I'm told that for every real-valued function $f$ there is an injective function $f'$ that respects the ordering given by $f$ in that $$f(a) < f(b) \implies f'(a) < f'(b)$$ and $$a \neq b \text{ and } f(a) = f(b) \;\; \implies \;\; f'(a) \neq f'(b)$$ Where $<$ is the usual less-than operation.
I'm told that the existence of such an $f'$ has something to do with the axiom of choice. If anyone could shed some light on this..
( The context of this problem is page 18 of https://www.cs.ox.ac.uk/files/3378/PRG56.pdf )
This is false. Let $f$ be any real function such that the preimage of each real has at least 2 points. Suppose there is some injective $f'$ for this $f$. For each real $r$, choose $a_r \neq b_r \in f^{-1}[\{r\}]$. Let $x_r$ be a rational between $f'(a_r)$ and $f'(b_r)$. Then $r \neq s$ implies $x_r \neq x_s$ which is impossible.