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Section 9.4 Complement, union, and intersection
First, it is often convenient to restrict the scope of the discussion.
universal set
a set that contains all objects currently under consideration
We will consider all of the following set operations to be performed within a universal set \(U \text{.}\) In particular, suppose \(A,B\subseteq U \text{.}\)
Subsection 9.4.1 Universal and relative complement
complement
the set of elements of
\(U \) thar are
not in
\(A \)
relative complement
if
\(A,B \subseteq U \text{,}\) the
complement of \(A \) in \(B \) is the set of elements of
\(B \) that are not in
\(A \)
\(\cmplmnt{A} \)
the complement of
\(A \) (in
\(U \) ), so that
\(\cmplmnt{A} = \setdef{ x \in U }{ x \notin A } \)
\(B \relcmplmnt A \)
the complement of
\(A \) in
\(B \text{,}\) so that
\(B \relcmplmnt A = \setdef{ x \in B }{ x \notin A } \)
Figure 9.4.1. Venn diagrams of universal and relative set complements.
Example 9.4.3 . Some examples of relative complement involving number sets.
Suppose \(B = \{ 1, 2, 3, 4, 5, 6 \} \) and \(A = \{ 1, 3, 5 \} \text{.}\) Then \(B \relcmplmnt A = \{ 2, 4, 6 \} \text{.}\)
The complement of the set of rational numbers \(\Q \) inside the set of real numbers \(\R \) is called the set of irrational numbers , and we write \(\I = \R \relcmplmnt \Q \) for this set. If you are thinking of real numbers in terms of their decimal expanions, the irrational numbers are precisely those that have nonterminating, nonrepeating decimal expansions.
Subsection 9.4.2 Union, intersection, and disjoint sets
union
the combined collection of all elements in a pair of sets
\(A \union B \)
the union of sets \(A \) and \(B \text{,}\) so that
\begin{equation*}
A \union B = \setdef{x\in U}{x\in A \text{ or } x\in B \text{ (or both)}}
\end{equation*}
intersection
the collection of only those elements common to a pair of sets
\(A \intersection B \)
the intersection of \(A \) and \(B \text{,}\) so that
\begin{equation*}
A \intersection B = \setdef{x\in U}{x\in A \text{ and } x\in B}
\end{equation*}
Figure 9.4.4. Venn diagrams of set union and intersection.
Example 9.4.6 .
For subsets \(A = \{1,2,3,4\} \) and \(B = \{3,4,5,6\} \) of \(\N \text{,}\) we have
\begin{align*}
A \union B \amp = \{1,2,3,4,5,6\}, \amp A \intersection B \amp = \{3,4\}.
\end{align*}
Example 9.4.7 .
Consider the following subsets of \(\N \text{.}\)
\begin{align*}
\mathscr{E} \amp = \setdef{n\in\N}{n\text{ even}} \amp
\mathscr{P} \amp = \setdef{n\in\N}{n\text{ prime, } n\ne 0} \\
\mathscr{O} \amp = \setdef{n\in\N}{n\text{ odd}} \amp
\mathscr{T} \amp = \setdef{3n}{n\in\N} = \{ 0,\, 3,\, 6,\, 9,\, \dotsc \}
\end{align*}
Then,
\begin{align*}
\mathscr{E} \union\mathscr{O} \amp = \N, \amp
\mathscr{E} \intersection \mathscr{P} \amp = \{2\}, \amp
\mathscr{E} \intersection \mathscr{T} \amp = \setdef{6n}{n\in\N}, \\
\mathscr{E} \intersection \mathscr{O} \amp = \emptyset, \amp
\mathscr{O} \intersection \mathscr{P} \amp = \mathscr{P} \relcmplmnt \{2\}, \amp
\mathscr{O} \intersection \mathscr{T} \amp = \setdef{6n+3}{n\in\N}.
\end{align*}
disjoint sets
sets that have no elements in common; that is, sets
\(A,B \) such that
\(A \intersection B = \emptyset \)
disjoint union
a union
\(A \union B \) where
\(A \) and
\(B \) are disjoint
\(A \disjunion B \)
the disjoint union of sets
\(A \) and
\(B \)
Figure 9.4.8. A Venn diagram of a disjoint set union.
Example 9.4.9 .
Sets
\(\mathscr{E},\mathscr{O} \) from
ExampleΒ 9.4.7 are disjoint, and
\(\N = \mathscr{E} \disjunion \mathscr{O} \text{.}\)
Subsection 9.4.3 Rules for set operations
Proposition 9.4.11 . Rules for Operations on Sets.
Suppose \(A,B,C \) are subsets of a universal set \(U \text{.}\) Then the following set equalities hold.
Properties of the universal set.
\(\displaystyle A \union U = U \)
\(\displaystyle A \intersection U = A \)
Properties of the empty set.
\(\displaystyle A \union \emptyset = A \)
\(\displaystyle A \intersection \emptyset = \emptyset \)
Duality of universal and empty sets.
\(\displaystyle \cmplmnt{U} = \emptyset \)
\(\displaystyle \cmplmnt{\emptyset} = U \)
Double complement.
\(\displaystyle \cmplmnt{(\cmplmnt{A})} = A \)
Idempotence.
\(\displaystyle A \union A = A \)
\(\displaystyle A \intersection A = A \)
Commutativity.
\(\displaystyle A \union B = B \union A \)
\(\displaystyle A \intersection B = B \intersection A \)
Associativity.
\(\displaystyle (A \union B) \union C = A \union (B \union C) \)
\(\displaystyle (A \intersection B) \intersection C = A \intersection (B \intersection C) \)
Distributivity.
\(\displaystyle A \intersection (B \union C) = (A \intersection B) \union (A \intersection C) \)
\(\displaystyle A \union (B \intersection C) = (A \union B) \intersection (A \union C) \)
\(\displaystyle (A \union B) \intersection C = (A \intersection C) \union (B \intersection C) \)
\(\displaystyle (A \intersection B) \union C = (A \union C) \intersection (B \union C) \)
DeMorganβs Laws.
\(\displaystyle \cmplmnt{(A \union B)} = \cmplmnt{A} \intersection \cmplmnt{B} \)
\(\displaystyle \cmplmnt{(A \intersection B)} = \cmplmnt{A} \union \cmplmnt{B} \)
Proof of RuleΒ 9.a.
Recall that to prove this set equality, we need to show both
\begin{align*}
\cmplmnt{(A \union B)} \amp \subseteq \cmplmnt{A} \intersection \cmplmnt{B} \text{,} \amp
\cmplmnt{A} \intersection \cmplmnt{B} \amp \subseteq \cmplmnt{(A \union B)} \text{.}
\end{align*}
Show \(\cmplmnt{(A \union B)} \subseteq \cmplmnt{A} \intersection \cmplmnt{B} \) .
We need to show
\begin{equation*}
x \in \cmplmnt{(A \union B)} \lgcimplies x \in \cmplmnt{A} \intersection \cmplmnt{B} \text{.}
\end{equation*}
If \(x \in \cmplmnt{(A \union B)} \) then by definition of complement, \(x\in U \) but \(x \notin A \union B \text{.}\) Then \(x \notin A \) must be true, since if \(x \) were in \(A \) then it would also be in \(A \union B \text{.}\) Similarly, \(x \notin B \) must also be true. Therefore, \(x \in \cmplmnt{A} \) and \(x\in\cmplmnt{B} \text{,}\) so that \(x \in \cmplmnt{A} \intersection \cmplmnt{B} \text{.}\)
Show \(\cmplmnt{A} \intersection \cmplmnt{B} \subseteq \cmplmnt{(A \union B)} \) .
We need to show
\begin{equation*}
x \in \cmplmnt{A} \intersection \cmplmnt{B} \lgcimplies x \in \cmplmnt{(A \union B)} \text{.}
\end{equation*}
If \(x \in \cmplmnt{A} \intersection \cmplmnt{B} \text{,}\) then by definition of intersection both \(x \in \cmplmnt{A} \) and \(x \in \cmplmnt{B} \) are true; that is, \(x \notin A \) and \(x \notin B \text{.}\) Since \(A \union B \) is all elements of \(U \) which are in one (or both) of \(A,B \text{,}\) we must have \(x \notin A \union B \text{.}\) Thus \(x \in \cmplmnt{(A \union B)} \text{.}\)
Proofs of the other rules.