# Superset, Power Set, Universal Set

**All Subsets of a Set**

The subsets of any set consists of all possible sets including its elements and the null set.

Let us understand with the help of an example.

**Example:** Find all the subsets of set A = {1,2,3,4}

Solution: Given, A = {1,2,3,4}

Subsets = {}

{1}, {2}, {3}, {4},

{1,2}, {1,3}, {1,4}, {2,3},{2,4}, {3,4},

{1,2,3}, {2,3,4}, {1,3,4}, {1,2,4}

{1,2,3,4}

**Power Set**

The power set is the set of all subsets that can be created from a given set. The cardinality of

the power set is 2 to the power of the given set’s cardinality.

Notation: P (set name)**Example**: A = {a, b, c} where |A| = 3

P (A) = {{a, b}, {a, c}, {b, c}, {a}, {b}, {c}, {a, b, c},{} } and |P (A)| = 8

Note: In general, if |A| = n, then |P (A) | = 2^{n}

**Superset **

If A ⊂ B and A ≠ B, this means that A is a proper subset of B. And is known as the superset of set A. For e.g. all natural numbers are integers. If N and Z represent the set of all the natural numbers and integers, respectively, then we can write that N ⊂ Z

Here, N is a proper subset of Z and Z is called the superset of N

### Universal Set

A set which contains all the sets relevant to a certain condition is called the universal set. It is the set of all possible values.

**Example:** If A = {1,2,3} and B {2,3,4,5}, then universal set here will be:

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

Let P be the set of prime numbers and let S = {t | 2^{t} – 1 is a prime}.

Prove that S ⊂ P.

Solution Now the equivalent contrapositive statement of x ∈ S ⇒ x ∈ P is x ∉ P ⇒

x ∉ S.

Now, we will prove the above contrapositive statement by contradiction method

Let x ∉ P

⇒ x is a composite number

Let us now assume that x ∈ S

⇒ 2^{x} – 1 = m (where m is a prime number)

⇒ 2^{x} = m + 1

Which is not true for all composite number, say for x = 4 because

2^{4} = 16 which can not be equal to the sum of any prime number m and 1.

Thus, we arrive at a contradiction

⇒ x ∉ S.

Thus, when x ∉ P, we arrive at x ∉ S

So S ⊂ P.