The title is kind of misleading because the task actually to show
Every monotonic increasing and bounded sequence $(x_n)_{n\in\mathbb{N}}$ is Cauchy
without knowing that:
- Every bounded non-empty set of real numbers has a least upper bound. (Supremum/Completeness Axiom)
- A sequence converges if and only if it is Cauchy. (Cauchy
Criterion) - Every monotonic increasing/decreasing, bounded and real
sequence converges to the supremum/infimum of the codomain (not sure if this is the right word).
However, what is allowed to use listed as well:
- A sequence is called covergent, if for $\forall\varepsilon>0\,\,\exists N\in\mathbb{N}$ so that $|\,a_n - a\,| < \varepsilon$ for $\forall n>N$. (Definition of Convergence)
- A sequence $(a'_k)_{k≥1}$ is called a subsequence of a sequence $(a_n)_{n≥1}$, if there is a monotonic increasing sequence $(n_k)_{k≥1}\in\mathbb{N}$ so that $a'_{k} = a_{n_{k}}$ for $\forall k≥1$. (Definition of a Subsequence)
- A sequence $(a_n)_{n≥1}$ is Cauchy, if for $\forall\varepsilon>0\,\,\exists N=N(\varepsilon)\in\mathbb{N}$ so that $|\,a_m - a_n\,| < \varepsilon$ for $\forall m,n>N$. (Definition of a Cauchy Sequence)
- (Hint) The sequence $(\varepsilon\cdot\ell)_{\ell\in\mathbb{N}}$ is unbounded for $\varepsilon>0$. (Archimedes Principle)
Would appreciate any help.
If $x_n$ is not Cauchy then an $\varepsilon>0$ can be chosen (fixed in the rest) for which, given any arbitrarily large $N$ there are $p,q \ge n$ for which $p<q$ and $x_q-x_p>\varepsilon.$
Now start with $N=1$ and choose $x_{n_1},\ x_{n_2}$ for which the difference of these is at least $\varepsilon$. Next use some $N'$ beyond either index $n_1,\ n_2$ and pick $N'<n_3<n_4$ for which $x_{n_4}-x_{n_3}>\varepsilon.$ Continue in this way to construct a subsequence.
That this subsequence diverges to $+\infty$ can be shown using the Archimedes principle, which you say can be used, since all the differences are nonnegative and there are infinitely many differences each greater than $\varepsilon$, a fixed positive number.