Problem 149 (difficulty: 7/10)

Does the ordered field of the rational functions satisfy the completeness theorem: all non-empty set has a supremum?

Solution:

No.

Denote by \(\displaystyle \R(x)\) the ordered field of the rational functions. Mapping the real numbers to the constant functions, \(\displaystyle \R\) can be considered as an ordered subfield of \(\displaystyle \R(x)\). We show that \(\displaystyle \R\) is nonempty, bounded from above but it has no smallest upper bound.

\(\displaystyle \R\) is obviously nonempty. The function \(\displaystyle x=\dfrac{x}1\in\R(x)\) is an upper bound of \(\displaystyle \R\) because for any \(\displaystyle a\in\R\) we have \(\displaystyle x-a=\dfrac{x-a}1>0\). Hence, \(\displaystyle \R\) is a nonempty subset of \(\displaystyle \R(x)\) and it is bounded from above.

Now we show that \(\displaystyle \R\) has no smallest upper bound. If \(\displaystyle K\in\R(x)\) is an upper bound then \(\displaystyle K-1\) is also an upper bound since for every \(\displaystyle a\in\R\) we have \(\displaystyle a+1\in\R\Rightarrow a+1\le K \Rightarrow a\le K\).

Give me another random problem!