Mapping (any, not only bijection) from $\mathbb{R}^2$ to $\mathbb{R}$ that preserves lexicographic order?

Question

Is there a (not necessarily bijective) mapping $f \colon \mathbb{R}^2 \rightarrow \mathbb{R}$ that preserves lexicographic order?

That's to say, we'd need to have $f(x_1, x_2) \leq f(y_1, y_2)$ iff $(x_1, x_2) \leq_{lex} (y_1, y_2)$.

What would be a formal argument that such an order does not exist?

Answer 1

There is no such map: the intervals

$$\big(f(\langle x,0\rangle),f(\langle x,1\rangle)\big)$$

for $x\in\Bbb R$ would have to be disjoint, which is impossible. Each would have to contain a different rational, but there are uncountably many intervals and only countably many rationals.