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).

## leave a comment

## 0 Comments