A representation of well-orderings? Is there a well-ordering $P$ of the set of real numbers $\mathbb{R}$ such that there is NO function $f: \mathbb{R}->\mathbb{R}$ satisfying the property:

for all $x,y \in \mathbb{R}$, $xPy$ iff $f(x)\geq f(y)$

We could also think of another well-ordered set $(X,P)$ for which there is no $f: X->\mathbb{R}$ that satisfies the property.
 A: This is basically answered at https://mathoverflow.net/questions/25100/order-types-of-positive-reals and is a standard fact in set theory.

Fact. A well ordering $\alpha$ can be embedded into the order $(\mathbb{R},<)$ if and only if $\alpha$ is countable.

First, if $\alpha$ is countable then $\alpha$ can be embedded in $(\mathbb{Q},<)$ by an argument of Cantor.
Second, if $\alpha$ embeds into $\mathbb{R}$, say $\alpha = \{ \alpha_i : i \in I\}$ and $f\colon \alpha \to \mathbb{R}$ is order preserving. Let $\alpha_i \in \alpha$ and let $\alpha_j$ be its successor in $\alpha$. Then there is a rational number between $f(\alpha_i)$ and $f(\alpha_j)$, and no other element of $\alpha$ is mapped into this interval. Thus we can make an injective map from $\alpha$ to $\mathbb{Q}$, which shows that $\alpha$ is countable.
Therefore, the answer to the question is that, in fact, there is absolutely no well ordering of $X = \mathbb{R}$ that can be embedded into the usual order on $\mathbb{R}$, and the same holds if $X$ is replaced with any uncountable well order: it will not embed in $\mathbb{R}$.
