Is completeness theorem satisfied?

Does the ordered field of the rational functions satisfy the completeness theorem: "all non - empty sets have a supremum")

Solution

The answer is negative. Denote by $\mathbb{R}(x)$ the ordered field of the rational functions. Mapping the real numbers to the constant functions , $\mathbb{R}$ can be considered as an ordered subfield of $\mathbb{R}(x)$. We will show that $\mathbb{R}$ is non empty , bounded from above but it has no smallest upper bound.

• $\mathbb{R}$ is obviously non empty. The function $x = \frac{x}{1} \in \mathbb{R}(x)$ is an upper bound of $\mathbb{R}$ because for any $a \in \mathbb{R}$ we have that $x-a = \frac{x-a}{1}>0$. Hence, $\mathbb{R}$ is a non empty set of $\mathbb{R}(x)$ and it bounded from above.
• What remains to be proven is that $\mathbb{R}$ has no smallest upper bound. If $K \in \mathbb{R}(x)$ is an upper bound then $K-1$ is also an upper bound since for every $a \in \mathbb{R}$ we have $a+1 \in \mathbb{R} \Rightarrow a+1 \leq K \Rightarrow a \leq K$.