# Power Set

In set theory, a power set is a set that contains all subsets of a given set, including the empty set and the set itself. Thus, a power set is also called a set of subjects, containing more elements than the original sets.

## Symbol

If S is a set, then the power set of set S is denoted by the notation P(S) or ${\mathcal{P}(S)}$ and expressed as:

P(S) = {x | x âŠ† S}

## Example

Let S = {a, b, c} be a given set.

Listing down the subsets of set S, we get:

{ }, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}

Thus, the power set of S; P(S) = {{ }, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}}

## Cardinality

If a set has â€˜nâ€™ elements, the total number of subsets for the set is given by 2n. Since a power set has all the subsets of the set, the cardinality of the power set is written by the notation:

|P(A)| = 2n, where n = the cardinality of the given set.

Again, considering the set S = {a, b, c}

Here, n = the cardinality of set S = |S| = 3

Thus, |P(S)| = 2n = 23 = 8.

Now, let us verify by listing down all the subsets of set S.

We already obtain P(s) = {{ }, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}}, which has 8 subsets.

Thus, |P(A)| = 8, verified.

### Proof

We use mathematical induction to prove the cardinality of a power set.

Base Step

Let us consider an empty set S; S = É¸ = { }

Here, the power set of S, P(S) = {É¸}

Also, the cardinality of the power set S = |P(S)| = 1 = 20

Inductive Step

Let S be a set with n number of elements, S = {x1, x2, â€¦, xn}, and |P(S)| = 2n

Now, considering another set T = {x1, x2, â€¦, xn, xn + 1}

The cardinality of the two sets S and T are |S| = n and |T| = n + 1

Now, T = A âˆª {an + 1}

Here, we conclude that every subset of set S is also a subset of set T, which means a subset of set T may or may not contain the element an+1

If the subset of set T does not contain the element an + 1, it is an element of set S.

Also, if the subset of T contains the element an + 1, then the element an + 1 is included in any of the 2n subsets of set S. Here, we conclude that set T has 2n subsets with the element an + 1

Thus, set T has 2n + 1 subsets with element an + 1 and 2n subsets without the element an + 1

Hence, the cardinality of the power set is proved.

## Power Set of Empty Set

An empty or null set has zero elements. It contains only the set itself as its subset. Thus, the power set of an empty set É¸ = { } has 20 = 1 element.

P(É¸) = {É¸} = {{ }}

## Power Set of Finite Set

The power set of a finite set is countable; thus, it is a finite set.

For example,

In A = {a, b, c}, the power sets are countable.

## Power Set of Infinite Set

An infinite set has an infinite number of elements. Thus, the power set of an infinite set has infinite subsets.

For example,

A set of integers has an infinite number of elements. In this case, the power set contains an infinite number of subsets.

## Solved Examples

If D = {5, 9, 13, 17}, find the power set and its cardinality.

Solution:

Given, D = {5, 9, 13, 17}, which has 4 elements.
Here, the cardinality of the P(D) = 24 = 16
The subsets of set D are É¸, {5}, {9}, {13}, {17}, {5, 9}, {5, 13}, {5, 17}, {9, 13}, {9, 17}, {13, 17}, {5, 9, 13}, {5, 9, 17}, {5, 13, 17}, {9, 13, 17}, {5, 9, 13, 17}
Thus, the power set D = {É¸, {5}, {9}, {13}, {17}, {5, 9}, {5, 13}, {5, 17}, {9, 13}, {9, 17}, {13, 17}, {5, 9, 13}, {5, 9, 17}, {5, 13, 17}, {9, 13, 17}, {5, 9, 13, 17}} and its cardinality is |P(D)| = 16

If A = {9, 8, 7, 6, u, v, w, x, y}, how many elements will its power set have?

Solution:

Given set A = {9, 8, 7, 6, u, v, w, x, y}, which has 9 elements.
Thus, its power set has 29 = 512 elements.

Determine the power set of M = {Monday, Friday}

Solution:

Given set M = {Monday, Friday}
The subsets of M are É¸, {Monday}, {Friday}, {Monday, Friday}
Thus, the power set of set M is P(M) = {É¸, {Monday}, {Friday}, {Monday, Friday}}