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?


For a more general construction : let $\Gamma$ be a totally ordered abelian group, and $\chi: \Gamma\to \{ \pm 1\}$ be a group morphism. 
Then we can form the Hahn series field $Q((\Gamma))$ as the set of formal series $\sum_{\gamma\in \Gamma} a_\gamma X^\gamma$ with $a_\gamma\in Q$ such that the support (ie $\{\gamma\in \Gamma\,|\, a_\gamma\neq 0\}$) is well-ordered. The product is then given by usual convolution, which is well-defined precisely thanks to the well-ordering condition on the support( this guarantees that the sums involved in the convolution product will be finite).
For instance, $\Gamma = \mathbb{Z}$ gives $Q((\Gamma)) = Q((X))$ the usual Laurent series field (the well-ordering condition just says that the series have bounded below degree terms).
Then for $f = \sum_{\gamma\in \Gamma} a_\gamma X^\gamma\in Q((\Gamma))$, we define $f\geqslant 0$ iff $\chi(\gamma)a_\gamma\geqslant 0$ in $Q$, where $\gamma$ is the least element of the support of $f$.
This gives a total ordering on $Q((\Gamma))$, and if you specialize to $\Gamma = \mathbb{Z}$ and $\chi:\mathbb{Z}\to \{\pm 1\}$ the reduction mod $2$ morphism, this gives an ordering on $Q((X))$ wich restricts to the ordering you want on $Q(X)$.
(I took this construction from Efrat's book "Valuations, orgerings, and Milnor K-Theory.)
