Least upper bound property implies Cauchy completeness Given an ordered field $F$, the following two statements are equivalent:


*

*$F$ has the Least-Upper-Bound property.

*$F$ is Archimedean and $F$ is "sequentially complete"/"Cauchy complete" (all Cauchy sequences in $F$ converge).
For some reason I have been unable to find a proof of this result on the web.
 A: First we will establish some intermediate results:
Theorem 1: Every Cauchy sequence is bounded.
Proof: Take $\epsilon=1$, then there is $N\in\mathbb{N}$ such that
$$m, n \geq N\quad\implies |x_m-x_n|<1$$
So that 
$$|x_m|<|x_N|+1\quad\quad\forall\;m\geq N$$
therefore
$$|x_n|\leq\max\lbrace |x_1|,\ldots,|x_N|,|x_N|+1\rbrace\quad\quad\forall\;n\in \mathbb{N}$$
Theorem 2: Every sequence has a monotone subsequence.
Proof: Lemma in Bolzano-Weierstrass
Theorem 3: Let $F$ be an ordered field with the LUB property, then every bounded and monotone sequence in $F$ is convergent.
Proof: WLOG lets assume the sequence is increasing. As it is bounded it has a supremum: $s=\sup_{n\in\mathbb{N}}\lbrace x_n\rbrace$ We claim that $s$ is the limit of the sequence.
Let $\epsilon>0$ then there is $N\in\mathbb{N}$ such that
$$s-\epsilon<x_N\leq x_{N+1}\leq \cdots \leq s$$
so that  for any $n\geq N$ we have
$$|x_n-s|<\epsilon$$
Which proves that $s$ is the limit.

To simplify discussion, we assume that the order field is $\mathbb{R}$.
For the first implication $(1)\implies (2)$:
To establish the Archimedean Property first note that the set of integers is not bounded above, if it were it would have a supremum. Let's call it $z$; then there would be an integer $n$ such that $z-1<n$ but then $z<n+1\in \mathbb{Z}$ This contradicts the fact that $z$ was the supremum.
Archimedean Property: For every real $x$ there is an integer $n$ such that $n>x$
Proof: If this is not the case, integers would be bounded by some $x$.
Theorems $(1)$, $(2)$ and $(3)$ above prove Cauchy completeness.

For the reverse implication $(2)\implies (1)$:
Theorem 4: Suppose $F$  has the Archimedean property, then every monotone bounded sequence is Cauchy.
Proof: WLOG take the sequence to be increasing. If $\lbrace x_n\rbrace_{n}$ is not Cauchy then there exists $\epsilon>0$ so that for every $N\in\mathbb{N}$
$$n>m\geq N\quad\implies x_n-x_m\geq \epsilon$$
We are going to extract a subsequence such that it is not bounded.
For $N=1$ choose ${n_1},\ {n_2}$ such that $x_{n_2}-x_{n_1}>\epsilon$
Now take $N^{\prime}>n_2$ and choose ${n_3},\ {n_4}$  such that  $x_{n_4}-x_{n_3}>\epsilon$. Continue in this way to construct a subsequence. Note that there are infintely many differences greater than $\epsilon$, so by the Archimedean Principle the subsequence diverges. This contradicts the fact that the sequence was bounded.
Observe that if $F$ is also Cauchy complete then every monotone bounded sequence is convergent.
We have established:
Cauchy completeness+Archimedean Property $\implies$ Convergence of every monotone and bounded sequence
The answer in this post proves: 
Convergence of every monotone and bounded sequence $\implies$ LUB Property
