Can any total ordering on $\Bbb R$ be mapped to the standard order?

Question

If I have a total ordering $\succeq$ on $\mathbb R$, is there guaranteed to be a function $g: \mathbb R \to \mathbb R$ such that $g(x) \geq g(y)$ if and only if $x \succeq y$? It seems to me that the answer should be yes, but this seems to cause an issue by allowing one to define a bijection from $\mathbb R$ to the natural numbers.

Answer 1

The natural total order $(\Bbb R,\le)$ has the property that every family of disjoint, non-sigleton and non-empty intervals is countable. For total orders this property is hereditary, which means that the order induced on any subset must have it too. The lexicographic order on $\Bbb R^2$ does not have this property, therefore it cannot be realised as a suborder of $(\Bbb R,\le)$. However, the lexicographic order is isomorphic to some order $(\Bbb R,\preceq)$, by using a bijection $f:\Bbb R\to\Bbb R^2$.