In the book of Real Algebraic Geometry by Bocknak-Coste-Roy, at page $7$, it is given that
There is exactly one ordering $\mathbb{R} (X)$ such that $X$ is positive and smaller than any positive real number.If $$P(X) = a_n > X^n + a_{n-1} X^{n-1} + ... + a_k X^k \quad with \quad a_k \not = 0,$$ then $P(X) > 0 $ for this ordering iff $a_k > 0$, and ...
So first of all, how can $<$ form a ordering, I mean is it not reflexive since for any $a_k$, $a_k \not > a_k$.
Plus, every source that I have looked says that the ordering is according to the leading coefficient, and not according to the constant term.
Secondly, what does the author mean by "X is positive and smaller than any positive real number" ,and how can we show that this ordering is unique since the author says "one ordering" ?
Edit: As is it point out in the comments, apparently $<$ should not be reflexive, but why ? isn't it in the definition of a total ordering ? if $< $ is not a total order, then what is it ?
Do you mean $$P(X)=a_nX^n+\cdots+a_kX^k?$$ I don't get your point about reflexivity. We should always have $f\not>f$.
If you order according to the leading coefficient, you get a different ordering, one where $X>a$ for all real $a$.
In this ordering $0<X<a$ for all real positive $a$. (When the authors say "$X$ is positive and smaller than any positive real number" they mean $X$ is positive and smaller than any positive real number, i.e., this). For example then, if $f=X^2-X$, then $$f=-X(1-X)<0$$ as $-X<0$ and $1-X>1-0.1>0$ etc. One can extend this argument to show that a polynomial with positive last coefficient is positive in this ordering.