Continuity as a Galaxy of Hyperreal Functions

In the present paper, the problem of defining continuity and scontinuity as a galaxy of hyperreal function is discussed. Our attempt is based on the fact that monads are subsets of some galaxies. New results are obtained, with nonstandard variables, related to a new extension of the continuity notion.


1-Introduction: -
The following definitions of nonstandard analysis will be needed in this paper.Every set or elements defined in a classical mathematics is called a standard.[2] Any set or formula which does not involve a new predicates "standard, infinitesimals, limited, unlimited…etc" is called internal, otherwise is called external.[2] , [4] A real number x is called unlimited if xr > for all real 0 r > otherwise called limited. [2] The set of all unlimited real numbers is denoted byR , and the set of all limited real numbers is denoted by R [2] A real number x is called infinitesimal if xr < for all positive standard real number r [4].
A real number x is called appreciable if it is neither unlimited nor infinitesimal, and the set of all positive appreciable numbers is denoted by A + .[4] Two real numbers x and y are said to be infinitely close if xy − is infinitesimal and is denoted by xy ≅ .[4] If x is a limited number in R , then it is infinitely close to a unique standard real number, this unique number is called the standard part of x or shadow of x denoted by () stx or 0 x , in this case we say that x is nearly standard [2][6].
Let f be an internal function, we say that f satisfies k-Lipschitz condition if ()() fxfykxy −<− for some standard constant k ∈N.[5] If E is a standard metric space, then E is complete if for each xE ∈ either x is nearly standard or there exists a standard number 0 r > , such that the ball (,) Bxr contains no standard elements. [2] Remark 1.1:

1.
From the definition of monad and galaxy we have: Using the definition of monad we obtain:

2-Galaxy Continuity
With nonstandard analysis the region of tangible elements is larger than that of standard analysis, therefore all problems that deal with such unusual elements take its frame space in nonstandard analysis, so the study of the behaviour of a function and its properties in nonstandard analysis give us a real phase and precision of the nature of it, which can never be imagined classically.
For functions that deal with only unlimited values or its values does not lie in the monad of any other points, the conventional and nonstandard definitions of continuity are meaningless and we are unable to say any thing about its continuity.In this paper we treat such problem by presenting a new version of continuity which includes all possible standard and nonstandard cases.

Definition 2.1:
Let : fXY → be an internal function then we say that f is , and is denoted by g-continuous.
The following examples give some relations between continuity, scontinuity, and g-continuity.
=≠ , then we have: 1-For all standard x , f is continuous, and neither s-continuous nor g-continuous.

2-
x , and otherwise it is not g-continuous.
3-f is not s-continuous, take . ii) , where 0 ε ≅ , is not continuous at 0 but it is g-continuous. iii) is not continuous at any real point while it is g-continuous everywhere.

iv) ()
, where n is standard positive integer, and 0 ε ≅ is g-continuous but not s-continuous.

2.
If f is s-continuous then f is g-continuous.

3.
If f satisfies k-Lipschitz condition then f is g-continuous. Proof: 1.
Let f be a standard continuous function at a standard point , thus f is g-continuous.

2.
It's proof is similar to the proof of (1).

3.
Let f be an internal function (standard or nonstandard) such that f satisfies k -Lipschitz condition and , xy ∈R such that xy ≅ , to prove that ( ) for all xy ≅ , where ε is an infinitesimal number, then f is g-continuous if and only if f is uniformly s-continuous.

Proof:
let f be g-continuous, then foe all , x xy ≅ we have ( ) for some standard numbers k and n .Thus ()() fxfy − is infinitesimal Therefore f is uniformly s-continuous.
The converse is obvious.Proof: Use Theorem 2.8 [3] to get that f is limited.
Therefore h is limited, Now, use Theorem 3.3 to obtain the result.Theorem 3.5 Let f be an infinitesimal valued function, then f is s-continuous function if and only if f is g-continuous.

Proof:
The forward way is direct.
For the backward direction, let f be g-continuous function, Therefore ( ) Thus f is s-continuous.

Theorem 3.6
Let f be an internal function, then f is g-continuous if and only if ( ) for all x , xy ≅ and α is appreciable.
ii. y is appreciable.
iii. y is unlimited.
For the first two possibilities define () o hxx = , then ()() hxhy − is limited, therefore h is g-continuous.For the third possibility, if xy ≅ , then x is also an unlimited number, if xy ≅ / we get a contradiction to our assumption.
In this case, define () ∈ , Since f satisfies k-Lipschitz condition then we get ()() fxfykxy −<− for some standard constant k ∈N , but xy ≅ , so ()() The converse of the above theorem is not true, for 1. and 2.see examples 2.2(iii) and 2.2(ii) respectively.For the third statement let that f is g-continuous and to show that it does not satisfy k-Lipschitz condition; Let 0 x = and y ε = where ε is an infinitesimal number, then xy ≅ and 0 xy −≅ , but there is no standard number k satisfies the inequality ()() fxfykxy −<− .