What is the correct way of adjoining an "infinite" point to a totally-ordered field, such that the end result is still totally-ordered? Let $Q$ denote a totally-ordered field. Now adjoin an infinite element $\infty$ to $Q$, by equipping the field of rational functions $Q(\infty)$ in the indeterminate $\infty$ with the least preorder $\lesssim$ such that the obvious axioms hold:


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*If $x \in Q$, then $x \lesssim \infty$.

*If $x,y \in Q$ and $x \leq y$, then $x \lesssim y$.

*If $x \lesssim y$, then $x+z \lesssim y+z$.

*If $0 \lesssim x$ and $0 \lesssim y$, then $0 \lesssim xy$.



Question. Is $\lesssim$ necessarily a total ordering of $Q(\infty)$? (i.e. antisymmetric and linear). And if not, what is the correct way of adjoining an infinite point to a totally-ordered field, such that the end result is still totally-ordered?


We give a model-theoretic version of the construction. Let $Q$ be an ordered field. Construct the language $L$ as follows. Among the non-logical symbols of $L$ are two binary function symbols (for addition and multiplication) and a binary relation symbol $\lt$. In addition, $L$ has a constant symbol $k_a$ for every element $a$ of $Q$, and an additional constant symbol $B$ (for "big").
We construct a theory $T$ over the language $L$ as follows:
(i) Among the axioms of $T$ are the usual axioms of the theory of ordered fields.
(ii) In addition, $T$ contains as axioms $k_a+k_b=k_{c}$ for every triple $(a,b,c)$ of elements of $Q$ such that $a+b=c$, and $k_a+k_b\ne k_{c}$ for every triple such that $a+b\ne c$. We also add similar axioms for multiplication, and for $\lt$. We also add $k_a\ne k_b$ for all pairs $(a,b)$ such that $a\ne b$. (We are adding the complete diagram of $Q$.)
(iii) Finally, we add $k_a\lt B$ for every $a\in Q$.
It is not hard to see that the above theory $T$ is consistent, since every finite subset of the axioms has $Q$ as a model if the $k_a$ and $B$ are interpreted suitably.
Thus by the Compactness Theorem, the theory $T$ has a model $M$. Any model of $T$ is an ordered field that has a subfield naturally isomorphic to $Q$. The interpretation of $B$ in $M$ is larger than every element of $Q$.
Remark:  The above may not look elementary if one has had no exposure to model theory. But if one has, the argument writes itself.
One can make a somewhat more explicit-looking construction by using a suitable ultrapower of $Q$.
