Dynkin System Closed under Set Difference with Subset

Theorem
Let $X$ be a set.

Let $\DD$ be a Dynkin system on $X$.

Let $D, E \in \DD$ and suppose that $E \subseteq D$.

Then the set difference $D \setminus E$ is also an element of $\DD$.

Proof
For brevity, write for example $E^\complement$ for $\relcomp X E = X \setminus E$.

We reason as follows:

Now this implies that $D \setminus E \in \DD$ $D^\complement \cup E \in \DD$.

It is already known that $D^\complement$ and $E$ are in $\DD$ by axiom $(2)$ for a Dynkin system.

Since $E \subseteq D$, it follows that $D^\complement \cap E = \O$, and thus Dynkin System Closed under Disjoint Union applies to give:


 * $D^\complement \cup E \in \DD$

which, combined with above reasoning, yields $D \setminus E \in \DD$.