Question

Is it true that for any set A and B, \[P(A) \cup P(B) = P(A \cup B)\] ? Justify your answer.

Answer 1

Hint:
Take the example: let there be some elements in set A and set B and then calculate the total subsets numbers for both the sets and then calculate the set union of A and B, and calculate their total proper subset. Hence, from there we can see the above answer.

Complete step by step solution:
Let the given condition be \[P(A) \cup P(B) = P(A \cup B)\]
Let the set \[A = \{ 0,1\} \] and set \[B = \{ 1,2\} \]
A union B will contain all the elements that are present in set A and B,
And so, \[A \cup B = \{ 0,1,2\} \]
The power set of a Set A is defined as the set of all subsets of the Set A including the Set itself and the null or empty set. It is denoted by $P(A)$ .
Hence, \[P(A) = \{ \phi ,\{ 0\} ,\{ 1\} ,\{ 0,1\} \} \]
And similarly \[P(B) = \{ \phi ,\{ 1\} ,\{ 2\} ,\{ 1,2\} \} \]
And so \[P(A \cup B) = \{ \phi ,\{ 0\} ,\{ 1\} ,\{ 2\} ,\{ 0,1\} ,\{ 0,2\} ,\{ 1,2\} ,\{ 0,1,2\} \} \]
Now calculating,
\[P(A) \cup P(B) = \{ \phi ,\{ 0\} ,\{ 1\} ,\{ 2\} ,\{ 0,1\} ,\{ 1,2\} \} \]
Hence, we can see that \[P(A) \cup P(B) \ne P(A \cup B)\]

So, the given statement is false.

Note:
In set theory, the power set (or power set) of a Set A is defined as the set of all subsets of the Set A including the Set itself and the null or empty set. It is denoted by $P(A)$. Basically, this set is the combination of all subsets including null set, of a given set.
In set theory, the union of a collection of sets is the set of all elements in the collection. It is one of the fundamental operations through which sets can be combined and related to each other.
Take union of sets without any mistake and also calculate their complete subsets without any error.
Hence we had justified the given statement by taking an example.