# 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?

• The rationals are specifically not Dedekind complete. – Arthur Jan 26 '16 at 22:30
• Dedekind completeness says: "A set $X$ has the least-upper-bound property if and only if every non-empty subset of $X$ with an upper bound has a supremum in $X$." Could you give me an example of rational subset which doesn't fit this property? – user285146 Jan 26 '16 at 22:31
• @user285146 $\{x\in\mathbb{Q}: x^2<2\}$. (More generally, $\{x\in\mathbb{Q}: x<\alpha\}$ for any fixed irrational $\alpha$.) – Noah Schweber Jan 26 '16 at 22:34
• How to prove that these don't have a supremum? – user285146 Jan 26 '16 at 22:37
• @user285146 See the first part of my answer. Does this help? – Noah Schweber Jan 26 '16 at 22:49

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 $\mathbb{S}$.

• Is there a way to prove its uniqueness without using the other definitions of reals? – user285146 Jan 26 '16 at 23:03
• @user285146 Yes, but (in my opinion) it's conceptually harder, even though the math is basically the same: given two ordered Dedekind complete fields $\mathbb{S}_1$ and $\mathbb{S}_2$, we construct an isomorphism between them in stages. First, we let $\mathbb{Q}_1$ and $\mathbb{Q}_2$ be their prime subfields; we show that each is just (isomorphic to) the rationals, so we get a field isomorphism $f: \mathbb{Q}_1\rightarrow\mathbb{Q}_2$. We then extend $f$ to a field homomorphism $F: \mathbb{S}_1\rightarrow\mathbb{S}_2$; (cont'd) – Noah Schweber Jan 26 '16 at 23:13
• this step requires us to show that $\mathbb{Q}_1$ is dense in $\mathbb{S}_1$. The argument is basically the same as the last bulletpoint in my answer: first show that $\mathbb{S}_1$ has no elements bigger than every rational, then show that this means that the rationals are dense in $\mathbb{S}_1$. Finally, apply the same argument to show that $F$ is surjective, so $\mathbb{S}_1\cong\mathbb{S}_2$. – Noah Schweber Jan 26 '16 at 23:14
• How do we know that the rational in your first example even exists? – user285146 Feb 12 '16 at 14:01