Give a small proof of the above?

If the range of a function ff is RR and f is strictly monotone in RR then f is continuous in RR

1 Answer
May 11, 2018

I'm going to assume that by "range" you are referring to a function's image and not the function's codomain.

A strictly monotonic function is either strictly non-decreasing or strictly non-increasing. We consider whether such a function can have discontinuities.

A monotonic function cannot have removable discontinuities because having one would imply that at the particular value x_0 at which the discontinuity occurs, f(x_0) is either greater or lesser than f(x_0 + delta) and f(x_0 - delta), where delta is sufficiently small. This implies either a change from increasing to decreasing or vice-versa.

A monotonic function cannot have an essential discontinuity because, if it did, then there would be some x_0 for which at least one of lim_(x->x_0^-) f(x) or lim_(x->x_0^+) f(x) is either infinite or undefined. This would imply a change from increasing to decreasing or vice-versa.

A monotonic function can have jump discontinuities because such continuities do not necessarily cause a change in the sign of f'(x). However, jump discontinuities necessary "skip" values in the image, hence the "jump" in "jump discontinuity". Such values can be recovered later if the function changes from increasing to decreasing or vice-versa, but such a function is not monotonic. Our assumption, however, is that the image for the function f: RR -> RR has an image of RR, so there cannot exist a jump discontinuity given our restraints.

These are the only types of discontinuities and our given constraints do not allow any of them to occur. Thus, our function is continuous.