Ordered pair

The order of pairing is important. The pair (a,b) and (b,a) are different. This ordering is required as there are real time situations, where order makes a difference. Consider for example, we are required to find the integers which can be formed from two integer subsets like {1,2,3} and {3,4,5}. Clearly, “13” and “31” represent different integers. We need to distinguish them. All pairs formed from two sets should be distinct.

Keeping this restriction in mind, let us work out an example to find ordered pairs formed from elements of two sets.

A = set of first letter of the names of cities = { N , D , H } A = set of first letter of the names of cities = { N , D , H }

B = set of numbers denoting flight numbers = { 001,002,003 } B = set of numbers denoting flight numbers = { 001,002,003 }

All possible ordered pairs formed from two sets are :

N , 001 , N , 002 , N , 003 , D , 001 , D , 002 , D , 003 , H , 001 , H , 002 , H , 003 N , 001 , N , 002 , N , 003 , D , 001 , D , 002 , D , 003 , H , 001 , H , 002 , H , 003

There are all together 9 ordered pairs. From this example, we can deduce a method for writing ordered pairs from two sets. We begin with the first elements of two sets. Progressively, we change the elements from the second set till it is exhausted, while keeping the elements from the first set unchanged. Then, we switch to “next” element from first set and start with “first” from the second set. Again, we change the elements from the second set progressively till it is exhausted, while keeping the elements from the first set unchanged. We continue in this manner till all elements from the first set is also exhausted.

From this discussion, it is also evident that two ordered pairs are equal if and only if the corresponding first and second elements are equal.

Cartesian product

The Cartesian product of two sets is defined in terms of ordered pairs.

Definition 1: Cartesian product

The Cartesian product of two non-empty sets “A” and “B” is the set of all ordered pairs of the elements from two sets.

We should emphasize the use of word “non-empty”, The Cartesian product of a non-empty set with an empty set is equal to empty set.

A × φ = φ A × φ = φ

On the other hand, if one of the sets is infinite, then resulting Cartesian product is also infinite.

We express the Cartesian product set in set building form as :

A × B = { x , y : x ∈ A , y ∈ B } A × B = { x , y : x ∈ A , y ∈ B }

Here, use of "," in the set builder form is equivalent to "and". Therefore, we can write Cartesian product of two sets also as :

A × B = { x , y : x ∈ A and y ∈ B } A × B = { x , y : x ∈ A and y ∈ B }

Further, we can emphasize two ways validity of the conditional statements as in the case of other set operators :

If x , y ∈ A × B ⇔ x ∈ A and y ∈ B If x , y ∈ A × B ⇔ x ∈ A and y ∈ B

Graphical representation

The ordered pairs can be represented in the form of tabular cells or points of intersection of perpendicular lines. The elements of one set are represented as rows, whereas elements of other set are represented as columns. Look at the representation of ordered pairs by points in the figure for the example given earlier.

Figure 1: The elements of one set are represented as rows, whereas elements of other set are represented as columns.

Cartesian product

Note that there are a total of 9 intersection points, corresponding to 9 ordered pairs.

Multiple products

Like other set operations, the product operation can also be applied to a series of sets in sequence. If A 1, A 2, … … . . , A n A 1, A 2, … … . . , A n is a finite family of sets, then their Cartesian product, one after another, is symbolically represented as :

A 1 × A 2 × … … … … … . × A n A 1 × A 2 × … … … … … . × A n

This product is set of group of ordered elements. Each group of ordered elements comprises of “n” elements. This is stated as :

A 1 × A 2 × … × A n = { x 1, x 2, … , x n : x 1 ∈ A 1, x 2 ∈ A 2, … , x n ∈ A n } A 1 × A 2 × … × A n = { x 1, x 2, … , x n : x 1 ∈ A 1, x 2 ∈ A 2, … , x n ∈ A n }

Cartesian Coordinate system

The Cartesian product, consisting of ordered triplets of real numbers, represents Cartesian three dimensional space.

R × R × R = { x , y , z : x , y , z ∈ R } R × R × R = { x , y , z : x , y , z ∈ R }

Each of the elements in the ordered triplet is a coordinate along an axis and each ordered triplet denotes a point in three dimensional coordinate space.

Figure 2: The coordinate of a point is an ordered tripplet.

Cartesian coordinate system

Similarly, the Cartesian product " R × R R × R " consisting of ordered pairs defines a Cartesian plane or Cartesian coordinates of two dimensions. It is for this reason that we call three dimensional rectangular coordinate system as Cartesian coordinate system.

Commutative property of Cartesian product

The Cartesian product is set of ordered pair. Now, the order of elements in the ordered pair depends on the position of sets across product sign. If sets "A" and "B" are unequal and non-empty sets, then :

A × B ≠ B × A A × B ≠ B × A

In general, any operation involving Cartesian product that changes the "order" in the "ordered pair" will yield different result.

However, if "A" and "B" are non-empty, but equal sets, then the significance of the order in the "ordered pair" is lost. We can use this fact to formulate a law to verify "equality of sets". Hence, if sets "A" and "B" are two non-empty sets and

A × B = B × A A × B = B × A

Then,

A = B A = B

It can also be verified that this condition is true other way also. If sets "A" and "B" are equal sets, then A × B = B × A A × B = B × A . The two way conditional statements can be symbolically represented with the help of two ways arrow,

A × B = B × A ⇔ A = B A × B = B × A ⇔ A = B

Distributive property of product operator

The distributive property of product operator holds for other set operators like union, intersection and difference operators. We write equations involving distribution of product operator for each of other operators as :

A × B ∪ C = A × B ∪ A × C A × B ∪ C = A × B ∪ A × C

A × B ∩ C = A × B ∩ A × C A × B ∩ C = A × B ∩ A × C

A × B − C = A × B − A × C A × B − C = A × B − A × C

Here, sets “A”,”B” and “C” are non-empty sets. In order to ascertain distributive property product operator over other set operators we need to check validity of the equations given above.

We can check these relations proceeding from the defining statements. For the time being, we reason that sequence of operation on either side of the equation does not affect the “order” in the “ordered pair”. Hence, distributive property should hold for product operator over three named operators. Let us check this with an example :

A = { a , b } , B = { 1,2 } a n d C = { 2,3 } A = { a , b } , B = { 1,2 } a n d C = { 2,3 }

1: For distribution over union operator

⇒ LHS = A × B ∪ C = { a , b } × { 1,2,3 } ⇒ LHS = A × B ∪ C = { a , b } × { 1,2,3 }

⇒ LHS = { a , 1 , a , 2 , a , 3 , b , 1 , b , 2 , b , 3 } ⇒ LHS = { a , 1 , a , 2 , a , 3 , b , 1 , b , 2 , b , 3 }

Similarly,

⇒ RHS = A × B ∪ A × C = { a , 1 , a , 2 , b , 1 , b , 2 } ∪ { a , 2 , a , 3 , b , 2 , b , 3 } ⇒ RHS = A × B ∪ A × C = { a , 1 , a , 2 , b , 1 , b , 2 } ∪ { a , 2 , a , 3 , b , 2 , b , 3 }

⇒ RHS = { a , 1 , a , 2 , a , 3 , b , 1 , b , 2 , b , 3 } ⇒ RHS = { a , 1 , a , 2 , a , 3 , b , 1 , b , 2 , b , 3 }

Hence,

⇒ A × B ∪ C = A × B ∪ A × C ⇒ A × B ∪ C = A × B ∪ A × C

2: For distribution over intersection operator

⇒ LHS = A × B ∩ C = { a , b } × { 2 } ⇒ LHS = A × B ∩ C = { a , b } × { 2 }

⇒ LHS = { a , 2 , b , 2 } ⇒ LHS = { a , 2 , b , 2 }

Similarly,

⇒ RHS = A × B ∩ A × C = { a , 1 , a , 2 , b , 1 , b , 2 } ∩ { a , 2 , a , 3 , b , 2 , b , 3 } ⇒ RHS = A × B ∩ A × C = { a , 1 , a , 2 , b , 1 , b , 2 } ∩ { a , 2 , a , 3 , b , 2 , b , 3 }

⇒ RHS = { a , 2 , b , 2 } ⇒ RHS = { a , 2 , b , 2 }

Hence,

⇒ A × B ∩ C = A × B ∩ A × C ⇒ A × B ∩ C = A × B ∩ A × C

3: For distribution over difference operator

⇒ LHS = A × B − C = { a , b } × { 1 } ⇒ LHS = A × B − C = { a , b } × { 1 }

⇒ LHS = { a , 1 , b , 1 } ⇒ LHS = { a , 1 , b , 1 }

Similarly,

⇒ RHS = A × B − A × C = { a , 1 , a , 2 , b , 1 , b , 2 } − { a , 2 , a , 3 , b , 2 , b , 3 } ⇒ RHS = A × B − A × C = { a , 1 , a , 2 , b , 1 , b , 2 } − { a , 2 , a , 3 , b , 2 , b , 3 }

⇒ RHS = { a , 1 , b , 1 } ⇒ RHS = { a , 1 , b , 1 }

Hence,

⇒ A × B − C = A × B − A × C ⇒ A × B − C = A × B − A × C