# Field generated by a class of subsets

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.

• 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. Nov 20 at 20:50
• @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. Nov 20 at 21:04