Question

How many elements has P(A), if \[A = \phi \]?

Answer 1

Hint: If we have a set ‘A’ with m elements we have the number of elements it has is \[n(A) = m\]. And Generally, if a set has n elements, the power set will have, \[{2^n}\] elements altogether.
Thus using these results we will solve our given problem.

Complete step by step solution: We know that if A is a set with m elements i.e., \[n(A) = m\],
Now, the power set of a set is the set of all subsets of A. now if a set has \[A = \{ a,b,c\} \] the power set of the set would be, \[P(A) = \{ \phi ,\{ a\} ,\{ b\} ,\{ c\} ,\{ a,b\} ,\{ a,c\} ,\{ b,c\} ,\{ a,b,c\} \} \] so, we have 8 elements in the set of P(A).
Generally, if a set has n elements, the power set will have, \[{2^n}\] elements altogether.
then, \[n\left[ {P\left( A \right)} \right] = {2^m}\], where m is number of elements in set A.
If \[A = \phi \], then the number of elements in the set is 0, so, \[n\left( A \right) = 0\]
\[\therefore n\left[ {P\left( A \right)} \right] = {2^0} = 1\], as \[{a^0} = 1\].

Hence, P(A) has one element.

Note: In mathematics, the power set (or power set) of any set S is the set of all subsets of S, including the empty set and S itself, variously denoted as P(S). P(S), or, identifying the power set of S with the set of all functions from S to a given set of two elements, \[{2^S}\]. The power set is closely related to the binomial theorem. The number of subsets with k elements in the power set of a set with n elements is given by the number of combinations, \[C\left( {n,\;k} \right),\] also called binomial coefficients.