Cardinality of Power Set of Finite Set/Proof 1

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:

So:

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:

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.