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Find the power set of A = { \[\phi \], { \[\phi \] } }.
(a) A
(b) { \[\phi \], { \[\phi \] }, A}
(c) { \[\phi \], { \[\phi \] }{ \[\phi \] }, A }
(d) None of these

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Last updated date: 22nd Jun 2024
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Answer
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Hint: To solve this problem we have to first know what is the power set. Power set is a set of all the subsets of the given set. So using this definition we will find all the subsets of the given set A and then combine them to form a set and that set will be the power set of A. After that we will match out which option satisfies our answer and find the correct answer.

Complete step by step answer:
We are given a set A = { \[\phi \], { \[\phi \] } },
Now we have to find the power set of this set A,
But before that we need to know that what do we mean by power set,
According to the definition of the power set, it is a set of all the subsets of the given set,
For example consider we have a set S = { a, b, c }
Now all the subsets of this set will be,
{}, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}
There are total 8 subsets hence power set of set S, P(S) is given as,
P(S) = { {}, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c} }
Now coming to the given question we have a set A given as,
A = { \[\phi \], { \[\phi \] } },
All the subsets of the A are given as,
\[\phi \], { \[\phi \] } , {{ \[\phi \] }}, { \[\phi \], { \[\phi \] } },
Hence we get power set of A, P(A) as,
P(A) = { \[\phi \], { \[\phi \] } , {{ \[\phi \] }}, { \[\phi \], { \[\phi \] } } }
We will substitute { \[\phi \], { \[\phi \] } } equal to A as options given have set A in it, so we get
P(A) = { \[\phi \], { \[\phi \] } , {{ \[\phi \] }}, A }

So, the correct answer is “Option D”.

Note: To solve this problem you should have prior knowledge about the power set otherwise you would not be able to solve this question. You can also note that if the given set has n elements in it then its power set will have ${{2}^{n}}$ elements in it. Using this you can eliminate the options which have more or less number of elements than ${{2}^{n}}$ and solve the question more easily.