Set Difference with Set Difference is Union of Set Difference with Intersection
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Theorem
Let $R, S, T$ be sets.
Then:
- $R \setminus \paren {S \setminus T} = \paren {R \setminus S} \cup \paren {R \cap T}$
where:
- $S \setminus T$ denotes set difference
- $S \cup T$ denotes set union
- $S \cap T$ denotes set intersection.
Illustration by Venn Diagram
Corollary
- $T \setminus \paren {S \setminus T} = T$
Proof
Consider $R, S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the universal set.
\(\ds R \setminus \paren {S \setminus T}\) | \(=\) | \(\ds R \cap \overline {\paren {S \cap \overline T} }\) | Set Difference as Intersection with Complement | |||||||||||
\(\ds \) | \(=\) | \(\ds R \cap \paren {\overline S \cup \overline {\paren {\overline T} } }\) | De Morgan's Laws | |||||||||||
\(\ds \) | \(=\) | \(\ds R \cap \paren {\overline S \cup T}\) | Complement of Complement | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {R \cap \overline S} \cup \paren {R \cap T}\) | Intersection Distributes over Union | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {R \setminus S} \cup \paren {R \cap T}\) | Set Difference as Intersection with Complement |
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Exercise $3.4 \ \text{(d)}$