# Necessity of Archimedean property in construction of Reals?

I am currently self-studying the construction of reals as equivalence classes of rationals. In it, I have read that the Archimedean property is a necessary assumption we have to make to construct $$\mathbb{R}$$. However, from what I have studied, I haven't been able to find out where this has been necessary.

As far as I can see, we can prove that Cauchy Completeness $$\implies$$ Least Upper Bound Property (for example, like in this Wikipedia entry) without the Archimedean Property, which should make Cauchy Completeness equivalent to the Least Upper Bound Property. So, am I missing out on something, or is the Archimedean property necessary for a different reason?

• The assertion "… and $\lvert A_n - B_n \rvert \to 0$ as $n \to \infty$" only follows if we have the Archimedean property. May 4, 2020 at 15:21
• I think when you say something is a "necessary assumption" that doesn't mean you need to know it is true in order to construct something. But rather if it were not true then construction can not be made.. If the Achemedian property is false on a set $X$ then $X$ can not be the reals. May 7, 2020 at 22:18
• The Levi-Civita field is a non-Archimedean valued field, that equipped with the metric $d(x,y)=|x-y|$ induced by the valuation $|\cdot|$, is Cauchy complete but does not satisfy the archimedean property nor the Least Upper Bound property. So yes, the Archimedean property is crucial in the construction of the real numbers. May 8, 2020 at 14:51

I assume you are constructing the set $$\mathbb {R}$$ from $$\mathbb {Q}$$ and each element of $$\mathbb{R}$$ is an equivalence class of Cauchy sequences of rationals.
When this is done you should know that $$\mathbb{Q}$$ possesses Archimedean property ie if $$a, b$$ are rationals with $$a>0$$ then there is a positive integer $$n$$ such that $$na>b$$. And this is a trivial property of set $$\mathbb {Q}$$ and it remains trivial even in $$\mathbb {R}$$ when we construct it from $$\mathbb {Q}$$.