Defining $\mathbb{R}$ as a field One of the definitions of real numbers I've encountered is "Dedekind complete totally ordered field". How to prove said field is unique?
The definition seems to be not complete too, since rational numbers seem to be a Dedekind complete totally ordered field too. Am I misunderstanding something, or is this really not a whole definition? What's the full one then - Dedekind complete totally ordered field with biggest cardinality, perhaps?
 A: Note that the rationals are not complete - e.g. $\{x\in\mathbb{Q}: x^2<2\}\cup\{x\in\mathbb{Q}: x<0\}$ is a set of rationals with an upper bound (say, 17) but no least upper bound.
This last point deserves a bit of explanation. Let $L=\{x\in\mathbb{Q}: x^2<2\}\cup\{x\in\mathbb{Q}: x< 0\}$ and suppose $\alpha$ is an upper bound of $L$, $\alpha\in\mathbb{Q}$. Then since $\sqrt{2}$ is irrational, there are two possibilities: 
$\alpha$ is too big: $\alpha^2>2$. Fix a positive rational $\epsilon$ such that $0<2\alpha\epsilon-\epsilon^2<\alpha^2-2$. Then $$(\alpha-\epsilon)^2=\alpha^2-(2\alpha\epsilon-\epsilon^2)>2,$$
 so $\alpha$ was not in fact a least upper bound of $L$.
$\alpha$ is too small: $\alpha^2<2$. A variation of the same argument will work here. We pick a rational $\epsilon$ such that $$0<2\alpha\epsilon+\epsilon^2<2-\alpha^2;$$ then $(\alpha+\epsilon)^2=\alpha^2+(2\alpha\epsilon+\epsilon^2)<\alpha^2+2-\alpha^2=2$.
(Alternately, both cases can be solved much faster if you know that $\sqrt{x}$ is increasing: given a putative rational least upper bound $\alpha$, pick $\beta$ some rational between $\alpha$ and $\sqrt{2}$.)

As to the proof of uniqueness, there's a few ways to do this. I like to start with the "usual" reals $\mathbb{R}$, that is, the Dedekind completion of $\mathbb{Q}$. Now suppose $\mathbb{S}$ is a Dedekind complete ordered field.


*

*First, show that $\mathbb{Q}$ embeds into $\mathbb{S}$.

*Next, use Dedekind completeness to extend the embedding $f: \mathbb{Q}\rightarrow\mathbb{S}$ to an embedding $F: \mathbb{R}\rightarrow\mathbb{S}$.

*Suppose $F$ is not surjective. Show that there is some $\alpha\in\mathbb{S}$ which is greater than $F(r)$ for every $r\in\mathbb{R}$.

*If $\alpha<F(r)$ for some $r\in\mathbb{R}$, call $\alpha$ finite. Assuming $F$ is not surjective, show that the finite elements of $\mathbb{S}$ are do not have a least upper bound (HINT: if $\beta$ is an upper bound, so is $\beta-1$). Together with the previous bullet point, this contradicts the Dedekind completeness of $\mathbb{S}$.
