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Last modified on June 27th, 2024

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.

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}**

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}}**

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

**|P(A)| = 2**** ^{n}**, 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)| = 2^{n} = 2^{3} = 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.

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 = 2^{0}

**Inductive Step**

Let S be a set with n number of elements, S = {x_{1}, x_{2}, â€¦, x_{n}}, and |P(S)| = 2^{n}

Now, considering another set T = {x_{1}, x_{2}, â€¦, x_{n}, x_{n + 1}}

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

Now, T = A âˆª {a_{n + 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 a_{n+1}

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

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

Thus, set T has 2^{n + 1} subsets with element a_{n + 1} and 2^{n} subsets without the element a_{n + 1}

Hence, the cardinality of the power set is proved.

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 2^{0} = 1 element.

P(É¸) = {É¸} = {{ }}

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.

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.

**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) = 2^{4} = 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 2^{9} = 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}}

Last modified on June 27th, 2024