I need to prove this:
If there is a definable bijection between $\varphi(C)$ and $\psi(C)$ then $RM(\varphi)= RM(\psi)$.
Where $C$ is the monster model. I can intuitively understand it, the Morley Rank is closely related to the amount of elements, if there is a bijection, between them it means they both have the same amount of elements so they have the same Morley Rank. Now, knowing this I have issues with the formal details, I'm not sure about how to start writing the proof formally, and would very much appreciate any input, both about how to tackle it and if my intuition is right. Thank you for your help
The most direct proof is likely by induction on Morley rank. It suffices to show, by induction on $\alpha$, that, given definable sets $X$ and $Y$ and $f:X \to Y$ a definable injection, if $RM(X) \geq \alpha$, then $RM(Y) \geq \alpha$.
Assume we know the claim for all $\beta < \alpha$, and we want to show the claim for $\alpha$. Assume $RM(X) \geq \alpha$. That is, for all $\beta < \alpha$, there are infinitely many disjoint definable subsets of $X$, each of $RM \geq \beta$. To show that $RM(Y) \geq \alpha$, we must show that, for all $\beta < \alpha$, there are infinitely many disjoint subsets of $Y$, each of $RM \geq \beta$. Fixing $\beta < \alpha$, we know we have $X_1,X_2,\ldots$ disjoint, definable subsets of $X$ with $RM \geq \beta$. Since $f$ is definable, $f(X_1), f(X_2), \ldots$ are definable, also. Since $f$ is injective, they are disjoint. By the inductive hypothesis, they have $RM \geq \beta$. We therefore have the desired conclusion.