# Give a small proof of the above?

## If the range of a function $f$ is $\mathbb{R}$ and $f$ is strictly monotone in $\mathbb{R}$ then $f$ is continuous in $\mathbb{R}$

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, $f \left({x}_{0}\right)$ is either greater or lesser than $f \left({x}_{0} + \delta\right)$ and $f \left({x}_{0} - \delta\right)$, 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 ${\lim}_{x \to {x}_{0}^{-}} f \left(x\right)$ or ${\lim}_{x \to {x}_{0}^{+}} f \left(x\right)$ 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 $f ' \left(x\right)$. 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 $f : \mathbb{R} \to \mathbb{R}$ has an image of $\mathbb{R}$, 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.