The empty set has only one subset, itself. Thus .
A set with a single element has two subsets, the empty set and the entire set. Thus .
A set with two elements has four subsets, and .
Similarly, for any finite set with elements, the power set has elements.
Note that for any nonnegative integer , and so for any finite set , (where absolute value signs here denote the cardinality of a set). The analogous result is also true for infinite sets (and thus for all sets): for any set , the cardinality of the power set is strictly larger than the cardinality of the set itself.
Now, note that by definition if and only if , so if and only if . This is a clear contradiction. Thus the bijection cannot really exist and so , as desired.
Size for Finite Sets
Either an element in the power set can have 0 elements, one element, ... , or n elements. There are ways to have no elements, ways to have one element, ... , and ways to have n elements. We add:
We proceed with induction.
Let be the set with elements. If , then is the empty set. Then
and has element.
Now let's say that the theorem stated above is true or n=k. We shall prove it for k+1.
Let's say that Q has k+1 elements.
In set Q, if we leave element x out, there will be elements in the power set. Now we include the sets that do include x. But that's just , since we are choosing either 0 1, ... or k elements to go with x. Therefore, if there are elements in the power set of a set that has k elements, then there are elements in the power set of a set that has k+1 elements.
Therefore, the number of elements in a power set of a set with n elements is .
We demonstrated in Method 2 that if S is the empty set, it works.
Now let's say that S has at least one element.
For an element x, it can be either in or out of a subset. Since there are n elements, and each different choice of in/out leads to a different subset, there are elements in the power sum.
- Power Set at Wolfram MathWorld.