I am going through Counter Examples of Analysis but I am having trouble understanding a claim it makes.
The book establishes that the set of rational functions defines an ordered field where the "positive" subset ($P$) is defined as all rational functions where the leading coefficients of the numerator and the denominator have the same sign.
The book further states that the ordered field not being dense means that "there are two distinct members of $\mathscr H$ having no rational number between them". It finally claims to show that the field mentioned above is not dense because "two distinct members of $\mathscr H$ having no rational number between them are any two distinct nonconstant polynomials with positive leading coefficients"
I am having trouble seeing how that statement is true. Suppose we take two elements $f = x$ and $g=x^3$ then it is straight forward to show that $h = x^2$ is in between $f$ and $g$
I will run through it in case I misunderstood something: $$f < g \quad since \quad g - f = x^3 - x \in P $$ $$f < h \quad since \quad h - f = x^2 - x \in P $$ $$h < g \quad since \quad g - h = x^3 - x^2 \in P $$
which means $f < h < g$.
So the questions are:
- Is the book right about their example? If so, what did I misunderstand?
- If not, is there a better counterexample or another proof to show that this field is not dense?