Closure of Intersection and Symmetric Difference imply Closure of Set Difference
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Theorem
Let $\RR$ be a system of sets such that for all $A, B \in \RR$:
- $(1): \quad A \cap B \in \RR$
- $(2): \quad A \symdif B \in \RR$
where $\cap$ denotes set intersection and $\symdif$ denotes set symmetric difference.
Then:
- $\forall A, B \in \RR: A \setminus B \in \RR$
where $\setminus$ denotes set difference.
Proof
Let $A, B \in \RR$.
From Set Difference as Symmetric Difference with Intersection:
- $A \symdif \paren {A \cap B} = A \setminus B$
By hypothesis:
- $A \cap B \in \RR$
and:
- $A \symdif \paren {A \cap B} \in \RR$
and so:
- $A \setminus B \in \RR$
$\blacksquare$