Closure of Union and Complement imply Closure of Set Difference

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Theorem

Let $\mathcal R$ be a system of sets on a universe $\mathbb U$ such that for all $A, B \in \mathcal R$:

$(1): \quad A \cup B \in \mathcal R$
$(2): \quad \map \complement A \in \mathcal R$

where $\cup$ denotes set union and $\complement$ denotes complement (relative to $\mathbb U$).


Then:

$\forall A, B \in \mathcal R: A \setminus B \in \mathcal R$

where $\setminus$ denotes set difference.


Proof

Let $A, B \in \mathcal R$.

\(\displaystyle A \setminus B\) \(=\) \(\displaystyle A \cap \map \complement B\) Set Difference as Intersection with Complement
\(\displaystyle \) \(=\) \(\displaystyle \map \complement {\map \complement A \cup B}\) De Morgan's Laws: Complement of Intersection

As both set union and complement are closed in $\mathcal R$ the result follows.

$\blacksquare$


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