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Properties of Set in Discrete mathematics

The sets can be described as a group of well-defined objects. For example, the example of a set can be described by the set of even numbers which is less than 20. There is one more example of the set, i.e., the set of natural numbers between 1 and 10. If we try to change the order to set’s elements, it will not make any affect on the set. If we replace one or more than one element of a set, in this case, the set will remain the same. Here we will learn about the important properties of sets. The following notation is used to describe the sets:

The above notation can be read as “X is a set which contains three elements a, b, and c”. We enclose the elements of a set with the help of curly braces “{ }”.

To understand the property of sets, we must be aware of the concepts named: union, intersection, null set, universal set. These concepts are described as follows:

Union:

If we have two sets and there is a union operation between them, it means that the resultant set will contain all the elements presented in these two sets. The symbol ‘U’ is used to represent the union. In general terms, the word ‘or’ is used to describe the union. Suppose there are two sets, X and Y. The union of these sets will be represented by set (X U Y) and can be read as X union Y. The resultant set will contain all the elements of set X and set Y. The resultant set will not contain repeated elements.

For example: In this example, we have two sets X and Y, where X = {3, 4, 6, 7}, and Y = {1, 2, 3, 4, 5}. Now have to determine the union of these two sets.

Solution:

Intersection:

If we have two sets and there is an intersection operation between them, it means that the resultant set will contain all the common elements presented in these two sets. The symbol ‘⋂’ is used to represent the intersection. In general terms, the word ‘and’ is used to describe the intersection. Suppose there are two sets, X and Y. The intersection of these sets will be represented by set (X ⋂ Y) and can be read as X intersection Y. The resultant set will contain the common elements of set X and set Y. The resultant set will not contain repeated elements.

For example: In this example, we have two sets X and Y, where X = {3, 4, 6, 7}, and Y = {1, 2, 3, 4, 5}. Now have to determine the intersection of these two sets.

Solution:

Null Set:

The null set is also known as the empty set. It is a type of set which does not contain any element. The symbol ∅ is used to represent the null symbol. So the null set is the only logical way through which a set can have nothing.

Universal set

The universal set is a type of set which contains elements of all the related sets. In the universal set, there is no repetition of an element. Suppose there are two sets X and Y where X is used to have all the even numbers, i.e., X = {2, 4, 6, 8,….}, and Y is used to have all the odd numbers, i.e., Y = {1, 3, 5, 7,…..}. So the universal set will have all the natural numbers, i.e., U = {1, 2, 3, 4, 5, 6, 7, 8,….}. There is one more example of a universal set, which is described as follows:

Suppose there are two sets X and Y where X = {0, 1, 2, 3}, and Y = {0, x, y, z}. Now the universal set which is associated with set X and Y are described as follows:

U = {0, 1, 2, 3, x, y, z}.

Basic properties of sets

Property 1: Commutative properties

According to the commutative property, if we change the order of elements of the sets, then that change does not affect the result of the operation, or the result of the operation is the same. For example: Suppose a set is represented with the help of ‘X’ and the operation is represented with the help of symbol ‘#’. If there are two elements x and y in a set, then the commutative property for the set X with the help of operation # is described as follows:

This expression shows that for set X, set Y, and operation #, the commutative property always holds. That means no matter what elements we pick from X and put it in place of x and y, the result of x # y and the result of y # x will always remain the same. The commutative property is satisfied by both the sets of union and intersection. So

Example: In this example, we have two sets, X and Y, where X = {l, m, n, o, p, q}, and Y = {m, n, o, r, s, t}. Here we have to verify the commutative property of union and intersection for these sets.

Solution:

The given sets X and Y will be known as commutative if and only if X ⋃ Y = Y ⋃ X, and X ⋂ Y = Y ⋂ X.

As we know that the sets X = {l, m, n, o, p, q}, and Y = {m, n, o, r, s, t}.

First, we will verify the commutative property for a union. After that, we will verify it for the intersection.

1. X ⋃ Y = Y ⋃ X.

X ⋃ Y = {l, m, n, o, p, q} ⋃ {m, n, o, r, s, t} = {l, m, n, o, p, q, r, s, t}

Y ⋃ X = {m, n, o, r, s, t} ⋃ {l, m, n, o, p, q} = {l, m, n, o, p, q, r, s, t}

So by seeing the above result, we can say that X ⋃ Y = Y ⋃ X.

2. X ⋂ Y = Y ⋂ X

X ⋂ Y = {l, m, n, o, p, q} ⋂ {m, n, o, r, s, t} = {m, n, o}

Y ⋂ X = {l, m, n, o, p, q} ⋂ {m, n, o, r, s, t} = {m, n, o}

So by seeing the above result, we can say that X ⋂ Y = Y ⋂ X.

Hence, the set X = {l, m, n, o, p, q}, and Y = {m, n, o, r, s, t} both are commutative.

Property 2: Associative property

According to the associative property, if the sets are grouped with the help of parenthesis, then the result will not be affected. This statement specifies that if we change the position of parenthesis in any given expression of sets, and these sets involve union and intersection operation, these changes will not affect the resultant set.

So the associative property is satisfied by both the sets of union and intersection. Suppose there are three finite sets, i.e., X, Y, and Z. These sets will be associative for the union or intersection if it satisfies the following property:

Example: In this example, we have two sets X and Y, and Z where X = {1, 2, 3, 4}, and Y = {3, 4, 5, 6}, and Z = {6, 7, 8}. Here we have to verify the associative property of union and intersection for these sets.

Solution:

The given sets X and Y, and Z will be known as associative if and only if (X ⋂ Y) ⋂ Z = X ⋂ (Y ⋂ Z), and (X ⋃ Y) ⋃ Z = X ⋃ (Y ⋃ Z).

As we know that the sets are X = {1, 2, 3, 4}, and Y = {3, 4, 5, 6}, and Z = {6, 7, 8}

First, we will verify the associative property for the union. After that, we will verify it for the intersection.

1. (X ⋃ Y) ⋃ Z = X ⋃ (Y ⋃ Z)

Here we will first take the left side of the equation and try to solve it like this:

(X ⋃ Y) = {1, 2, 3, 4} ⋃ {3, 4, 5, 6} = {1, 2, 3, 4, 5, 6}

(X ⋃ Y) ⋃ Z = {1, 2, 3, 4, 5, 6} ⋃ {6, 7, 8} = {1, 2, 3, 4, 5, 6, 7, 8}

Now we will take the right side of the equation and try to solve it like this:

(Y ⋃ Z) = {3, 4, 5, 6} ⋃ {6, 7, 8} = {3, 4, 5, 6, 7, 8}

X ⋃ (Y ⋃ Z) = {1, 2, 3, 4} ⋃ {3, 4, 5, 6, 7, 8} = {1, 2, 3, 4, 5, 6, 7, 8}

So by seeing the above result, we can say that (X ⋃ Y) ⋃ Z = X ⋃ (Y ⋃ Z)

2. (X ⋂ Y) ⋂ Z = X ⋂ (Y ⋂ Z)

Here we will first take the left side of the equation and try to solve it like this:

(X ⋂ Y) = {1, 2, 3, 4} ⋂ {3, 4, 5, 6} = {3, 4}

(X ⋂ Y) ⋂ Z = {3, 4} ⋂ {6, 7, 8} = {ϕ}

Now we will take the right side of the equation and try to solve it like this:

(Y ⋂ Z) = {3, 4, 5, 6} ⋂ {6, 7, 8} = {6}

X ⋂ (Y ⋂ Z) = {1, 2, 3, 4} ⋂ {6} = {ϕ}

So by seeing the above result, we can say that (X ⋂ Y) ⋂ Z = X ⋂ (Y ⋂ Z).

Hence, the sets X = {1, 2, 3, 4}, and Y = {3, 4, 5, 6}, and Z = {6, 7, 8} are associative.

Property 3: Distributive Property

According to distributive property, if we are doing a union of a set with the intersection of two other sets, it will generate the same result if we do a union of the original set with both the other two sets separately, and after that, we do the intersection of a result of these separate union.

The same things will happen at the intersection. That means if we are doing intersection of a set with the union of two other sets, it will generate the same result if we do the intersection of an original set with both the other two sets separately, and after that, we do a union of the result of these separate intersection. So the distributive property is satisfied by both the sets of union and intersection. Suppose we have three sets: X, Y, and Z. These sets will be distributive for the union or intersection if it satisfies the following property:

Example: In this example, we have three sets X and Y, and Z where X = {0, 1, 2, 3, 4}, and Y = {1, -2, 3, 4, 5, 6}, and Z = {2, 4, 6, 7}. Here we have to verify the distributive property of union and intersection for these set.

Solution:

The given sets X and Y, and Z will be known as distributive if and only if X ⋃ (Y ⋂ Z) = (X ⋃ Y) ⋂ (X ⋃ Z), and X ⋂ (Y ⋃ Z) = (X ⋂ Y) ⋃ (X ⋂ Z)

As we know that the sets are X = {0, 1, 2, 3, 4}, and Y = {1, -2, 3, 4, 5, 6}, and Z = {2, 4, 6, 7}.

First, we will verify the distributive property for the union. After that, we will verify it for the intersection.

1. X ⋃ (Y ⋂ Z) = (X ⋃ Y) ⋂ (X ⋃ Z)

Here we will first take the left side of the equation and try to solve it like this:

(Y ⋂ Z) = {1, -2, 3, 4, 5, 6} ⋂ {2, 4, 6, 7} = {4, 6}

X ⋃ (Y ⋂ Z) = {0, 1, 2, 3, 4} ⋃ {4, 6} = {0, 1, 2, 3, 4, 6}

Now we will take the right side of the equation and try to solve it like this:

(X ⋃ Y) = {0, 1, 2, 3, 4} ⋃ {1, -2, 3, 4, 5, 6} = {-2, 0, 1, 2, 3, 4, 5, 6}

(X ⋃ Z) = {0, 1, 2, 3, 4} ⋃ {2, 4, 6, 7} = {0, 1, 2, 3, 4, 6, 7}

(X ⋃ Y) ⋂ (X ⋃ Z) = {-2, 0, 1, 2, 3, 4, 5, 6} ⋂ {0, 1, 2, 3, 4, 6, 7} = {0, 1, 2, 3, 4, 6}

So by seeing the above result, we can say that X ⋃ (Y ⋂ Z) = (X ⋃ Y) ⋂ (X ⋃ Z)

2. X ⋂ (Y ⋃ Z) = (X ⋂ Y) ⋃ (X ⋂ Z)

Here we will first take the left side of the equation and try to solve it like this:

(Y ⋃ Z) = {1, -2, 3, 4, 5, 6} ⋃ {2, 4, 6, 7} = {-2, 1, 2, 3, 4, 5, 6, 7}

X ⋂ (Y ⋃ Z) = {0, 1, 2, 3, 4} ⋂ {-2, 1, 2, 3, 4, 5, 6, 7} = {1, 2, 3, 4}

Now we will take the right side of the equation and try to solve it like this:

(X ⋂ Y) = {0, 1, 2, 3, 4} ⋂ {1, -2, 3, 4, 5, 6} = {1, 2, 3, 4}

(X ⋂ Z) = {0, 1, 2, 3, 4} ⋂ {2, 4, 6, 7} = {2, 4}

(X ⋂ Y) ⋃ (X ⋂ Z) = {1, 2, 3, 4} ⋃ {2, 4} = {1, 2, 3, 4}

So by seeing the above result, we can say that X ⋂ (Y ⋃ Z) = (X ⋂ Y) ⋃ (X ⋂ Z)

Hence, the sets X = {0, 1, 2, 3, 4}, and Y = {1, -2, 3, 4, 5, 6}, and Z = {2, 4, 6, 7} are distributive.

Property 4: Identity Property

In the identity property, 0 and 1 are used for addition and multiplication. Same as ∅ is used for the union, and U is used for the intersection. If there is a union of any set with the null set, then the result will be the set itself. The null set or empty set will not contain any element.

Same as if there is an intersection of any set with the universal set, then the result will be the set itself. So, the identity property is satisfied by both the sets of union and intersection. Suppose there is a set X. This set will be identity for the union or intersection if it satisfies the following property:

Example: In this example, we have a set X where X = {1, 2, 3, 4}. Here we have to verify the identity property of union and intersection for this set.

Solution:

The given sets X will be known as the identity if and only if X ⋃ ∅ = X, and X ⋂ U = X.

As we know that the set X = {1, 2, 3, 4}. So, U = {1, 2, 3, 4}, and ∅ = {}.

First, we will verify the identity property for the union. After that, we will verify it for the intersection.

1. X ⋃ ∅ = X

X ⋃ ∅ = {1, 2, 3, 4} ⋃ {}

= {1, 2, 3, 4} = X

So by seeing the above result, we can say that X ⋃ ∅ = X

2. X ⋂ U = X

X ⋂ U = {1, 2, 3, 4} ⋂ {1, 2, 3, 4}

= {1, 2, 3, 4}

So by seeing the above result, we can say that X ⋂ U = X.

Hence, the set X = {1, 2, 3, 4} has identity.

Property 5: Complement Property

According to the complement property, if there is a union of a set with the complement of same set, then the result will be the universal set. The universal set is a type of set which contains elements of all the related sets. In the universal set, there is no repetition of an element.

The result will be different in the case of intersections. If there is an intersection of a set with the complement of same set, then the result will be an empty set. So, the complement property is satisfied by both the sets of union and intersection. Suppose there is a set X and its complement XC. This set will be known as a complement for the union or intersection if it satisfies the following property:

Example: In this example, we have a set X where X = {1, 2, 3}. Here we have to verify the complement property of union and intersection for this set.

Solution:

The given sets X will be known as a complement if and only if X ⋃ X^C = U, and X ⋂ X^C = ∅.

As we know that the set X = {1, 2, 3}. So, U = {1, 2, 3, 4, 5}, and X^C = {4, 5}.

First, we will verify the complement property for the union. After that, we will verify it for the intersection.

1. X ⋃ X^C = U

X ⋃ X^C = {1, 2, 3} ⋃ {4, 5}

= {1, 2, 3, 4, 5} = U

So by seeing the above result, we can say that X ⋃ X = U.

2. X ⋂ X^C = ∅

X ⋂ XC = {1, 2, 3} ⋂ {4, 5}

= ∅

So by seeing the above result, we can say that X ⋂ X = ∅.

Hence, the set X = {1, 2, 3} is complement.

Property 6: Idempotent Property

According to the idempotent property, if there is a union or intersection of any set with the same set, then the result will be the set itself. So the idempotent property is satisfied by both the sets of union and intersection. Suppose there is a set X. This set will be idempotent for the union or intersection if it satisfies the following property:

So we can say that if two identical sets contain the same elements, then the result will be the original elements of the set.

Example: In this example, we have a set X where X = {1, 2, 3, 4}. Here we have to verify the idempotent property of union and intersection for this set.

Solution:

The given sets X will be known as idempotent if and only if X ⋂ X = X, and X ⋃ X = X.

As we know that the set X = {1, 2, 3, 4}

First, we will verify the idempotent property for the union. After that, we will verify it for the intersection.

1. X ⋃ X = X

X ⋃ X = {1, 2, 3, 4} ⋃ {1, 2, 3, 4}

= {1, 2, 3, 4} = X

So by seeing the above result, we can say that X ⋃ X = X.

2. X ⋂ X = X

X ⋂ X = {1, 2, 3, 4} ⋂ {1, 2, 3, 4}

= {1, 2, 3, 4} = X

So by seeing the above result, we can say that X ⋂ X = X.

Hence, the set X = {1, 2, 3, 4} is idempotent.

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