The existence of the empty set ∅\emptyset means that any operation defined on sets must take it into account. Surprisingly, this simplifies the description of a set's power set. To begin with, the only subset of the empty set is… the empty set itself. The idea is then to describe what happens when an element x is added to a set E to form a set F=E∪{x}F=E\cup\{x\}, assuming that we already know the power set of E, namely P(E)\mathcal{P}(E). If a subset of F contains x, it is obtained by adding x to an element of P(E)\mathcal{P}(E); otherwise, it is an element of P(E)\mathcal{P}(E). This describes every subset of F. Adding one element to E therefore doubles the number of its subsets. Since the empty set has only one subset, induction shows that if E has a finite number n of elements, then P(E)\mathcal{P}(E) has 2n elements. This method also lets us find the power set of any finite set. Let's see how it works in a simple case.
To obtain the power set of {1, 2, 3}, we start with P(∅)=∅\mathcal{P}(\emptyset)=\emptyset and then consider {1}=∅∪{1}\{1\}=\emptyset\cup\{1\}, which gives
P({1})={∅,{1}}.\mathcal{P}(\{1\})=\{\emptyset,\{1\}\}.
Next, we consider {1,2}={1}∪{2}\{1,2\}=\{1\}\cup\{2\}, which gives
P({1,2})={∅,{1},{2},{1,2}}\mathcal{P}(\{1,2\})=\{\emptyset,\{1\},\{2\},\{1,2\}\}