# Cardinality of Power Set of Finite Set/Proof 1

## Theorem

Let $S$ be a set such that:

$\card S = n$

where $\card S$ denotes the cardinality of $S$,

Then:

$\card {\powerset S} = 2^n$

where $\powerset S$ denotes the power set of $S$.

## Proof

Let $T = \set {0, 1}$.

For each $A \in \powerset S$, we consider the characteristic function $\chi_A: S \to T$ defined as:

$\forall x \in S: \map {\chi_A} x = \begin{cases} 1 & : x \in A \\ 0 & : x \notin A \end{cases}$

Now consider the mapping $f: \powerset S \to T^S$:

$\forall A \in \powerset S: \map f A = \chi_A$

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

Also, consider the mapping $g: T^S \to \powerset S$:

$\forall \phi \in T^S: \map g \phi = \phi^{-1} \sqbrk {\set 1}$

where $\phi^{-1} \sqbrk {\set 1}$ is the preimage of $\set 1$ under $\phi$.

Consider the characteristic function of $\phi^{-1} \sqbrk {\set 1}$, denoted $\chi_{\phi^{-1} \sqbrk {\set 1} }$.

We have:

 $\ds \forall x \in S: \,$ $\ds \map {\chi_{\phi^{-1} \sqbrk {\set 1} } } x$ $=$ $\ds \begin{cases} 1 & : x \in \phi^{-1} \sqbrk {\set 1} \\ 0 & : x \notin \phi^{-1} \sqbrk {\set 1} \end{cases}$ $\ds$ $=$ $\ds \begin{cases} 1 & : \map \phi x = 1 \\ 0 & : \map \phi x = 0 \end{cases}$ $\ds$ $=$ $\ds \map \phi x$

So:

 $\ds \forall \phi \in T^S: \,$ $\ds \map {\paren {f \circ g} } \phi$ $=$ $\ds \map f {\phi^{-1} } {\sqbrk {\set 1} }$ $\ds$ $=$ $\ds \chi_{\phi^{-1} \sqbrk {\set 1} }$ $\ds$ $=$ $\ds \phi$

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

So far so good. Now we consider the preimage of $\set 1$ under $\chi_A$:

$\chi_A^{-1} \sqbrk {\set 1} = A$

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

Thus:

 $\ds \forall A \in \powerset S: \,$ $\ds \map {\paren {g \circ f} } A$ $=$ $\ds \map g {\map f A}$ $\ds$ $=$ $\ds \map g {\chi_A}$ $\ds$ $=$ $\ds \chi_A^{-1} \sqbrk {\set 1}$ $\ds$ $=$ $\ds A$

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

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.

$\blacksquare$