Proof. Suppose, for example, that is an increasing, bounded sequence; that is, suppose that we have
for some (fixed) number and all . Then, in particular, is a (nonempty) bounded set.
It follows that has a least upper bound (or supremum). In other words, there is an upper bound for the set, which we will again call , satisfying whenever is any other upper bound for . We will show that converges to .
To this end, suppose that we are handed a small positive number , and consider the number . Since < , we know that can't be an upper bound for . Thus, there is some such that
Since our sequence is increasing, this means that we have
for all > . In particular,
for all > . Hence, converges to .
Finally, if we're given a decreasing, bounded sequence , just apply the first part of the proof to the increasing sequence .