Is there a bijective mapping $f \colon \mathbb{R}^n \rightarrow \mathbb{R}$ that preserves lexicographic order?
That's to say, we'd need to have $f(x_1, \dots, x_n) \leq f(y_1, \dots, y_n)$ iff $(x_1, \dots, x_n) \leq_{lex} (y_1, \dots, y_n)$.
In the same ballpark: what would be a formal argument that such an order exists, or not (or some hint to how the formal argument would look like)?
Even in the case $n=2$, you would have to be able to fit uncountably many disjoint nondegenerate intervals into the real line. Indeed, because $f$ is order-preserving and bijective, it must be the case that $f(\{x\}\times[a,b])=[f(a),f(b)]$ and likewise for the corresponding open intervals.
Since every open interval contains a rational number, we cannot have uncountably many such disjoint intervals.
(après MPW)