Cardinality of Power Set of Finite Set

Theorem
Let $$S$$ be a set such that $$\left|{S}\right| = n$$.

Then $$\left|{\mathcal{P} \left({S}\right)}\right| = 2^n$$.

It can be seen that the power set's alternative notation $2^S$ is indeed appropriate.

However, because of possible confusion over the conventional meaning of $$2^n$$, its use is deprecated.

Proof 1
Let $$T = \left\{{0, 1}\right\}$$.

Let $$\chi: S \to T$$ be defined as:


 * $$\forall A \subseteq S: \chi \left({A}\right) = \chi_A$$ where:



\forall x \in S: \chi_A \left({x}\right) = \begin{cases} 1 & : x \in A \\ 0 & : x \notin A \end{cases} $$

It is clear that $$\chi: A \to \chi_A$$ is a bijection from $$\mathcal{P} \left({S}\right) \to T^S$$.

Thus by Cardinality of Set of All Mappings the result follows.

Proof 2
We can see that enumerating the subsets of $$S$$ is equivalent to counting all of the ways of selecting $$k$$ out of the $$n$$ elements of $$S$$ with $$k = 0, 1, \ldots, n$$.

In other words the number we are looking for is:


 * $$\left|{\mathcal{P} \left({S}\right)}\right| = \sum_{k=0}^{n}{{n}\choose{k}}$$

But from the binomial theorem:


 * $$(x+y)^n = \sum_{k=0}^{n}{{n}\choose{k}}x^{n-k}y^k$$

it follows that:
 * $$(1+1)^n = \sum_{k=0}^{n}{{n}\choose{k}}(1)^{n-k}(1)^k = \sum_{k=0}^{n}{{n}\choose{k}} = 2^n = \left|{\mathcal{P} \left({S}\right)}\right|$$

See Sum of Binomial Coefficients for Given n.