In my analysis class, our professor showed us a "proof" that shows that every ordered field $ \mathbb{F}$ thats also complete holds the Archimedean property. The proof was trivial, and consisted of showing that for any $x\in \mathbb{F}$ there exists an $ n \in \mathbb{N}$ such that $x<n$. While I was reviewing my journal I came upon this proof and felt that this was wrong since we're assuming that we can compare the elements in $\mathbb{F}$ to the natural numbers. I haven't seen fields and rings yet in school, only groups and vector spaces. However I'm aware that there are fields that consists of elements other than numbers, so I'm thinking that there should be a field where this doesn't hold....Unfortunately the class is very boring and non rigorous, so the only example I have thought of so far is the the field constructed by Dedekind cuts since it represents $ \mathbb{R}$ but I'm not completely sure since thats something I self learned through Rudins book and might not understand Dedekind cuts as I deeply as I probably should.

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    $\begingroup$ An odered field has a multiplicative unit say $e,$ and then a copy of the naturals in the field is the set consisting of $e,e+e,e+e+e,\cdots.$ $\endgroup$ – coffeemath Feb 9 at 7:37
  • $\begingroup$ In an ordered field you can compare elements with one another - that is what the ordering enables you to do. $\endgroup$ – Mark Bennet Feb 9 at 8:26
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    $\begingroup$ By completeness, I guess you mean that any non-empty set that is upper bounded has a supremum. If by completeness you mean that every Cauchy sequence is convergent, be aware that there are ordered fields that are complete in the latter sense and does not satisfy the Archimedean property. For example, the Levi-Civita field. $\endgroup$ – Chilote Feb 9 at 19:15

Lemma: Every ordered field contains a copy of the natural numbers, all of which are positive.

Proof: We know that $1$ (the multiplicative identity) is not equal to $0$ - that's a standard field axiom. Since this is an ordered field, $1$ is either positive or negative. But then, $1\cdot 1=1$ is either the product of two positive numbers or the product of two negative numbers. Either way, it's positive.

From $1>0$, we get $2=1+1>1>0$, $3=2+1>2>0$, and so on. By induction, the natural numbers (defined as $n$-fold sums of $1$) are all positive. Since they're positive, they're not equal to zero. Also, as a strictly increasing sequence, none of them is equal to any other. That's it.

With this lemma, it makes sense to talk about the Archimedean property in any ordered field, and to prove the property if we also have completeness.


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