from Theorem 4.7which are characterized in Proposition4.3.

Corollary 4.8. Let X be a completely regular Hausdorff space andp∈ βX.

Then the maximal ideal M^Qp is fixed if and only if M^Qp = MQ_x for some x∈X.

Proof. WheneverM^Qp =M_Qx, clearly M^Qp is fixed. Conversely, letM^Qp be fixed. Then there is x ∈ X such that x ∈ Z(f) for each f ∈ M^Qp. Using Proposition 3.4, f(Qx) = 0, ∀f ∈ M^Qp. This means that M^Qp ⊆ MQ_x and

henceM^Qp=MQ_x because M^Qp is maximal.

Remark 4.9. As we observed in Theorem4.7,M^Qp = (S

q∈Q^pM^q)∩C_e(X) for eachp∈βXwhile for fixed maximal idealsMQ_x, we haveMQ_x = (T

y∈QxMy)∩

Ce(X). The reason is that the equivalence classes Q^p’s are constructed by a bit different relation which is defined preceding Lemma4.5and in contrast to Qx’s, they are not necessarily the quasi components ofβX. In factQ^pfor each p∈βX is contained in the quasicomponent ofpin βX. To see this letq∈Q^p but not in the quasicomponent of p in βX. Then there is a clopen set G in βX such thatp ∈G but q /∈G. Now define g ∈C(βX) with g(G) = 0 and g(βX \G) = 1. Clearly g|_X ∈ C_e^∗(X) and g|^β_X = g separates p and q, i.e., q /∈Q^p.

Moreover, as the representation ofM^Qp’s, we may wrightMQ_x = (S

y∈QxMx)∩

C_e(X), in fact M_Qx =M_y∩C_e(X) for each y ∈Q_x. To this end, whenever f ∈C_e(X), then fory∈Q_x, we havef(y) = 0 if and only iff(Q_x) =f(Q_y) = 0 by proposition3.4.

Since every element of βXz is of the form ¯σ(p), p ∈ βX, the ideal O^¯σ(p) is defined as usual by the set {fz ∈ C(Xz) : ¯σ(p)∈ intβX_zclβX_zZ(fz)}. The related ideals in Ce(X) are ϕ⁻¹(O^σ(p)¯ ), p ∈ βX and we are to characterize them by the following lemma and corollary.

Lemma 4.10. Let f ∈Ce(X)and p∈βX. Thenσ(p)¯ ∈intβX_zclβX_zZ(fz)if and only ifQ^p⊆intβXclβXZ(f)

Proof. Let ¯σ(p) ∈ int_βXzcl_βXzZ(f_z). Then there exists g_z ∈ C^∗(X_z) (g ∈ C_e^∗(X)) such that g^β_z(¯σ(p)) = 1 and cozg^β_z ⊆ intβX_zclβX_zZ(fz). Therefore cozgz⊆Z(fz) which implies thatfzgz= 0 whencef g= 0. On the other hand, sinceg=gz◦σonX, we haveg^β=g_z^β◦σ¯ and henceg^β(p) = 1. Nowf g= 0 implies that clβXZ(f)∪clβXZ(g) = βX or equivalently, βX\clβXZ(g) ⊆ clβXZ(f). Using the equivalence relation defined preceding Lemma 4.5, for

everyq∈Q^p we haveg^β(q) =g^β(p) = 1. Hence

Q^p⊆βX\clβXZ(g)⊆intβXclβXZ(f).

For the converse, suppose that Q^p ⊆intβXclβXZ(f). By a similar argument in the proof of Lemma 4.5, there exists an e-continuous function h∈C_e^∗(X) such that h^β(p) = 1 and βX\intβXclβXZ(f) ⊆ clβXZ(h). Thus cozh^β ⊆ intβXclβXZ(f), whence cozh ⊆ Z(f) and hence f h = 0 which implies that f_zh_z = 0. Moreover, since h^β =h^β_z◦σ¯ and h^β(p) = 1, we haveh^β_z(¯σ(p)) = 1 which means that hz ∈/ M^σ(p)¯ . Therefore fz ∈ O^¯σ(p) by 7.12(b) in [4], so

¯

σ(p)∈intβX_zclβX_zZ(fz).

Now, by the above lemma and the isomorphism ϕ : Ce(X) → C(Xz), we have the following corollary.

Corollary 4.11. For each p∈βX,

O^Qp:=ϕ⁻¹(O^¯σ(p)) ={f ∈C_e(X) :Q^p⊆int_βXcl_βXZ(f)}.

Corollary 4.12. An ideal I inC_e(X)is contained in a unique maximal ideal M^Qp if and only ifO^Qp⊆I.

Using Theorem4.1, to study the algebraic properties of the ringsC_e(X), the spaceX may be considered as a zero-dimensional space. The following propo- sition also states thatC_e(X) (C_e^∗(X)) determines the topology of a Hausdorff X if and only if it is zero-dimensional.

Proposition 4.13. Let X be a Hausdorff space. ThenX is zero-dimensional if and only if its topology coincides with the weak topology induced by Ce(X) (C_e^∗(X)).

Proof. WheneverXis a zero-dimensional space, thenCe(X) =C(X) by Propo- sition3.7. On the other handX is completely regular, so its topology coincides with the weak topology induced byC(X) =Ce(X) (C^∗(X) =C_e^∗(X)) by The- orem 3.6 in [4]. Conversely suppose that the topology onX coincides with the weak topology induced by Ce(X). Then the collection B ={f⁻¹(a, b) : f ∈ Ce(X), a, b∈R}is a base forX. But eachf⁻¹(a, b) is a union ofe-open subsets ofX by Proposition3.2and this shows that the spaceX has a base consisting of e-open sets, so X is an e-space. On the other hand, using Theorem 3.7 in [4] the spaceX is completely regular and henceX will be zero-dimensional by

Proposition2.7.

Remark 4.14. If we consider X to be zero-dimensional, then C_e(X) =C(X) and every two different points p, q ∈ βX can be separated by f^β, for some f ∈ C^∗(X) =C_e^∗(X). This means thatQ^p is singleton for eachp∈ βX and each maximal idealsM^Qp ofCe(X) is exactly M^p.

The rest of this section is devoted to characterization of real maximal ideals of C_e(X). IfR is a commutative ring which contains the real field R, then a maximal ideal M ofR is said to be real whenever _M^R 'R. In C(X) an ideal I is a real maximal ideal if and only if for eachf ∈C(X),f −r∈I for some

r∈R; see Corollary 3.3 in [2]. From 8B.2 in [4], for everyf ∈C(X), we denote υfX ={p∈βX:f^∗(p)6=∞}, wheref^∗ :βX →R^∗ is the Stone extension of f from X to the one-point compactificationR^∗ =R∪ {∞}. The realcompact spaces between X and βX are precisely the spacesυ_C⁰X =T

f∈C⁰υ_fX for a subsetC⁰ofC(X). The spaceυ_C(X)Xis called the Hewitt realcompactification ofX which is the smallest; see 8B.3 in [4].

Theorem 4.15. Let X be a completely regular Hausdorff space and p∈βX.

Then the following statements are equivalent.

(1) The maximal idealM^Qp of Ce(X)is real.

(2) Q^p∩υ_Ce(X)X 6=∅.

(3) p∈υ_Ce(X)X, or equivalently Q^p⊆υ_Ce(X)X.

Proof. First note thatM^Qp is real if and only ifϕ(M^Qp) is real. On the other hand, Lemma4.6, implies thatϕ(M^Qp) =M^σ(p)¯ and therefore,M^Qp is real if and only if ¯σ(p)∈υX_z.

(1)⇒(2). Suppose thatQ^p∩υ_Ce(X)X =∅. Let q∈Q^p be arbitrary. Then by the definition ofυ_Ce(X)X, there exists an e-continuous functionf ∈C_e(X) such thatf^∗(q) =∞. Using Theorem 7.6 (a) in [4],|M^q(f)| is infinitely large and therefore by Theorem 5.7 in [4] we infer that the zero-setsZ_n ={x∈X :

|f(x)| ≥n},n∈Nbelong toZ[M^q]. Thus Lemma4.6implies that σ(Zn) ={y∈Xz:|fz(y)| ≥n} ∈Z[M^σ(p)¯ ],

and hence by Theorems 5.7 and 7.6(a) in [4] we conclude that f_z^∗(¯σ(q)) = f_z^∗(¯σ(p)) = ∞. This is a contradiction by the preceding argument at the beginig of the proof; see also Theorem 8.4 in [4].

(2)⇒(3). Letq∈Q^p∩υ_Ce(X)X butp /∈υ_Ce(X)X. Then by the definition of υ_Ce(X)X, there exists ane-continuous function f ∈Ce(X) such that f^∗(p) =

∞. Since q ∈ υ_Ce(X)X, so f^∗(q) = r, for some real number r ∈ R. Let g =r∨f ∧(|r|+ 1), then g ∈ C_e^∗(X) and g^β(q) = f^∗(q) = r. But q ∈ Q^p and by the definition of Q^p we have g^β(p) = g^β(q) = r, which implies that f^∗(p) =r, a contradiction.

(3)⇒(1). Using the argument at the beginning of the proof, it is enough to show that ¯σ(p)∈υXz. To do this, we need to prove that, f_z^∗(¯σ(p))6=∞, for everyfz∈C(Xz); see Theorem 8.4 in [4]. In fact sincef_z^∗◦σ¯ agree withf^∗on X , we havef_z^∗◦σ¯=f^∗. Now (f_z^∗◦σ)(p) =¯ f^∗(p) implies thatf_z^∗(¯σ(p))6=∞

becausef^∗(p)6=∞and we are done.

Using Theorem 3.1 in [2] and our Theorem4.15, we have also the following elementwise characterization of real maximal ideals ofCe(X).

Proposition 4.16. Let Ibe an ideal ofCe(X). Then the following statements are equivalent.

(1) The ideal I is a real maximal ideal ofCe(X).

(2) For eachf ∈Ce(X), there isr∈Rsuch thatf−r∈I.

(3) There existsp∈βX such that for eachf ∈Ce(X),Q^p∩clβXf⁻¹(r)6=

∅, for somer∈R.

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