Consider a universal set consisting of all the integers

N = {0, 1, -1, 2, -2, 3, -3, …}.

The following are all examples of ordered pairs:

(1, -1)	(0, 12)	(12, 0)
(5, 12)	(3, 4)	(-11, -13)
(-10, 100)	(0, 12)	(0, -9)

Let’s refer to the upper-middle ordered pair (0, 12) as (a, b). We then have

a = 0 and b = 12.

Furthermore, let’s refer to the lower-middle ordered pair (0, 12) as (c, d), then

c = 0 and d = 12.

Based on the above definition of equal, since we have that

a	=	c	=	0
b	=	d	=	12

We have that (0, 12) = (0, 12). This might be obvious, but we also wanted to demonstrate using the actual definition, by comparing each element in order for equality.

Now let’s compare the ordered pairs (0, 12) and (0, -9). First, let’s use (a, b) and (c, d) to refer to these ordered pairs as

(0, 12)	=	(a, b)
(0, -9)	=	(c, d)

We have that

a = c = 0

but

b = 12 and d = -9.

Since 12 ≠ -9, we also have that b ≠ d. And since b ≠ d, we have that

(0, 12) ≠ (0, -9).

For similar reasons, we also have that

(0, 12) ≠ (12, 0).

None of the following n-tuples are equal to each other:

(1, 2, 3)	(2, 1, 3)	(3, 1, 2)
(1, 3, 2)	(2, 3, 1)	(3, 2, 1)

However, all of the following sets are equal to each other:

{1, 2, 3}	{2, 1, 3}	{3, 1, 2}
{1, 3, 2}	{2, 3, 1}	{3, 2, 1}

The following n-tuples are not equal:

(1, 2, 3)

(1, 1, 2, 3)

Even though the elements are the same, and the 1s appear before the 2, which appears before the 3, there are a different number of 1s. The first n-tuple has only one 1, whereas the second n-tuple has two 1s. Remember, since order matters, repetition will also matter, and the order in which repeated elements appear also matters.

Remember, equality of n-tuples depends on having the same number of elements, and having equal corresponding elements.

Suppose we have

A = {1, 2}

B = {a, b, c}

where a, b, and c are not variables, but simply the first, second, and third letters of the English alphabet. Then we have that

A ✕ B = { (1, a), (1, b), (1, c), (2, a), (2, b), (2, c) }.

Furthermore, we have that

B ✕ A = { (a, 1), (a, 2), (b, 1), (b, 2), (c, 1), (c, 2) }.

Often times, we may want to stylize how we write out the sets in order to make them easier to visually parse through. As such, we can rewrite both Cartesian Products like so:

Consider the following sets:

A = {1, 2}

B = {a, b}

C = {α, β}

Then we have that

A ✕ B ✕ C = { (1, a, α), (1, a, β), (1, b, α), (1, b, β), (2, a, α), (2, a, β), (2, b, α), (2, b, β) }

Consider the set

Z = {1, a, α}.

Then we have that

Z ✕ Z = { (1, 1), (1, a), (1, α), (a, 1), (a, a), (a, α), (α, 1), (α, a), (α, α) }

Consider the set

A = {1, a}.

We have that

A3

=

A ✕ A ✕ A

=

{(1, 1, 1), (1, 1, a), (1, a, 1), (1, a, a), (a, 1, 1), (a, 1, a), (a, a, 1), (a, a, a)}

Reconsider the sets

A = {1, 2}

B = {a, b}

C = {α, β}

From Example 4.1.4, we know that

A ✕ B ✕ C = { (1, a, α), (1, a, β), (1, b, α), (1, b, β), (2, a, α), (2, a, β), (2, b, α), (2, b, β) }.

To evaluate (A ✕ B) ✕ C, let’s compute A ✕ B first, since it is the part in parenthesis:

A ✕ B = { (1, a), (1, b), (2, a), (2, b) }.

Now, to compute (A ✕ B) ✕ C, recognize that it is the same as computing { (1, a), (1, b), (2, a), (2, b) } ✕ {α, β}:

(A ✕ B) ✕ C	=	{ (1, a), (1, b), (2, a), (2, b) } ✕ {α, β}
	=	{ ( (1, a), α), ( (1, a), β), ( (1, b), α), ( (1, b), β), ( (2, a), α), ( (2, a), β), ( (2, b), α), ( (2, b), β) }

Notice that each ordered pair in the set (A ✕ B) ✕ C contains an ordered pair as an element. Just as sets can contain other sets as elements, ordered pairs can contain other ordered pairs as elements as well. Let’s rewrite the set (A ✕ B) ✕ C to make the elements even more obvious:

{	(	(1, a)	,	α	)
	(	(1, a)	,	β	)
	(	(1, b)	,	α	)
	(	(1, b)	,	β	)
	(	(2, a)	,	α	)
	(	(2, a)	,	β	)
	(	(2, b)	,	α	)
	(	(2, b)	,	β	)	}

Going further, we can even construct sets that contain ordered pairs as elements, and we can construct ordered pairs that contain sets as elements.

Of course, the same idea applies when computing A ✕ (B ✕ C):

A ✕ (B ✕ C)	=	{1, 2} ✕ { (a, α), (a, β), (b, α), (b, β) }
	=	{ (1, (a, α)), (1, (a, β)), (1, (b, α)), (1, (b, β)), (2, (a, α)), (2, (a, β)), (2, (b, α)), (2, (b, β)) }

There are some clear similarities between A ✕ B ✕ C, (A ✕ B) ✕ C, and A ✕ (B ✕ C). One similarity is that if removed the inner parenthesis, all of these sets would be equal to each other. Of course when parenthesized, the inner ordered pair may be the first or second element within the outer ordered pairs.

Another difference worth pointing out is that A ✕ B ✕ C is a set of ordered triples. The sets (A ✕ B) ✕ C and A ✕ (B ✕ C) are sets of ordered pairs.

Suppose we had the sets

Ψ = { {1, 2}, (3, 4) }

Ω = { {a, b}, (c, d) }

One way we could compute this Cartesian Product is to tabulate everything:

Ψ ✕ Ω
	{a, b}	(c, d)
{1, 2}	( {1, 2}, {a, b} )	( {1, 2}, (c, d) )
(3, 4)	( (3, 4), {a, b} )	( (3, 4), (c, d) )

So then we have that

Ψ ✕ Ω = { ( {1, 2}, {a, b} ), ( {1, 2}, (c, d) ), ( (3, 4), {a, b} ), ( (3, 4), (c, d) ) }.

It can be a little confusing to deal with all of the parenthesis () and braces {}. Another way to compute the above Cartesian Product is to replace each of {1, 2}, (3, 4), {a, b}, (c, d) with some symbol of our choosing. We can create any arbitrary symbol using our imagination, or use existing symbols, but we must remain consistent. Let’s use the following symbols simply for the sake of variety:

★	=	{1, 2}
✦	=	(3, 4)
♫	=	{a, b}
♥	=	(c, d)

Now we can simply use those simpler symbols in place of the sets and ordered pairs originally used:

Ψ ✕ Ω	=	{ ★, ✦ } ✕ { ♫, ♥ }
	=	{ (★, ♫), (★, ♥), (✦, ♫), (✦, ♥) }

Even though it would be technically correct to leave the above answer as is (because it is known what ★, ✦, ♫, ♥ represent), it’s good etiquette to replace all of those symbols with their original counterparts:

Either way, we get the set

Ψ ✕ Ω = { ( {1, 2}, {a, b} ), ( {1, 2}, (c, d) ), ( (3, 4), {a, b} ), ( (3, 4), (c, d) ) }.