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
.