Let \(A\) and \(B\) be sets. The union of \(A\) and \(B\text{,}\) denoted \(A \cup B\text{,}\) is the set defined by
\begin{equation*}
A \cup B = \{x | x\in A \text{ or } x\in B\}\text{.}
\end{equation*}
The intersection of \(A\) and \(B\text{,}\) denoted \(A \cap B\text{,}\) is the set defined by
\begin{equation*}
A \cap B = \{x | x\in A \text{ and }x\in B\}\text{.}
\end{equation*}
Theorem3.17.
Let \(A\text{,}\)\(B\) and \(C\) be sets.
\(A \cap B \subseteq A\) and \(A \cap B \subseteq B\text{.}\) If \(X\) is a set such that \(X \subseteq A\) and \(X \subseteq B\text{,}\) then \(X \subseteq A \cap B\text{.}\)
\(A \subseteq A \cup B\) and \(B \subseteq A \cup B\text{.}\) If \(Y\) is a set such that \(A \subseteq Y\) and \(B \subseteq Y\text{,}\) then \(A \cup B \subseteq Y\) .
Commutative Laws.
\(A \cup B = B \cup A\) and \(A \cap B = B \cap A\text{.}\)
Associative Laws.
\((A\cup B)\cup C = A\cup (B\cup C)\) and \((A\cap B)\cap C = A\cap (B\cap C)\text{.}\)
Distributive Laws.
\(A \cap (B \cup C) = (A \cap B) \cup (A \cap C)\) and \(A \cup (B \cap C) = (A \cup B) \cap (A \cup C)\text{.}\)
Identity Laws.
\(A \cup\emptyset = A\) and \(A \cap\emptyset = \emptyset\text{.}\)
Idempotent Laws.
\(A \cup A = A\) and \(A \cap A = A\text{.}\)
Absorbtion Laws.
\(A \cup (A \cap B) = A\) and \(A \cap (A \cup B) = A\text{.}\)
If \(A \subseteq B\text{,}\) then \(A \cup C \subseteq B \cup C\) and \(A \cap C \subseteq B \cap C\text{.}\)
Definition3.18.
Let \(A\) and \(B\) be sets. The sets \(A\) and \(B\) are disjoint if \(A \cap B = \emptyset\text{.}\)
Example3.19.
Let \(\E\) be the set of even integers, let \(\O\) be the set of odd integers and let \(\P\) be the set of prime numbers. Then \(\E\) and \(\O\) are disjoint, whereas \(\E\) and \(\P\) are not disjoint (because \(\E \cap\P = \{2\}\)).
Definition3.20.
Let \(A\) and \(B\) be sets. The set difference (also called the difference) of \(A\) and \(B\text{,}\) denoted \(A \sm B\text{,}\) is the set defined by \(A \sm B = \{x | x\in A \text{ and }x \not\in B\}\text{.}\)
Convention3.21.
Some books use the notation \(A - B\) instead of \(A \sm B\text{.}\) Though the notation \(A - B\) may seem intuitive, there are situations where it can become misleading or confusing.
Theorem3.22.
Let \(A\text{,}\)\(B\) and \(C\) be sets.
\(A \sm B \subseteq A\text{.}\)
\((A \sm B) \cap B = \emptyset\text{.}\)
\(A \sm B = \emptyset\) if and only if \(A \subseteq B\text{.}\)
\(B \sm (B \sm A) = A\) if and only if \(A \subseteq B\text{.}\)
If \(A \subseteq B\text{,}\) then \(A \sm C = A \cap (B \sm C)\text{.}\)
If \(A \subseteq B\text{,}\) then \(C \sm A ⊇ C \sm B\text{.}\)
\(C \sm (A \cup B) = (C \sm A) \cap (C \sm B)\) and \(C \sm (A \cap B) = (C \sm A) \cup (C \sm B)\) (DeMorgan's Laws).
Definition3.23.
Let \(A\) and \(B\) be sets. The product (also called the Cartesian product) of \(A\) and \(B\text{,}\) denoted \(A \times B\text{,}\) is the set
\begin{equation*}
A \times B = \{(a, b) | a\in A and b\in B\}\text{,}
\end{equation*}
where \((a, b)\) denotes an ordered pair.
Example3.24.
We can think of \(\R^2\text{,}\) which is defined in terms of ordered pairs of real numbers, as \(\R^2 = \R \times\R\text{.}\) Similarly, we think of \(\R^n\) as
Let \(A\text{,}\)\(B\) and \(C\) be sets. Suppose that \(C \subset A \cup B\text{,}\) and that \(C \cap A = \emptyset\text{.}\) Prove that \(C \subseteq B\text{.}\)
For real numbers \(a\text{,}\)\(b\) and \(c\text{,}\) we know that \(a - (b - c) = (a - b) + c\text{.}\) Let \(A\text{,}\)\(B\) and \(C\) be sets.
Suppose that \(C \subseteq A\text{.}\) Prove that \(A \sm (B \sm C) = (A \sm B) \cup C\text{.}\)
Does \(A \sm (B \sm C) = (A \sm B) \cup C\) hold for all sets \(A\text{,}\)\(B\) and \(C\text{?}\) Prove or give a counterexample for this formula. If the formula is false, find and prove a modification of this formula that holds for all sets.
