The power set of a set *A* is the collection of all subsets of A. When working with a finite set with *n* elements, one question that we might ask is, “How many elements are there in the power set of *A* ?” We will see that the answer to this question is 2^{n} and prove mathematically why this is true.

### Observation of the Pattern

We will look for a pattern by observing the number of elements in the power set of *A*, where *A* has *n* elements:

- If
*A*= { } (the empty set), then*A*has no elements but*P (A)*= { { } }, a set with one element. - If
*A*= {a}, then*A*has one element and*P (A)*= { { }, {a}}, a set with two elements. - If
*A*= {a, b}, then*A*has two elements and*P (A)*= { { }, {a}, {b}, {a,b}}, a set with two elements.

In all of these situations, it is straightforward to see for sets with a small number of elements that if there is a finite number of *n* elements in *A*, then the power set *P* (*A*) has 2*n* elements. But does this pattern continue? Just because a pattern is true for *n* = 0, 1, and 2 doesn’t necessarily mean that the pattern is true for higher values of *n*.

But this pattern does continue. To show that this is indeed the case, we will use proof by induction.

### Proof by Induction

Proof by induction is useful for proving statements concerning all of the natural numbers. We achieve this in two steps. For the first step, we anchor our proof by showing a true statement for the first value of *n* that we wish to consider.

The second step of our proof is to assume that the statement holds for *n* = *k*, and the show that this implies the statement holds for *n* = *k* + 1.

### Another Observation

To help in our proof, we will need another observation. From the examples above, we can see that P({a}) is a subset of P({a, b}). The subsets of {a} form exactly half of the subsets of {a, b}.

We can obtain all of the subsets of {a, b} by adding the element b to each of the subsets of {a}. This set addition is accomplished by means of the set operation of union:

- Empty Set U {b} = {b}
- {a} U {b} = {a, b}

These are the two new elements in P({a, b}) that were not elements of P({a}).

We see a similar occurrence for P({a, b, c}). We start with the four sets of P({a, b}), and to each of these we add the element c:

- Empty Set U {c} = {c}
- {a} U {c} = {a, c}
- {b} U {c} = {b, c}
- {a, b} U {c} = {a, b, c}

And so we end up with a total of eight elements in P({a, b, c}).

### The Proof

We are now ready to prove the statement, “If the set *A* contains *n* elements, then the power set *P( A)* has 2^{n} elements.”

We begin by noting that the proof by induction has already been anchored for the cases *n* = 0, 1, 2 and 3. We suppose by induction that the statement holds for *k*. Now let the set *A* contain *n* + 1 elements. We can write *A* = *B* U {x}, and consider how to form subsets of *A*.

We take all elements of *P(B)*, and by the inductive hypothesis, there are 2^{n} of these. Then we add the element x to each of these subsets of *B*, resulting in another 2^{n} subsets of *B*. This exhausts the list of subsets of *B*, and so the total is 2^{n} + 2^{n} = 2(2^{n}) = 2^{n + 1} elements of the power set of *A*.