Question

Is it true that for any sets A & B, P(A) ∪ P(B) = P(A∪B)? Justify your answer.

Answer 1

Hint: In the above problem, P (A) represents the power set of A. Let us assume that n (A) = {0, 1}, n (B) = {1, 2} then n (A∪B) = {0, 1, 2}. P (A) = {∅, {0}, {1}, {0, 1}}. Similarly write P(B), P(A∪B) then put these values in this expression P(A) ∪ P(B) = P(A∪B) and see whether it is holding true or not.

Complete step-by-step answer:
Let us assume that n (A) = {0, 1}, n (B) = {1, 2} and n (A∪B) = {0, 1, 2} where “n” represents number of elements.

Power set of A is written as P (A) and is equal to {∅, {0}, {1}, {0, 1}}. Similarly, P (B) = {∅, {1},
{2}, {1, 2}} and P (A∪B) = {∅, {0}, {1}, {2}, {0, 1}, {1, 2}, {0, 2}, {0, 1, 2}}.

Now, finding P (A) ∪ P (B) we get:

P (A) = {∅, {0}, {1}, {0, 1}}

P (B) = {∅, {1}, {2}, {1, 2}}

P (A) ∪ P (B) = {∅, {0}, {1}, {2}, {0, 1}, {1, 2}}

And P (A∪B) = {∅, {0}, {1}, {2}, {0, 1}, {1, 2}, {0, 2}, {0, 1, 2}}.

Now, on comparing P (A) ∪ P (B) and P (A∪B) from above we can see that P (A∪B) has two extra sets ({0, 2} and {0, 1, 2} as compared to P (A) ∪ P (B).

So, from the above discussion, we can say that P (A) ∪ P (B) ≠ P (A∪B).

Hence, it is not true that P (A) ∪ P (B) = P (A∪B).

Note: As we have written, the power set in the above solution. So, in the below we are going to show how to write a power set.

n (A) = {0, 1}

Power set of any set contains all the possible subsets.

We are showing the subsets of set A or n (A) = {0, 1}

The subsets consists of a null set {∅}, subsets comprising elements taking one at a time {0},

{1} and subsets comprising elements taking two at a time {0, 1}.

So, P (A) = {∅, {0}, {1}, {0, 1}}.

Similarly, you can find the power set of all the sets.