# 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.

$\blacksquare$

## Sources

- 1970: Avner Friedman:
*Foundations of Modern Analysis*... (previous) ... (next): $\S 1.1$: Rings and Algebras