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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?

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    $\begingroup$ The assertion "… and $\lvert A_n - B_n \rvert \to 0$ as $n \to \infty$" only follows if we have the Archimedean property. $\endgroup$ – Daniel Fischer May 4 '20 at 15:21
  • $\begingroup$ 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. $\endgroup$ – fleablood May 7 '20 at 22:18
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    $\begingroup$ 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. $\endgroup$ – Chilote May 8 '20 at 14:51
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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} $.

Only when the set of real numbers is presented to us axiomatically this Archimedean property acquires some sort of a non-trivial nature because then it can not be proved without the use of completeness axiom.

My own confusion regarding this was sorted out long back in this thread on mathoverflow.

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I think that you are missing nothing. The Archimedian property is a consequence of the least upper bound property. So, if you somehow define the reals in some way which allows you to prove that the least upper bound property holds, then automatically the Archimedian property will hold too.

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  • $\begingroup$ .... which would mean if the archimedian property were false on a set, (and you had a counter example), you could prove the set does not have the lub property and therefore can not be the reals. $\endgroup$ – fleablood May 7 '20 at 22:22

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