Now that we have defined sets, let's remind ourselves that we already know of a few: \(\mathbb{N} \subseteq \mathbb{Z} \subseteq \mathbb{Q} \subseteq \mathbb{R} \subseteq \mathbb{C}\text{,}\) the natural numbers, integers, rational numbers, real numbers and complex numbers. Having defined sets, we also want to know how we can combine them to form new sets, which is the purpose of this section.
Given two sets \(A,B\) there are three basic binary operations we can perform.
It is straightforward to see that union and intersection are each commutative and associative since disjunction (\(\vee\)) and conjunction (\(\wedge\)) are. Moreover, they distribute over each other for the same reason. In addition, you can think of set difference as being analogous to negation (\(\neg\)), a point which will be made clearer later. These facts are encapsulated in the following theorem which I encourage you to prove for yourself using if-and-only-if proofs for set equality.
The binary operations union and intersection are commutative, associative and distribute over each other. That is,
\begin{align*} A \cup B = B \cup A \quad&\text{and}\quad A \cap B = B \cap A,\\ (A \cup B) \cup C = A \cup (B \cup C) \quad&\text{and}\quad (A \cap B) \cap C = A \cap (B \cap C),\\ (A \cup B) \cap C = (A \cap C) \cup (B \cap C) \quad&\text{and}\quad (A \cap B) \cup C = (A \cup C) \cap (B \cup C). \end{align*}Moreover, with the difference they satisfy a sort of De Morgan's Laws:
\begin{align*} A \setminus (B \cup C) &= (A \setminus B) \cap (A \setminus C),\\ A \setminus (B \cap C) &= (A \setminus B) \cup (A \setminus C). \end{align*}In addition, the set difference satisfies a sort of double negation elimination, namely,
\begin{equation*} A \setminus \big( A \setminus ( A \setminus B) \big) = A \setminus B. \end{equation*}Sets \(A,B\) are said to be disjoint if their intersection is empty, i.e., \(A \cap B = \emptyset\text{.}\)
If there is a universe of discourse \(U\) which is specified, and a set \(A\) of elements from this universe, we can talk about the complement of \(A\text{,}\) denoted \(A^c\text{,}\) which is defined as \(U \setminus A\text{.}\)
The complement satisfies a few nice properties, and it really acts like negation. So, for example, \((A^c)^c = A\) and it satisfies De Morgan's Laws by virtue of Theorem 2.2.2, namely,
\begin{gather*} (A \cup B)^c = A^c \cap B^c,\\ (A \cap B)^c = A^c \cup B^c. \end{gather*}In the theorem below we collect a few more facts about set operations, particularly how they interact with the subset relation.
For any sets \(A,B\text{,}\)
One other way to make new sets is to make ordered tuples. Remember, sets are unordered, but it is very useful to have objects with order to them. For example, when we considered points in the plane in algebra and calculus, we used to represent them by a pair of real numbers, like \((a,b)\) where \(a\) is the horizontal offset from the origin, and \(b\) is the vertical offset. It is clear that \((0,1) \not= (1,0)\text{,}\) so order matters.
For sets \(A,B\) we can form their Cartesian product (or just product) which consists of all ordered pairs where the first component is an element of \(A\) and the second component is an element of \(B\text{.}\) Symbollically,
\begin{equation*} A \times B := \{ (a,b) \mid a \in A, b \in B \}. \end{equation*}Note that your book calls the Cartesian product the “cross product.” This is simply incorrect; no one refers to it this way. Cross products are either the operation on three (or seven) dimensional vectors you learned in Calculus, or they are much more complicated objects that involve group actions (you are not supposed to know what a group action is). Just never use the term cross product in place of Cartesian product.
Of course, we can form multiple Cartesian products to get order triples, or more generally, order \(n\)-tuples. For example,
\begin{equation*} A \times B \times C = \{ (a,b,c) \mid a \in A, b \in B, c \in C \}. \end{equation*}When we take the Cartesian product of a set with itself, we use exponential notation for convenience, i.e.,
\begin{equation*} \underbrace{A \times A \times \cdots \times A}_{n} = A^n. \end{equation*}This is precisely why we use the notation \(\mathbb{R}^n\) in linear algebra to denote collection of vectors (they are ordered \(n\)-tuples of real numbers).
To maintain a grasp of Cartesian products, you should keep in mind the analogy of rectangles. Consider the open intervals \(A = (2,3)\) and \(B = (4,7)\text{.}\) Then since \(A,B \subseteq \mathbb{R}\text{,}\) we see that \(A \times B \subseteq \mathbb{R}^2\text{,}\) that is, it lies in the plane, so we can visualize it! In particular, \(A \times B\) is the (open) rectangle in the plane whose \(x\)-values are between \(2\) and \(3\) and whose \(y\)-values are between \(6\) and \(7\text{.}\) So, when you have some property about Cartesian products and you want to see if and/or why it is true, imagine first that your Cartesian products are rectangles in the plane. Hopefully this helps your intuition for the following theorem.
Suppose \(A,B,C,D\) are sets. Then