Dynkin System Closed under Set Difference with Subset

Theorem
Let $X$ be a set, and let $\mathcal D$ be a Dynkin system on $X$.

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

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

Proof
For brevity, write for example $E^c$ for $\complement_X \left({E}\right) = X \setminus E$.

We reason as follows:

Now this implies that $D \setminus E \in \mathcal D$ $D^c \cup E \in \mathcal D$.

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

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


 * $D^c \cup E \in \mathcal D$

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