SHI Yanwei , MA Chunhui

(1. School of Intelligent Science and Information Engineering, Xi’an Peihua University, Xi’an 710125, China; 2. School of Science, Xi’an University of Architecture and Technology, Xi’an 710055, China )

Abstract： This paper studies *-topology T* and s-topology Ts in polysaturated nonstandard model, which are induced by metric space on two nonstandard sets. In order to construct *-topology T*, the set of finite points is introduced. It shows that every internal set is compact, every open set is saturated and standard part mapping is continuous. Finally, the closure operator, interior operator and mapping * are discussed under s-topology Ts.

Key Words： finite point; saturated set; standard part mapping; functor

Foundationitem： Supported by Scientific Research Project of Xi’an Peihua University (PHKT17001) and Basic Research Fund of Xi’an University of Architecture and Technology( JC1709)

Abouttheauthor： SHI Yanwei(1980—), ORCID∶http: // orcid.org/0000-0001-6056-3014 , female, master, associate professor, the fields of interest are general topology and nonstandard analysis, E-mail: shi_yanwei@163.com.

DOI： 10.3785/j.issn.1008-9497.2018.04.005

0 Introduction and preliminaries

In 1960, nonstandard analysis was founded by American mathematical logician ROBINSON[1]. As a new branch of mathematics and new mathematical method, nonstandard analysis was a new mathematical theory that used nonstandard models to study various mathematical problems. In nonstandard analysis, nonstandard topology was an important reaching field．

As usual, there were two ways to study nonstandard topology. One was to research the general topological space[1-2]or uniform space[3-4]by nonstandard analysis methods, the other was directly to study the topology on a nonstandard set[5-6]. In this paper, the latter will be shown. On two nonstandard sets, Fin(*X) and*X，*-topology T*ands-topology Tsinduced by metric space (X,ρ) are studied in polysaturated nonstandard model. In *-topological space (Fin(*X), T*), the following conclusions are proved: (1)Every internal set is compact; (2) Every open set is saturated; (3) Standard part mappingstis continuous. Ins-topological space (*X, Ts), the relations, between in (X, T ) and in (*X, T*), of closure operators and interior operators are discussed, respectively. And it is obtained that mapping * between two categories is a functor．

Firstly, some basic concepts and conclusions are recalled. For details, [1, 2, 5, 7] can be referenced. In this paper,Xis any non-empty set.N,RandR+denote the sets of the natural, real and positive real numbers, respectively．

Definition1The mappingρ:X×X→Ris called a metric onXif for anyx,y,z∈X,

(1)ρ(x,y) ≥ 0, andρ(x,y) = 0 if and only ifx=y;

(2)ρ(x,y) =ρ(y,x);

(3)ρ(x,y) ≤ρ(x,z) +ρ(z,y)．

The pair (X,ρ) is called a metric space．

For eachx∈Xandr∈R+,B(x,r) = {y∈X|ρ(x,y)

B= {B(x,r) |x∈X，r∈R+}.

As we all known, B is a topological base onX. The topology generated by B is denoted as T．

About nonstandard analysis, letSbe an individual set. The superstructureV(S) can be inductively defined as follows:

V0(S)=S，Vn+1(S)=Vn(S)∪P(Vn(S)),n∈N，

where P (Vn(S)) is the power set ofVn(S), i.e. the collection of all subsets ofVn(S).

Suppose thatX∪R∈V(S). It can be easily obtained that mathematical objects used in this paper are all in superstructureV(S), such as open sets, topology, metric, functions onXetc. Given thatV(*S) is polysaturated nonstandard model ofV(S)．

Lemma1(Saturation principle)[6]V(*S) is polysaturated nonstandard model ofV(S) if and only if for every internal family of sets with the finite intersection property has a non-empty intersection．

1 *-Topology on Fin(*X )

In this section, a kind of topology called *-topology on Fin(*X) will be constructed, and some properties about it will be discussed．

By transfer principle, mapping*ρ:*X×*X→*Rsatisfies conditions (1)～(3) in definition 1. Since the value of*ρis in hyper-real*R, the nonstandard extension (*X,*ρ) is a hyper-metric space．

Definition2Let (X,ρ) be a metric space．

(1) A pointa∈*Xis said to be finite in (*X,*ρ) if*ρ(x,a) is finite for somex∈X. And Fin(*X) denotes the set of all finite points in (*X,*ρ)．

(2) A pointa∈*Xis said to be near-standard in (*X,*ρ) if*ρ(x,a) is infinitesimal for somex∈X. Andns(*X) denotes the set of all near-standard points in (*X,*ρ)．

Since 0 is infinitesimal and infinitesimal is finite, every standard point is near-standard point and near-standard point is finite point. Thus,X?ns(*X) ? Fin(*X) ?*X．

Consider the copy B of collectionBof all open balls，

B={*B(x,r) |B(x,r) ∈B};

Since

∪ B =∪{*B(x,r) |B(x,r) ∈B} = Fin(*X ), B can be a subbase for a topology on Fin(*X)．

Definition3The topology generated by B is called *-topology on Fin(*X), and denoted by T*．

Theorem1Let (X,ρ) be a metric space. Then

(1) For every internal subsetA? Fin(*X),Ais compact in Fin(*X) under T*.

(2)Xis dense in Fin(*X) under T*．

(2) LetGbe arbitrary non-empty T*-open set in Fin(*X). For eacha∈G, by definition of T*, there existx∈Xandr∈R+such thata∈*B(x,r) ?G. So, there isx∈Xsuch thatx∈G, i.e.X∩G≠ ? for every non-emptyG∈ T*. The proof is finished．

Definition4Let (X,ρ) be a metric space. For eacha∈*X, the set

m(a) = ∩{*B(x,r) ∈ B |a∈*B(x,r)}

is called the monad of pointa．

A subsetA?*Xis said to be saturated ifm(a) ?Afor alla∈A．

It can be easily to see that

(1)m(a) = ∩ {*G|a∈*G,G∈ T} =

{b∈*X|*ρ(a,b) is infinitesimal}.

(2) A is saturated if and only if

A= ∪ {m(a) |a∈A}．

Theorem2Let (X,ρ) be a metric space.Gis saturated for everyG∈ T*．

ProofLetG∈ T*. For eacha∈G, there arex∈Xandr∈R+such thata∈*B(x,r) ?G. Sincem(a) ?*B(x,r),m(a) ?G. Hence,Gis saturated．

Define mappingst:ns(*X) →X, for eacha∈ns(*X)，st(a) =xif and only if*ρ(x,a) is infinitesimal.

As we all known, metric space is also Hausdorff space. For metric space (X,ρ), mappingstis well-defined. It is clear thatst(a) =xif and only ifa∈m(x). Generally, for anyA?ns(*X), define

st[A] = {x∈X|st(a) =x，a∈A}.

IfA= {a},st({a}) is also denoted asst(a)．

Definition5Mappingstdefined above is called standard part mapping．

Lemma2[6]Let (X,ρ) be a metric space. For anyA?X,clX(A) =st[*A], whereclX(A) is the closure ofAinXunder T ．

Theorem3Standard part mappingst:ns(*X)→Xis continuous with respect to T*and T ．

ProofFor eacha∈ns(*X), letst(a) =x∈X,G,H∈ T such thatx∈G∩HandclX(G) ?H．Thenm(x) ?*G. Hence*Gis a T*-open neighborhood ofainns(*X). By lemma 2,x=st(a) ∈st[*G]=clX(G) ?H. Therefore, mappingstis continuous．

2 s-Topology on *X

At last,s-topology Tswill be constructed on*X, where (X,ρ) is a metric space. Some properties will be shown．

Let (X,ρ) be a metric space. Consider the following copyTof topology T generated by B, which is the collection of all open balls in (X,ρ)．

T= {*G|G∈ T },

althoughTis not a topology, it forms a base for a topology on*Xsince*X∈T．

Definition6The topology generated byTis calleds-topology on*X, and denoted by Ts．

Now, three topological spaces have been obtained, (X, T ), (Fin(*X), T*) and (*X, Ts) are all induced by metric space (X,ρ). Since B is base of T , it is easily to see that

(X, T ) ≤ (Fin(*X ), T*) ≤ (*X, Ts),

which relation ≤ is partial order relation between topological subspace．

Theorem4Let (X,ρ) be a metric space．

(1) Every internal subsetA?*Xis compact under Ts.

(2)Xis dense in*Xunder Ts．

ProofThe proof is similar to theorem 1．

Theorem5For everyA?X，

(1)*(clX(A)) =cl*X(*A)， whereclXandcl*Xare the closure operators in (X, T ) and(*X, Ts)， respectively．

(2)*(intX(A)) = int*X(*A), where intXand int*Xare the interior operators in (X, T ) and(*X, Ts), respectively．

ProofOnly (2) is proved．

*(intX(A)) ? ∪{*G|*(intX(A)) ?*G,G∈ T } =

∪{*G | intX(A) ?G,G∈ T }=

∪{*G|A?G,G∈ T } = ∪{*G|*A?

*G,G∈ T }=int*X(*A)．

Conversely, by transfer principle,A? intX(A) implies*A?*(intX(A)). Since intX(A) ∈ T implies*(intX(A)) ∈ Ts, int*X(*A) ?

*(intX(A)). The proof is finished．

In the following conclusion, the mapping * will be a functor between two categories C and Cs．

Let (X,ρ) be a metric space. The topologies T and Tsare generated byBandT, respectively. Two categories C and Csare defined as follows:

(1) The objects of C and CsareXand*X, respectively.

(2) The homomorphism of C and Csare continuous functions on topological spaces (X, T ) and on (*X, Ts), respectively．

Theorem6Let C and Csbe two categories defined above. Mapping * : C→ Csis a functor．

ProofStep1Clearly, for eachx∈X,*(x)=

*x∈*X．

Step2*fis a function on*Xfor every functionfonX. In fact, sincefis a function onX, the following sentence holds:

[?z∈X][[〈x,z〉∈f] → [z=y]]].

By transfer principle，

[?z∈*X][[〈x,z〉∈*f] → [z=y]]].

That is,*fis a function on*X．

Step3Let functionfbe continuous on (X, T). For anyG∈ T ,*G∈ T . Then*f-1[*G] =*(f-1[G]) ∈ T . Function*f: (*X, Ts) →

(*Y, Ts) is continuous, since T are bases for Ts．

Step4For any continuous functionsfandgonX, by step 3,*f，*gand*(f°g) are all continuous on*X, sincef°gis continuous onX. Then by definition off°g, the following sentence holds:

[?z∈X][[〈x,z〉∈g] ∧ [〈z,y〉∈f]]].

By transfer principle，

*(f°g)] ? [?z∈*X][[〈x,z〉∈*g] ∧

[〈z,y〉∈*f]]].

That is,*(f°g) =*f°*g．

Step5For identity 1XonX, the following sentence holds：

[?x∈X][〈x,x〉∈ 1X].

By transfer principle，

[?x∈*X][〈x,x〉∈*(1X)]．

That is,*(1X) = 1*X．

So, mapping * : C → Csis a functor．