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Let $A_1,...,A_n$ be subsets of a set $\Omega$. For all $a \in \{0,1 \}^n$, define

\begin{equation} \label{eu_eqn} F(a) := \bigcap^n_{i=1}A_i^{(a_i)} , \end{equation} where $A^{(0)}:= A_i$ and $A^{(1)}:= A_i^c$.

I need to prove that the collection $\{F(a): a \in \{0,1 \}^n \}$ is a partition of $\Omega$. However, I fail to see why this collection is a partition, because a partition never contains the empty set and in general the sets $A_1,...,A_n$ can be disjoint, right? So in that case I can easily find an $a$ such that $F(a) = \emptyset$, namely $a := (0,...,0)$. I have already shown that $\Omega = \bigcup_{a \in \{0,1 \}^n} F(a)$ but in order for it to be a partition, it cannot have the empty set. Am I missing something? Or maybe I didn't understand the notation of some of these objects.

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    $\begingroup$ Some will define a partition of a set $X$ to be a set $\{X_i\}_{i\in I}$ where $X_i\cap X_j=\emptyset$ for $i\neq j$ and $\cup_{i\in I}X_i=X$, so with this definition the empty set is allowed to be a part in the partition, so perhaps just take that as a part of the definition. $\endgroup$
    – Steven Creech
    Nov 20 at 20:50
  • $\begingroup$ @StevenCreech Oh wow, I didn't know that alternative definition of partition, tyvm! I recently joined a new university and I noticed they are using a ton of different notations/conventions. $\endgroup$
    – Einsteinwasmyfather
    Nov 20 at 21:04

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