## Example 6.2.8(d):

_{n}} is a Cauchy sequence, is the sequence { f( x

_{n}) } also Cauchy ?

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

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

- | x
_{n}- x_{j}| < if n, j > N

But then, by uniform continuity, we have that

- | f( x
_{n}) - f(x_{j}) | < if n, j > N

Therefore, the sequence { f( x_{n})
} 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 { x_{n}}
that converges to a fixed point inside (0, 1), the same argument
as above would to show that { f( x_{n})
} 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.