Closure of Union and Complement imply Closure of Set Difference
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Theorem
Let $\RR$ be a system of sets on a universe $\mathbb U$ such that for all $A, B \in \RR$:
- $(1): \quad A \cup B \in \RR$
- $(2): \quad \map \complement A \in \RR$
where $\cup$ denotes set union and $\complement$ denotes complement (relative to $\mathbb U$).
Then:
- $\forall A, B \in \RR: A \setminus B \in \RR$
where $\setminus$ denotes set difference.
Proof
Let $A, B \in \RR$.
\(\ds A \setminus B\) | \(=\) | \(\ds A \cap \map \complement B\) | Set Difference as Intersection with Complement | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \complement {\map \complement A \cup B}\) | De Morgan's Laws: Complement of Intersection |
As both set union and complement are closed in $\RR$ the result follows.
$\blacksquare$
Sources
- 1970: Avner Friedman: Foundations of Modern Analysis ... (previous) ... (next): $\S 1.1$: Rings and Algebras