# Real numbers/Continuous functions/Motivation/Introduction/Section

We denote the distance between two real numbers and by . For a given function

one may ask whether it is possible to control the distance in the image (the target) by controlling the distance in the domain. Let and let be its image. One would like that for points which are close to , also their image points are close to . Already linear functions with different slopes show that the measure for being close in the image and in the domain can not be the same. A reasonable question is whether for any desired exactness in the image there exists at all an exactness in the domain which ensures that the values of the function are lying inside the desired exactness. To make this intuitive idea more precise, let be given. This represents an accuracy for the target. The question is whether one can find a (representing an accuracy for the domain) such that for all fulfilling , also the relation holds. This idea leads to the concept of a continuous function.

Let be a subset,

a
function,
and
a point. We say that is
*continuous*
in the point , if for every
,
there exists a
,
such that for all
fulfilling
,
the estimate
holds. We say that
*continuous*,
if it is continuous in every point

stands for the set of definition of the function. Typically, is just , or an interval, or with finitely many points removed. Instead of working with the real numbers and , one might also work just with the unit fractions and .

A constant function

is continuous. For every given , one can choose an arbitrary , since

holds anyway.

The identity

is also continuous. For every given , one can take , yielding the tautology: If

then

holds.

We consider the function

given by

This function is not continuous in . For and every positive , there exists a negative number such that . But for such a number we have .

The following statement relates continuity to convergent sequences.

Let be a subset,

a function and

a point. Then the following statements are equivalent.- is continuous in the point .
- For every convergent sequence in with , also the image sequence is convergent with limit .

Suppose that (1) is fulfilled and let be a sequence in converging to . We have to show that

holds. To show this, let be given. Due to (1), there exists a fulfilling the estimation property (from the definition of continuity) and because of the convergence of to there exists a natural number such that for all the estimate

holds. By the choice of we have

so that the image sequence converges to .

Suppose now that (2) is fulfilled. We assume that is not continuous. Then there exists an such that for all there exist elements such that their distance to is at most , but such that the distance of their value to is larger than . This holds in particular for every , . This means that for every natural number , there exists a with

This sequence converges to , but the image sequence does not converge to , since the distance of its members to is always at least . This contradicts condition (2).