There is a useful sequential characterization of continuity in metric spaces. Let be a map between metric spaces. If for every convergent sequence in we have in , then is continuous. And the converse is true as well.

Uniformly continuous functions also have a useful property related to sequences: if is uniformly continuous and is a Cauchy sequence in , then is a Cauchy sequence in . However, this property does not characterize uniform continuity. For example, if , then Cauchy sequences are the same as convergent sequences, and therefore any continuous function preserves the Cauchy-ness of sequences—it does not have to be uniformly continuous.

Let us say that two sequences and are **equivalent** if the distance from to tends to zero. The **sequential characterization of uniform continuity** is: is uniformly continuous if and only if for any two equivalent sequences and in , their images and are equivalent in . The proof of this claim is straightforward.

In the special case when is a constant sequence, the sequential characterization of uniform continuity reduces to the sequential characterization of continuity.

A typical example of the use of this characterization is the proof that a continuous function on a compact set is uniformly continuous: pick two equivalent sequences with non-equivalent images, pass to suitable subsequences, get a contradiction with continuity.

Here is a different example. To state it, introduce the notation .

**Removability Theorem**. Let be continuous. Suppose that there exists such that for every , the restriction of to is uniformly continuous. Then is uniformly continuous on .

This is a removability result because from having a certain property on subsets of we get it on all of . To demonstrate its use, let with the standard metric, , and . The uniform continuity of on follows immediately from the derivative being bounded on that set (so, is Lipschitz continuous there). By the removability theorem, is uniformly continuous on .

Before proving the theorem, let us restate the sequential characterization in an equivalent form (up to passing to subsequences): is uniformly continuous if and only if for any two equivalent sequences and there exist equivalent subsequences and , with the same choice of indices in both.

**Proof of the theorem**. Suppose and are equivalent sequences in . If , then as well, and the continuity of at implies that both and converge to , hence are equivalent sequences. If , then by passing to a subsequence we can achieve for some constant . By the triangle inequality, for sufficiently large we have . Since is uniformly continuous on , it follows that and are equivalent.