Definition. Let be metric spaces and a function. We say is Cauchy regular provided that any Cauchy sequence in , is a Cauchy sequence in .
Note that if is uniformly continuous on , then it is Cauchy regular. Indeed, let be a Cauchy sequence in and let be given. Since is uniformly continuous, there exists such that implies .
Since is Cauchy, there exists such that implies . So . Hence is Cauchy in .
One may ask whether a Cauchy regular function is continuous. It is clearly true.
Proposition. Let be metric spaces
and a Cauchy regular function. Then is continuous
Proof. Suppose not. There exists such that is not continuous at . Then there exists such that for any , there exists such that but .
By taking , we can choose in such that in . Now define . Since is convergent, is Cauchy. So is Cauchy. But
shows is not Cauchy, a contradiction.
So Cauchy regular is continuous. Here we note that we didn’t used any kind of continuity.
Recall the Heine-Cantor theorem. If is compact and is continuous, then is uniformly continuous. One may ask what condition on gurantees a continuous function is Cauchy regular. In fact, if we impose is complete, this is enough.
Proposition. Let be continuous. If is complete, then is Cauchy regular. In this case, continuity and Cauchy regular is
Proof. Let be a Cauchy sequence in . By completeness of , there is such that in .
Then triangluar inequality shows
Since is continuous at , there exists such that implies
Hence there exists such that implies
So is Cauchy regular.
Then is continuous . But this function is not Cauchy regular since it cannot be extended to as a continuous function. The example does not contradict our previous theorem since is not complete.
There is a continuous function which is Cauchy regular, but not uniformly continuous. Consider defined by . Then it is Cauchy regular, but not uniformly continuous.
Definition. A metric space is Cauchy precompact if every sequence admits a Cauchy subsequence.
Proposition. If is Cauhcy-precompact and is Cauchy regular, then is uniformly continuous.
Proof. Suppose not. Then there exists and sequences , such that
but . Since is Cauchy precompact, there exists a Cauchy subsequence of . Again by Cauchy-precompactness, there exists a Cauchy subsequence of . Define
by , ,. One may check it is Cauchy sequence. Then if is odd, then
But is Cauchy, a contradiction. Hence is uniformly continuous on .