Uniqueness of the real line

Question

A few days ago, I came across this question in a review queue. I tried my luck at it. Here is what I did:

1. If I want a homomorphism (isomorphism, but even just homomorphism) $f:\mathbb{R}\to F$, then I'll need 0 and 1 to map to themselves. From this, via homomorphism properties, I conclude $\mathbb{Q}$ has to map to itself.
2. $F$ is ordered, so it has an order topology which is easily seen to be Hausdorff. So I can take the identity of $\mathbb{Q}$ and extend it to $f:\mathbb{R}\to F$ via sequential continuity, i.e. if $x_n\to x$ and $x_n\in\mathbb{Q}$, I want $f(x_n)\to f(x)$ in $F$. This is a well-defined extension.
3. $f$ is order preserving. Naturally, I am assuming "ordered subfield" means the operations and order of $F$ restrict the the normal operations and order of $\mathbb{Q}$. With that, that it preserves the order of $\mathbb{Q}$ is obvious. So let e.g. $a\in\mathbb{R}\smallsetminus\mathbb{Q}$. $\mathbb{Q}$ is dense, which is what allows $f$ to be defined on all $\mathbb{R}$, so there exists $a_n\to a$ a sequence of rationals. Naturally, $f(a)=\lim f(a_n)$. $a_n$ has to eventually be in every open set around $a$, including $(a-\frac d2,a+\frac d2)$, where $d=b-a$. So eventually $a_n<b$, and $f(a_n)<f(b)$ since $b$ is rational. But then $f(a)>f(b)$ is a contradiction. So $f(a)\leq f(b)$. We need injectivity to conclude strict preserving of the order. Oh with similar arguments we conclude for the case $a,b$ both irrational. I'm having problems recalling my proof of this injectivity…
4. $f$ is a homomorphism. On the rationals it is evident, so let $a,b$ be irrational. Then $a_n\to a,b_n\to b$ exist in the rationals. $a_nb_n\to ab$, so $f(ab)=\lim f(a_nb_n)=\lim(f(a_n)f(b_n))=\lim f(a_n)\cdot\lim f(b_n)=f(a)f(b)$. Similarly you prove additive linearity.

So modulo recalling the proof of injectivity, we have proven that any complete ordered field containing $\mathbb{Q}$ as an ordered subfield must contain an isomoprhic copy of $\mathbb{R}$, which is homeomorphic to $\mathbb{R}$ if equipped with the restricted order topology, as one can easily verify since sequential continuity, which is the basis for the definition of $f$, implies continuity in first countable spaces and the order topology is first countable by taking intervals of size $\frac1n$ as local bases. Completeness of $F$ was of course used when defining $f$ via limits, to ensure those limits existed: the sequences of rationals are Cauchy both in $\mathbb{R}$ and in $F$, so the images converge by completeness of $F$.

We are left with proving $f(\mathbb{R})=F$, i.e. $f$ is surjective. I tried by contradiction, and assumed there was $x\in F\smallsetminus f(\mathbb{R})$. I concluded all real multiples of $x$ were of that kind, i.e. not reals, except for $0x=0$. But I can't see how to go on from there.

Can you help me? Is there a more algebraic and less topological way, perhaps an easier way, to do this?

Answer 1

To be clear, let's say that in an ordered field $F$ and a subfield $K$ of $F$, a sequence $u: \mathbb{N} \rightarrow F$ is $K$-Cauchy if $\forall \varepsilon \in K, \varepsilon > 0 \rightarrow \exists N \in \mathbb{N}, \forall n,p \in \mathbb{N}, n,p \geq N \rightarrow |u_n-u_p| < \varepsilon$.

Your theorem (more preceisely part 2 of your proof) holds if completeness is interpreted as the convergence (in the order topology) of every $\mathbb{Q}$-Cauchy sequence of $F$, but not if means convergence of every $F$-Cauchy sequence. Informally, in the latter case, when $F$ is not archimedean - yes, non archimedean ordered fields do exist -, being $F$-Cauchy is much more restricting than being $\mathbb{Q}$-Cauchy because terms of the sequence have to be infinitesimally close to each other.

To end your proof, you could start by proving that $F$ is archimedean, because then its smallest subfield $\mathbb{Q}$ would be dense in it and this would yeild surjectivity. To prove this, assume there is $\omega \in F$ greater than every integer and prove that $(\frac{1}{n+1})_{n \in \mathbb{N}}$ is $\mathbb{Q}$-Cauchy in $F$ (this is clear) but it doesn't converge in the order topology (here use $\omega$).

You can first prove the more general fact that every convergent sequence in the order topology is $F$-Cauchy.