Order-preserving isomorphism between $\mathbb{R}^n$ and $\mathbb{R}$

Question

Suppose we have a linearly ordered group over $\mathbb Z^n$ where the ordering goes left-to-right, i.e. when deciding if $(x_1,x_2,\dots)<(y_1,y_2,\dots)$ we first check if $x_1< y_1$, if it is then $X< Y$. If they are the same, we compare $x_2< y_2$ and so on.

I believe there is an isomorphism to a subset of $\mathbb Q$ which preserves this ordering. Namely, a Godel numbering $(x_1,x_2,\dots)\mapsto 2^{x_1}3^{x_2}\dots$ is an isomorphism which has the same ordering if we first use the 2-adic norm, then the 3-adic norm etc.

I think a slightly more convoluted isomorphism exists even if we're dealing with $\mathbb Q^n$. However, I'm not sure if this is true if we're dealing with $\mathbb R^n$. So I ask:

For any group over $\mathbb R^n$ which uses the left-to-right ordering, is there an equivalent group over $\mathbb R$?

Answer 1

No. Note that an order-isomorphism will be an injective continuous function when both are considered in the order topology. Consider then $\Bbb R^2$ ordered as you describe, and in particular consider the images of the connected sets $\{x\}\times\Bbb R$ of $\Bbb R^2$ under any continuous function. The continuous image of a connected set is connected, so for $f$ to be injective, we will require for the images of those connected sets to be disjoint, uncountable connected subsets of $\Bbb R$. The interiors of these connected sets would be disjoint open intervals or rays, but there would be uncountably many of these, which is impossible, since $\Bbb R$ has a countable dense subset.

Incidentally, I'm not sure what your rational isomorphism is supposed to do, but I don't think it works. However, as a countable dense linear order with no endpoints, $\Bbb Q^n$ is order isomorphic to $\Bbb Q$.

Answer 2

An ordered abelian group $(G,+,<)$ is Archimedean if for all $x,y \in G$ with $x > 0$, there is a positive integer $n$ with $nx > y$.

Famously the ordered group $(\mathbb{R},+,<)$ is Archimedean. But for $N > 1$, $\mathbb{R}^N$ endowed with the lexicographic ordering is not: for all $n \in \mathbb{Z}^+$, we have $n (0,1,0,\ldots,0) < (1,0,\ldots,0)$. So they cannot be order isomorphic.

It is a theorem of Holder that an ordered abelian group is Archimedean iff it can be embedded in $\mathbb{R}$: see $\S$ 17.2 of these notes for a proof.

These ideas can be pushed much further: a subgroup $H$ of an ordered abelian group $(G,+,<)$ is convex if for all $x < y< z \in G$, if $x,z \in H$ then also $y \in H$.

The family of convex subgroups of an ordered abelian group is linearly ordered under inclusion (Proposition 17.10 of loc. cit.). So the order isomorphism type of this set is an invariant of $(G,+,<)$, say $r(G)$. When $r(G)$ is finite we identify it with a natural number and call it the rank of $G$. Then:

$\bullet$ An ordered group has rank at most $1$ iff it is Archimedean (Proposition 17.11 of loc. cit.).

$\bullet$ The rank of $\mathbb{R}^N$ is $N$. (Exercise! This refines the current question.)