Question

Give a small proof of the above?

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

Answer 1

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.