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.