# Heine–Cantor theorem

In mathematics, the **Heine–Cantor theorem**, named after Eduard Heine and Georg Cantor, states that if *f* : *M* → *N* is a continuous function between two metric spaces, and *M* is compact, then *f* is uniformly continuous. An important special case is that every continuous function from a bounded closed interval to the real numbers is uniformly continuous.

## Proof

Suppose that *M* and *N* are two metric spaces with metrics *d _{M}* and

*d*, respectively. Suppose further that is continuous, and that

_{N}*M*is compact. We want to show that

*f*is uniformly continuous, that is, for every there exists such that for all points

*x*,

*y*in the domain

*M*, implies that .

Fix some positive . Then by continuity, for any point *x* in our domain *M*, there exists a positive real number such that when *y* is within of *x*.

Let *U _{x}* be the open -neighborhood of

*x*, i.e. the set

Since each point *x* is contained in its own *U _{x}*, we find that the collection is an open cover of

*M*. Since

*M*is compact, this cover has a finite subcover. That subcover must be of the form

for some finite set of points . Each of these open sets has an associated radius . Let us now define , i.e. the minimum radius of these open sets. Since we have a finite number of positive radii, this number is well-defined and positive. We may now show that this works for the definition of uniform continuity.

Suppose that for any two *x,y* in *M*. Since the sets form an open (sub)cover of our space *M*, we know that *x* must lie within one of them, say . Then we have that . The Triangle Inequality then implies that

implying that *x* and *y* are both at most away from *x _{i}*. By definition of , this implies that and are both less than . Applying the Triangle Inequality then yields the desired

For an alternative proof in the case of *M* = [*a*, *b*] a closed interval, see the article on non-standard calculus.