Dynkin System Closed under Set Difference with Subset

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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:

\(\ds D \setminus E\) \(=\) \(\ds D \cap E^\complement\) Set Difference as Intersection with Relative Complement
\(\ds \) \(=\) \(\ds \paren {D^\complement \cup E}^\complement\) De Morgan's Laws: Complement of Union, Relative Complement of Relative Complement

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


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 $D \setminus E \in \DD$.

$\blacksquare$


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