The standard counterexample is the field $F$ of finitely-tailed Laurent series over $\mathbb{R}$: series of the form $\sum a_n t^n$ where $a_n \in \mathbb{R}$, exponents $n \in \mathbb{Z}$, but with only finitely many negative exponents (i.e., $a_n = 0$ for all but finitely many negative integers $n$.)
Addition and multiplication are defined just like for power series; you should verify that the "finitely many negative exponents" condition is essential to ensuring that multiplication of finitely-tailed Laurent series is well-defined. Check that non-zero elements of $F$ have multiplicative inverses: a series that starts with $a_n t^n$ has an inverse that starts with $a_n^{-1} t^{-n}$. (Here, and below, when I say that a nonzero element of $F$ starts with $a_n t^n$, I mean that $n$ is the lowest integer for which $a_n \ne 0$.)
Ordering is defined as follows: An element is positive if its starting term has positive coefficient. It's easy to check: (1) for every nonzero $x \in F$, exactly one of $x$ and $-x$ is positive; (2) the sum and product of 2 positive elements is positive.
So $F$ is an ordered field. Note that $F$ is not Archimedean: $t^{-1}$ is bigger than every integer. The field $F$ also does not satisfy the least upper bound property: $t^{-1}$ is an upper bound for the integers, but there is no least upper bound (check that if $x$ is such an upper bound, then $x/2$ is a smaller one.)
To discuss Cauchy completeness, we have to put a metric on $F$, as follows. Pick any positive real number $q>1$. If a nonzero element $x \in F$ starts with $a_n t^n$, then define $|x| = q^{-n}$ (and of course define $|0|=0$), and then define the distance between $x$ and $y$ to be $|x-y|$. Check that this is a metric space. (Caution: This doesn't extend the existing metric on $\mathbb{R}$; all real numbers have the same size under this metric.) In fact, it makes $F$ into a valued field; we have $|xy| = |x||y|$.
This is a "non-Archimedean" metric because it satisfies the strong triangle inequality $d(x,z) \le \max(d(x,y), d(y,z))$. For a non-Archimedean metric, a sequence is Cauchy iff $|x_n - x_{n+1}| \to 0$ as $n \to \infty$.
To prove that $F$ is complete under this metric, note that for a Cauchy sequence in $F$, the coefficients for a fixed exponent $t^n$ must be eventually constant; otherwise the difference between consecutive terms could never be smaller than $q^{-n}$. Call this constant $a_n$; then the Cauchy sequence converges to $\sum a_n t^n$.
First: the "least upper bound" and "greatest lower bound" principles are not at all about ordered fields, they are about partially ordered sets; an ordered field is necessarily a partially ordered set, but there are plenty of partially ordered sets that satisfy the least upper bound and greatest lower bound properties without being fields.
Second: You are trying to prove the least upper bound principle, assuming the greatest lower bound principle holds. Let's be clear about what these principles are:
We say a partially ordered set $(X,\leq)$ satisfies the Greatest Lower Bound Principle if and only if every nonempty subset $P$ of $X$ that has a lower bound has a greatest lower bound; that is, if there exists $x\in X$ such that $x\leq p$ for all $p\in P$, then there exists $g \in X$ such that:
- $g\leq p$ for all $p\in P$; and
- if $x\in X$ is such that $x\leq p$ for all $p\in P$, then $x\leq g$.
As for the Least Upper Bound Principle, we have the dual definition:
We say a partially ordered set $(X,\leq)$ satisfies the Least Upper Bound Principle if and only if every nonempty subset $P$ of $X$ that has an upper bound has a least upper bound; that is, if there exists $x\in X$ such that $p\leq x$ for all $p\in P$, then there exists $\ell\in X$ such that:
- $p\leq \ell$ for all $p\in P$; and
- if $x\in X$ is such that $p\leq x$ for all $p\in P$, then $\ell\leq x$.
Now, let's look at your argument. You are assuming that your field $F$ satisfies the Greatest Lower Bound Principle, and you take a subset $P$ of $F$. You assume it is nonempty, and that there is an element $k$ that is an upper bound. You incorrectly assert that this element $k$ will be in $P$: you don't know that, and you don't need that.
Then you consider the set $-P=\{-p \mid p\in P\}$, and note that $-k$ is a lower bound for $-P$ (you incorrectly assert it is a greatest lower bound; you don't know that). You then deduce that $-P$ has a greatest lower bound (correct). And you stop there. But you need to show that $P$ has a least upper bound, and you have not done so. (Of course, you probably meant to take the greatest lower bound $\ell$ of $-P$, and then show that $-\ell$ is a least upper bound for $P$; but you have not done so). So your argument is incomplete.
Now, there is nothing wrong with the idea behind your argument (provided you fix the errors and fill in the gaps). The reason you may not want to do this is that the fact that the Least Upper Bound Principle and the Greatest Lower Bound Principles are equivalent is true for any partially ordered set. In a partially ordered set, the argument that relies on "multiplying by $-1$" will not in general work, because that operation will not make sense. So rather than use a proof that only works for a certain kind of partially ordered set, we can use a proof which is not any more complicated, but which works in general. And moreover, a proof that highlights that the least upper bound of a set is in fact a greatest lower bound of a different set (and likewise for greatest lower bounds).
Namely:
Suppose $X$ satisfies the greatest lower bound principle, and we want to prove it satisfies the least upper bound principle. Let $P$ be a nonempty set that has at least one upper bound. We need to show it has a least upper bound. Let $B=\{b\in X\mid b\text{ is an upper bound for }P\}$. By assumption, $B$ is not empty. Moreover, since $P$ is not empty, there is a $p\in P$, and so $p\leq b$ for all $b\in B$. That means that $B$ is a nonempty set that is bounded below. By the Greatest Lower Bound Principle, $B$ has a Greatest Lower Bound; call it $g$. We claim that $g$ is also a least upper bound for $P$.
First, note that if $p\in P$ we have $p\leq b$ for all $b\in B$. By the second property of the greatest lower bound, we conclude that $p\leq g$. Therefore, for every $p\in P$ we have $p\leq g$, so $g$ satisfies the first property necessary to be the least upper bound of $P$. To verify it satisfies the second property, if $x\in X$ is such that $p\leq x$ for all $p\in P$, then $x\in B$ by definition. And since $g$ is the greatest lower bound of $B$, then the first property of the greatest lower bound tells us that $g\leq b$ for all $b\in B$, and in particular that $g\leq x$. Thus, if $x$ is an upper bound for $P$, then $g\leq x$. This verifies that $g$ satisfies the second property of the least upper bound for $P$. We conclude that $g$ is indeed the least upper bound for $P$.
So we have shown that if $X$ satisfies the greatest lower bound principle, then any nonempty subset that is bounded above has a least upper bound; that is, $X$ also satisfies the least upper bound principle.
A symmetric argument (or applying the above argument to the partially ordered set $(X,\leq^{\rm op})$) shows that if $X$ satisfies the least upper bound principle, then it satisfies the greatest lower bound principle. $\Box$
Once you fill in all the details into your argument, you'll see that it is no shorter and no less difficult than the above one. But the above one has the virtue of not needing any of the field properties that you use (that $a\leq b$ if and only if $-b\leq -a$, for example), only the properties of order, of upper and lower bounds, and the corresponding principle. So it is a nicer proofs, because it assumes less, concludes the same, and is not any harder or longer than the proof you suggest.
Best Answer
The field with two elements is not an ordered field; 1+1 is 0 in that field, but it would have to be greater than 0 in an ordered field. Every ordered field has characteristic 0, actually. See http://en.wikipedia.org/wiki/Ordered_field .