# Which of the Completeness Properties of $\mathbb{R}$ imply the Archimedean Property of $\mathbb{R}$?

Completeness Properties of $\mathbb{R}$: Least Upper Bound Property, Monotone Convergence Theorem, Nested Intervals Theorem, Bolzano Weierstrass Theorem, Cauchy Criterion.

Archimedean Property: $\forall x\in \mathbb{R}\forall \epsilon >0\exists n\in \mathbb{N}:n\epsilon >x$

I can show that LUB implies the Archimedean Property but what about the rest of these properties? Please provide proofs (even hints) or counterexamples.

EDIT: It was shown by Isaac Solomon that the Bolzano-Weierstrass implies the Archimedean Property.

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For Bolzano-Weierstrass $\to$ Archimedean Property: Take $\epsilon > 0$, and consider the sequence $\{n\epsilon: n \in \mathbb{N}\}$. If the Archimedean Property fails, then this sequence is bounded, so that it has a convergent subsequence. However, it is easy to see that this sequence does not have a convergent subsequence.
For Monotone Convergence Theorem $\to$ Archimedean Property: Taking the same sequence $\{n\epsilon: n \in \mathbb{N}\}$, it is easy to see that it is monotone. If the Archimedean Property fails, then this sequence is bounded, so by the MCT, it would have a finite limit, which is not the case.
A sequence $x_{n}$ converges to a limit $L$ if for every $\delta > 0$ there exist $N \in \mathbb{N}$ such that for $n \geq N$, $|x_{n} - L| < \delta$. However, consider our sequence $\{n\epsilon\}$, and take $\delta < \epsilon$. No matter how large you make $N$, $(N+1)\epsilon - N\epsilon = \epsilon > \delta$, and so this sequence cannot converge. – Isaac Solomon Mar 28 '12 at 17:29