Ordering of $\mathbb{R}[X]$ such that $0 < X < a$ for all positive numbers $a$

Question

If $\mathbb{R}[X]$ denotes the field of real valued rational functions, then according to Bochnak, Coste and Roy's Real Algebraic Geometry, the unique ordering for which $X > 0$ and $X < a$ for all $a > 0$ is given by, for $a_k \neq 0$ and $n \geq k$, $$P(X) = a_n X^n + \dots + a_k X^k > 0 \text{ if and only if } a_k > 0$$

and we say $P(X)/Q(X) > 0$ if and only if $P(X)Q(X) > 0$.

This defines a total order of the field of real valued rational functions.

What I am trying to prove is that this is necessarily the only one for which we have $X > 0$ and $X < a$ for all positive numbers $a$.

It is enough to show, as $X^k > 0$, that we must have $a_n X^n + \dots + a_1 X + a_0 > 0$ if $a_0 > 0$. I was able to show this for the $n=1$ case, but for $n > 1$, I am unable to do so.

Answer 1

Let $\alpha>0$ and $n\ge 1$. We have

• if $\beta\le 0$, then $\beta X^n \le 0 < \alpha$;
• if $\beta>0$, then $\beta X^n< \beta X^{n-1}<\dots< \beta X < \beta\left(\dfrac\alpha\beta\right) = \alpha$;

therefore $\alpha>\beta X^n$ $\forall \beta$.

In particular, if $a_0>0$, then $\dfrac{a_0}{n}>-a_k X^k~\forall k\ge 1$, so $a_0>-a_1X-a_2X^2-\dots-a_nX^n$ and $$a_nX^n+\dots+a_2X^2+a_1X+a_0>0$$ as we wanted.

The reverse implication is not true, as pointed out by a comment.