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:
- $\left({R \setminus S}\right) \cap T = \left({R \cap T}\right) \setminus \left({S \cap T}\right)$
- $R \cap \left({S \setminus T}\right) = \left({R \cap S}\right) \setminus \left({R \cap T}\right)$
where:
- $R \setminus S$ denotes set difference
- $R \cap T$ denotes set intersection.
Proof
\(\ds \left({R \cap T}\right) \setminus \left({S \cap T}\right)\) | \(=\) | \(\ds \left({\left({R \cap T}\right) \setminus S}\right) \cup \left({\left({R \cap T}\right) \setminus T}\right)\) | De Morgan's Laws: Difference with Intersection | |||||||||||
\(\ds \) | \(=\) | \(\ds \left({\left({R \cap T}\right) \setminus S}\right) \cup \varnothing\) | Set Difference of Intersection with Set is Empty Set | |||||||||||
\(\ds \) | \(=\) | \(\ds \left({R \cap T}\right) \setminus S\) | Union with Empty Set | |||||||||||
\(\ds \) | \(=\) | \(\ds \left({R \setminus S}\right) \cap T\) | Intersection with Set Difference is Set Difference with Intersection |
$\Box$
Then:
\(\ds R \cap \left({S \setminus T}\right)\) | \(=\) | \(\ds \left({S \setminus T}\right) \cap R\) | Intersection is Commutative | |||||||||||
\(\ds \) | \(=\) | \(\ds \left({S \cap R}\right) \setminus \left({T \cap R}\right)\) | from above | |||||||||||
\(\ds \) | \(=\) | \(\ds \left({R \cap S}\right) \setminus \left({R \cap T}\right)\) | 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 $1.2 \ \text{(g)}$