Set Intersection Distributes over Set Difference
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Theorem
Set intersection is distributive over set difference.
Let $R, S, T$ be sets.
Then:
- $\paren {R \setminus S} \cap T = \paren {R \cap T} \setminus \paren {S \cap T}$
- $R \cap \paren {S \setminus T} = \paren {R \cap S} \setminus \paren {R \cap T}$
where:
- $R \setminus S$ denotes set difference
- $R \cap T$ denotes set intersection.
Proof
\(\ds \paren {R \cap T} \setminus \paren {S \cap T}\) | \(=\) | \(\ds \paren {\paren {R \cap T} \setminus S} \cup \paren {\paren {R \cap T} \setminus T}\) | De Morgan's Laws: Difference with Intersection | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\paren {R \cap T} \setminus S} \cup \O\) | Set Difference of Intersection with Set is Empty Set | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {R \cap T} \setminus S\) | Union with Empty Set | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {R \setminus S} \cap T\) | Intersection with Set Difference is Set Difference with Intersection |
$\Box$
Then:
\(\ds R \cap \paren {S \setminus T}\) | \(=\) | \(\ds \paren {S \setminus T} \cap R\) | Intersection is Commutative | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {S \cap R} \setminus \paren {T \cap R}\) | from above | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {R \cap S} \setminus \paren {R \cap T}\) | Intersection is Commutative |
$\blacksquare$
Sources
- 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 5$: Complements and Powers
- 1968: A.N. Kolmogorov and S.V. Fomin: Introductory Real Analysis ... (previous) ... (next): $\S 1$: Sets and Functions: Problem $4 \ \text{(a)}$
- 2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 1$: Fundamental Concepts: Exercise $1.2 \ \text{(g)}$