Homeomorphism between the Euclidean and a "modified" Euclidean topological space

Question

The topology of Euclidean space is defined by the Euclidean metric, which in turn is derived from the set of real numbers. Furthermore, the real numbers are defined via Dedekind cuts of the rational numbers. My question is: if one were to define an alternative set of "rational numbers" (i.e. a set of rational expressions with irrational numerators/denominators) from which to derive the real numbers via the Dedekind approach, could one ensure that the resulting topology would be homeomorphic to the usual one?

Answer 1

In response to the comments: Cantor proved that any two countable dense linear orders without endpoints are order-isomorphic. In particular, they have isomorphic Dedekind completions (where the Dedekind completion of a linear order $L$ is the set of nonempty proper initial segments of $L$ ordered by inclusion). So as clarified by the comments the answer to your question is yes.

• Incidentally, it's worth noting that more complicated things happen when we look at uncountable linear orders. For example, since $\mathbb{R}$ is second-countable the linear order $\omega_1$ does not order-embed into $\mathbb{R}$ (even if $\vert\mathbb{R}\vert$ is much larger than $\aleph_1$). See also the discussion here.