## Section6.2Cartesian Products

A Cartesian product of two sets is a new set that is constructed from the two sets. In order to define Cartesian products, we need to define a mathematical object called an ordered pair.

### Definition6.2.1.

An ordered pair is an ordered list of two mathematical objects, $$a$$ and $$b\text{,}$$ written as $$(a,b)\text{.}$$ The objects in an ordered pair are called components. The object $$a$$ is the first component of $$(a,b)\text{,}$$ and the object $$b$$ is the second component of $$(a,b)\text{.}$$

### Definition6.2.2.

Let $$A$$ and $$B$$ be sets. The Cartesian product of $$A$$ and $$B\text{,}$$ denoted $$A \times B\text{,}$$ is the set of ordered pairs $$(a,b)\text{,}$$ where $$a \in A$$ and $$b \in B\text{.}$$

The Cartesian product of two sets $$A$$ and $$B\text{,}$$ formulated in set-builder notation, is

\begin{equation*} A \times B := \set{(a,b) \mid a \in A \text{ and } b \in B}\text{.} \end{equation*}

Complete the definition of Cartesian products in Checkpoint 6.2.3.

Complete the following:

Let $$A$$ and $$B$$ be sets.

The

• select

• intersection

• union

• difference

• sum

• Cartesian product

• complement

of the sets $$A$$ and $$B\text{,}$$ denoted by $$A\times B\text{,}$$ is the set of all
• select

• unordered pairs

• ordered pairs

• sets

• numbers

$$(a,b)$$ where $$a$$ is
• select

• related to

• an element of

• a subsets of

• not related to

• not an element of

• a proper subset of

• equal to

• not equal to

the set $$A$$
• select

• and

• or

$$b$$ is
• select

• related to

• an element of

• a subsets of

• not related to

• not an element of

• a proper subset of

• equal to

• not equal to

the set $$B\text{.}$$

$$\text{Cartesian product}$$

$$\text{ordered pairs}$$

$$\text{an element of}$$

$$\text{and}$$

$$\text{an element of}$$

To form the Cartesian product $$A \times B\text{,}$$ we pair each element of $$A\text{,}$$ placed in the first component of the ordered pair, with each element of $$B\text{,}$$ placed in the second component of the ordered pair.

Let $$A = \set{0,1}\text{,}$$ and let $$B = \set{4,5,6}\text{.}$$ Then,

\begin{equation*} A \times B = \set{(0, 4), (0, 5), (0, 6), (1, 4), (1, 5), (1, 6)}\text{,} \end{equation*}

and

\begin{equation*} B \times A = \set{(4, 0), (4, 1), (5, 0), (5, 1), (6, 0), (6,1)}\text{.} \end{equation*}

Let $$A=\{1,2,3\}$$ and let $$B=\{-50\}\text{.}$$ Give the set $$A\times B$$ in roster form.

Solution.

The set $$A\times B$$ contains all ordered pairs whose first entry is an element of the set $$A$$ and whose second entry is an element of the set $$B\text{.}$$ We write ordered pairs whose first entry is $$c$$ and whose second entry is $$d$$ as $$(c,d)\text{.}$$ We get

\begin{equation*} A\times B= \{(1,-50),(2,-50),(3,-50)\} \end{equation*}

In the next problem a Cartesian product is given in set builder notation.

Let $$A = \{12,13,34\}\text{.}$$ Give $$\{(a,a \fmod 5)\mid a \in A\}$$ in roster form.

Solution.

We find all pairs whose first entry is an element $$a$$ of the set $$A$$ and whose second entry is $$a \fmod 5\text{.}$$ We get

\begin{align*} \{(a,a \fmod 5) \mid a \in A\} \amp = \{(12,12 \fmod 5), (13,13 \fmod 5), (34,34\fmod 5)\}\\ \amp =\{(12,2), (13,3), (34,4)\}\text{.} \end{align*}

In Checkpoint 6.2.7 determine all the elements of a Cartesian product that is given in set-builder notation.

Let $$A = \lbrace 5, 6, 7, 8 \rbrace\text{.}$$ Give the set in roster form.

$$\lbrace ( a, 3\cdot a ) \mid a \in A \rbrace =$$$$\lbrace$$$$\rbrace$$

[although it would be mathematically correct to list elements multiple times, this problem is marked wrong if you do so.]

$$\left(5,15\right), \left(6,18\right), \left(7,21\right), \left(8,24\right)$$

To determine when two Cartesian products are equal we need to know when two ordered pairs are equal.

### Definition6.2.8.

Let $$A$$ and $$B$$ be sets. Saying that $$(a,b)\in A\times B$$ and $$(c,d)\in A\times B$$ are equal means that $$a=c$$ and $$b=d\text{.}$$

When $$(a,b)$$ and $$(c,d)$$ are equal, we write $$(a,b)=(c,d)\text{.}$$

When $$(a,b)$$ and $$(c,d)$$ are not equal, we write $$(a,b)\ne(c,d)\text{.}$$

So, two ordered pairs are equal if they have matching first components and matching second components. The fact that the elements $$(a,b)$$ of $$A \times B$$ are called ordered pairs indicates that we must pay attention to order for Cartesian products. In comparison, recall that the order of the elements in a set given in roster form does not matter. (See Example 5.3.8.)

We have the equality of sets $$\{1,2\} = \{2,1\}\text{.}$$ However, as ordered pairs, $$(1,2)\neq(2,1)\text{.}$$

Since the empty set $$\emptyset$$ does not contain any elements, there are no elements to be placed into the second component of the Cartesian product $$A\times \emptyset\text{.}$$ So, we have that $$A\times\emptyset=\emptyset$$ for any set $$A\text{.}$$ Similarly, $$\emptyset \times B = \emptyset$$ for any set $$B\text{.}$$

In Checkpoint 6.2.10 find all elements of a Cartesian product.

Let $$A = \lbrace 5, 6, 7\rbrace$$ and let $$B = \lbrace 1, 2\rbrace\text{.}$$ Give the set in roster form.

$$A \times B = \lbrace$$$$\rbrace$$

$$\left(5,1\right), \left(5,2\right), \left(6,1\right), \left(6,2\right), \left(7,1\right), \left(7,2\right)$$