Proof
Let
denote a bound from above, so that
holds for all
. We assume that
is not a Cauchy sequence. Then there exists some
such that for every
, there exist indices
fulfilling
.
Because of the monotonicity, there is also for every
an
with
.
Hence, we can define inductively an increasing sequence of natural numbers satisfying
-
-
and so on. On the other hand, there exists, due to the
axiom of Archimedes,
some
with
-
![{\displaystyle {}k\epsilon >b-x_{n_{0}}\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c690efa32dbb84a3b6d5f7eb7b0a596032a303ed)
The sum of the first
differences of the
subsequence
,
,
is
![{\displaystyle {}{\begin{aligned}x_{n_{k}}-x_{n_{0}}&={\left(x_{n_{k}}-x_{n_{k-1}}\right)}+{\left(x_{n_{k-1}}-x_{n_{k-2}}\right)}+\cdots +{\left(x_{n_{2}}-x_{n_{1}}\right)}+{\left(x_{n_{1}}-x_{n_{0}}\right)}\\&\geq k\epsilon \\&>b-x_{n_{0}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6083aa3d190af3ed3e9175ae8df5ce1a70734bc5)
This implies
,
contradicting the condition that
is an upper bound for the sequence.