Cantor-Bendixson rank for a formula is defined as (in my notation) for $\varphi(x)$ in fixed $L$ and complete theory $T$,
(i)$CB(\varphi(x))\geq 0$ if $\phi$ is consistent with $T$
(ii)$CB(\varphi(x))\geq\alpha$ if $CB(\varphi(x))\geq\beta$ for all $\beta<\alpha$ ($\alpha$ is limit ordinal)
(iii)$CB(\varphi(x))\geq\alpha +1$ if there's a pairwise inconsistent $(\psi_i |i<\omega)$ such that $CB(\varphi(x) \wedge \psi_i (x)\geq\alpha$
(iv) $CB(\varphi)=\alpha$ iff $CB(\varphi)\geq\alpha\wedge CB(\varphi)\ngeq\alpha +1$
(v)$CB(p):=min\{\varphi | \varphi\in p \}$ for $p$ (partial) type of T
It is known that
(i)$p\models q \Rightarrow CB(p)\leq CB(q)$ for $p,q$ (partial) types of T
(ii)$CB(\varphi\vee\psi)=max\{CB(\varphi),CB(\psi)\}$
(iii)For type $p$, $\exists q\in S_n(T)$ such that $p\subseteq q \wedge CB(p)=CB(q)$
Here, $S_n(T)=S_n(\emptyset)$ is a stone space consisting of complete $n$-types of T.
I'm proving $CB(\varphi)=\alpha$ iff $\{p\in S_n (T) | \varphi\in p\wedge CB(p)\geq\alpha\}$ is nomempty finite. I can do $\Leftarrow$ part, but it is not easy to prove $\Rightarrow$ part. I proved above set is not empty from (iii) property so I assumed contradiction so that there is infinite $p_i$'s containing $\varphi$ and $CB(p_i)\geq\alpha$.
I aimed to show there is a pairwise inconsistent $(\varphi_i | i<\omega)$ having $CB(\varphi_i)\geq\alpha$ for $\varphi_i\in p_i$. I could prove for every $n<\omega$, there's pairwise inconsistent $\varphi_0$,...,$\varphi_n$ with $\varphi_i\in p_i$ for $i\leq n$ only. I tried to derive infinite pairwise inconsistent formulas but I couldn't do from this. I'm stuck here and don't know how to derive such formulas. Any hint will be thankful.
I finally proved this. since $S_n(T)$ is Hausdorff, we can take disjoint open sets $O_{\varphi_{i}}=\{p\in S_n(T) | \varphi_i\in p\}$ containing each $p_i$. then $\varphi_i$'s are desired ones for $CB(\varphi)\geq\alpha+1$ which yields contradiction.