# $C_n:=A_n\cap (A_1\cup\cdots\cup A_{n-1})^c$ pairwise disjoint?

Let $\Omega$ be a set and $A_1,\ldots\in Pot(\Omega)$.

Why are the sets $C_n:=A_n\cap (A_1\cup\cdots\cup A_{n-1})^c$ pairwise disjoint?

I've tried to write it like $C_n\cap C_m=\bigcap_{k=1}^{n-1}(A_n\setminus A_k)\cap\bigcap_{k=1}^{m-1}(A_m\setminus A_k)$ but it didn't get me anywhere.

How can you verify it?

Say $m<n$. Note that $$C_m \cap C_n= A_m \cap \left( \cap_{i=1}^{m-1}A_i^c\right) \cap A_n \cap \left( \cap_{i=1}^{n-1}A_i^c\right).$$
If $x\in C_m \cap C_n$, then $x\in A_m$ and $x\in \cap_{i=1}^{n-1}A_i^c$. But ($m<n$) $$\cap_{i=1}^{n-1}A_i^c \subseteq A_m^c.$$.
Consider $C_n$ and $C_{m}$ with $m>n$. If $x\in C_n$, then $x\in A_n$. In particular, $x\in A_1\cup\ldots\cup A_{m-1}$ (here, we have used $m>n$) so $x\not\in C_m$. So $C_n\cap C_m=\emptyset$.
If $m \neq n$, then either $m > n$ or $m<n$. Assume without loss of generality that $m<n$. Then $C_m = A_m \cap A_1^{c} \cap \cdots \cap A_{m-1}^{c}$ is a subset of $A_m$, and $C_n = A_n \cap A_1^{c} \cap \cdots \cap A_{m}^{c} \cap \cdots \cap A_{n-1}^{c}$ is a subset of $A_m^{c}$. Since $A_m$ and $A_m^{c}$ are disjoint, so are $C_m$ and $C_n$ as their respective subsets. Hope this helps!