Consider a map where and are metric spaces.

The standard (wrong) definition says: is uniformly continuous if for every there exists such that .

According to this definition, every function is uniformly continuous. What exactly is uniform about the function ? I don't see it.

The right definition is: is uniformly continuous if there exists a function such that and for all .

What is the difference, you ask? For nice spaces, such as intervals, there is none. But for uglier spaces: highly nonconvex, disconnected, etc the second definition is more restrictive, or, the way I put it, the first definition is overly permissive (e.g., toward the function ).

There is a parallel with the definition of a quasisymmetric map, which is the version of “uniformly continuous” for relative distances. This definition was worked out after the concept of a general metric space was well understood, which is why mathematicians were able to get it right. Uniform continuity predates metric spaces: it was defined for functions on an interval, and then the same definition was applied to more general spaces whether it fit or not.

Here is a statement which is true with the right definition of uniform continuity and false with the wrong one.

**Theorem.** For a metric space , the following are equivalent:

- is compact
- every continuous function is uniformly continuous
- every map , for every metric space , is uniformly continuous

**Proof.** . Given , consider the set . This is a closed subset of the compact set , hence compact. Since the map is continuous, it follows that is a bounded subset of . Define . One can show that , for example, by covering the diagonal of with appropriate open neighborhoods.

is trivial.

. We must show that if is either not complete or not totally bounded, then there exists a continuous function which is not uniformly continuous. First, suppose is not complete. Pick where is the completion of . The function is continuous but not bounded on . Since the right definition of uniform continuity requires to be bounded on bounded sets, is not uniformly continuous.

Next, suppose that is not totally bounded. Then there exists and an infinite sequence such that whenever . Let . This function is continuous on but is not uniformly continuous.

**QED**

At a first glance one might think than the difference between the two notions boils down to preserving bounded sets. This is not so, since any function is bounded on bounded subsets of , yet not every such function is uniformly continuous according to the right definition.

When needs to be emphasized, one can say that is -continuous. In particular, is Lipschitz continuous if it is -continuous with for some . Also, is -Hölder continuous if it is -continuous with as . (I don’t think many people really want to impose the Hölder condition at large distances.) Finally, is Dini-continuous if it is -continuous with .