The ordered field of rational functions does not have the LUB property, can someone give an example?

Question

S is a set of some rational functionals and is bounded above, but S has no least upper bound.

Can someone give an example of S?

Answer 1

We assume that you mean rational functions with real coefficients. One common way to define order is to say that $r_1(x)\lt r_2(x)$ if for all large enough real $b$ we have $r_1(b)\lt r_2(b)$.

Let $S=\{1,2,3,4,\dots\}$. This set is bounded above, for example by the polynomial $x$, but has no least upper bound.

Outline of proof: Suppose to the contrary that $r(x)=\frac{p(x)}{q(x)}$ is a least upper bound for $S$, where $p$ and $q$ are polynomials, without loss of generality with positive lead coefficients

If $p$ has degree less than the degree of $q$, then $p/q$ is not an upper bound for $S$.

If $p$ and $q$ have the same degree, then the ratio has limit say $a$ as $x\to\infty$. Since there are integers greater than $a$, $p/q$ is not an upper bound for $S$.

Finally, if $p$ has degree greater than the degree of $q$, then $\frac{p(x)}{2q(x)}$ is an upper bound for $S$ which is less than $p/q$.

Answer 2

Here's another, more conceptual example (albeit without the proof) that shows the property fails even if we define a (partial) order by $r_1()\lt r_2()$ iff $\forall x r_1(x)\lt r_2(x)$: let $p_0(x)=1$, $p_1(x)=1+x^2$, $p_2(x)=1+x^2+\frac{x^4}{2}$, $p_3(x)=1+x^2+\frac{x^4}{2}+\frac{x^6}{3!}$, $p_n(x)=\sum_{i=0}^n\frac{x^{2i}}{i!}$; then we have $p_i()\lt p_{i+1}()$ (since each term in the sum is manifestly positive) but there is no LUB in the space of rational functions. (The LUB in the space of analytic functions is the function $e^{x^2}$, for which the $p_i()$ are of course the partial Taylor expansions.)