What is wrong with this reasoning for empty class? Let $\{A_i\}$ be empty class. I want to show $(\cup_{i}A_i)^{\prime}=\cap_{i}A_i^{\prime}$. Since $(\cup_{i}A_i)^{\prime}=U$ where $U$ represents universal set what we need to show is that $U=\cap_{i}A_i^{\prime}$. Since $\{A_i\}$ is empty class it contains no set, then it is complement contains every subset of $U$, but then $\emptyset \in A_i^{\prime}$ then $\cap_{i}A_i^{\prime}=\emptyset$. Thanks for any help.

$\forall x\in U$ we can show that $x\in\cap_{\alpha\in\emptyset} A_{\alpha}$. Just see that: $$x\in\cap_{\alpha\in\emptyset}A_{\alpha}\equiv(x\in A_{\alpha}, \forall \alpha\in\emptyset )\equiv(\alpha\in\emptyset\longrightarrow x\in A_{\alpha})$$
