Cartesian Products

Definition. Let S and T be sets. The Cartesian product of S and T is the set $S \times T$ consisting of all ordered pairs $(s, t)$ , where $s
   \in S$ and $t \in T$ .

Ordered pairs are characterized by the following property: $(a, b) =
   (c, d)$ if and only if

$$a = c \quad\hbox{and}\quad b = d.$$

Remarks. (a) $S \times T$ is not the same as $T \times S$ unless $S = T$ .

(b) You can define an ordered pair using sets. For example, the ordered pair $(x, y)$ can be defined as the set $\{x, \{x, y\}\}$ .


Example. Let $S = \{a, b, c\}$ and $T = \{1, 2\}$ . List the elements of $S \times T$ and sketch the set.

$$S \times T = \{(a, 1), (a, 2), (b,1 ), (b, 2), (c, 1), (c, 2)\}.$$

Notice that S and T are not subsets of $S \times T$ . There are subset which "look like" S and T; for example, here's a subset that "looks like" S:

$$U = \{(a, 1), (b, 1), (c, 1)\}.$$

But this is not S: The elements of S are a, b, and c, whereas the elements of the subset U are pairs.

Here's a picture of $S
   \times T$ . The elements are points in the grid:

$$\hbox{\epsfysize=1.75 in \epsffile{cartesian-products-1.eps}}\quad\halmos$$


$\real \times \real$ consists of all pairs $(x, y)$ , where $x,
   y \in \real$ . This is the same thing as the the x-y-plane:

$$\hbox{\epsfysize=1.75 in \epsffile{cartesian-products-2.eps}}\quad\halmos$$

Example. Consider the following subset of $\real \times \real$ :

$$S = \{(2 x, 5 x) \mid x \in \real\}.$$

(a) Prove that $(-14, -35)
   \in S$ .

(b) Prove that $(18, 50)
   \notin S$ .

(a)

$$(-14, -35) = (2 \cdot (-7), 5 \cdot (-7)) \in S.\quad\halmos$$

(b) Suppose $(18, 50) \in
   S$ . Then for some $x \in \real$ , I have

$$(18, 50) = (2 x, 5 x).$$

Equating the first components, I get $2 x = 18$ , so $x =
   9$ . But equating the second components, I get $5 x =
   50$ , so $x = 10$ . This is a contradiction, so $(18, 50) \notin S$ .


Example. $\integer \times \integer$ is the set of pairs $(m,
   n)$ of integers. Consider the following subsets of $\integer \times \integer$ :

$$A = \{(3 n + 5, 9 n + 10) \mid n \in \integer\} \quad\hbox{and}\quad B = \{(s, t) in \integer \times \integer \mid s + t \quad\hbox{is odd}\}.$$

Let $(3 n + 5, 9 n + 10) \in
   A$ . Then

$$(3 n + 5) + (9 n + 10) = 12 n + 15 = 2(6 n + 7) + 1.$$


You can take the product of more than 2 sets --- even an infinite number of sets, though I won't consider infinite products here.

For example, $\integer
   \times \integer \times \integer$ consists of ordered triples $(a, b, c)$ , where a, b, and c are integers.

Example. Consider the following subset of $\integer \times \integer \times
   \integer$ :

$$W = \{(a, b, a + 2 b) \mid a, b \in integer\}.$$

(a) Show that $(-3, 5, 7)
   \in W$ .

(b) Show that $(2, -4, 6)
   \notin W$ .

(a)

$$(-3, 5, 7) = (-3, 5, -3 + 2 \cdot 5) \in W.\quad\halmos$$

(b) Suppose $(2, -4, 6) \in
   W$ . Then for some integers a and b, I have

$$(2, -4, 6) = (a, b, a + 2 b).$$

Equating components, I get three equations:

$$a = 2, \quad b = -4, \quad a + 2 b = 6.$$

But substituting $a = 2$ and $b = -4$ into $a + 2
   b$ gives

$$a + 2 b = 2 + 2 \cdot (-4) = -6 \ne 6.$$

This contradiction proves that $(2, -4, 6) \notin W$ .


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