Cardinality of Power Set of Finite Set

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

where $$\left|{S}\right|$$ denotes the cardinality of $$S$$,

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

where $$\mathcal P \left({S}\right)$$ denotes the power set of $$S$$.

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

For each $$A \in \mathcal P \left({S}\right)$$, we consider the characteristic function $$\chi_A: S \to T$$ defined as:


 * $$\forall x \in S: \chi_A \left({x}\right) =

\begin{cases} 1 & : x \in A \\ 0 & : x \notin A \end{cases}$$

Now consider the mapping $$f: \mathcal P \left({S}\right) \to T^S$$:
 * $$\forall A \in \mathcal P \left({S}\right): f \left({A}\right) = \chi_A$$

where $$T^S$$ is the set of all mappings from $$S$$ to $$T$$.

Also, consider the mapping $$g: T^S \to P \left({S}\right)$$:
 * $$\forall \phi \in T^S: g \left({\phi}\right) = \phi^{-1} \left({\left\{{1}\right\}}\right)$$

Note that $$g$$ is itself a mapping from a set of mappings: $$\phi: S \to T$$ is itself a mapping.

Consider the characteristic function of $$\phi^{-1} \left({\left\{{1}\right\}}\right)$$, denoted $$\chi_{\phi^{-1} \left({\left\{{1}\right\}}\right)} \left({x}\right)$$.

We have:

$$ $$ $$

So:

$$ $$ $$

So $$f \circ g = I_{T^S}$$, that is, the identity mapping on $$T^S$$.

So far so good. Now we consider:
 * $$\chi_A^{-1} \left({\left\{{1}\right\}}\right) = A$$

from the definition of the characteristic function $$\chi_A$$ above.

So:

$$ $$ $$ $$

So $$g \circ f = I_{\mathcal P \left({S}\right)}$$, that is, the identity mapping on $$\mathcal P \left({S}\right)$$.

It follows from Bijection iff Left and Right Inverse that $$f$$ and $$g$$ are bijections.

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:


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

But from the binomial theorem:


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

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

See Sum of Binomial Coefficients for Given n.

Special Case
This formula even works when $$S = \varnothing$$.

Clearly:
 * $$\mathcal P \left({\varnothing}\right) = \left\{{\varnothing}\right\}$$

has one element, that is, $$\varnothing$$.

So:
 * $$\left|{\mathcal P \left({\varnothing}\right)}\right| = \left|{\left\{{\varnothing}\right\}}\right| = 1 = 2^{0}$$

thus confirming that the main result still holds when $$\left|{S}\right| = 0$$.