Example: If f(x) is uniformly continuous in R, and { } is a Cauchy sequence, is the sequence { f( ) } also Cauchy ?

Note that this is not true if the function is only assumed to be continuous (as the example f(x) = 1 / x on the interval (0, 1) illustrates). But if f is uniformly continuous, we have:

• given any > 0 there exists > 0 such that if | t - s | < then | f(t) - f(s) | <

If { } is a Cauchy sequence, there exists an integer N such that

• | - | < if n, j > N

But then, by uniform continuity, we have that

• | f( ) - f( ) | < if n, j > N

Therefore, the sequence { f( ) } is again a Cauchy sequence.

Why will this argument will not work for a continuous function ? As an example, consider again the function f(x) = 1/x. It is continuous in the open interval (0, 1). Therefore, if we consider a sequence { } that converges to a fixed point inside (0, 1), the same argument as above would to show that { f( ) } is also Cauchy. We can not use this argument, however, for a sequences that converge to 0, since f is not even defined at zero. On the other hand, we can find Cauchy sequences which converge to 0, without ever having to refer to the value of the function at the limit of the sequence.