The intersection of sets P and Q is written P ∩ Q; it contains the elements common to both sets. (p.96)
Set B is a subset of set A (B ⊂ A or B ⊆ A) when all elements of B are found in A; then A ∩ B = B. (p.98)
An empty set has no element and is written ∅ or { }. If two sets share no element, their intersection is an empty set. (p.99)
The complement of an intersection, (A ∩ B)′, refers to all elements in ξ that are not in the intersection of A and B. (p.100)
Intersection problems: draw a Venn diagram, fill in the intersection first, then compute the “only” regions by subtraction (e.g. n(B only) = n(B) − n(B ∩ H)). (p.103)
Three-set problems: fill the centre (all three sets) first, then the regions in exactly two sets, then those in one set only; elements in no set = n(ξ) − total in the circles. (p.104)
4.2 Union of Sets
The union of sets P and Q is written P ∪ Q; it contains all the elements in set P, set Q, or both. (p.106)
The complement of a union, (A ∪ B)′, refers to all elements that are not in set A and not in set B. (p.110)
Union problems: n(A ∪ B) = n(A) + n(B) − n(A ∩ B); elements in neither set = n(ξ) − n(A ∪ B). (p.113)
Three-set problems: draw a Venn diagram of the three sets and compute each region; n(A ∪ B ∪ C)′ = n(ξ) − sum of all regions in the circles. (p.114)
4.3 Combined Operations on Sets
Combined operations involve both intersection (∩) and union (∪) at the same time. (p.116)
Combined operations are solved from left to right; operations in brackets are carried out first. (p.116)
To shade (P ∪ Q) ∩ R: shade P ∪ Q, shade R, then take the common region of both. (p.117)
To shade Q ∪ (P ∩ R): shade Q and shade P ∩ R; the answer is all regions covered by either. (p.117)
For a set A in the universal set, the complement A′ means all elements not in A. (p.118)
Complement of combined operations is solved left to right, with brackets first; e.g. (A ∪ B)′ ∩ C. (p.118)
“Common region” is used for ∩; “all regions covered by either” is used for ∪. (p.119)
Problem solving: draw a Venn diagram of all given information and compute each region, e.g. one competition only = sum of the “only” regions. (p.121)