# Power set

The power set of a given set $S$ is the set $\mathcal{P}(S)$ of all subsets of that set. It is also sometimes denoted by $2^S$.

## Examples

The empty set has only one subset, itself. Thus $\mathcal{P}(\emptyset) = \{\emptyset\}$.

A set $\{a\}$ with a single element has two subsets, the empty set and the entire set. Thus $\mathcal{P}(\{a\}) = \{\emptyset, \{a\}\}$.

A set $\{a, b\}$ with two elements has four subsets, and $\mathcal{P}(\{a, b\}) = \{\emptyset, \{a\}, \{b\}, \{a, b\}\}$.

Similarly, for any finite set with $n$ elements, the power set has $2^n$ elements.

## Size Comparison

Note that for any nonnegative integer $n$, $2^n > n$ and so for any finite set $S$, $|\mathcal P (S)| > |S|$ (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 $S$, the cardinality $|\mathcal P (S)|$ of the power set is strictly larger than the cardinality $|S|$ of the set itself.

### Proof

There is a natural injection $S \hookrightarrow \mathcal P (S)$ taking $x \mapsto \{x\}$, so $|S| \leq |\mathcal P(S)|$. Suppose for the sake of contradiction that $|S| = |\mathcal P(S)|$. Then there is a bijection $f: \mathcal P(S) \to S$. Let $T \subset S$ be defined by $T = \{x \in S \;|\; x \not\in f(x) \}$. Then $T \in \mathcal P(S)$ and since $f$ is a bijection, $\exists y\in S \;|\; T = f(y)$.

Now, note that $y \in T$ by definition if and only if $y \not\in f(y)$, so $y \in T$ if and only if $y \not \in T$. This is a clear contradiction. Thus the bijection $f$ cannot really exist and $|\mathcal P (S)| \neq |S|$ so $|\mathcal P(S)| > |S|$, as desired.

Note that this proof does not rely upon either the Continuum Hypothesis or the Axiom of Choice. It is a good example of a diagonal argument, a method pioneered by the mathematician Georg Cantor.

## Size for Finite Sets

The number of elements in a power set of a set with n elements is $2^n$ for all finite sets. This can be proven in a number of ways:

### Method 1

Either an element in the power set can have 0 elements, one element, ... , or n elements. There are $\binom{n}{0}$ ways to have no elements, $\binom{n}{1}$ ways to have one element, ... , and $\binom{n}{n}$ ways to have n elements. We add: $\binom{n}{0}+\binom{n}{1}+\cdots+\binom{n}{n}=2^n$

as desired.

### Method 2

We proceed with induction.

Let $S$ be the set with $n$ elements. If $n=0$, then $S$ is the empty set. Then $P(S)=\{\emptyset \}$

and has $2^0=1$ 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 $2^k$ elements in the power set. Now we include the sets that do include x. But that's just $2^k$, since we are choosing either 0 1, ... or k elements to go with x. Therefore, if there are $2^k$ elements in the power set of a set that has k elements, then there are $2^{k+1}$ 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 $2^n$.

### Method 3

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 $2^n$ elements in the power sum.