It is a theorem that every $\omega$-stable first-order theory is totally transcendental. The only proof I have seen of this theorem relies on the following fact:
Let $T$ be a first-order theory, $\mathbb{M}\models T$ a monster model, and $\text{RM}$ the Morley-rank function for $\mathbb{M}$. If $\phi$ is an ${\mathcal{L}}_{\mathbb{M}}$-formula with $\text{RM}(\phi)=\infty$, then there is an ${\mathcal{L}}_{\mathbb{M}}$-formula $\psi$ such that $\text{RM}(\phi\wedge\psi)=\text{RM}(\phi\wedge\neg\psi)=\infty$.
In other words, we can split any definable set of infinite Morley rank into two definable sets of infinite Morley rank. Intuitively this is plausible, but I don't see how to prove it. (The closest I can come is that for any ordinal $\alpha$, we can split a definable set of infinite Morley rank into a definable set of infinite Morley rank and a definable set of rank $\alpha$.) Marker's book and page 10 of these notes both argue as follows:
Let $\beta=\text{sup}\big\{\text{RM}(\psi):\psi\text{ an }{\mathcal{L}}_{\mathbb{M}}\text{-formula and RM}(\psi)<\infty\big\}$. Because $\text{RM}(\phi)=\infty\geq\beta+2$, we can find an ${\mathcal{L}}_{M}$-formula $\psi$ such that $\text{RM}(\phi\wedge\psi)\geq\beta+1$ and $\text{RM}(\phi\wedge\neg\psi)\geq\beta+1$. Then $\text{RM}(\phi\wedge\psi)=\text{RM}(\phi\wedge\neg\psi)=\infty$.
But unless we assume that $\beta<\infty$, the second sentence of this argument appears to beg the question. For if $\beta=\infty$, then second sentence simply reasserts that there is a $\psi$ with the desired property.
My question is, just how do we know that if $\text{RM}(\phi)=\infty$, then there is a $\psi$ with $\text{RM}(\phi\wedge\psi)=\text{RM}(\phi\wedge\neg\psi)=\infty$? Am I not understanding the quoted paragraph?