Write down the power set of A = {3, {4, 5}}.

Hint: First, we will see how many elements the original set has and then use that in the formula for the number of elements in the power set to find out the number of elements in the final set. After that, we know that $\varphi $ will be one of the elements and so on.

Complete step-by-step answer:
We see that we are provided with the set A = {3, {4, 5}}.
We can clearly observe that A consists of two elements which are 3 and {4, 5}.
Now, we know that if we have a set X with n elements, then P(X) has ${2^n}$ elements.
Now, put in n = 2 for the given set.
We have with us: n(A) = 2.
$\therefore n(P(A)) = {2^2} = 2 \times 2 = 4$.
$\therefore $ The final set must have 4 elements.
Now, we know that the $A \subset P(A)$.
\[ \Rightarrow P\left( A \right) = \{ \varphi ,3,\{ 4,5\} ,\{ 3,\{ 4,5\} \} \} \].

$\therefore $ P(A) basically consists of four elements namely $\varphi $ , 3, {4, 5} and {3, {4, 5}}.

Note: The students must wonder why it is important to calculate the number of elements in the set P(A). It just gives us an indication to when to stop thinking of anymore elements. For example: if the original set has 4 or more elements, we may tend to forget the elements in the power set because there are too many. So, to keep a check on whether we are right or incomplete, we will have an easier way to know so.
A power set of A that is P(A) is the set of all the possible subsets of A. So, we have clubbed in basically all the subsets of A together in one set.
The students must notice that we used the symbol $\varphi $ which is called phi, we may use { } instead of that because both the symbols represent empty or null.
Let us ponder over, why a power set has ${2^n}$ elements. Basically every element has a choice whether to be in an element in the power set or not. So, every element of the set gets two choices, thus resulting in ${2^n}$.