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Mathematics · Ch 1 — Sets

Union and Intersection of Sets

1.6

Union and Intersection of Sets

Given two sets, the two most fundamental ways of combining them are union and

intersection.

Union. The union of two sets AA and BB, written A∪BA \cup B, is the set of all

elements that belong to AA, or to BB, or to both:

A∪B={x:x∈A or x∈B}.A \cup B = \{x : x \in A \text{ or } x \in B\}.

Here "or" is used in the inclusive sense (as always in mathematics) — an element that

belongs to both sets is still only listed once in the union, since a set never

repeats an element.

Intersection. The intersection of AA and BB, written A∩BA \cap B, is the set of

all elements that belong to both AA and BB simultaneously:

A∩B={x:x∈A and x∈B}.A \cap B = \{x : x \in A \text{ and } x \in B\}.

Disjoint sets. If A∩B=∅A \cap B = \varnothing — that is, AA and BB share no elements

at all — then AA and BB are called disjoint sets.

Worked illustration. Let A={1,2,3,4,5}A = \{1, 2, 3, 4, 5\} and B={4,5,6,7,8}B = \{4, 5, 6, 7, 8\}. Then

A∪B={1,2,3,4,5,6,7,8},A∩B={4,5}.A \cup B = \{1, 2, 3, 4, 5, 6, 7, 8\}, \qquad A \cap B = \{4, 5\}.

Since A∩B≠∅A \cap B \ne \varnothing, these two sets are not disjoint.

Basic properties. For any sets AA, BB, CC:

  • Commutative laws: A∪B=B∪AA \cup B = B \cup A, and A∩B=B∩AA \cap B = B \cap A.
  • Associative laws: (A∪B)∪C=A∪(B∪C)(A \cup B) \cup C = A \cup (B \cup C), and similarly for ∩\cap.
  • Idempotent laws: A∪A=AA \cup A = A, A∩A=AA \cap A = A.
  • A∪∅=AA \cup \varnothing = A and A∩∅=∅A \cap \varnothing = \varnothing.
  • If A⊆BA \subseteq B, then A∪B=BA \cup B = B and A∩B=AA \cap B = A (drawing this as a Venn diagram — a smaller circle entirely inside a bigger one — makes both facts immediate: the "union" of the two circles is just the bigger one, and their "overlap" is the whole of the smaller one).

Counting formula (finite sets). For two finite sets,

n(A∪B)=n(A)+n(B)−n(A∩B).n(A \cup B) = n(A) + n(B) - n(A \cap B).

The subtraction corrects for double-counting: elements in A∩BA \cap B get counted once

in n(A)n(A) and once again in n(B)n(B), so one copy must be removed. …