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I was seeing this post and its nice answer. I asked myself if I could have found a more "crazy" example. I was not very succesful. So I formalized what I was trying to achieve:

Question: Let $(R,+,\cdot,0,1,<)$ be a totally ordered commutative ring without infinities/infinitesimals, i.e. $$\forall x\in R:\exists n\in\Bbb N:\underbrace{1+\cdots+1}_{n}>|x|., \qquad\forall x\in R-\{0\}:\exists n\in\Bbb N:\underbrace{|x+\cdots+x|}_{n}>1.$$ Is it true that this already suffices to make $R$ isomorphic to a subring of $\Bbb R$?

This would show that there are no "crazy" examples.

My approaches mostly started from $\Bbb R$ by adding new elements, but it seems any such additional element led to infinities/infinitesimals. On the other hand I do not even have an idea on how to show that $R$ might be embedded into $\Bbb R$ in general.

Update: I found the Hahn embedding theorem to be very relevant for my question.

M. Winter
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This is indeed true. Such a totally ordered ring is called Archimedean. Here's how you embed an Archimedean ordered ring $R$ into $\mathbb{R}$. For any positive $r\in R$, consider the set $S_r$ of positive rational numbers $\frac{a}{b}$ (with $a,b\in\mathbb{Z}_+$) such that $br\geq a$. Your second assumption implies $S_r$ is nonempty, and your first assumption implies $S_r$ is bounded above. Define $f(r)=\sup S_r\in\mathbb{R}$. Also define $f(0)=0$, and $f(r)=-f(-r)$ if $r<0$. (Morally, $S_r$ is the set of positive rational numbers less than or equal to $r$, so we are sending $r>0$ to the real number which defines the same cut in the rational numbers as $r$.)

This defines a map $f:R\to\mathbb{R}$, and it is straightforward to check that this map is an order-preserving homomorphism. It is also injective, since $f(r)$ is by definition always positive if $r>0$ and negative if $r<0$.

(Alternatively, to avoid separate cases based on signs in the definition of $f$, you could define $S_r$ to be the set of $\frac{a}{b}$ with $a,b\in\mathbb{Z}$ and $b>0$ such that $br\geq a$ and define $f(r)=\sup S_r$ for any $r\in R$. This probably makes it a bit easier to check that $f$ is a homomorphism. You still have to separate into cases to prove the supremum exists, though: if $r>0$, then $S_r$ is obviously nonempty but you must use the non-existence of infinite elements to prove it is bounded above, while if $r<0$, then $S_r$ is obviously bounded above but you must use the non-existence of infinite elements to prove it is nonempty. You then have to use the non-existence of infinitesimal elements to prove $f(r)>0$ when $r>0$ and $f(r)<0$ when $r<0$, so that $f$ is injective.)

Eric Wofsey
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  • This is amazing. Thank you! – M. Winter Oct 15 '17 at 21:44
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    I am correct that we never used the ring-multiplication, hence this result even holds for ordered groups? – M. Winter Oct 15 '17 at 22:38
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    Yes. More precisely, it shows that if $G$ is an Archimedean totally ordered abelian group and $u\in G$ is positive, then there exists an embedding $f:G\to\mathbb{R}$ such that $f(u)=1$, by letting $u$ play the role of $1$ in this argument. Here, in the absence of a chosen unit element, "Archimedean" means that if $x$ and $y$ are any two positive elements there exists $n\in\mathbb{N}$ such that $nx>y$. – Eric Wofsey Oct 15 '17 at 22:43