Consider the set $E$ and its power set $2^E$. We say that a collection of subsets of $E$, denoted $D \subset 2^E$, is a d-system if
- $E \in D$,
- $A,B \in D \wedge B \subset A \implies A \setminus B \in D$, and
- $A_n \subset D \wedge A_n \uparrow A \implies A \in D$.
where $A_n$ is an increasing sequence of sets such that $\bigcup_{i=1}^\infty A_i = A$. It seems to me that condition 2 is redundant, since if $A \in D$ and $B \subset A$, then $A \setminus B \subset A \in D$. I think that condition 2 exists so that the set $A \setminus B$ is explicitly included as part of $D \subset 2^E$. That is, rather than writing $D = \{E,A,B\}$, where $A \setminus B$ is hidden inside $A$, we instead write $D = \{E,A,B,A \setminus B\}$. However, I'm not sure about this reasoning.