Prerequisite – Introduction of Set theory, Set Operations (Set theory)

For a given set S, **Power set** P(S) or 2^S represents the set containing all possible subsets of S as its elements. For example,

S = {1, 2, 3}

P(S) = {ɸ, {1}, {2}, {3} {1,2}, {1,3}, {2,3}, {1,2,3}}

**Number of Elements in Power Set –**

For a given set S with n elements, number of elements in P(S) is 2^n. As each element has two possibilities (present or absent}, possible subsets are 2×2×2.. n times = 2^n. Therefore, power set contains 2^n elements.

**Note –**

- Power set of a finite set is finite.
- Set S is an element of power set of S which can be written as S ɛ P(S).
- Empty Set ɸ is an element of power set of S which can be written as ɸ ɛ P(S).
- Empty set ɸ is subset of power set of S which can be written as ɸ ⊂ P(S).

Let us discuss the questions based on power set.

Q1. The cardinality of the power set of {0, 1, 2 . . ., 10} is _________.

(A) 1024

(B) 1023

(C) 2048

(D) 2043

Solution: The cardinality of a set is the number of elements contained. For a set S with n elements, its power set contains 2^n elements. For n = 11, size of power set is 2^11 = 2048.

Q2. For a set A, the power set of A is denoted by 2^A. If A = {5, {6}, {7}}, which of the following options are True.

I.Φ ϵ 2 AII.Φ⊆ 2 AIII.{5,{6}} ϵ 2 AIV.{5,{6}} ⊆ 2 A

(A) I and III only

(B) II and III only

(C) I, II and III only

(D) I, II and IV only

Explanation: The set A has 5, {6}, {7} has its elements. Therefore, the power set of A is:

2^S = {ɸ, {5}, {{6}}, {{7}}, {5,{6}}, {5,{7}}, {{6},{7}}, {5,{6},{7}} }

Statement I is true as we can see ɸ is an element of 2^S.

Statement II is true as empty set ɸ is subset of every set.

Statement III is true as {5,{6}} is an element of 2^S.

However, statement IV is not true as 5,{6}} is an element of 2^S not subset.

Therefore, correct option is (C).

Q3. Let P(S) denotes the power set of set S. Which of the following is always true?

(a)P(P(S))=P(S)(b)P(S) ∩ P(P(S)) = { Φ }(c)P(S) ∩ S = P(S)(d)S ∉ P(S)

(A) a

(B) b

(C) c

(D) d

Solution: Let us assume set S ={1, 2}. Therefore P(S) = { ɸ, {1}, {2}, {1,2}}

Option (a) is false as P(S) has 2^2 = 4 elements and P(P(S)) has 2^4 = 16 elements and they are not equivalent.

Option (b) is true as intersection of S and P(S) is empty set.

Option (c) is false as intersection of S and P(S) is empty set.

Option (d) is false as S is an element of P(S).

**Countable set and its power set –**

A set is called countable when its element can be counted. A countable set can be finite or infinite.

For example, set S1 = {a, e, i, o, u} representing vowels is a countably finite set. However, S2 = {1, 2, 3……} representing set of natural numbers is a countably infinite set.

**Note –**

- Power set of countably finite set is finite and hence countable.

For example, set S1 representing vowels has 5 elements and its power set contains 2^5 = 32 elements. Therefore, it is finite and hence countable. - Power set of countably infinite set is uncountable.

For example, set S2 representing set of natural numbers is countably infinite. However, its power set is uncountable.

**Uncountable set and its power set –**

A set is called uncountable when its element can’t be counted. An uncountable set can be always infinite.

For example, set S3 containing all fractional numbers between 1 and 10 is uncountable.

**Note –**

- Power set of uncountable set is always uncountable.

For example, set S3 representing all fractional numbers between 1 and 10 is uncountable. Therefore, power set of uncountable set is also uncountable.

Let us discuss gate questions on this.

Q4. Let ∑ be a finite non-empty alphabet and let 2^∑* be the power set of ∑*. Which of the following is true?

(A).Both 2^∑* and ∑* are countable(B).2^∑* is countable and ∑* is uncountable(C).2^∑* is uncountable and ∑* is countable(D).Both 2^∑* and ∑* are uncountable

Solution: Let ∑ = {a, b}

then ∑* = { ε, a, b, aa, ba, bb, ……………….}.

As we can see, ∑* is countably infinite and hence countable. But power set of countably infinite set is uncountable.

Therefore, 2^∑* is uncountable. So, the correct option is (C).

This article is contributed by **Sonal Tuteja**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.